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Chapter 1

Assessed Problems

There are three assessed problem sets. You will be assigned groups to work with. The general idea forthese problem sheets is to allow more time for thinking, discussion and research. In some cases you mayfind that there is more than one way to answer a question. Occasionally you will be asked things youcannot answer solely using what we have done in the lectures - you will have to consult a book or theweb. On occasion you might need to find out additional facts in order to answer a question.

1.1 Assessed Problem Set 1: Statistics

Bring a complete set of solutions to these questions to the first problems class. You may work on yourproblems in the group assigned to you.

The questions

1. FelonyListen to the 7 August 2009 episode of the BBC programme More Or Less. It can be found on thefollowing webpage:

http://news.bbc.co.uk/1/hi/programmes/more or less/8200392.stm#august

The last item (beginning at 21 minutes and 45 seconds) in the news piece refers to a public consul-tation on the DNA data base. In the electronic version of this booklet, the above link is clickable.Marks: Total 10.

(a) What are the reasons why the vice-president of the Royal Statistical Society, Prof. Sheila Bird,describes this paper as a “statistical felony”? Marks: 2.Two main reasons: The Home Office only looked at rearrest rates, not convictions. The

sample size was too small to establish a statistically signifcant result.

(b) Consider the two groups compared by the Home Office in the article. Suppose there was a5% difference in their rearrest rates. With a group size of 250 each, could such a differencebe considered significant? Take the rearrest rates to be 0.35 for one group and 0.4 for theother (note that this is a 5% difference). You will need to look up how to combine errors orstandard deviatons and how many standard deviations a result would have to vary from theexpectation to be considered significant in most scientific fields. Marks: 4.The standard deviation for each groups is found using the Binomial distribution standard

deviation, σ =√Np(1− p) with N = 250, p = 0.4 or p = 0.35. In the case where p = 0.4,

then σ0.4 =√

250 · 0.4 · 0.6 = 7.7. As a percentage of the group size this is therefore σ0.4 =7.7/250% = 3%. Similarly for the other group, σ0.35 = 7.2 = 3%. The total standard deviationon the difference between the rates is σ =

√0.032 + 0.032 = 0.04. So a difference of 5% is just

above 1σ. Normally a 3σ difference is required to be considered a significant result.

1

http://news.bbc.co.uk/1/hi/programmes/more_or_less/8200392.stm#august

2 CHAPTER 1. ASSESSED PROBLEMS

(c) How significant would a 5% difference be for the suggested sample size of about 1500 vs 1500?Marks: 4.Same calculation as above with N = 1500, or simply scale the result by

√250/1500 = 0.4,

so σ1500 = 1.6% and a 5% difference would now correspond to 3σ, approximately 99.5%C.L..Much more significant.

2. Binomial Distribution

Marks: Total 15.

(a) Under what conditions does the binomial distribution apply?Marks: 2To use the Binomial distribution one needs a finite number of trials and only two outcomes,

success or failure.

(b) What is the probability of obtaining 4 or more heads in 20 tosses of a coin? Marks: 5

P (x ≥ 4; 12, 0.5) = 1− P (x < 4; 100, 0.5) ,

= 1− (P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3)) ,

= 1−(9.54× 10−7 + 1.91× 10−5 + 1.81× 10−4 + 1.09× 10−3

),

= 1− 1.29× 10−3 ,

= 0.9987 .

(c) In the final of a tournament, there are two teams. They play a series of games, and the matchis decided when one team wins 4 games. What is the probability that the series will last 4games? 5 games? 6 games? Assume that the teams are evenly matched. Marks: 8This is a tricky application of the binomial distribution. If you can follow the logic of this

solution, you have a good understanding of the distribution, to this point.

We start by defining a success as a win by the team that ultimately becomes the world serieschampion. For the purpose of this analysis, we assume that the teams are evenly matched.Therefore, the probability that a particular team wins a particular game is 0.5.

Let’s look first at the simplest case. What is the probability that the series lasts only 4 games.This can occur if one team wins the first 4 games. The probability of a team A winning 4games in a row is:

P (4; 4, 0.5) =

(44

)(0.54 · 0.50

),

= 0.0625 .

Similarly, when we compute the probability of team B winning 4 games in a row, we findthat it is also 0.0625. Therefore, probability that the series ends in four games would be0.0625 + 0.0625 = 0.125; since the series would end if either team A or team B won 4 gamesin a row.

Now let’s tackle the question of finding probability that the series ends in 5 games. The trickin finding this solution is to recognise that the series can only end in 5 games, if one team haswon 3 out of the first 4 games. So let’s first find the probability that team A wins exactly 3 ofthe first 4 games.

P (4; 4, 0.5) =

(43

)(0.53 · 0.51

),

= 0.25 .

The next part is more tricky still. Given that team A has won 3 of the first 4 games, team Ahas a 50/50 chance of winning the fifth game to end the series. Therefore, the probability of

1.1. ASSESSED PROBLEM SET 1: STATISTICS 3

team A winning the series in 5 games is 0.25 · 0.50 = 0.125. Since team B could also win theseries in 5 games, the probability that the series ends in 5 games would be 0.125+0.125 = 0.25.

The rest of the problem would be solved in the same way. You should find that the probabilityof the series ending in 6 games is 0.3125.

3. Poisson Distribution

Marks: Total 7.

(a) Under what conditions does the Poisson distribution apply? Marks: 2A continuous spectrum of “trials”, with only success or failure outcomes.

(b) The average number of giant massive huge laser bolts of death shot by the rage filled eyes ofthe ancient sea god and scourge of humanity known only as Cthulhu, is 2 per second. What isthe probability that exactly 3 homes will each be blasted into annihilation in the next secondby the master of the ocean? Marks: 5This is a Poisson experiment in which we know the following: λ = 2; since 2 laser bolts are

fired per second, on average. r = 3; since we want to find the likelihood that 3 homes will bedestroyed. We plug these values into the Poisson formula as follows:

P (r;λ) =e−λλr

r!,

P (3; 2) = 2.71828−2 · 233! ,

= 0.13534 · 86 ,

= 0.180 .

Thus, the probability of 3 destroyed homes in the next second is 0.180.

4. OutbreakAssume 0.15% of population in the UK carry a newly discovered and dangerous virus. Immediatelya test is developed by the governement. The test has a 99% chance of correctly detecting the virusin a person who has it and a 0.5% chance of falsely detecting it in a person that does not haveit. Calculate the probability that a person that tests positive for the virus, does indeed carry thevirus. How good is the test? Marks: 5.

P (positive test) = P (positive test|not carry)P (not carry) + P (positive test|carry) · P (carry)

= 0.005 · 0.9985 + 0.99 · 0.0015

= 0.65%

P (carry|positive test) = P (positive test|carry) · P (carry)

P (positive test)

= 0.99 · 0.0015

0.0065= 0.23

= 23%

The test is useless.

5. Which distribution?Which distribution best describes the following: Marks: Total 5.

(a) The number of flashes of lightening in one hour during a thunderstorm. Marks: 1Poisson

(b) The number of detected Higgs events at the LHC in one year of operation. Marks: 1.Poisson, or Binomial if you assume a fixed number of proton-proton collisions which can eitherresult in a Higgs or not.


(c) The number of cars failing an MOT test out of a batch of 100. Marks: 1.Binomial

(d) The weight of individual sugar cubes at the end of a sugar cube production line. Marks: 1.Gauss

(e) The number of grains of sand in 1kg of sand. Marks: 1.Gauss

6. Super SymmetryA lot of particle physics research is concerned with looking for cracks and flaws in the (so far verysuccessful) Standard Model of particle physics and searching for new, unknown particles. This isgenerally done by looking for the stable decay products of the new particle of interest. The massof the hypothetical mother particle is reconstructed from the decay kinematics. For backgroundevents this generally results in a random result, distributed more-or-less evenly throughout the massrange, while a real unstable particle would produce a peak around its actual mass. Marks: Total:6.

(a) The year is 2016. In the search for Supersymmetry at the LHC, researchers expect (aftercleaning up their data sample with careful selection criteria), within a given mass-window,100 events from background and known Standard Model processes. They find 120. Is thissignificant? To estimate how much evidence against the Standard Model this constitutes,calculate the probability of finding 120 or more events when one expects 100. Marks: 4.

• We use the Poisson distribution, and σ =√N = 10

• This is a 2σ effect,

• What needs to be calculated is the probability to get a 2σ increase or more:

1−2ˆ

−∞

e−x2/2dx = 1− 97.72% = 2.28%

If you integrated a Poisson, that is of course OK, too; it should have given the same result(or very similar).

(b) Now you learn that in fact they have not been looking in only one mass window, but they havelooked at 32 mass windows, all chosen such that one expects about 100 events from known,Standard Model process in each window. How significant is the above result, now? To answerthis question, calculate how likely it is to have at least one bin amongst the 32 that has 120events or more due to statistical fluctuations. Marks: 2.The probability to find at least one such fluctuation is 1 minus the probability to see no such

fluctuation: 1− (97.72%)32 = 52%

1.2. ASSESSED PROBLEM SET 2 5

1.2 Assessed Problem Set 2

Marks: 77 in total.

1. Paths & FieldsRead the full question before you start to make sure you leave enough space in your diagrams toanswer all parts. Marks: 20 in total.

(a) Sketch the following vector field in the x− y plane (i.e. z = 0):

~G =1√

x2 + y2

−yx0

Marks: 4.See plots below. The important features: The magnitude of the vectors does not change. Check

∣∣∣~G∣∣∣ =

√~G · ~G =

√G2x +G2

y +G2z =

√x2

x2 + y2+

y2

x2 + y2= 1

The vector field is symmetric with respect to rotation around the z-axis, all vectors are tan-gential to circles centred at 0. Notice that this is in fact the vector field of ~eφ.

(b) Without calculation, add to the drawing you made for ~G one simple path (A simple path isone where, as you traverse it, you go through each point exactly once, it is a path that doesnot cross itself) with different start and end points where:

• the line integral is clearly positive.

• the line integral is clearly negative.

• the line integral is clearly zero.

Marks: 6, 2 for each path.

• Paths parallel to vectors are positive.

• Paths anti-parallel to vectors are negative.

• Lines perpendicular to vectors are zero.

• In the example below, a more complicated path has a positive and part that cancels it out.

Of course you didn’t have to provide 8 examples, like below, but only three. This is just foradditional illustration.


Describe in a few words why each of the paths results in either a positive, negative or zero


line integral. For one extra mark, make the zero line integral path go through the points (1,0)and (0,1) though the path does not need to start or end there.

(c) Draw the closed path described by x2 + y2 = 14 and z = 0. Calculate the line integral of ~G

counter-clockwise around this closed loop. Marks: 7, 1 point for the sketch and 6 points forcalculating the line integral.The path looks like this

Definitely better to use polar/cylindrical co-ordinates. Can use either the usual i, j, k basis or


cylindrical. Here, I’ll do the calculation for both:

~r(t) =

12 cos t12 sin t

0

=

1

2~er

d~r

dt=

1

2

− sin tcos t

0

=

1

2~eφ

~G(~r(t)) =

− sin tcos t

0

= ~eφ

ˆ

circle

~G · d~r =

2πˆ

0

1

2

− sin tcos t

0

·

− sin tcos t

0

dt =

2πˆ

0

1

2~eφ · ~eφ dt

=1

2

2πˆ

0

(sin2 t+ cos2 t+ 0

)dφ =

2πˆ

0

1

2dφ

=1

2

2πˆ

0

1 dφ

= π

(d) Draw the path described by (x − 1)2 + (y − 1)2 = 14 and z = 0. How would you parametrise

this path? Marks: 3, 1 for the sketch and 2 for a correct parametrisation.

Same as before, plus

110

~r(t) =

12 cos t12 sin t

0

+

110

with 0 ≤ t < 2π

Or something equivalent. Looks like this (I’ve picked arbitrarily a positive orientation, which,in my parametrisation, would correspond to going from t = 0 to t = 2π).


The path (x–1)2 + (y–1)2 = 1/4

x

y

1–

–

1

–0.5

–

0.5

2. Multiple Choice. Marks: 10.Which of these expressions represent a vector field (v), scalar field (s) or something nonsensical(n)? Write the question numbers and next to them your answer as v, s, or n. You need to give

the answer only, justification is unnecessary. ~A, ~B, ~C are generic vector fields, with no particularmeaning attached i.e. ~B is not meant to represent a magnetic field. Marking: +1/2 for a correctanswer, -1/2 points for a wrong answer, 0 points for no answer ( The overall mark for the questionwill not be negative - the minimum overall mark is 0).

(a) ~A · ~Bs

(b) ~C ×(~A× ~B

)

v

(c)(~A× ~B

)× C

v

(d) ~C ×(~A · ~B

)

n

(e) ~C ·(~A× ~B

)

s

(f) ~A× ~Bv

(g) ~B × ~Bv

(h) ~0× ~Bv

(i)

abc

·

hij

s

(j)

ahbicj

v

(k) ~∇× ~Av

(l) ~∇ · ~As

(m) ~∇ ~An

(n) ~∇φv

(o) ~∇×(~∇ · ~A

)

n

(p) ~∇ ·(~∇× ~A

)

s

(q) ~∇×(~∇× ~A

)

v

(r)‚

~A · d~Ss

(s)˝

~∇φdVv

(t)¸~A · d~l

s


3. More Multiple Choice:Which of these expressions are obviously nonsensical and which are valid? In the expressions below,a, b, c, d, e, f, g, h, i, j, k are all simple numbers. Marks: 5, ±1 for each question.

(a)

abc

·

hij

= ah+ bi+ cj

Valid

(b)

abc

·

hij

=

ahbicj

n

(c)

abc

·

hij

= (ah+ bi+ cj, 0, 0)

n

(d)

abc

×

hij

= g

n

(e)

abc

×

hij

=

def

Valid

4. Potentials & Fields Marks: 20.

(a) Express the following potential in cylindrical coordinates. Marks: 2.

φ(x, y, z) =1

2

(x2 + y2

)− 1√

x2 + y2 + z2

φ(ρ, φ, z) = φ(ρ, z) =1

2ρ2 − 1√

ρ2 + z2

(b) Draw the equipotential lines for the above potential in the x− y plane and in the x− z plane.Marks: 8, 4 for each plot.Should reflect the basic structure and symmetries of the field. Many sensible choices are

possible for the values of φ for which to draw the lines. Mark generously. Let’s start withthe x-y plane. There

φ(x, y, 0) =1

2

(x2 + y2

)− 1√

x2 + y2=

1

2ρ2 − 1

ρ

Clearly lines of constant φ are circles - the only issue is the spacing. Trying to solve thisfor ρ, I got some irritating cubic equation, which I did not fancy solving. Instead, I typedφ(ρ) = 1

2ρ2 − 1

ρ into a graphing programme - and then you can read off the ρ values thatcorrespond to an equal spacing in φ, e.g.:


φ(ρ

)fo

rz

=0

0.5 1.0 1.5 2.0 2.5 3.0 3.5

�8

�6

�4

�2

2

4

6

−→ ρNOTE THAT THE GRAPH ABOVE IS NOT THE SOLUTION TO THE QUESTION. It’sjust a step towards it - that’s only necessary because it’s to tricky to sovle φ = 1

2ρ2− 1

ρ for ρ. I

get (solving it numerically - but you can also use the plot, that you could have drawn, or sometrial-and-error method. Would all be fine)φ = ... ρ = ...

-8 0.12-6 0.17-4 0.25-2 0.470 1.262 2.214 2.956 3.548 4.06

Of course you could have chosen different values for φ, which is fine as long as they areequidistant. This is what I get in the x− y plane:

−→y

�8�6

�4

�2

0

2

4

6

8

8

8

8

�4 �2 0 2 4�4

�2

0

2

4

−→ x(of course for your assignment you do not have to provide a separate contour plot w/o vectorfield - one single plot with both contours and the field is fine)


Let’s now move to the x-z plane. Here

φ(x, y = 0, z) =1

2x2 − 1√

x2 + z2

The best way to draw contour lines is to solve the equation for one of the parameters - z asa function of x seems the easiest. Then you can draw z vs x curves for several values of φ.But before doing this, let’s take a moment to think about what structure we expect to see. Thewhole thing is cleary symmetric under a change of sign of x or z or both. For very large xthe dependence on z will disappear and we’ll get 1

2x2, which would be lines parallel to z with

decreasing spacing. For very small x and z, you’d expect the term 1√x2+z2

to dominate, so we

get circles.

Let’s solve it:

φ =1

2x2 − 1√

x2 + z2

1

2x2 − φ =

1√x2 + z2

(1

2x2 − φ

)2

=1

x2 + z2

1(

12x

2 − φ)2 − x2 = z2

z = ±(

1(

12x

2 − φ)2 − x2

) 12

With this we should be able to draw a few contour lines by calculating for each φ we considera few points (x, z) that we then connect to countour lines - more efficiently since we know thatthe field is symmetric about the x and the z axis, so once we have drawn the contour on thefirst quadrant, we know how it looks like in the remaining quadrants.

This is what I get from Mathematica:

−→z

�8�6�4

�2

0 0

2

2

4

4

6

6

8

8

�4 �2 0 2 4

�4

�2

0

2

4

−→ x(c) Calculate the force field. Sketch that in the same two plots. Marks: 6, 2 for the correct field

~F , and 2 for each correct sketch of ~F .


A correct sketch means that the structure and symmetries should be apparent and the vectorsshould be consistent with the potential - perpendicular to the lines, and longer where equipo-tential lines are closer together. First φ:

−~∇φ = −1

2

∂(x2+y2)∂x

∂(x2+y2)∂y

∂(x2+y2)∂z

−

∂ 1√x2+y2+z2

∂x∂ 1√

x2+y2+z2

∂y∂ 1√

x2+y2+z2

∂z

=

−x−y0

+

1

(x2 + y2 + z2)(3/2)

xyz

The plots should look like this (of course if you draw them by hand - which is the idea - it isOK to draw fewer arrows):

x− y, z = 0 x− z, y = 0

−→y

�8�6

�4

�2

0

2

4

6

8

8

8

8

�4 �2 0 2 4�4

�2

0

2

4

−→ x

−→z

�8�6�4

�2

0 0

2

2

4

4

6

6

8

8

�4 �2 0 2 4

�4

�2

0

2

4

−→ x

(d) Can you imagine a physical setup that would produce such a potential? Marks: 4.

Something like this:

no friction

gravity

spring

5. For which of the fields below does there exist a potential? If there is a potential, find it. Forthis question, use any of the methods described in the lecture - or the “Murder Mystery Method”described at this link:


http://www.math.oregonstate.edu/bridge/papers/mmm.pdf.Make sure that it is clear how you arrived at your result. Marks: 8.

(a) This field:

~F =

αxβyγ

where α, β, γ are constants. Marks: 6.

• Does it have a potential? One way to show this is to show that

∂Fi∂xj

=∂Fj∂xi

for all i, j

on a simply connected region.

– Is the area simply connected? Since the force field is well defined everywhere, the regionmust be simply connected.

– The condition above means we need to test

∂(γ)∂y −

∂(βy)∂z

−∂(γ)∂x + ∂(αx)

∂z∂(βy)∂x −

∂(αx)∂y

=

000

Or one can simply find the potential and show that it works (i.e. that ~∇φ = ~F ), then youdo not have to do this part.

• What is it’s potential? You are allowed to guess it and then verify that the guess iscorrect. However, you could also use the line-integral approach: First calculate the lineintegral from some fixed point ~r0 to an arbitrary point ~r1, along a convenient path. We’lluse a straight line, and also start at the origin, ~r0 = ~0. Let’s do the line integral:

– The path connecting

000

and

x1

y1

z1

in a straight line is:

~γ =

x1ty1tz1t

with 0 < t < 1

– The line element

d~r =d~r

dtdt =

x1

y1

z1

– The field along the path:

~F (~r(t)) =

αx1tβy1tγ


Putting it all together for the line integral:

I =

ˆ

γ

~F · d~r

=

1ˆ

0

αx1tβy1tγ

·

x1

y1

z1

dt

=

1ˆ

0

(αx2

1t+ βy21t+ γz1t

)dt

=1

2

(αx2

1 + βy21

)+ γz1

So, our potential is (not forgetting the minus sign):

V (x, y, z) = −1

2

(αx2 + βy2

)+ γz1

Cross check:

−~∇V = −

∂(− 12αx

2− 12βy

2−γz1)∂x

∂(− 12αx

2− 12βy

2−γz1)∂y

∂(− 12αx

2− 12βy

2−γz1)∂z

=

αxβyγ

= ~F

Yes, it’s the right potential. You have to do either this cross check (that I recommend

anyway), or calculate ~∇× ~F to make sure that ~F really has a potential. Otherwise whatyou’ve calculated could just be the line-integral of a field w/o a potential. Which one you

do is up to you. The advantage of checking ~∇× ~F = ~0 is that you can do that before youstart.

(b) The field:

~H =

−yxz

Marks: 2.

~∇× ~H =

∂z∂y − ∂x

∂z

− ∂z∂x + ∂(−y)

∂z∂x∂x −

∂(−y)∂y

=

002

6= ~0

So it has no potential. We can stop here.

6. Flat SurfacesMarks: 6.

(a) The differential area element in cartesian co-ordinates is dx dy. Show geometrically (by drawingthe appropriate differential area element, with labels) that the differential area element in polarcoordinates is r dr dφ. Marks: 2.


Jonas Rademacker Mathematical Physics 202, part A.2: Vector Calculus

Differential Area Element in Polar Co-ordinates

rr+dr

φ

φ+dφ

!"#!$%&'(')*"'$"+%&

,)*"!!-+&./+.0

,#(!-+&.0

#!$"+%&

• Sidelength:

• purple side: dr

• red side: r dφ

• Area: dA = r dφ dr

r dφ

dr

dφ

(b) In the lectures, we discussed how to calculate the vectorial surface element for general surfaces(curved or flat). Use the same formalism for the x − y plane. The x − y plane is given by

~S(x, y) =

xy0

in cartesian coordinates, and by ~S(rφ) =

r cosφr sinφ

0

in polar coordinates.

Calculate for both cases∣∣∣d~S∣∣∣. Compare with the result from the previous question. Marks: 4,

2 for each d~S.They should make a comment that the answer is the same as the geometrical method.

• Cartesian:

d~S =

xy0

∂~S

∂x=

100

∂~S

∂y=

010

d~S =∂~S

∂x× ∂~S

∂ydx dy

=

100

×

010

dx dy

=

001

dx dy

∣∣∣d~S∣∣∣ = dx dy


• Polar

d~S =

r cosφr sinφ

0

∂~S

∂r=

cosφsinφ

0

∂~S

∂φ=

−r sinφr cosφ

0

d~S =∂~S

∂r× ∂~S

∂φdr dφ

=

cosφsinφ

0

×

−r sinφr cosφ

0

dr dφ

=

00

r(cos2 φ+ sin2 φ)

dr dφ

=

001

r dr dφ

∣∣∣d~S∣∣∣ = r dr dφ

So they match - it’s OK if you haven’t got any more to say about this. The moral of the storyis that the d~S formalism automatically takes care of getting the area element right in differentcoordinate systems. The only difference to the area element you calculated before is that d~Salso has a direction.

7. Calculate the flux of the following field

~F =

−x−y−1

through the upper (z > 0) half of the unit sphere centered at the origin. Marks: 8.

There are different ways to do this - since it is a sphere, it makes sense to use spherical polarcoordinates. And that will work. However, the true symmetry of the problem, including the field ~F ,is cylindrical. So it is actually better to use cylindrical coordinates. I’ll do both. And at the end, Ishow a “clever” solution, that makes it even easier.

Option 1: cylindrical coordinates

Cylindrical co-ordinates can sometimes be very useful even when integrating over a sphere. Whileyour sphere is clearly spherically symmetric it exhibits also some part of cylindrical symmetry (fullcylindrical symmetry would be symmetry under rotation around the z-axis, plus symmetry undertranslation along z - a sphere clearly shows the former, and not the latter). Often the problem seenas a whole (which will include a field) is not spherically symmetric, but is still symmetric underrotations around the z-axis - in such cases it might be easier to use cylindrical coordinates even


with spheres. Anyway, it’s a good exercise, and here it does turn out to be easier. Recall that theposition vector in cylindrical co-ordinates is:

xyz

=

ρ cosφρ sinφz

where, remember, this ρ is the distance from the z−axis, a different quantity than the r in sphericalcoordinates. The cartesian equation for a sphere

x2 + y2 + z2 = R2

in cylindrical coordinates becomes:

ρ2 + z2 = R2

where in our case R = 1, but we keep it as a general parameter for now. We have a choice whichpair of co-ordinates to pick to parameterise the surface. I can eliminate either ρ or z using thisequation. It turns out, eliminating ρ =

√R2 − z2 and parametersing the sphere in terms of z and

φ is easier than using ρ and φ, not least because for R, which is positive by definition, we have aunique solution to ρ2 + z2 = R2 (see the solution in cartesian coordinates, next, for an illustrationwhy this matters).

A sphere in cylindrical coordinates, parameterised in terms of z and φ is therefore (simply replacing

ρ with√z2 −R2 in

ρ cosφρ sinφz

):

~S =

√z2 −R2 cosφ√z2 −R2 sinφ

z

∂~S

∂z=

−z√R2−z2 cosφ−z√R2−z2 sinφ

1

∂~S

∂φ=

−√R2 − z2 sinφ√R2 − z2 cosφ

0

d~S =∂~S

∂φ× ∂~S

∂zdz dφ

d~S =


0

×


1

dz dφ

=

√R2 − z2 cosφ√R2 − z2 sinφ

z

dz dφ

Now the integrationPut “sphere into the field”:

~F (~S(z, φ)) = −


1


Now all into the integral

¨

hemisphere

~F · d~S = −ˆ 2π

0

ˆ R

0


1

·


z

dz dφ

= −ˆ 2π

0

ˆ R

0

((R2 − z2) cos2 φ+ (R2 − z2) sin2 φ+ z

)dz dφ

= −ˆ 2π

0

ˆ R

0

((R2 − z2)(cos2 φ+ sin2 φ) + z

)dz dφ

= −ˆ 2π

0

ˆ R

0

(R2 − z2 + z) dz dφ

= −2π

ˆ R

0

(R2 − z2 + z) dz

= −2π

[R2z − 1

3z3 +

1

2z2

]R

0

= −2π

(R3 − 1

3R3 +

1

2R2

)

putting in R=1

= −2π

(1− 1

3+

1

2

)

= −2π6− 2 + 3

6

= −2π7

6

= −7π

3

Using cylindrical basis vectors:

~S = ~r = ρ~eρ + z~ez

d~S =(√

R2 − z2~eρ + z~ez

)dz dφ

~F = −√R2 − z2~eρ + ~ez

¨

hemisphere

~F · d~S

= −¨

hemisphere

(√R2 − z2~eρ + ~ez

)·(√

R2 − z2~eρ + z~ez

)dz dφ

= −¨

hemisphere

(R2 − z2 + z

)dz dφ

as before...

= −7π

3


Option 2: Spherical coordinates (a bit more complicated but still betterthan cartesian, and if in doubt, probably usually the better choice if youintegrate of some part of a sphere)

Parameterise the sphere: The general co-ordinate vector on spherical coordinates is

r sin θ cosφr sin θ sinφr cos θ

In terms of this, the equation for a sphere becomes very simple!

x2 + y2 + z2 = R2

∣∣∣∣∣∣

r sin θ cosφr sin θ sinφr cos θ

∣∣∣∣∣∣

2

= R2

r2 = R2

That’s it - with just replace the parameter r with the fixed length R = 1 and parameterise the spherein terms of θ, φ.

~S(θ, φ) =

sin θ cosφsin θ sinφ

cos θ

partial derivatives

∂

∂φ~S

=

− sin θ sinφsin θ cosφ

0

∂

∂θ~S =

cos θ cosφ− cos θ sinφ− sin θ

d~S =∂

∂θ~S × ∂

∂φ~S =

sin2 θ cosφsin2 θ sinφsin θ cos θ

dθ dφ

=∂

∂θ~S × ∂

∂φ~S =


cos θ

sin θdθ dφ

Put the “sphere into the field”

~F (~S) =

− sin θ cosφ− sin θ sinφ−1


Now put it all in:

¨

hemisphere

~F · d~S =

2πˆ

φ=0

π/2ˆ

θ=0

− sin θ cosφ− sin θ sinφ−1

·


cos θ

sin θ dθ dφ

=

2πˆ

φ=0

π/2ˆ

θ=0

(− sin2 θ cos2 φ− sin2 θ sin2 φ− cos θ

)· sin θ dθ dφ

= −2πˆ

φ=0

π/2ˆ

θ=0

(sin2 θ

(cos2 φ+ sin2 φ

)+ cos θ

)sin θ dθ dφ

= −2πˆ

φ=0

π/2ˆ

θ=0

sin3 θ + sin θ cos θ dθ dφ

= −2π

π/2ˆ

θ=0

(sin3 θ + sin θ cos θ

)dθ

= −2π

([− cos θ +

1

3cos3 θ

]π/2

0

+

[1

2sin2 θ

]π/2

0

)

= −2π

(1− 1

3+

1

2

)

= −2π6− 2 + 3

6

= −7π

3

We used for this: ˆsin3 ax dx = −1

acos ax+

1

3acos3 ax

Option 3: Clever Method

Here, we use the divergence theorem. Now the divergence theorem only applies to closed surfaces,and this is an open surface. Still, it will allow us to simplify the problem, and it works like this:The upper hemisphere plus the unit disk in the x− y plane form, together, a closed surface. Let ususe the following notation:

• Ih ≡˜

upper hemisphere

~F · d~S = integral over upper hemisphere, radius R = 1 (with surface normal

such that it has a positive z component)

• Id ≡˜disk

~F · d~S = integral over unit disk in x− y plane (with surface normal in −z direction)

• It = Ih + Id the integral over the closed surface put together from the upper hemisphere andthe disk. It’s important that I get the orientation of the surfaces consistent with each other -pointing outwards.

Now let’s calculate It and Id, from which we can get Ih = It−Id. This can be easier than calculatingIh directly, as we can use the divergence theorem for It, and then only have to calculate Id. Param-eterising a disk is a bit easier than parameterising a sphere, so we have a simpler surface (althoughit is perfectly possible to integrate directly over the hemisphere, as shown previously). Also, the z


component of ~F is particularly simple, and that’s the one that the disk in the x− y plane will pickout. Let’s start with It

It =

‹

disk+hemisphere

~F · d~S

=

˚

upper half−ball

~∇ · ~F dV

=

˚

upper half−ball

(−2)dV

= −2 · 1

2· (Volume of ball)

= −2 · 1

2· 4

3π

= −4

3π

Now Id

Id =

¨

disk

~F · d~S

Paramaterise surface (disk)

~S =

ρ cosφρ sinφ

0

So our parameters are ρ, going from 0 to 1, and φ, going from 0 to 2π. Let’s find d~S:

d~S =∂~S

∂φ× ∂~S

∂ρdφ dρ

=

−ρ sinφρ cosφ

0

×

cosφsinφ

0

dφ dρ

=

00

−ρ sin2 φ− ρ cos2 φ

dφ dρ

=

00−ρ

dφ dρ

Check orientation: We wanted it to point in −z direction, and (since ρ ≥ 0), it does. Good!

Otherwise we’d just have taken the negative of the d~S we found. Now put ~S into the field

~F =

−x−y−1

~F (~S(ρ, φ)) =

−ρ cosφ−ρ sinφ−1


Finally, do the integral:

Id =

¨

disk

~F · d~S

=

2πˆ

0

1ˆ

0

−ρ cosφ−ρ sinφ−1

·

00−ρ

dρ dφ

=

2πˆ

0

1ˆ

0

ρ dρ dφ

= 2π

1ˆ

0

ρ dρ

= 2π · 1

2= π

Therefore, we finally get

Ih = It − Id= −4

3π − π

= −7

3π


1.3 Assessed Problem Set 3

The deadline for this work is Wednesday in week 8 of first term. As usual, you can hand work ingroups of up to 4 students. Please write everyone’s name and student ID clearly onto the sheet.Note the competition in Q1. Marks: 99 total

1) Marks: 25 in total for this question. What is the (classical, time-averaged) electric field causedby a charged (quantum-mechanical) particle trapped in a spherical infinite potential well ofradius R = 10−14m, in the groundstate? In which way could this be result be relevant?Competition: The first fully correct submitted answer to this question wins a prize (onefor each member of the group). Over the course of the week, I will publish successivelycomprehensive hints how to solve this.

The Normalisation Factor.

First we calculate the normalisation factor C:

2πˆ

0

π̂

0

R̂

0

|ψ(r, θ, φ)|2 r2 sin θ dr dθ dφ = 1

2πˆ

0

π̂

0

R̂

0

C2 sin2(kr)

k2r2r2 sin θ dr dθ dφ = 1

C2

k2

2πˆ

0

π̂

0

R̂

0

sin2(kr) sin θ dr dθ dφ = 1

C2

k24π

R̂

0

sin2(kr) dr = 1

Using the standard integralˆ

sin2(ax) dx =1

2x− 1

4asin(2ax),

we see that,

C2

k24π

(1

2R− 1

4ksin(2kR)

)= 1

with k = π/R:

R2C2

π24π

(1

2R− R

4πsin(2π)

)= 1

2R3C2

π= 1

C =

√π

2R3=k3/2

√2π

And hence

ψ(r, θ, φ) =

{ √π

2R3sin krkr if r < R

0 else=

{k3/2√

2πsin krkr if r < R

0 else


The Field.

The probability density above, multiplied by the charge of the charged particle, can be seen as a(time-averaged) charge density. Since it is clearly spherically symmetric (only depends on r),we can use the usual symmetry trick to calculate the electric field. Gauss’ divergence theoremtells us: ˚

V

~∇ · ~E dV =

‹

∂V

~E · d~S

The trick is now to take a spherical volume. Because of the symmetry of the physics causingthe field, the ~E field must also possess that symmetry. This means that

• The field must be radial: ~E = Er~er

• Its strength can only depend on r: ~E = Er(r)~er

For the surface integral over the sphere, this means the ~E and d~S are always parallel, and fora given radius (i.e. a given sphere), Er(r) is constant. Therefore, for a sphere of radius r0

(using the symbol r0 to distinguish it from the radius R of the potential well, as well as theintegration variable r): ‹

∂Sphere

~E · d~S =

‹

∂Sphere

Er~er · d~S

= Er(r0)

‹

∂Sphere

~er · d~S

= Er(r0)

‹

∂Sphere

∣∣∣d~S∣∣∣

= Er(r0)4πr20

The other side of the equation is (using q for the charge of the trapped particle), for r0 ≤ R:˚

V

~∇ · ~E dV =

˚

V

ρ

ε0dV

=

2πˆ

0

π̂

0

r0ˆ

0

q

ε0|ψ(r, θ, φ)|2 r2 sin θ dr dθ dφ

=

2πˆ

0

π̂

0

r0ˆ

0

q

ε0C2 sin2(kr)

krr2 sin θ dr dθ dφ

=

2πˆ

0

π̂

0

r0ˆ

0

q

ε0C2 sin2(kr)


=q

ε0

C2

k24π

(1

2r0 −

1

4ksin(2kr0)

)

=q

ε0

k3

2π2k24π

(1

2r0 −

1

4ksin(2kr0)

)

=k

π2

(1

2r0 −

1

4ksin(2kr0)

)

=q

ε0

(r0

R− 1

2πsin(

2πr0

R

))


Hence, for r0 ≤ R:

Einsider (r0)4πr2

0 =q

ε0

(r0

R− 1

2πsin(

2πr0

R

))

Einsider (r0) =

q

4πε0r20

(r0

R− 1

2πsin(

2πr0

R

))

Now for r0 ≥ R, the same logic applies, except that we stop integrating at r0 = R when thecharge density drops to zero:˚

V

~∇ · ~E dV =

˚

V

ρ

ε0dV

=

2πˆ

0

π̂

0

r0ˆ

0

q

ε0|ψ(r, θ, φ)|2 r2 sin θ dr dθ dφ

=

2πˆ

0

π̂

0

R̂

0

q

ε0C2 sin2(kr)


=q

ε0

Leading, for r0 ≥ R, to

Eoutsider (r0)4πr2

0 =q

ε0

Eoutsider (r0) =

q

4πε0r20

Note that for r0 = R, this is the as the previously calculated Einside, as one would expect - wewant a continuous function.

The total result is for Er is:

Er(r0) =

{q

4πε0r20

(r0R − 1

2π sin(2π r0R

))for r0 ≤ R

q4πε0r20

otherwise

where, for aesthetic reasons, and since there is no ambiguity about its meaning, we can nowdrop the subscript and substitute r for r0:

Er(r) =

{q

4πε0r2

(rR − 1

2π sin(2π r

R

))for r ≤ R

q4πε0r2

otherwise

The electric field (a vector field) is therefore:

~E(r) =

{q

4πε0r2

(rR − 1

2π sin(2π r

R

))~er for r ≤ R

q4πε0r2

~er otherwise

=

{q

4πε0r2

(rR − 1

2π sin(2π r

R

))~r|~r| for r ≤ R

q4πε0r2

~r|~r| otherwise

2) Marks: 38 in total for this question. Consider the following fields:

~G =

x3 + yzy3 + xzxy

~H =

y2

x2

1


a) Show that ~G has sources, while ~H does not. Marks: 4, 2 for each field.The question is whether their divergence is zero (no sources) or not (sources).

~∇ · ~G =

∂∂x∂∂y∂∂z

·

x3 + yzy3 + xzxy

= 3(x2 + y2) = 3ρ2

So ~G does have sources.

~∇ · ~H =

∂∂x∂∂y∂∂z

·

y2

x2

1

= 0

~H does not have any sources.

b) Which of the above fields (if any) is conservative? Marks: 4, 2 for each field.Both fields are defined for all values of x, y, z, so we do not need to worry whether

the region where they are defined is simply connected. (Don’t forget that this would beimportant if we had fields that are undefined for some values.)

~∇× ~G =

∂∂x∂∂y∂∂z

×

x3 + yzy3 + xzxy

=

∂(xy)∂y −

∂(y3+xz)∂z

−∂(xy)∂x + ∂(x3+yz)

∂z∂(y3+xz)

∂x − ∂(x3+yz)∂y

= ~0

So ~G is conservative because its curl is ~0 and it is defined over a simply connected region.Now let’s look at ~H:

~∇× ~H =

∂∂x∂∂y∂∂z

×

y2

x2

1

=

∂1∂y − ∂x2

∂z

− ∂1∂x + ∂y2

∂z∂x2

∂x −∂y2

∂y

=

00

2x− 2y

This is not ~0 and therefore ~H is not conservative.

c) Find the potential for the conservative field. Marks: 10.

We are looking for a φ such that ~G = −~∇φ. There are different ways of finding it.


Guessing is one way. Using a line integral another. But the easiest way is probably this:

∂φ

∂x= −x3 − yz

φ = −1

4x4 − xyz + f(y, z)

∂φ

∂y= y3 + xz

∂(− 14x

4 − xyz + f(y, z))

∂y= y3 + xz

−xz ∂(f(y, z))

∂y= −y3 − xz

∂(f(y, z))

∂y= −y3

f(y, z) = −1

4y4 + g(z)

φ = −1

4x4 − xyz − 1

4y4 + g(z)

∂φ

∂z= −xy

∂(− 14 (x4 + y4)− xyz − 1

4y4 + g(z))

∂z= −xy

−xy +∂g(z)

∂z= −xy

∂g(z)

∂z= 0

φ = −1

4(x4 + y4)− xyz + C

where C is a constant. A possible solution is therefore

φ = −1

4

(x4 + y4

)− xyz

To be sure, we can try it out:

−

∂φ∂x∂φ∂y∂φ∂z

=

∂( 14 (x4+y4+)+xyz)

∂x∂( 1

4 (x4+y4+)+xyz)

∂y∂( 1

4 (x4+y4+)+xyz)

∂z

=

x3 + yzy3 + xzxy

= ~G

So this method gave the correct answer.

d) Calculate the flux of ~G and of ~H through a closed cylindrical surface of radius R, centeredaround the z axis and extending from z = 0 to z = L. Marks: 10.Marks: 10 points in total, 4 for field ~H and 6 for field ~G. This is a clear case for using

the divergence theorem, as a sphere is a closed surface. So we use

‹

∂V

~F · d~S =

˚

V

∂∂x∂∂y∂∂z

· ~F dV


For ~H this is trivial, as we just calculated its divergence and it is zero, therefore

‹

∂V

~H · d~S =

˚

V

∂∂x∂∂y∂∂z

· ~H dV =

˚

V

0 dV = 0

Next the slightly more complicated ~G

‹

∂V

~G · d~S =

˚

V

∂∂x∂∂y∂∂z

· ~GdV

=

˚

V

3ρ2 dV

=

2πˆ

0

L̂

0

R̂

0

3ρ2 ρ dρ dz dφ

= 3

2πˆ

0

L̂

0

R̂

0

ρ3 dρ dz dφ

= 6πL

R̂

0

ρ3dρ

=3

2πLR4

e) Calculate the flux of ~H through the surface, shown in the figure below.


x

y

z

This surface is bound by a circle of radius R in the x− y plane (shown as a red [or greyin black and white printouts] circle) and is otherwise irregularly shaped. Calculate the

flux of ~H through that surface. Marks: 10.~∇· ~H = 0 and ~H is defined over the entire simply connected region, so we can use Stokes’

theorem. We know that for ~H, the surface integral only depends on the boundary, so weget the same surface integral if we integrate over any surface that has the same boundaryas the gnome hat. The obvious choice is the disk of radius R in the x− y plane.

• Parametrise the disk:

~S =

xy0

=

ρ cosφρ sinφ

0

with ρ ∈ [0, 1] and φ ∈ [0, 2π].


• Find d~S

∂

∂ρ~S =

cosφsinφ

0

∂

∂φ~S =

−ρ sinφρ cosφ

0

d~S =∂

∂ρ~S × ∂

∂φ~S dρ dφ

=

00ρ

dρ dφ

• Everything into integral:¨

complicated surface

~H(~S) · d~S =

¨

disk

~H(~S) · d~S

=

R̂

0

2πˆ

0

~H(~S)

00ρ

dρ dφ

=

R̂

0

2πˆ

0

ρ2 sin2 φρ2 cos2 φ

1

·

00ρ

dρ dφ

=

R̂

0

2πˆ

0

ρ dρ dφ

= πR2

3) Marks: 30. Consider the following vector field, which represents the vector potential of themagnetic field outside an infinitely long solenoid of radius a:

~A =Bz2

a2

x2 + y2

−yx0

.

i. Show that the corresponding magnetic field is zero. Marks: 4

~∇× ~A =a2Bz

2

∂0∂y −

∂(x/(x2+y2))∂z

− ∂0∂x + ∂(−y/(x2+y2))

∂z∂(x/(x2+y2))

∂x − ∂(−y/(x2+y2))∂y

=a2Bz

2

00

(x2+y2)−2x2

(x2+y2)2 + (x2+y2)−2y2

(x2+y2)2

=a2Bz

2

00

2x2+2y2−2x2−2y2

(x2+y2)2

=

000


ii. To obtain the flux of the magnetic field through the solenoid, calculate the integral of~A along a closed path around the solenoid. Use a circular path in the x− y plane withradius R > a. Do the line-integral explicitly, do not use Stokes’ theorem. Marks: 10.

• Path: γ(t) =

R cos tR sin t

0

, t from 0 to 2π.

• γ̇(t) =

−R sin tR cos t

0

• Put the path into the field

~A(γ(t)) =a2Bz

2

1

R2

−R sin tR cos t

0

• The integral

I =a2Bz

2

2πˆ

0

1

R2

−R sin tR cos t

0

·

−R sin tR cos t

0

dt

=a2Bz

2

2πˆ

0

1 dt

= πa2Bz

iii. Given part (a), the non-zero result that you should have got in part (b) seems at firstsight to contradict Stokes’ theorem. Why does it not contradict Stokes’ theorem afterall? Marks: 5.It is not a simply connected region, because x = y = 0 is missing.

iv. What is the magnetic field inside the solenoid (you can use your knowledge that the fieldinside a solenoid is constant)? Marks: 5.

Taking a cross section through the solenoid (so the surface element ~S points upwards inthe z-direction), the flux is

¨

B1

B2

B3

· d~S = πa2Bz

B3 = Bz

For symmetry reasons, B1 = B2 = 0, and hence

~Binside =

00Bz

v. Find the magnetic vector potential for the entire field. Make sure that it is continuous -so your task is to find a vector potential (through making sensible guesses that can then

be refined) which gives you the correct field inside, and has the same value for ~A wherethe two potentials meet meet at r = a. You can assume that the walls of the solenoidare infinitely thin. Marks: 6.


So at x2+y2 = a2 we want Ainside(x2+y2 = a2) = Aoutside(x2+y2 = a2) = Bz2

−yx0

and in fact, this is already the answer:

Ainside =Bz2

−yx0

We can try it out:

~∇× ~A =Bz2

∂0∂y − ∂x

∂z

− ∂0∂x + ∂(−y)

∂z∂x∂x −

∂(−y)∂y

=Bz2

002

=

00Bz

as required.

4) Show that if ~P and ~Q are both gradient fields, i.e. ~P = ~∇φ and ~Q = ~∇ψ, then

~R ≡ ~P × ~Q

has no sources. Marks: 6.

~∇ ·(~P × ~Q

)= ~∇ ·

(~P × ~Q

)

= ~∇ ·(↓~P × ~Q

)+ ~∇ ·

(~P×

↓~Q

)

= ~Q ·(~∇×

↓~P

)− ~P ·

(~∇×

↓~Q

)

= ~Q ·(~∇× ~P

)− ~P ·

(~∇× ~Q

)

= ~Q ·~0− ~P ·~0 = 0

Chapter 2

Practice Problems Week 1

2.1 Math basics

2.1.1 Derivatives

(a) f(x) = (cosx− sinx) · ex. Show that dfdx = −2 sinx ex

Use the product rule: (uv)′ = (u′)v + u(v′) with u = (cosx− sinx) and v = ex. Thederivatives of u and v:u′ = (− sinx− cosx) and v′ = ex. Plugging this in:dfdx = (− sinx− cosx) ex + (cosx− sinx) ex = −2 sinx ex.

(b) g(t) = 2eat arcsin t. What is dgdt ?

2eax(a · arcsinx+ 1√

1−x2

)

(c) g(t) = 2eat arcsin t. What is dgdt (0)?

2

(d) g(a) = 2eat arcsin t. What is dgda?

2teat arcsin t

(e) y(θ) = ln θ√θ

What is dydθ ?

2−ln θ2θ√θ

(f) y(x) = ln√

4x− x2. Calculate y′.

y = ln(4x− x2

) 12 = 1

2 ln(4x− x2

)

y′ = 2−x4x−x2

2.1.2 Integrals

Recall the difference between indefinite integrals (=reverse differentiation) and definite integrals:

• Calculate the indefinite integral´x dx.

12x

2 + C. Note that the notation´f(x)dx sometimes means the indefinite integral / anti-

derivative of f(x), and sometimes means the (definite) integral over all values, i.e. usually−∞ and +∞ (for angles 0, . . . , 2π).

35

36 CHAPTER 2. PRACTICE PROBLEMS WEEK 1

• Calculateb́

a

x dx.

12b

2 − 12a

2 (definite integral)

• Calculate2́

0

x dx.

2 (definite integral)

(a) Calculate the indefinite integral I =´

(−2 sinx ex) dx.Follows directly from question (a) on derivatives: I = (cosx− sinx) · ex + C

(b) Calculate1́

0

x(1+x2)2

dx Hint: Substitute u = 1 + x2.

dudx = 2x, dx = du

2x

I =

1ˆ

0

x

(1 + x2)2

=

2ˆ

u=1

x

u2

du

2x

=1

2

2ˆ

1

1

u2

=1

2

[− 1

u

]2

1

=1

4

(c) Calculate´

e2x

1+ex dx. (Hint: substitute u = 1 + ex)dudx = ex, dx = 1

ex dx

With this: I =éx

u du =ú−1u du =

´ (1− 1

u

)du = u− ln |u|+ C

Substitute back: u = 1 + ex

I = (1 + ex)− ln |1 + ex|+ C = ex − ln (1 + ex) +K with K = 1 + C

(d) Calculate I =´

(1 + 2x) e−x dx. Hint: Use integration by parts:ú ·v′ dx = u ·v−

ú′ ·v dx

Use u = (1 + 2x), v′ = e−x, which gives v = −e−x and u′ = 2. HenceI = −

´(1 + 2x) e−x dx = (1 + 2x)(−e−x)−

´2(−e−x)dx = −(3 + 2x) · e−x + C.

(e) Calculate I =1́

x=0

−x+1ý=0

x2y2dy dx

2.2. STATISTICS 37

Step one: integrate inner integral (y):

−x+1ˆ

y=0

x2y2dy = x2

−x+1ˆ

y=0

y2dy

= x2

[1

3y3

]−x+1

y=0

= x2

(1

3(−x+ 1)

3 − 0

)

=1

3x2 (−x+ 1)

3

=1

3

(−x5 + 3x4 − 3x3 + x2

)

Now the outer integral (x):

I =1

3

1ˆ

0

(−x5 + 3x4 − 3x3 + x2

)

=1

3

ˆ [1

6(−x6) +

1

5· 3x5 1

4· (−3)x4 +

1

3· x3

]1

0

=1

180

2.2 Statistics

(a) Calculate the mean, variance and standard deviation of the following histogram:

-3 -2 -1 0 1 2 30

1

2

3

4

5

6

7

gaussgauss


• mean:

x =1

Nevents

Nevents∑

i=0

xi (for individual events with values {xi})

≈ 1

Nevents

Nbins∑

j=0

fj · xbin j(for binned data, where fj is the number of events inthe bin with center xbin j)

=1

10· (7 · 0 + 3 · (−1))

= −0.3

• variance:

V =1

Nevents

Nevents∑

i=0

(xi − x)2 (for individual events with values {xi})

≈ 1

Nevents

Nbins∑

j=0

fj · (xbin j − x)2 (for binned data, where fj is the number of events inthe bin with center xbin j)

=1

10·(7 · ((−0.3)− 0)2 + 3 · ((−0.3)− (−1))2

)

= 0.21

• standard deviation:

σ =√V = 0.46

(b) (Supplementary) What is the 3rd central moment of the histogram in the previous question?What would it be for a completely symmetric distribution?

c3 =1

Nevents

Nevents∑

i=0

(xi − x)3 (for individual events with values {xi}

≈ 1

Nevents

Nbins∑

j=0

fj · (xbin j − x)3 (for binned data, where fj is the number of events inthe bin with center xbin j)

=1

10·(7 · (0− (−0.3))3 + 3 · ((−1)− (−0.3))3

)

=1

10·(7 · (0.3)3 + 3 · (−0.7)3

)

= −0.084

For a symmetric distribution, all odd moments (including this one) would be 0.

(c) Estimate the standard deviation σ of the following histogram, assuming it is roughly Gaussian(as many distributions in practice are), and using

FWHM ≈ 2.35σ.

2.2. STATISTICS 39

70 80 90 100 110 120 1300

50

100

150

200

250

300

gaussgauss

Generated with σ = 10.

(d) (Supplementary) A person is driving a car to a location and back. On the outward journey,the person averages 20 miles per gallon, on the way back this becomes 30 miles per gallon.Calculate the harmonic mean of the two numbers, and show that this is the correct numberto use if you want to know the miles per gallon for the combined trip.

• Length of each trip: s

• miles per gallon uphill gu

• miles per gallon downhill gd

• total gallons: gt = s/gu + s/gd

• total miles per gallon

2s/gt = 2s/(s/gu + s/gd) =1

12

(1gu

+ 1gd

)

• so it is the harmonic mean of the mpg for each trip.

(e) (Supplementary) In the question above, what would have been the appropriate kind of averageif you had measured the performance of the car in litres per 100 kilometres rather than milesper gallon?

• l = liters per km (you can convert this into liters per 100 km easily, if you want).

• total liters: lt = slu + sld (now s is in kilometers)

• total liters per km s(lu + ld)/(2s) = 12 (lu + ld)

• ... so for that it’s the arithmetic mean.

Chapter 3


3.1 Theoretical Distributions

3.1.1 Coins

Level: EasyRecall the distribution of obtaining NH heads when flipping a coin 4 times? If you can’t, look itup in the lecture notes.

P (NH) =

(1

2

)4(4NH

)

P (0) = 1/16

P (1) = 1/4

P (2) = 3/8

P (3) = 1/4

P (4) = 1/16

(a) We call the mean of a predicted or theoretical distribution the “expectation value”. What isthe expectation value 〈NH〉 and the standard deviation σNH , of the above distribution?Note that the factor 1∑

P (NH i)that appeared in the original answer is superfluous (but not

wrong), because our probability function is already normalised, so that∑

all possible NH i

P (NH i) =

1. Check this yourself.Mean:

〈NH〉 =∑

P (NH i)NH i

=1

1

∑P (NH i)NH i

=1

1

(0 · 1

16+ 1 · 1

4+ 2 · 3

8+ 3 · 1

4+ 4 · 1

16

)

= 2

41


Variance:

V =∑

P (NH i) (NH i − 〈NH〉)2

=1

1

∑P (NH i) (NH i − 〈NH〉)2

= (0− 2)2 · 1

16+ (1− 2)2 · 1

4+ (2− 2)2 · 3

8+ (3− 2)2 · 1

4+ (4− 2)2 · 1

16= 1

Standard deviation σ =√V = 1. Also possible to use

V =⟨N2H

⟩− 〈NH〉2

Try this yourself.

3.1.2 PDFs

Even PDF’s

Level: Easy-MediumShow that the expectation value of any even PDF is 0. An even function is one that has f(x) =f(−x).

〈x〉 =

∞̂

−∞

x f(x)dx

=

0ˆ

−∞

x f(x)dx+

∞̂

0

x f(x)dx

in first integral, u = −x

= −0ˆ

∞

(−u) f(−u)du+

∞̂

0

x f(x)dx

=

0ˆ

∞

u f(−u)du+

∞̂

0

x f(x)dx

=

0ˆ

∞

u f(u)du+

∞̂

0

x f(x)dx

= −∞̂

0

u f(u)du+

∞̂

0

x f(x)dx

= −∞̂

0

x f(x)dx+

∞̂

0

x f(x)dx

= 0 QED

3.1. THEORETICAL DISTRIBUTIONS 43

Flat PDF’s

Level: MediumShow that the standard deviation of a flat distribution is

σflat =width√

12

Hints: Your distribution is:

F (x) =

{1/(b− a) if x ∈ [a, b]0 else

}

with width = (b − a). Calculate Vx =⟨x2⟩− 〈x〉2. It’s obvious what 〈x〉 should be, but do check

that you can calculate it.

〈x〉 =

ˆxF (x)dx

=

bˆ

a

x/(b− a)dx

=

[1

2x2

]b

a

/(b− a)

=1

2

b2 − a2

b− a=

1

2

(b− a)(b+ a)

b− a=

1

2(a+ b)

as expected. Now⟨x2⟩

⟨x2⟩

=

ˆx2F (x)dx

=

bˆ

a

x2/(b− a)dx

=

[1

3x3

]b

a

/(b− a)

=1

3

b3 − a3

b− a

=1

3

(b− a)3 − 3(ba2 − b2a)

b− a

=1

3

(b− a)3 − 3ba(a− b)b− a

=1

3

((b− a)2 + 3ab

)


and finally V

⟨x2⟩− 〈x〉2 =

1

3(b− a)2 + ab− 1

4(a+ b)2

=1

3(b− a)2 + ab− 1

4(a+ b)2

=1

3(a2 + b2)− 2

3ab+ ab− 1

4(a2 + b2)− 1

2ab

=1

12(a2 + b2)− 1

6ab

=1

12(a− b)2

QED

Decay

Level: MediumThe probability density for an unstable nucleus that exits at time t0 = 0 to decay at a later time tis given by

P (t) =1

τe−t/τ

(a) Show that this distribution is properly normalised. Keep in mind that the particle cannothave decayed in the past, so P (t) = 0 for t < 0.What needs to be shown is:

+∞ˆ

−∞

P (t) dt = 1

Keeping in mind that P (t) = 0 for t < 0 this is:

+∞ˆ

0

1

τe−t/τ dt = 1

Calculating the integral:

+∞ˆ

0

1

τe−t/τ dt =

1

τ

[−1

τe−t/τ

]∞

0

= −(e−∞ − e0

)

= −(0− 1)

= 1 QED

(b) What is the cumulative probability distribution for this probability density?Same calculation as above, except that you replace the upper limit with some variable, say t0

(in fact, it would have been more efficient to do this calculation first and then let t0 go to ∞


to answer the previous question):

F (t0) ≡t0ˆ

−∞

P (t) dt

=

t0ˆ

0

1

τe−t/τ dt

=1

τ

[−1

τe−t/τ

]t0

0

= −(e−t0/τ − e0

)

= 1− e−t0/τ

(c) What is the expectation value and the standard deviation of P (t)?

• Expectation value:

〈t〉 =

t0ˆ

0

t1

τe−t/τ dt

There are now several ways of solving this. I’ll show two. First: standard method usingpartial integration:

〈t〉 =1

τ

∞̂

0

te−t/τ dt

=1

τ

[t(−τ)e−t/τ

]∞0− 1

τ

∞̂

0

(−τ)e−t/τ dt

=1

τ

(0 + τ

[(−τ)e−t/τ

]∞0

)

=1

τ

(−τ2 (0− 1)

)

= τ


Alternative way: Define −1τ = a. Then

〈t〉 =1

τ

∞̂

0

te−t/τ dt

=1

τ

∞̂

0

teat dt

=1

τ

d

da

∞̂

0

eat dt

=1

τ

d

da

[1

aeat]∞

0

=1

τ

d

da

1

a(0− 1)

= −1

τ

d

da

1

a

=1

τ

1

a2

=1

ττ2

= τ

• Expectation value of x2, i.e.⟨x2⟩

〈t〉 =

t0ˆ

0

t21

τe−t/τ dt

Using partial integration:

〈t〉 =1

τ

∞̂

0

t2e−t/τ dt

=1

τ

[t2(−τ)e−t/τ

]∞0− 1

τ

∞̂

0

(−τ)2te−t/τ dt

= 0 + 2τ1

τ

∞̂

0

te−t/τ dt

using previous result for integral:

= 2ττ = 2τ2


Alternative way: Define −1τ = a. Then

〈t〉 =1

τ

∞̂

0

t2e−t/τ dt

=1

τ

∞̂

0

t2eat dt

= −1

τ

d2

da2

1

a

=1

τ

d

da

1

a2

=1

τ

d

da

−2

a3

=1

τ(−2)(−τ)3

= 2τ2

• So the variance is V =⟨t2⟩− 〈t〉2 = τ2 and the standard deviation σ =

√V = τ .

(d) Find the co-ordinate transformation that transforms a flat distribution between 0 and 1 (which is the PDF in question 3.1.2 with a = 0, b = 1) to the above exponential. If we call theflatly-distributed random variable x, you want to find t(x) such that

P (t) dt = P (x) dx

where P (t) = 1τ e−t/τ and P (x) = 1 within 0, 1 and P (x) = 0 elsewhere. The way to solve

this is to first find the inverse transformation, x(t), and then solve this for t to find what youreally want, t(x). The first step is to put in P (t) and P (x) and re-arrange things a bit.

dx

dt=

1

τe−t/τ

You should be able to take it from here. This will give you x(t). Now solve for t to get t(x).Use your integration constants wisely - so that you map a distribution between 0 and 1 to adistribution between 0 and ∞.

dx

dt=

1

τe−t/τ

x(t) = −e−t/τ + C

C − x = e−t/τ

log(C − x)) = − tτ

t(x) = −τ log(C − x)

This is my co-ordinate transformation, except that I still have the undetermined integrationconstant C in there. This is determined by the fact that I want t to be between 0 and ∞,therefore C−x must be between 0 and 1. Given that x varies between 0 and 1, for C−x ∈ [0, 1]we require C = 1. So finally we get:

t(x) = −τ log(1− x)

You can try it out based on your random number generator you used for the assignment - foryour flatly-distributed x, just calculate −τ log(1− x) for some value of τ and see how this isdistributed.


Gaussian

Level: Medium to HardThe Gaussian PDF is given by:

g(x;µ, σ) =1√2πσ

e−(x−µ)2

2σ2

This notation means that g(x;µ, σ) is a function of x, with the parameters µ, σ. You could alsosimply write g(x).

You can assume that this is properly normalised, i.e.+∞́

−∞g(x) dx = 1 (you might or might not need

this information).

(a) Show that all integrals over g(x;µ, σ) can be expressed in terms of the gaussian with µ = 0and σ = 1:

bˆ

a

1√2πσ

e−(x−µ)2

2σ2 dx =

(b−µ)/σˆ

(a−µ)/σ

1√2πe−

u2

2 du

simple co-ordinate transformation u = x−µσ .

(b) Show that its mean is µ. Hint: Use the result from previous question and an appropriateco-ordinate transformation.

∞̂

−∞

x1√2πσ

e−(x−µ)2

2σ2 dx =

∞̂

−∞

(σ · u+ µ)1√2πe−

u2

2 du = 0 + µ = µ

(c) Advanced question: Show that the standard deviation of the Gaussian is (as the variablename suggests) σ

V =⟨x2⟩− 〈x〉2. We know 〈x〉. Now calculate

⟨x2⟩:

⟨x2⟩

=

∞̂

−∞

x2 1√2πσ

e−(x−µ)2

2σ2 dx

Now do co-ordinate transformation to make it look easier:

u =x− µσ

dx = σ du

With this:

⟨x2⟩

=1√2π

∞̂

−∞

(σ · u+ µ)2e−u2

2

=1√2π

∞̂

−∞

(σ2u2 + 2σµu+ µ2)e−u2

2 du

Term by term:

•1√2π

∞̂

−∞

2σµue−u2

2 du

this vanishes as it is a odd function over a symmetric interval.


•1√2π

∞̂

−∞

µ2e−u2

2 du = µ2 1√2π

∞̂

−∞

e−u2

2 du = µ2

due to normalisation.

• To solve the remaining term we use a trick. We introduce a dummy parameter a andexpress the integral as the derivative of a at a = 1:

d

da

∣∣∣a=1

(−2e−au2

2 ) = u2e−u2

2

Where the dda

∣∣∣a=1

means “first calculate the derivative, and then evaluate it at a = 1.”

With this:

1√2πσ2

∞̂

−∞

u2e−u2

2 du = −21√2πσ2 d

da

∣∣∣a=1

∞̂

−∞

e−au2

2 du

coordinate transformation: v =√au

= −21√2πσ2 d

da

∣∣∣a=1

1√a

∞̂

−∞

e−v2

2 dv

= −2σ2 d

da

∣∣∣a=1

1√a

1

= −2σ2(−1

2)

= σ2

So we have ⟨x2⟩

= σ2 + µ2

and consequentlyV =

⟨x2⟩− 〈x〉2 = σ2 + µ2 − µ2 = σ2

and σ =√V = σ, as expected.

Convolution

Level: Hard.(Supplementary) Note: This question could help you with (but is not required for) assessed prob-lems. Consider the random variable z = x+ y, where x is distributed according to the PDF Px andy is distributed according to the PDF Py. The PDF of z is given by:

Pz(z) =

∞̂

−∞

Px(x)Py(z − x)dx =

∞̂

−∞

Py(x)Px(z − x)dx

If x and y are both distributed according to a flat distribution between 0 and 1 (see question 3.1.2),what is Pz?

Pz(z) =

∞̂

−∞

Px(x)Py(z − x)dx

=

1ˆ

0

Py(z − x)dx


• For z ≤ 0 P (z) = 0

• for z ≥ 2 P (z) = 0

• Else: u=z-x x=z-u

Pz(z) =

1ˆ

0

Py(z − x)dx

= −z−1ˆ

z

Py(u)du

=

zˆ

z−1

Py(u)du

=

min(z,1)ˆ

max(z−1,0)

Py(u)du

The result depends on the explicit form of the integration limits, which are ulower = max(z −1, 0) and uupper = min(z, 1).

– For z < 0, the integration limits are ulower = uupper = 0 and hence Pz(z) = 0

– for z > 1, the integration limits are ulower = uupper = 1 and hence also Pz(z) = 0

– For 0 < z < 1: ulower = 0, uupper = z and therefore Pz(z) = z

– For 1 < z < 2: ulower = z, uupper = 1 and therefore Pz(z) = 1− z.

So the solution is a triangle that peaks at P (z = 1) = 1.

3.1.3 Life Insurance

Level: Easy-Medium once you’ve done the Binomial distribution (Wed)

If the probability that any person thirty years old will be dead within a year is p = 0.01, find theprobability that out of a group of 10 such people, one will find after one year that:Clearly a case for the binomial distribution.

P (r) =

(nr

)pr(1− p)n−r =

n!

(n− r)! r!pr(1− p)n−r =

n · (n− 1) · . . . · (n− r + 1)

r!pr(1− p)n−r

with n = 10.You’ll find the the Poisson with λ = n · p = 0.1 will give you a decent approximation.

(a) none have died

P (0) = (1− p)10 = 90.44%

(b) exactly one will have died

P (1) = n · p(1− p)n−1 = 9.04%


(c) not more than one will have died

P (0) + P (1) = 99.48%

(d) more than one will have died

1− (P (0) + P (1)) = 0.52%

(e) at least one with have died

1− P (0) = 9.56%

3.1.4 Parity Bit

Level: Medium (once you’ve done the binomial distribution)If the probability that a single bit (which can have the values “I” or “0”) is transmitted wrongly(i.e. what is sent as “I” arrives as “0” and vice versa) over the internet is p = 10−6 (This is a madeup probability!).

(a) What is the probability that an ASCII character of 7 bits will have at least one wrong bit?Binomial:

P (r) =

(nr

)pr(1−p)n−r =

n!

(n− r)! r!pr(1−p)n−r =

n · (n− 1) · . . . · (n− r + 1)

r!pr(1−p)n−r

with n = 7, p = 10−6

P (at least 1) = 1− P (0) = 1− (1− p)7 = 7.00 · 10−6

If your calculator gave you 0, it hasn’t got enough significant digits for this calculation. Thenuse

P (at least 1) = 1− P (0) = 1.0− (1− p)7 = 1− (1− 7p+O(p2)) ≈ 7p = 7.00 · 10−6

where we’ve used that p� 1

(b) How many characters do you expect to be wrong in a 1MB = 220bytes of text? (1 byte 6= 1bit)

• Single byte: Pbyte = 7 · 10−6

• 220 bytes:〈Nwrong〉 = 220 · 7 · 10−6 = 7.3

(c) One way to reduce the error rate in such transmissions is the so-called parity bit. The senderadds to each 7-bit ASCII character one more bit. This is done such that the over-all number ofbits sent to “I” is even. So the 7-bit character “0III000” would have a parity bit of “I” added,giving “0III000I”, while “0II0000” would have a “0” added, giving “0II00000”. The recipientwill check the number of “I” bits and request re-transmission if it finds an odd number. Thismethod will find single-bit failures. How many wrong characters to you expect in a 1MB text,now?


Now we have again a binomial distribution, however instead of n = 7 we have n = 8. Thismethod can detect single bit failures (and any odd number of bit failures), but not 2-bit failures(or any even number of bit failures). We can safely ignore any larger contribution from higher-order terms such as three or four bit failures since they are suppressed relative to the “leadingorder” term of 2-bit failures by a factor of p = 7 · 10−6, so they are a correction somewhere inthe 5th significant digit of our result. (For the same reason we don’t have to worry about thedetails of re-transmission, we can safely assume that a character once identified as wronglytransmitted, is correctly transmitted the next time - the correction due to this relative to thenumber we will calculate is again less than 1/100,000, so somewhere in the 5th significantdigit)

• So, while in principle we would calculate

P (2, 4, 6, 8 out of 8 bits wrong; for n=8)

• It is really sufficient to calculate

P (2 bits wrong; n=8) =

(82

)p2(1− p)6 = 2.80 · 10−11

• So now the expected number of wrong bytes in a 220 byte text is:

〈Nwrong〉 = 2.9 · 10−5

So it’s a huge effect - it turns something quite useless (any major text - and any long computerprogramme - would have errors in it) to something that’s quite OK. I suspect that even theerror rate we calculated with the parity bit is still unacceptably high and the actual error rateis much lower.

3.2 Extra Statistics Questions

These are for additional practice

3.2.1 Probabilities

(a) Suppose you throw two dice, what is the probability that the dice add up to 4 given that thefirst shows a 2?If the first die shows a 2, then for the dice to add up to 4 the second die must also show a 2.

So the probability that the dice add up to 4 given that the first shows a 2 is 1/6. (Q& A fromhttp://philosophy.hku.hk/think/stat)

(b) Suppose you throw two dice, what is the probability that the dice add up to 4 and that thefirst shows a 2?Note that we cannot use the simplified multiplication rule here, because the dice adding up

to 4 is not independent of the first die showing a 2. So we need to use the full multiplicationrule. This tells us that probability that the first die shows a 2 and the dice add up to 4 isgiven by the probability that the first die shows a 2, multiplied by the probability that thedice add up to 4 given that the first die shows a 2. This is 1/6 ×1/6 = 1/36. (Q&A fromhttp://philosophy.hku.hk/think/stat)

3.2. EXTRA STATISTICS QUESTIONS 53

3.2.2 Combinatorics

(a) (Supplementary) The Zimbabwean Cabinet has 30 ministries to be divided between threeparties: Zanu-PF (14), Main MDC (13), and MDC (Mutambara) (3).

i. How many ways are there to allocate the various ministries to the different parties, giventhese constraints?How many ways are there to put 30 things (ministries) into 3 boxes (parties) each withn1 = 14, n2 = 13, n3 = 3 places?

N =n!

n1!n2!n3!=

30!

14! · 13! · 3!= 8.14 · 1010

ii. If neither Robert Mugabe’s Zanu-PF nor the opposition (Main MDC and MDC (Mu-tambara)) are allowed to have both the defence and home affairs (which includes police)ministries, how many ways are there now to divide the ministries between the parties?

• Number of ways in which Mugabe’s Zanu PF can have both:

NRM =n!

n1!n2!n3!=

28!

12! · 13! · 3!= 1.7 · 1010

• Number of ways in which the Main MDC can have both:

NMDC−main =n!

n1!n2!n3!=

28!

14! · 11! · 3!= 1.5 · 1010

• Number of ways in which the MDC-Mutaba can have both:

NMDC−mutaba =n!

n1!n2!n3!=

28!

14! · 13! · 1!= 5.6 · 108

• Number of ways in which the Main MDC and MDC-Mutaba can each have exactlyone

NMDC−shared = 2 · n!

n1!n2!n3!=

28!

14! · 12! · 2!= 3.7 · 109

• So, total:

Nconstraint = N −NRM −NMDC−main−NMDC−mutaba−NMDC−shared = 4.6 · 1010

3.2.3 Molecules (Supplementary)

Level: Medium-Hard

i. According to the equipartition theorem, the average kinetic Energy of gas molecules attemperature T is Ekin =32kBT(Look up the equipartition theorem if necessary and don’t forget that the movement ofthe particle corresponds to 3 degrees of freedom. Also never forget the equipartitiontheorem again). What is the mean speed squared, |v|2, for a given molecular mass m?3kTm


ii. From Maxwell’s speed distribution, we know that the mean speed of a molecule at

temperature T is |v| =√

8kBTπm . Compare this to the previous result. Why is it OK to

calculate the |v|2 from the kinetic energy Ekin, while it is not OK to calculate |v| fromit?Because the mean is “linear”, i.e. it behaves like this:

λx+ µy = λx+ µy

where λ and µ are fixed numbers, and x and y represent data from two different datasets. This of course also implies that

λx = λx.

There is a linear relationship between E and v2 (which is E = 12m(v2), i.e. in the

formalism above, λ = 12m and x = v2). Therefore

E =1

2mv2

However, the relationship between v and v2 is v =√v2, which is not linear. Hence, in

general v 6=√v2 (of course one cannot exclude the possibility that by pure chance the

two numbers turn out to be the same).

iii. Show that if, and only if the fluctuations about the mean |v|2 are small, then |v| ≈√|v|2.

Instructions:

A. Obviously, v =√v2. Express v2 as the |v|2 + ∆v2. Perform a Taylor expansion of

v =√|v|2 around |v|2 = |v|2 up to second order (i.e. get something that looks like

a+ b∆v2 + c(∆v2

)2)

Taylor expansion:

f(x0 + ∆x) = f(x0) +df

dx

∣∣∣∣∣x=x0

·∆x+1

2

d2f

dx2

∣∣∣∣∣x=x0

· (∆x)2 + . . .

Here we have x = v2 and x0 = x = v2, and f(x) =√x, and therefore

f(x) =√x (with x = v2)

df

dx=

1

2

1√x

d2f

dx2= −1

4

1

x3/2

Plugging it all in:

v =√x0 +

1

2√x0

∆x− 1

8 (x0)3/2

(∆x)2

=√x+

1

2√x

∆x− 1

8 (x)3/2

(∆x)2

=

√|v|2 +

1

2

√|v|2

∆|v|2 − 1

8(|v|2)3/2

(∆|v|2

)2

3.2. EXTRA STATISTICS QUESTIONS 55

B. Now take the mean of that. Note that ∆|v|2 is the 1st central moment of |v|2 - what

happens to that term? What does the (∆|v|2)2

term represent?

v =

√|v|2 +

1√|v|2

∆|v|2 − 1

8(|v|2)3/2

(∆|v|2)2

=

√|v|2 +

1√|v|2

∆|v|2 − 1

8(|v|2)3/2

(∆|v|2)2

Let’s look at each term with a bar over it in turn:

√v2 =

√v2 just a constant

∆|v|2 = |v|2 − |v|2

= |v|2 − |v|2= |v|2 − |v|2= 0

(∆|v|2)2

=(|v|2 − |v|2

)2

=1

N

N∑

i=1

(|v|2 − |v|2

)2

= V ar(|v|2) = σ2|v|2

Putting these results back in:

v ≈√|v|2 − 1

8(|v|2)3/2

(∆|v|2)2

=

√|v|2 − 1

8(|v|2)3/2

Var(|v|2)

=

√|v|2 − 1

8

√|v|2

Var(|v|2)

|v|2

C. Hence, if the Variance of |v|2 is small compared to |v|2, and only then, is |v| ≈√|v|2

an acceptable approximation.

Chapter 4


4.1 Vector Basics

4.1.1 What form is it?

We denote vectors with arrows on top, like this: ~A. Everything else is a scalar.

(a) Which of the expressions below is

• a vector

• a scalar (simply a number)

• nonsense?

Remember that the dot, ·, between two vectors is the dot-product. The same symbol betweenscalars is the usual product between to numbers, and between scalars and vectors it is normalscalar multiplication of vectors (which changes the length, but not the direction of vectors). Inthe latter two cases you can omit the dot - you cannot between vectors, to avoid confusion withthe cross product. The cross, ×, indicates the cross-product and only makes sense betweenvectors.

i. λ ~Avector

ii. ~A× ~Bvector

iii. ~A · ~Bscalar

iv. ~C ·(~A× ~B

)

scalar

v.(~C · ~A

)× ~B

nonsense

vi. ~C · ~A× ~B

scalar (This is interpreted as ~C ·(~A× ~B

))

vii.(~C · ~A

)~B

vector

57


viii. ~A× ~B × ~Cvector

ix. Why is the following statement nonsensical? ~A× ~B + ~A · ~BOne can only add vectors and vectors, as well as scalars and scalars, but never vectorsand scalars.

(b) Which of the below is zero (either the scalar-zero 0 or the zero vector ~0), which are non zero,

which can be either depending on the value of ~A and ~B. You should assume that neither ~Anor ~B themselves are ~0:

i. ~A · ~Anon-zero

ii. ~A× ~A~0

iii.(~A× ~B

)· ~A

It’s 0 because ~A × ~B is perpendicular to ~A and the dot-product between perpendicularvectors is 0. This can also be seen like this: the triple-dot product between three vectorsrepresents the volume spanned by the three. Here we only have two vectors - they alwayslie in one plane, so the volume is 0.

iv.(~A× ~B

)× ~A

Could go either way. Definitely not 0 if ~A× ~B 6= ~0.

v. λ ~A+ µ~B for µ 6= 0, λ 6= 0Can be either. Definitely non-zero only if ~A and ~B are linearly independent - but wedon’t know that.

vi. ~A× ~B +(~A · ~B

)~B Hint: The answer to this one is not “can be either”.

Cannot be 0 because

• ~A× ~B and ~B are linearly independent.

• Given that ~A and ~B are each non-zero, ~A × ~B can only be zero if ~A and ~B areparallel, while ~A · ~B can only be zero if they are perpendicular. Both cannot be thecase at the same time.

• So we have the sum of two linearly independent vectors, at least one of which isnon-zero, so the result is non-zero.

4.1.2 Vector operations

i. Calculate

123

×

321

−48−4

ii. Draw the vectors ~A =

120

and ~B =

210

. These “span” a parallelogram. What

is its area?

area =∣∣∣ ~A× ~B

∣∣∣ =

∣∣∣∣∣∣

00−3

∣∣∣∣∣∣

= 3

4.1. VECTOR BASICS 59

iii. Calculate ~A · ~B for ~A =

012

and ~B =

140

012

·

140

= 0 · 1 + 1 · 4 + 2 · 0 = 4

iv. Calculate the length of ~A and ~B in the previous questionslength = square-root of dot-product with itself:

∣∣∣ ~A∣∣∣ =√

0 · 0 + 1 · 1 + 2 cot 2 =√

5 ≈ 2.23

same for ~B: ∣∣∣ ~A∣∣∣ =√

1 · 1 + 4 · 4 + 0 cot 0 =√

17 ≈ 4.1

v. Calculate the angle between ~A and ~B from the previous two questions.Let’s call the angle θ. With this:

~A · ~B =∣∣∣ ~A∣∣∣∣∣∣ ~B∣∣∣ cos θ

Therefore

cos θ =22√5√

17= 0.43

andθ = 1.1

(which is 64o).

vi. What this the unit vector in the direction of

340

?

We divide the vector by it’s length to get the unit vector:

~u =

340

1∣∣∣∣∣∣

340

∣∣∣∣∣∣

where ∣∣∣∣∣∣

340

∣∣∣∣∣∣

=

√√√√√

340

·

340

=

√25 = 5

So

~u =

3/54/50

vii. How much work is done when applying a constant force ~F =

123

N to move an

object in a straight line from point ~A =

101112

m to point ~B =

101113

m?


Work = the force in direction of path times the path length. Usually this requiresintegration but since the path is straight and the force is constant, we can just multiplythings together:

W = ~F ·(~B − ~A

)=

123

N ·

001

m = 3J

4.2 Other Math Basics

(a) (First Year Revision) Use a Taylor/McLauren expansion to show that

1

1 + x≈ 1− x+ x2

f(x) =

∞∑

n=0

1

n!

dnf(x)

dxn

∣∣∣x=0

xn ≈ f(0) +df(x)

dx

∣∣∣x=0

x +1

2

d2f(x)

dx2

∣∣∣x=0

x2

with dfdx = −1

(1+x)2 and with d2fdx2 = 2

(1+x)3

f(x) ≈ f(0) + (−1)·x+1

2· 1 · x2 = 1− x+ x2

(b) (First Year Revision) Use a Taylor/McLauren expansion to show that, for values of x near 1,

ln(x) ≈ −3

2+ 2x− 1

2x2

Hint: Expand around x0 = 1.

f(x) =

∞∑

n=0

1

n!

dnf(x)

dxn

∣∣∣x=x0

(x−x0)n ≈ f(x0) +df(x)

dx

∣∣∣x=x0

(x−x0)2 +1

2

d2f(x)

dx2

∣∣∣x=x0

(x−x0)2

with dfdx = 1

x , d2fdx2 = −1

x2 ,

ln(x) ≈ (x− 1)− 1

2(x− 1)2 = x− 1− 1

2x2 − 1

2+ x =

−3

2+ 2x− 1

2x2

(c) Calculate the following integral

¨

unit disk

(x2 + y2)dx dy

over the unit disk. Hint: Think carefully about the integration limits, especially if you do itin cartesian co-ordinates. It is easier to change the co-ordinate system to one that is more

4.3. STATISTICS 61

suitable for a disk.Polar co-ordinates: dx dy = r dr dφ.

¨

unit disk

(x2 + y2)dx dy =

¨

unit disk

r3dr dφ

=

2πˆ

0

1ˆ

0

r3dr dφ

=

2πˆ

0

1

4[r3]10 dφ

=

2πˆ

0

1

41̇ dφ

=1

4[φ]2π0

=π

2

4.3 Statistics

4.3.1 Combining Errors (Supplementary)

(a) (Supplementary) If we have measurements of the mass and velocity of an object with un-certainties σm and σv respectively, assuming the measurements are uncorrelated, what’s theuncertainty on the object’s kinetic energy?

E =1

2mv2

∂E

∂m=

1

2v2

∂E

∂v= mv

σ2E =

1

4v4σ2

m +m2v2σ2v

σE =

√1

4v4σ2

m +m2v2σ2v

(b) (Supplementary) In a particle detector we can measure the speed of a particle using a RICHdetector, and the momentum of a particle by looking how it bends in a magnetic field. Wecan combine the two to extract its mass (ignoring relativistic effects, thus p = mv). Thecorrelation coefficient between speed and momentum errors is ρvm. What then is the erroron the mass?


m =p

v∂m

∂p=

1

v

∂m

∂v= − p

v2

σ2m =

1

v2σ2v +

p2

v4σ2p + 2ρvmσvσm

σm =

√1

v2σ2v +

p2

v4σ2p−2

p

v3ρvmσvσm

4.3.2 2 (Supplementary)

centre of bin x1 = 1 x2 = 2 x3 = 3 x4 = 4 x5 = 5Number of events Ni 7 13 32 43 56

(Supplementary) For the above data, all the bins have the same width. Fit a straight line goingthrough the origin to the data given above, i.e. you fit:

f(x) = ax

(a) Find the best fit for a? Here is a step by step guide:

i. Write down the expression for the χ2. Use σi ≈√N (rather than σi =

√f(xi)), which

makes things significantly easier.

n∑

i=1

(f(xi)−Ni)2/Ni =

n∑

i=1

(axi −Ni)2/Ni

where n is the number of bins, here, n = 5.

ii. To find the minimum χ2, differentiate the above answer with respect to a, set this to 0and solve for a

0 =d(χ2)

da

=d

da

n∑

i=1

(axi −Ni)2/Ni

=

n∑

i=1

2xi (axi −Ni) /Ni

0 =

n∑

i=1

(ax2i

Ni− xi

)

∑xi = a

n∑

i=1

x2i

Ni

a =

∑xi

n∑i=1

x2i

Ni

=15

1.55= 9.7

4.4. ADDITIONAL QUESTIONS 63

(b) What is the χ2 of your fit. What is the number of degrees of freedom? Do you think a straightline through the origin describes the data well?

χ2 =

n∑

i=1

(f(xi)−Ni)2/Ni = 1.02 + 3.10 + 0.28 + 0.43 + 1.04 = 5.87

Number of degrees of freedom = number of bins minus number of fit parameters = 4. Soχ2/ndof = 1.47. So this somewhat larger than the expected value of ≈ 1, but there is also asmall number of degrees of freedom, so larger fluctuations are to be expected. If you look upa tabulated (or electronic) χ2 probability table, you’ll find that the probability to get a χ2 thisbad or worse for 4 bins is 21%, so this is very reasonable and no cause for concern. Also, thisdata was generated according to the function we used to fit it (f(x) = ax), with a = 10.

4.4 Additional Questions

4.4.1 CLT

A random variable X is the sum of N independent random variables yi with i = 1, . . . , N :

X =

N∑

i=1

yi

(a) Prove that

〈X〉 =∑〈yi〉

(use the rules for the linearity of expectation values)

X =∑

i

yi

〈X〉 =

⟨∑

i

yi

⟩

〈X〉 =∑

i

〈yi〉

(b) (Relevant, but beyond exam level) Prove that the standard deviation of X is given by

σx =√∑

σ2(yi)

where σ(yi) is the standard deviation of the random variable yi. Hint: Start with

V (X) =⟨

(X − 〈X〉)2⟩


V (X) =⟨

(X − 〈X〉)2⟩

=

⟨(∑

i

yi −∑

i

〈yi〉)2⟩

=

⟨(∑

i

(yi − 〈yi〉))2⟩

=

⟨(∑

i

(yi − 〈yi〉))∑

j

(yj − 〈yj〉)

⟩

=

⟨∑

ij

(yi − 〈yi〉) ((yj − 〈yj〉))⟩

=∑

ij

〈 (yi − 〈yi〉) ((yj − 〈yj〉)) 〉

=∑

ij

cov(yi, yj)

We stated that all the yi are independent - so the covariance be-tween yi and yj vanishes, unless of course i = j:

=∑

i

cov(yi, yi)

The covariance of a parameter with itself is simply the variance

=∑

i

Vi

σX =√Vx =

√∑

i

σ2y i QED

4.4.2 Breit Wigner (Supplementary)

The Breit Wigner function describes the probability density of finding that a short-lived particlewith mean mass M0 and lifetime τ = ~

Γ c2 actually has the mass M .

BM0(M) =1

2π

Γ

(M −M0)2 + (Γ/2)2

An example for M0 = 10 and Γ = 1 (arbitrary units, usually one would use MeV or GeV), is givenbelow:

4.4. ADDITIONAL QUESTIONS 65

7 8 9 10 11 12 130

0.1

0.2

0.3

0.4

0.5

0.6

1/(2*TMath::Pi())*1/((10-x)^2+1/4)

(a) Show that the maximum of the Breit Wigner is at M = M0

dB

dM= − 1

2π

Γ(

(M −M0)2

+ (Γ/2)2)2 · 2 (M −M0)

dB

dM

∣∣∣∣∣M=Mmax

= 0

⇔ (M −M0) = 0

⇔Mmax = M0

We won’t bother with the 2nd derivative here, because it is pretty obvious that this is a maxi-mum.

(b) Show that the FWHM of the Breit Wigner is Γ

B(Mmax) =1

2π

Γ

Γ2/4

B(Mmax +1

2Γ) = B(Mmax −

1

2Γ) =

1

2π

Γ

Γ2/4 + Γ2/4=

1

2B(Mmax) QED

(c) Using that the expectation value of an even PDF is 0, and assuming that the Breit Wignerfunction as given is properly normalised, show that the expectation value of the Breit Wigneris M0.For clarity we can define,

A ≡ M0

BA(M) =1

2π

Γ

(M −A)2 + (Γ/2)2

B0(M) =1

2π

Γ

M2 + (Γ/2)2

With this notation, we want to show that 〈BA(M)〉 = A. Clearly, the function B0 defined


above is an even function of M , B0(M) = B0(−M).

〈M〉 =

∞̂

−∞

MBA(M)dM

=

∞̂

−∞

M1

2π

Γ

(M −A)2 + (Γ/2)2dM

change integration parameter: u = M −A

=

∞̂

−∞

(A− u)1

2π

Γ

(u)2 + (Γ/2)2du

= A

∞̂

−∞

1

2π

Γ

(u)2 + (Γ/2)2du−

∞̂

−∞

u1

2π

Γ

(u)2 + (Γ/2)2du

= A · 1 + 〈B0〉= A QED

(d) Show that the Breit-Wigner function has no standard deviation (the relevant integral di-verges).We know that 〈M〉 is well defined so we expect to find that that

⟨M2⟩

isn’t:

⟨M2⟩

=Γ

2π

ˆM2

(M −A)2 + (Γ/2)2

≥ Γ

2π

ˆM2

M2 + (Γ/2)2dM

=Γ

2π

ˆM2 + (Γ/2)2

M2 + (Γ/2)2dM − Γ

2π

ˆ(Γ/2)2

M2 + (Γ/2)2dM

=Γ

2π

∞̂

−∞

dM − (Γ/2)2

=Γ

2π· ”2 · ∞”− (Γ/2)2

= ”∞”

Chapter 5


5.1 Lines and Gradients

5.1.1 d~r

Calculate d~r and ds = |d~r| expressed in

(a) polar co-ordinates with the polar basis ~er, ~eφ

(b) cylindrical coordinates with the the cylindrical basis ~eρ, ~eφ, ~ez,

(c) spherical polar coordinates with the basis ~er, ~eθ, ~eφ,

Hint: in any orthonormal basis, ~u1, ~u2, ~u3, any vector ~A and be expressed as

~A =(~A · ~u1

)~u1+

(~A · ~u2

)~u2+

(~A · ~u3

)~u3

Recall the expressions for the different basis vectors from the lectures and use the formula above.

One way to approach this (there are others) is to first express d~r and ~er, . . . in the cartesian

basis vectors ~ı,~,~k and the appropriate curvilinear coordinates. Then calculate the dot products.Compare your result with those in the lecture notes.

d~r ds = |d~r| =∣∣d~rdt

∣∣ dt

polar dr ~er + r dφ~eφ

√(drdt

)2+ r2

(dφdt

)2

dt

cylindrical dρ~eρ + ρ dφ~eφ + dz ~ez

√(dρdt

)2

+ ρ2(dφdt

)2

+(dzdt

)2dt

spherical dr ~er + r dθ ~eθ + r sin θ dφ~eφ

√(drdt

)2+ r2

(dθdt

)2+ r2 sin2 θ

(dφdt

)2

dt

67


5.1.2 Multiple Choice Path & Field Guessing (Supplementary, but goodfor general understanding)

Consider the following vector fields and paths:1 2

A

B

A

B

3 4

AB B

(a) Which of the 8 closed paths is positive, negative, zero? Hint: three of the paths are zero(one might not be obvious).

i. Neither paths is zero.

ii. Neither path A or path B are zero.

iii. Both path A and path B are zero.

iv. the red path is not-zero. Path B is zero - not obvious, but you can check - any path thatdoes not enclose 0 (where the field is not defined) is zero, which is because ~∇× ~K = ~0,

5.1. LINES AND GRADIENTS 69

and this means for path on simply connected regions (i.e. they must not enclose the 0where the field is undefined), all closed line integrals are zero. Graphically, this can beseen as follows. The top and bottom of path B cancel. The left and right do to, but thisis not obvious, since the arrows are smaller at the right vertical part of the path than atthe left. However they are also more parallel to the path, enhancing this contributionsin exactly the right way. (As you certainly remember, the path integral is essentiallyan infinite sum of a dot products between field vectors and infinitly small path elements,so, since it’s a dot product, only the components of the field vectors parallel to the pathsegments count.) Of course you can’t see, just by looking at this path, that this is zero.You needed to use the information that there are exactly 3 zero paths. After removing thetwo obvious zero ones, amongst what’s left, you can deduce that this is the best candidate.

(b) Which field matches which plot?

~F =

(1− e−y/a

0

), ~G =

(−yx

), ~H =

(b0

), ~K =

1

x2 + y2

(−yx

)

where a > 0, b > 0

(a) ~F

(b) ~G

(c) ~H

(d) ~K

5.1.3 Three Paths

For

~F =

3x2 + 67y−14yz20xz2

Evaluate ˆ

γ

~F · d~s

from (0, 0, 0) to (1, 1, 1) along the following paths γ:

(a) x = t, y = t2, z = t3

• Step 1: Parameterise pathHere the path is already given in parameter form:

~γ(t) =

tt2

t3

Part of this step is also to find the boundaries of the integral given your parameterisation,i.e. over which values of t to integrate. We are supposed to integrate between (0, 0, 0)and (1, 1, 1), which corresponds to tmin = 0 and tmax = 1.


• Step 2: Vectorial line element

d~s =d~γ

dtdt =

12t3t2

dt

(Notation warning: Sometimes we use d~s with a small s, and sometimes d~r for the lineelement. Both are common notations and mean the same thing.)

• Step 3: Fill it all in

– Put the path into the field:

~F (~γ(t)) =

3(γx(t))2 + 67(γy(t))2

−14γy(t)γz(t)20γx(t)(γz(t))

2

=

70t2

−14t5

20t7

– Put this into the integral expression

ˆ

γ

~F · d~s =

tmaxˆ

tmin

~F (~γ(t)) · d~γdtdt

=

1ˆ

0

(70t2 · 1− 14t5 · 2t+ 20t7 · 3t2

)dt

=

1ˆ

0

(70t2 − 28t6 + 20t9

)dt

=70

3− 28

7+

20

10

= 251

3

(b) The straight lines from (0,0,0) to (1,0.0), then to (1,1,0), and then to (1,1,1).

• Step 1: Parameterise path. We now have a path that is the combination of 3 straightsections, that we’ll parameterise separately.

~γ1(t) =

t00

for 0 < t < 1

~γ2(t) =

1t0

for 0 < t < 1

~γ3(t) =

11t

for 0 < t < 1

• Step 2:

d~γ1

dtdt =

100

dt,

d~γ2

dtdt =

010

dt,

d~γ3

dtdt =

001

dt

• Step 3:


i. Put the path into the field:

~F (~γ1(t)) =

3t2

00

, ~F (~γ2(t)) =

3 + 67t00

, ~F (~γ3(t)) =

70−14t20t2

ii. All into (three) integrals. Total will be I = I1 + I2 + I3 with:

I1 =

ˆ

γ1

~F · d~s =

1ˆ

0

3t2

00

·

100

dt =

1ˆ

0

3t2 dt = 1

I2 =

ˆ

γ2

~F · d~s =

1ˆ

0

3 + 67t00

·

010

dt = 0

I3 =

ˆ

γ3

~F · d~s =

1ˆ

0

70−14t20t2

·

001

dt =

1ˆ

0

20t2 · 1 dt =20

3

So the result is ˆ

γ=γ1→γ2→γ3

~F · d~s = I1 + I2 + I3 =23

3

(c) The straight line joining (0, 0, 0) and (1, 1, 1).

• Step 1:

~γ(t) =

ttt

with 0 < t < 1

• Step 2:

d~s =d~γ

dtdt =

111

dt

• Step 3:

i. Put the path into the field:

~F (~γ(t)) =

3t2 + 67t−14t2

20t3

ii. All of the above into the integral:

ˆ

γ

~F · d~s =

1ˆ

0

3t2 + 67t−14t2

20t3

·

111

dt

=

1ˆ

0

(3t2 + 67t− 14t2 + 20t3

)dt

=

1ˆ

0

(67t− 11t2 + 20t3

)dt

=67

2− 11

3+

20

4=

418

12= 34

5

6


(d) Is this vector field conservative?No. If it were, all line integrals between the same two points would have given the sameanswer.

5.1.4 Conservative Fields

(a) i. Show that

~F =

2xy + z3

x2

3xz2

is a conservative force fieldIn Vector’s Lecture 3, slide 47, we learnt that the necessary and sufficient condition for

a conservative vector field is:

∂Fi∂xj

=∂Fj∂xi

for all i, j

on a simply connected region. In 3-D this is specifically:

∂Fz∂y

=∂Fy∂z

and∂Fx∂z

=∂Fz∂x

and∂Fx∂y

=∂Fy∂x

on a simply connected region in 3 dimensions. We now know that this condition can bewritten as

~∇× ~F = ~0

or, equivalently

~∇× ~F =

∂∂x∂∂y∂∂z

× ~F = ~0

on a simply connected region in 3D.

• Are we dealing with a simply connected region? Since ~F has no singularities (no valuefor which it is infinite or undefined, such a 1

x2 of log x which both are undefined atx = 0), it is defined on all space, without anything missing, thus the region is simplyconnected. You need to look out for entire lines or planes missing (single missingpoints would have been OK in 3 dimensions - see the discussion in slides 49-58 inVector Lecture 3).

• Let’s now calculate ~∇× ~F and see if it’s zero:

∂∂x∂∂y∂∂z

× ~F =

∂∂y (3xz2)− ∂

∂z (x2)

−(∂∂x (3xz2)− ∂

∂z (2xy + z3))

∂∂x (x2)− ∂

∂y (2xy + z3)

=

0− 0−(3z2 − (0 + 3z2)

)

2x− (2x+ 0)

=

000

So ~F is a gradient field and it has a potential.

ii. Find the scalar potentialUse the step-by-step guide on page 60 and 61 of the lecture notes for lecture 3.

• Step 1: Check if ~F has a potential. We’ve done that, we’ve shown that ~∇× ~F = ~0and that the region is simply connected.


• Step 2: Calculate a convenient line integral from some point ~r0 to an arbitrary point~r1:

V (~r1) = −ˆ

path from ~r0 to ~r1

~F · d~s

We’ll choose as our path the straight line between the origin, ~r0 =

000

, and

~r1 =

x1

y1

z1

.

– Path integral step 1: Parameterise Path:

~γ = (~r1 − ~r0) t = ~r1t,=

x1ty1tz1t

with 0 < t < 1

– Path integral step 2: d~s

d~s =d~γ

dtdt = ~r1 dt =

x1

y1

z1

dt

– Path integral step 3: Fill it all in

A. Put the path into the field ~F :

~F (~γ(t)) =

2x1ty1t+ (z1t)3

(x1t)2

3(x1t)(z1t)2

=

2x1y1t2 + z3

1t3

x21t

2

3x1z21t

3

B. Put it all into the integral

ˆ

γ

~F · d~s =

1ˆ

0

2x1y1t2 + z3

1t3

x21t

2

3x1z21t

3

·

x1

y1

z1

dt

=

1ˆ

0

((2x2

1y1t2 + x1z

31t

3) + (x21y1t

2) + (3x1z31t

3))dt

=2

3x2

1y1 +1

4x1z

31 +

1

3x2

1y1 +3

4x1z

31

= x21y1 + x1z

31

So our result is (not forgetting the minus sign):

V (x1, y1, z1) = −(x2

1y1 + x1z31

)

The subscript 1 was only added to avoid possible confusion with integration variables. We may nowdrop it:

V (x, y, z) = −(x2y + xz3

)

This is only determined up to a constant, so if C is a constant, then

V ′(x, y, z) = −(x2y + xz3

)+ C

is also a solution.


• Final Check: A highly recommended extra step: Check that the result is correct:

−gradV = −

∂∂x (−(x2y + xz3))∂∂y (−(x2y + xz3))∂∂z (−(x2y + xz3))

=

3xy + z3

x2

3xz2

= ~F (x, y, z)

So all is fine!

iii. (Supplementary) Find the work done, in arbitrary units, in moving an object in this fieldfrom (1,−2, 1) to (3, 1, 4)This is simply the difference in the potential at these two points:

V (from)− V (to) = V (1,−2, 1)− V (2, 1, 4)

= −(12 · (−2) + 1 · 13

)−(−(32 · 1 + 3 · 43

))

= 1− (−201) = 202

So, the work done by the force field on our imaginary object as rolls down the hill (itends up on a lower potential) is 202 units. So its kinetic energy would be 1

2mv2 =

202 energy units if it started at rest.

(b) In two dimensions:

i. show that

~F =

(αxβy

)

is a conservative fieldWe need to show that

∂Fi∂xj

=∂Fj∂xi

for all i, j

on a simply connected region.

• Simply connected? Since the force field is defined everywhere in our 2-D space, theregion is simply connected.

• In two dimensions, the condition above means we need to test

∂Fx∂y

=∂Fy∂x

So, for our field:

∂Fx∂y

=∂Fy∂x

⇔ ∂(αx)

∂y=

∂(βy)

∂x⇔ 0 = 0

Since 0 = 0 is evidently true, the field is conservative.

ii. Calculate its potential φWe use the same method in the previous question - we calculate the line integral from

some fixed ~r0 to an arbitrary point ~r1, along a convenient path. We’ll use again a straightline, and also start again at the origin, ~r0 = ~0. Let’s do the line integral:

• The path connecting

(00

)and

(x1

y1

)in a straight line is:

~γ =

(x1ty1t

)with 0 < t < 1


• The line element

d~s =d~γ

dtdt =

(x1

y1

)

• The field along the path:

~F (~γ(t)) =

(αx1tβy1t

)

Putting it all together for the line integral:

I =

ˆ

γ

~F · d~s

=

1ˆ

0

(αx1tβy1t

)·(x1

y1

)dt

=

1ˆ

0

(αx2

1t+ βy21t)dt

=1

2

(αx2

1 + βy21

)

So, our potential is (not forgetting the minus sign):

V (x, y) = −1

2

(αx2 + βy2

)

iii. Parameterise the equipotential lines, i.e. find an expression

~̀(t) =

(x(t)y(t)

)

such that φ(~̀(t)) = const. The section in the lecture notes on parameterising pathsmight be helpful for this.Equipotential lines are given by the condition

−1

2

(αx2 + βy2

)= C

For the next step, you have two options. Either you stick with Cartesian co-ordinates,or you go to polar co-ordinates.

• Option 1: Cartesian.

– Solve for one of the parameters, say y:

y2 =2(−C)− αx2

β

y1,2 = ±√−2C − αx2

β

Giving two solutions, one for positive or zero y, one for negative y.

– Parameterise the path in terms of x:

~̀1(t) =

(x√

−2C−αx2

β

)for y ≥ 0

~̀2(t) =

(x

−√−2C−αx2

β

)for y < 0


• Polar co-ordinates, by brute force: Put into the equation

−1

2

(αx2 + βy2

)= C

the polar co-ordinates for x and y, which are x = r cosφ and y = r sinφ:

−1

2

(αr2 cos2 φ+ βr2 sin2 φ

)= C

Now solve for one of the parameters to reduce the number of variables to 1 (as neededfor a path, which is, after all, 1-dimensional). We’ll pick r.

r2(α cos2 φ+ β sin2 φ

)= −2C

r2 =−2C

α cos2 φ+ β sin2 φ

r =

√−2C


This is now a unique solution, since r is positive by definition. Now put this intoour path vector expressed in terms of polar parameters:

~̀ =

(r cosφr sinφ

)

=

(cosφsinφ

)√ −2C


Now, this is a perfectly fine solution to the problem, even if it is perhaps a bit ugly.But it has the advantage over the previous, Cartesian solution (also perfectly valid),that we only need to parameterise one path, not two.

• Polar co-ordinates elegant. Re-write our equation:

−1

2

(αx2 + βy2

)= C

in a more standardised form:x2

a2+y2

b2= 1

with a2 = −C2α and b2 = −C

2β . You might recognise this as the equation for an ellipse.The parameter form of an ellipse is given by

~̀=

(a cos tb sin t

)

It is easy to show that this fulfils the above equation.

iv. (Relevant but beyond exam level) Using your parameterisation, prove that the equipo-tential lines are always perpendicular to the force field.We need to show that the angle between the force (given) and the tangent to our param-

eterised equipotential lines is 900.

• Step 1: Tangent vector ∝ “speed” of particle = d~̀

dt where you need to replace t withwhatever parameter you chose (could be x or φ).

– For the Cartesian solution (we chose x as our parameter):

d~̀1dx

=

(1

−αβx√

−βC+αx2

)

d~̀2dx

=

(1

αβx√

−βC+αx2

)


– For the “elegant” polar solution

(a cos tb sin t

)(it’s all the same principle, just

more painful for the messy solution):

~̀̇1 =

(−a sin tb cos t

)

To prove that the force is perpendicular to the equipotential lines, you need to showthat

~F (x, y) · d~̀

dt= 0

– Cartesian: We want to evaluate ~F for a value on a given equipotential line - andthen take the scalar product of the tangent at this point with ~F . Let’s start withthe upper part of the equipotential line, i.e. the one for y ≥ 1:

~F (~̀1) =

(αx

β√−2C−αx2

β

)

The dot product:

~F (~̀1) · d`1dx

=

(αx

β√−2C−αx2

β

)·(

1

−αβx√

−βC+αx2

)

= αx− αx = 0

Similarly for `2:

~F (~̀2) · d`2dx

=

(αx

β −√−2C−αx2

β

)·(

1αβx√

−βC+αx2

)

= αx− αx = 0 QED

– For the

(a cos tb sin t

)parameterisation:

~F (~̀(t)) =

(α`xβ`y

)

=

(αa cos tβb sin t

)

with α =1

2a2, β =

1

2b2

~F (~̀(t)) =1

2

(1a cos t1b sin t

)

the dot product:

~F (~̀(t)) · d~̀

dt=

1

2

(1a cos t1b sin t

)·(−asintb cos t

)

=1

2(− cos t sin t+ sin t cos t) = 0 QED

v. Sketch the field and equipotential lines for α = 2 and β = 12 . (Equipotential lines are

drawn for equal spacings in potential, i.e. for φ0, φ0+∆φ, φ0+2∆φ for constant, sensiblychosen ∆φ.)Mathematica made this plot. Note that the arrows are drawn such that their base is at


the point where they are evaluated. So whenever you see an arrow base on the read line,you should find that the two are perpendicular.

y

!1.5 !1.0 !0.5 0.0 0.5 1.0 1.5!1.5

!1.0

!0.5

0.0

0.5

1.0

1.5

x

5.2 Multiple Integrals Review

Consider the 2-D region B enclosed by the following lines:

• y = x

• xy = 1

• y = 2

Hint: Draw this before you continue.

5.2. MULTIPLE INTEGRALS REVIEW 79

(a) Calculate the area of the enclosed region

A =

¨

B

dx dy

A =

y=2ˆ

y=1

x=yˆ

x= 1y

dx dy

=

2ˆ

1

(y − 1

y

)dy

=

[1

2y2 − ln y

]2

1

= 2− ln 2− 1

2+ 0

=3

2− ln 2

(b) Calculate the volume enclosed by this area and the surface defined by z = y2

x2

V =

¨

B

y2

x2dx dy


A =

y=2ˆ

y=1

x=yˆ

x= 1y

y2

x2dx dy

=

y=2ˆ

y=1

y2

x=yˆ

x= 1y

1

x2dx dy

=

2ˆ

1

y2

[− 1

x

]x=y

x=1/y

dy

=

2ˆ

1

y2

(y − 1

y

)dy

=

2ˆ

1

(y3 − y

)dy

=

[1

4y4 − 1

2y2

]2

1

= 4− 2− 1

4+

1

2

=9

4

Chapter 6


6.1 Surface Integrals

6.1.1 Find d~S

Find d~S for each of the following surfaces. Sketch the intersections of the surfaces with the x − yplane, the x−z plane and the y−z plane. Add the projection of d~S at several points on the surfaceto your graph.

Try to draw a full 3-D graph of the surfaces and d~S.

(Of course, d~S is infinitesimally small - draw something that is visible and proportional to d~S.)

(a) z = 2− x− y(b) z2 + x2 = 1

(c) z = x2 + y2

(d) z = xy

(e)(xa

)2+(yb

)2+(zc

)2= 1

(f) Example: x2 + y2 + z2 = R2 (solutions for d~S using various approaches below.)

• Using spherical co-ordinates Reminder: ~r in spherical coordinates is xyz

=

r sin θ cosφr sin θ sinφr cosφ

translate the equation into spherical co-ordinate:

R2 = x2 + y2 + z2

R2 = (r sin θ cosφ)2 + (r sin θ sinφ)2 + (r cos θ)2

R2 = r2

This eliminates one parameter - the co-ordinate r is replaced with the constant R.

~S(r, θ, φ) = R


cos θ

partial derivatives:

∂

∂φ~S = R


0

,∂

∂θ~S = R

cos θ cosφcos θ sinφ− sin θ

81


Finally calculate d~S:

d~S =∂

∂θ~S× ∂

∂φ~S dθ dφ =


× − sin θ sinφ

sin θ cosφ0

R2dθ dφ =

sin2 θ cosφsin2 θ sinφsin θ cos θ

R2dθ dφ

Check correct orientation. Pick an easy point, e.g.: θ = π/2, φ = 0, i.e. where the sphere intersectsthe positive x axis. For d~S to point outwards, given that the centre of the sphere is at (0,0,0), it

should to point along +x at this point i.e along

+|a|00

where |a| is any positive number. Check:

d~S(θ = π/2, φ = 0) =

100

R2dθ dφ

Which is OK. (Note: if, unlike here, the surface is not closed, the orientation is your choice. Thereare two things to consider though: If you are using Stoke’s theorem the orientation of the surface isrelated to the direction in which you go through the circular path by the right-hand-rule. And if youhave split your surface into several parts, the orientations should be consistent with each other - sothe direction of the surface normal should be the same where the two component surfaces touch.)

• Cylindrical co-ordinates can sometimes be very useful even when integrating over a sphere. Whileyour sphere is clearly spherically symmetric, it is also cylindrically symmetric. Often the problemseen as a whole (which will include a field) is not spherically symmetric, but it might still exhibitcylindrical symmetry, in which case it tends to be easier to use cylindrical coordinates even withspheres. Anyway, it’s a good exercise. Reminder: the position vector in cylindrical co-ordinates is:

xyz

=

ρ cosφρ sinφz

where, remember, this ρ is the distance from the z−axis, a different quantity than the ρ in sphericalcoordinates. The equation for a sphere

x2 + y2 + z2 = R2

becomes in cylindrical coordinates:

ρ2 + z2 = R2

We have a choice which pair of co-ordinates to pick to parameterise the surface. I can eliminate eitherρ or z using this equation. It turns out, eliminating ρ =

√R2 − z2 and parameterising the sphere in

terms of z and φ is easier than using ρ and φ, not least because for ρ, which is positive by definition,we have a unique solution to ρ2 + z2 = R2 (see the solution in cartesian coordinates, next, for anillustration why this matters). A sphere in cylindrical coordinates, parameterised in terms of z and

φ is therefore (simply replacing ρ with√z2 −R2 in

ρ cosφρ sinφz

):

~S =


z

6.1. SURFACE INTEGRALS 83

∂~S

∂z=

−z√R2−z2

cosφ

−z√R2−z2

sinφ

1

∂~S

∂φ=


0

d~S =

∂~S

∂φ× ∂~S

∂zdz dφ

d~S =


0

×

−z√R2−z2

cosφ

−z√R2−z2

sinφ

1

dz dφ

=


z

dz dφ

Does d~S point outwards? Let’s check for z = 0 and φ = 0, when it should point towards +x:

d~S(z = 0, φ = 0) =

R00

dz dφ

and it does, all fine. Note that∣∣∣d~S∣∣∣ for a sphere is particularly simple in cylindrical co-ordinate:∣∣∣d~S∣∣∣ = Rdz dφ.

• Using cartesian coordinates (this is not recommended, just shown as an illustration that it can bedone, and that it is complex.)

solve for one of the co-ordinates, say z

z = ±√R2 − x2 − y2

2 solutions:

~S1(x, y) =

xy√

R2 − x2 − y2

for z ≥ 0

~S2(x, y) =

xy

−√R2 − x2 − y2

for z < 0

calculate partial derivatives:

∂

∂x~S1 =

10−x√

R2−x2−y2

,∂

∂y~S1 =

01−y√

R2−x2−y2

∂

∂x~S2 =

10x√

R2−x2−y2

,∂

∂y~S2 =

01y√

R2−x2−y2

d~S1 =∂

∂x~S1 ×

∂

∂y~S1 dx dy =

x√

R2−x2−y2

y√R2−x2−y2

1

dx dy

d~S2 =∂

∂x~S2 ×

∂

∂y~S2 dx dy =

−x√

R2−x2−y2

−y√R2−x2−y2

1

dx dy


Since it’s a closed surface, we should check that d~S points outwards. Obviously the case for d~S1, butnot for d~S2. Check the z-direction with to convince yourself - for d~S to point outwards, it shouldpoint downwards (to negative z) for the lower (z < 0) half of the sphere and upwards (positive z) forthe upper (z ≥ 0) part. So d~S1 is fine, but we need to take the negative of d~S2. Final result:

d~S1 =

x√

R2−x2−y2

y√R2−x2−y2

1

dx dy

d~S2 correct orientation =

x√

R2−x2−y2

−√R2−x2−y2

−1

dx dy

The task is to

(a) Parameterise the surface.

(b) Calculate, from that, d~S, which is

d~S(u, v) = ±∂~S

∂u× ∂~S

∂vdu dv

(c) If the orientation matters, check that you got it right - if not, swap the sign of d~S. Theorientation matters...

i. ... when the surface is close. d~S should point outwards.

ii. ... when you are using Stoke’s theorem. The right hand rule relates surface direction andline integral.

iii. ... if you split up your surface, d~S must be consistent between the different parts.

iv. ... whenever you care if something flowing in a given direction through the surface shouldbe a positive or negative flux.

v. ...not at all when you are not calculating the flux of a vector field through the surface,but the surface integral of a scalar function:¨

S

ρ∣∣∣d~S∣∣∣

I have added a section with quite a detailed explanation on parameterising surfaces to the end ofthis problem set. Read this if you get stuck. But the most important thing is to practice.

Solutions to question 6.1.1

(a) i. The surface is already given in a convenient form:

z = 2− x− y

solved for one of the parameters (z). So we stick with cartesian co-ordinates and replace

z in

xyz

with 2− x− y:

~S(x, y) =

xy

2− x− y


ii.

∂~S

∂x=

10−1

∂~S

∂y=

01−1

∂~S

∂x× ∂~S

∂y=

111

d~S =

111

dx dy

(b) This:z2 + x2 = 1

defines a cylinder centred at the y axis. We can use cartesian co-ordinates, or, recognisingthe symmetry around the y-axis, do something clever.

• Cartesian

i. In cartesian coordinates, we get two solutions for z (or x):

z1,2 = ±√

1− x2

and will therefore have to split the surface into two parts:

S1(x, y) =

xy√

1− x2

S2(x, y) =

xy

−√

1− x2

ii.

∂~S1,2

∂x=

10

∓ x√1−x2

∂~S1,2

∂y=

010

∂~S1,2

∂x× ∂~S1,2

∂y=

± x√

1−x2

01

d~S1,2 =

± x√

1−x2

01

dx dz

d~S1 =

x√1−x2

01

dx dz

d~S2 =

− x√

1−x2

01

dx dz


Now that we have two surfaces, we need to check that we have a consistentorientation between the two surfaces. The borderline is at z = 0. A point on theborderline between the two surfaces is, x = 1, y = 0. Unfortunately (and a good

reason to think about better parameterisations), d~S is not defined there, since its xcomponent goes to ±∞. But we can still check the consistency of the two vectorialsurface elements by checking what happens as we go closer and closer to that border.In fact we find that our surface orientations are not consistent, since

limx→1

d~S1 →

+∞01

(6.1)

limx→1

d~S2 →

−∞

01

(6.2)

Therefore we need to turn one of them around (say d~S2). So, our final result is

d~S1 =

x√1−x2

01

dx dz

d~S2 consistent orientation =

x√1−x2

0−1

dx dz

• The same thing in more suitable co-ordinates:

i. To make use of the symmetry, we need to use something like cylindrical co-ordinates(after all, we are dealing with a cylinder), but in such a way that the y axis becomesthe centre of the cylinder, and not, as usual, the z axis. With this:

~S(φ, y) =

cosφy

sinφ

ii.

∂~S

∂φ=

− sinφ

0cosφ

∂~S

∂y=

010

∂~S

∂φ× ∂~S

∂y=

− cosφ

0− sinφ

d~S =

− cosφ

0− sinφ

dφ dy

(c) This equationz = x2 + y2

describes a paraboloid centred around the z-axis. The most suitable co-ordinate system wouldbe cylindrical co-ordinates, since they reflect the symmetry of the problem. But cartesian willwork very well, too. What to choose depends on what integral you’ll want to do (e.g. ifyou know you want to integrate over a vector field that’s also got cylindrical symmetry, usescylindrical co-ordinates). Here we show both solutions.


• Cartesian:

i. We can write down the parameterised surface straight away, since it has already beensolved for one parameter:

~S =

xy

x2 + y2

ii.

∂~S

∂x=

10

2x

∂~S

∂y=

012y

∂~S

∂x× ∂~S

∂y=

10

2x

×

012y

=

−2x−2y

1

d~S =

−2x−2y

1

dx dy

• Cylindrical

i.

~S =

ρ cosφρ sinφρ2

ii.

∂~S

∂φ=

−ρ sinφρ cosφ

0

∂~S

∂ρ=

cosφsinφ2ρ

∂~S

∂ρ× ∂~S

∂φ=

cosφsinφ2ρ

×

−ρ sinφρ cosφ

0

=

−2ρ2 cosφ−2ρ2 sinφ

ρ

d~S =

−2ρ2 cosφ−2ρ2 sinφ

ρ

dρ dφ

(d) This:

z = xy

gives some kind of wavy surface that is quite close to the x − y plane near x = y = 0 anddeparts from that more and more the further we get away from x = y = 0. The surface isgiven as a “graph”, so it’s easy to parameterise:

~S =

xyxy


And d~S is

d~S =∂~S

∂x× ∂~S

∂xdx dy

=

10y

×

01x

dx, dy

d~S =

−y−x1

dx dy

This d~S points in positive z direction, −d~S that points downwards would also have been apossible solution. However, if the orientation matters (as it does when using Stokes’ theorem,or when a particular orientation is given in the question), you need to check which optionpoints in the right direction and pick the appropriate one.

(e) This (xa

)2

+(yb

)2

+(zc

)2

= 1

describes an ellipsoid.

i. We could again work through the step-by-step programme, using cartesian co-ordinates.However we’ve done this often enough, and we can find a easier way. The thing to realiseis that, if we change co-ordinates from x, y, z to x′, y′, z′ like this

x′ = x/a

y′ = y/b

z′ = z/c

we get the equation of a unit sphere:

(x′)2 + (y′)2 + (z′)2 = 1

in these new co-ordinates. We know how to parameterise a unit sphere, i.e. a sphere ofradius 1 (or any radius R). If you have forgotten, see the discussion in the model answerto question f . So we use the following parameterisation:

x′ = sin θ cosφ

y′ = sin θ sinφ

z′ = cos θ

and hence

x = a sin θ cosφ

y = b sin θ sinφ

z = c cos θ

So our parameterisation is:

~S(θ, φ) =

a sin θ cosφb sin θ sinφc cos θ

If that seems too easy to be true: try it out and check that it fulfils:

(xa

)2

+(yb

)2

+(zc

)2

= 1


ii.

∂~S

∂φ=

−a sin θ sinφb sin θ cosφ

0

∂~S

∂θ=

a cos θ cosφb cos θ sinφ−c sin θ

∂~S

∂φ× ∂~S

∂θ=

−bc sin2 θ cosφ−ac sin2 θ sinφ−ab sin θ cos θ

d~S =

−bc sin2 θ cosφ−ac sin2 θ sinφ−ab sin θ cos θ

dφ dθ

Now we have to check the orientation of d~S because we are dealing with a closed surface,so we want d~S to point outwards. Evaluate d~S at a convenient location. First attempt:try where the ellipsoid cuts through the positive z axis, at θ = 0, where we get

d~S(θ = 0, φ = 0) =

000

We see that it didn’t work - you cannot determine the orientation of ~0. We must tryanother one: this time we can try where it hits the positive x axis, i.e. θ = π/2 andφ = 0:

d~S(θ = π/2, φ = 0) =

−bc

00

So there it points in the negative x direction, which is cleary inwards. We want it to beoutwards, so we need to multiply the d~S we got by −1. Our final result is:

d~Spointing outwards =

bc sin2 θ cosφac sin2 θ sinφab sin θ cos θ

6.1.2 Mass (Relevant but beyond exam)

The distribution of mass on the following sphere shell

x2 + y2 + z2 = R2

is given by

σ(x, y, z) =σ0

R2

(x2 + y2

)

Find an expression in terms of σ0 and R for the total mass of the shell.A sphere by itself is most easily parameterised in spherical coordinates, however, this does not actu-

ally reflect the symmetry of the entire problem - the mass distribution is not spherically symmetric,as it changes as a function of x2 + y2. Because the mass distribution depends only on the distanceto the z axis, this problem has cylindrical symmetry. Since many of you will have solved this in inspherical coordinates (and that works just fine its only slightly more complicated), I give you thesolution in both systems.


• Cylindrical

(a) The equation for a spherex2 + y2 + z2 = R2

becomes in cylindrical coordinates:

r2 + z2 = R2

where, remember, this r is the distance from the z−axis, a different quantity than the rin spherical coordinates. We have a choice which two co-ordinates to pick to parametersthe sphere. I tried two options, z and φ as well as r and φ. It turns out, that z and φ iseasier (even though σ depends on r2).

A sphere in cylindrical coordinates, parameterised in terms of z and φ:

~S =


z

(b) Now find d~S:

∂~S

∂z=


1

∂~S

∂φ=


0

d~S =∂~S

∂φ× ∂~S

∂zdz dφ

d~S =


0

×


1

dz dφ

=


z

dz dφ

∣∣∣d~S∣∣∣ =

√R2 − z2 + z2 dz dφ

= Rdz dφ

(c) Now do the integral. The limits are −R < z < R and 0 < φ < 2π

2πˆ

0

R̂

−R

σ(z, φ)∣∣∣d~S∣∣∣ =

2πˆ

0

R̂

−R

σ(z, φ)Rdz dφ

=

2πˆ

0

R̂

−R

σ0

R2

(R2 − z2

)Rdz dφ

= 2πσ0

R2

R3

R̂

−R

dz −RR̂

−R

z2 dz

= 2πσ0

R2

(2R4 − 2

3R4

)

=8π

3σ0R

2


• The same in spherical coordinates

(a) Find parameterisation: We’ve done this often enough for a sphere, in spherical coordi-nates the answer is:

~S(θ, φ) =

R sin θ cosφR sin θ sinφR cos θ

Since we want to integrate over the entire shell, the integration limits are

0 < θ < π, 0 < φ < 2π

(b) Find |d~S| The same way in the previous questions, we get the differential, vectorialsurface element.

d~S =

R2 sin2 θ cosφR2 sin2 θ sinφR2 sin θ cos θ

Now we want to integrate over a scalar function, not a vector field, so we need |d~S|:

d~S2 = R4(sin4 θ cos2 φ+ sin4 θ sin2 φ+ sin2 θ cos2 θ

)(dθ dφ)2

= R4(sin4 θ

(cos2 φ+ sin2 φ

)+ sin2 θ cos2 θ

)(dθ dφ)2

= R4(sin4 θ + sin2 θ cos2 θ

)

= R4 sin2 θ(sin2θ + cos2 θ

)(dθ dφ)2

= R4 sin2 θ (dθ dφ)2

∣∣∣d~S∣∣∣ = R2 sin θ dθ dφ

Note that we should in principle worry about the sign of sin θ, after all we want∣∣∣d~S∣∣∣, a

positive value. But since θ only goes from 0 to π, sin θ is positive anyway, so no need toworry any further.

(c) Fill it all in

i. The function ρ on the surface:

ρ =σ0

R2

(x2 + y2

)

=σ0

R2

(R2 sin2 θ cos2 φ+R2 sin2 θ sin2 φ

)

= σ0 sin2 θ


ii. Everything into the integral:

I =

¨

S

ρ∣∣∣d~S∣∣∣

=

θ=πˆ

θ=0

φ=2πˆ

φ=0

(σ0 sin2 θ) R2 sin θ dφ dθ

= 2πR2σ0

θ=πˆ

θ=0

sin3 θ dθ

= 2πR2σ0

[− cos θ +

1

3cos3 θ

]θ=π

θ=0

= 2πR2σ0

((1− 1

3

)−(−1 +

1

3

))

= 2πR2σ0

(2− 2

3

)

=8π

3R2σ0

Same result as before, fortunately.

6.1.3 Flux through a disk

(a) Calculate the flux of the field ~C

~C =

001

through the unit-disk in the x − y plane (the plane is defined by z = 0, and the integrationlimits by x2 + y2 < 1, which are easier in polar or cylindrical coordinates).

i. Parameterise the disk:

~S =

r cosφr sinφ

0

rmin = 0, rmax = 1, φmin = 0, φmax = 2π

ii. Find d~S (see above)

d~S =

001

r dr dφ

iii. Put it all in


A. Put the surface into the field

~C(x, y, z) =

001

just a reminder, nothing done, yet.

Now replace all instances of x with the x-componentof ~S, all instances of y with the y-componenent of~S and all instances of z with the z-component of ~S(trivial, here):

~C(~S) =

001

B. Everything into integral:¨

complicated surface

~C(~S) · d~S =

¨

disk

~C(~S) · d~S

=

¨

disk

001

·

001

r dr dφ

=

¨

disk

(0 · 0 + 0 · 0 + 1 · 1) r dr dφ

=

¨

disk

1 r dr dφ

So it’s just the area of the disk, which you probablyknow - we’ll calculate it anyway:

=

2πˆ

0

R̂

0

r dr dφ

= 2π

R̂

0

r dr

= 2π

[1

2r2

]R

0

= 2π1

2R2

= π R2

(unit circle, so R=1):

= π

(b) Evaluate ¨~G · d~S

through the disk(x− 2)2 + (y − 3)2 < 9, z = 0

where

~G =

02yz

2− z2


Hint: the answer is 18πLet’s go:

i. Parameterise the disk (same as before, only now it’s not centred at (0, 0), but at (2, 3);the radius is 3.

~S =

r cosφ+ 2r sinφ+ 3

0


ii. Find d~S

∂

∂r~S =

cosφsinφ

0

∂

∂φ~S =

−r sinφr cosφ

0

d~S =∂

∂r~S × ∂

∂φ~S dr dφ

=

00r

dr dφ

=

001

r dr dφ

iii. Put it all in

A. Surface into field

~G(x, y, z) =

02yz

2− z2


Now replace all instances of x with the x-componentof ~S, all instances of y with the y-componenent of~S and all instances of z with the z-component of ~S

(since d~S is

001

we kow that only the z compo-

nent matters, but I’ll do them all, here):

~G(~S) = ~G(Sx, Sy, Sz)

= ~G(r cosφ+ 2, r sinφ+ 3, 0)

=

02 · (r sinφ+ 3) · 0

2− 02

=

002


B. Everything into integral:

¨

disk

~G(~S) · d~S

=

¨

disk

002

·

001

r dr dφ

=

¨

disk

2 r dr dφ

= 2

3ˆ

0

2πˆ

0

r dr dφ

= 4π

3ˆ

0

r dr

= 2π32

= 18π

6.1.4 Flux through a triangular plane section

Evaluate ¨

S

~A · d~S

where

~A =

18z−123y

and S is the part of the plane

2x+ 3y + 6z = 12

which is located in the first octant, i.e. where x > 0, y > 0 and z > 0.

(a) Parameterise the plane:

xy

2− 13x− 1

2y

Part of the task in parametersing the plane is to figure out the parameter limits. We’re toldto integrate over the section of the plane in the first octant, i.e. x and y are to be chosen suchthat all x, y, z are positive. The lower limits are easy:

xmin = ymin = 0

Quickly check that this leads to a legal value for z:

z(x = 0, y = 0) = 2− 0− 0 = 2


that’s positive, so that’s OK. The upper limits are defined by the requirement that z should bepositive, i.e.:

z > 0

2− 1

3x− 1

2y > 0

−x > −6 +3

2y

x < 6− 3

2y

So the upper limit of x in terms y is

xmax = 6− 3

2y

The upper limit on y derives from the requirement, that xmax > xmin, hence

6− 3

2y > xmin

6− 3

2y > 0

y < 4

So we haveymax = 4

(b) Find d~S

∂~S

∂x=

10− 1

3

∂~S

∂y=

01− 1

2

∂~S

∂x× ∂~S

∂y=

13121

d~S =

13121

dx dy

(c) Fill it all in

i. Put the surface into field:

~A =

18z−123y

=

18(2− 1

3x− 12y)

−123y

=

36− 6x− 9y−123y


ii. Put it all into the integral

I =

¨

plane section

~A · d~S

=

4ˆ

0

6− 32yˆ

0

~A ·

13121

dx dy

=

4ˆ

0

6− 32yˆ

0

36− 6x− 9y−123y

·

13121

dx dy

=

4ˆ

0

6− 32yˆ

0

(12− 2x− 3y − 6 + 3y) dx dy

=

4ˆ

0

6− 32yˆ

0

(6− 2x) dx dy

=

4ˆ

0

[6x− x2

]x=6− 32y

x=0dy

=

4ˆ

0

(36− 9y −

(36 +

9

4y2 − 18y

))dy

=

4ˆ

0

(9y − 9

4y2

)dy

=

[9

2y2 − 3

4y3

]y=4

y=0

=9

2· 16− 3

4· 64

= 72− 48

= 24

Chapter 7


7.1 Divergence

7.1.1 Sum and product rule for divergence (Supplementary)

Prove

(a)~∇ ·(~A+ ~B

)= ~∇ · ~A+ ~∇ · ~B

Step 1: Evaluate~∇ · ~A+ ~∇ · ~B

~∇ · ~A+ ~∇ · ~B

=

∂∂x∂∂y∂∂z

·

AxAyAz

+

∂∂x∂∂y∂∂z

·

BxByBz

=∂

∂xAx +

∂

∂xBx +

∂

∂yAy +

∂

∂yBy +

∂

∂zAz +

∂

∂zBz

Step 2: Evaluate~∇ ·(~A+ ~B

)

~∇ ·(~A+ ~B

)

=

∂∂x∂∂y∂∂z

·

Ax +BxAy +ByAz +Bz

=∂

∂x(Ax +Bx) +

∂

∂y(Ay +By) +

∂

∂z(Az +Bz)

=∂

∂xAx +

∂

∂xBx +

∂

∂yAy +

∂

∂yBy +

∂

∂zAz +

∂

∂zBz

as we’ve shown above:

= ~∇ · ~A+ ~∇ · ~B QED

99


(b)~∇ ·(φ ~A

)=(~∇φ)· ~A+ φ

(~∇ · ~A

)

~∇ ·(φ ~A)

=

∂∂x∂∂y∂∂z

·

φAxφAyφAz

=∂

∂x(φAx) +

∂

∂y(φAy) +

∂

∂z(φAz)

=

(∂

∂xφ

)Ax + φ

∂

∂xAx +

(∂

∂yφ

)Ay + φ

∂

∂yAy +

(∂

∂zφ

)Az + φ

∂

∂zAz

=

(∂

∂xφ

)Ax +

(∂

∂yφ

)Ay +

(∂

∂zφ

)Az + φ

∂

∂xAx + φ

∂

∂yAy + φ

∂

∂zAz

=

(∂

∂xφ

)Ax +

(∂

∂yφ

)Ay +

(∂

∂zφ

)Az + φ

(∂

∂xAx +

∂

∂yAy +

∂

∂zAz

)

=

∂∂x∂∂y∂∂z

φ

· ~A + φ

∂∂x∂∂y∂∂z

· ~A

=(~∇φ)· ~A + φ

(~∇ · ~A

)QED

7.1.2 Divergence of 1/r2 field

Prove:

~∇ ·(~r

r3

)= 0

(for r 6= 0)

• In spherical polars: The field in spherical polars is:

~r

r3=

1

r2~er

So, it’s θ and φ components are both 0, and its r component is 1r2 , which depends on r only.

Remembering that

~∇ · ~F =1

r2

∂r2Fr∂r

+ partial derivatives wrt θ, φ

we find

~∇ ·(~r

r3

)= ~∇ ·

(1

r2~er

)

=1

r2

∂(r2(

1r2

))

∂r

=1

r2

∂1

∂r= 0

7.1. DIVERGENCE 101

• Cartesian Coordinates:

~∇ · ~rr3

=

∂∂x∂∂y∂∂z

·

1

(x2 + y2 + z2)3/2

xyz

=∂

∂x

x

(x2 + y2 + z2)3/2

+∂

∂y

y

(x2 + y2 + z2)3/2

+∂

∂z

z

(x2 + y2 + z2)3/2

=x2 + y2 + z2 − 3x2

(x2 + y2 + z2)5/2

+x2 + y2 + z2 − 3y2

(x2 + y2 + z2)5/2

+x2 + y2 + z2 − 3z2

(x2 + y2 + z2)5/2

= 0

7.1.3 Incompressible fluid (Supplementary)

(a) Using the continuity equation, show that for the velocity field ~v of an incompressible liquid,~∇ · ~v = 0The continuity equation holds for fluids - i.e. we are assuming that we don’t create fluid from

nothing or lose any fluid into nothing:

dρ

dt+ ~∇ · ~ = 0

Incompressible fluid = density ρ is constant, hence

dρ

dt= 0

and consequently

~∇ · ~ = 0

(b) Given this velocity field:

~v =

x+ 3yy − 2zx+ az

For which value of a is ~∇ · ~v = 0?

~∇ · ~v = 0

∂∂x∂∂y∂∂z

·

x+ 3yy − 2zx+ az

= 0

∂

∂x(x+ 3y) +

∂

∂y(y − 2z) +

∂

∂z(x+ az) = 0

1 + 1 + a = 0

a = −2


7.2 Curl

(a) If

~A =

xz3

−2yz3yz4

find ~∇× ~A.

~∇× ~A =

∂∂x∂∂y∂∂z

×

AxAyAz

=

∂Az∂y −

∂Ay∂z

−∂Az∂x + ∂Ax∂z

∂Ay∂x − ∂Ax

∂y

=

∂(3yz4)∂y − ∂(−2yz)

∂z

−∂(3yz4)∂x + ∂(xz3)

∂z∂(−2yz)∂x − ∂(xz3)

∂y

=

3z4 + 2y3xz2

0

(b)

~A =

x2y−2xz2yz

Calculate ~∇×(~∇× ~A

)

~∇×(~∇× ~A

)= ~∇×

~∇×

x2y−2xz2yz

= ~∇×

∂(2yz)∂y − ∂(−2xz)

∂z

−∂(2yz)∂x + ∂(x2y)

∂z∂(−2xz)∂x − ∂(x2y)

∂y

= ~∇×

2x+ 2z0

−x2 − 2z

=

∂(−x2−2z)∂y − ∂0

∂z

−∂(−x2−2z)∂x + ∂(2x+2z)

∂z∂0∂x −

∂(2x+2z)∂y

=

02x+ 2

0

7.3. INTEGRALS AND INTEGRAL THEOREMS 103

7.3 Integrals and Integral Theorems

Hint: Make sure you use the right integral theorem to simplify the problem where possible.

(a) Evaluate

‹~F · d~S

where

~F =

4xz−y2

yz

and the surface is the surface of the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.

Use the divergence theorem:

‹

∂V

~F · d~S =

˚

V

(~∇ · ~F

)dV

What is ~∇ · ~F?

~∇ · ~F =

∂∂x∂∂y∂∂z

·

4xz−y2

yz

= 4z − 2y + y

= 4z − y

For the integration, we’ll use cartesian co-ordinates. Integration boundaries are easy for thecube:

xmin = 0 , xmax = 1

ymin = 0 , ymax = 1

zmin = 0 , zmax = 1


So, the integral is:

I =

"

∂V

~F · d~S

=

˚

V

~∇ · ~F dV

=

˚

V

(4z − y) dV

=

1ˆ

0

1ˆ

0

1ˆ

0

(4z − y) dx dy dz

=

1ˆ

0

1ˆ

0

1(̇4z − y) dy dz

=

1ˆ

0

(4z − 1

2) dz

= (2− 1

2)

=3

2

(b) Evaluate ‹∂V

~r · d~S

for and arbitrary volume V of size Vo (i.e V is the collection of points constituting the volume,while Vo is its size, for example in litres, pints of m3). The vector ~r is the position vector, i.e.

~r =

xyz

, and ∂V is the surface enclosing the volume V .

~∇ · ~r =

∂∂x∂∂y∂∂z

·

xyz

= 3

Hence

I =

"

∂V

~r · d~S

=

˚

V

~∇ · ~r dV

= 3

˚

V

dV

= 3Vo


(c) (Supplementary) An electrostatic field is given by

~E = λ

yzxzxy

where λ is a constant. Use Gauss’ law to find the total charge enclosed by the surface consistingof S1, the hemisphere

z =√R2 − x2 − y2

and S2, its circular base in the xy plane.Gauss law: "

~E · d~S =Q

ε0

where Q is the total charge inside the volume. Using the divergence theorem will usually makethings easier: ‹

∂V

~F · d~S =

˚

V

(~∇ · ~F

)dV

Calculate the divergence of ~E:

~∇ · ~E = λ

∂∂x∂∂y∂∂z

·

yzxzxy

=∂yz

∂x+∂xz

∂y+∂xy

∂z= 0 + 0 + 0

= 0

So

Q = ε0

"

∂V

~E · d~S

= ε0

˚

V

~∇ · ~E dV

= 0

(d) Verify the Divergence Theorem for the example of the vector field

~A =

4x−2y2

z2

and the volume V bound by

x2 + y2 = 4, z = 0, and z = 3

i.e. a cylinder section, centred at the z-axis with radius 2, and with lids at z = 0 and z = 3.Verify the divergence theorem by calculating

‹

∂V

~A · d~S


i. by not using the divergence theorem, i.e. calculating the surface integral directly. Notethat you are dealing with three surfaces: The cylinder wall, and two “lids” at z = 0 andz = 3.

A. Parameterise the surface. We have three surfaces - the cylinder wall (S1), the base(S2) and the lid (S3). They are given by

~S1(φ, z) =

2 cosφ2 sinφz

, 0 < φ < 2π, 0 < z < 3

~S2(r, φ) =

r cosφr sinφ

0

, 0 < φ < 2π, 0 < r < 2

~S3(r, φ) =

r cosφr sinφ

3

, 0 < φ < 2π, 0 < r < 2

B. Find d~S1, d~S2, d~S3:

• Surface 1

∂

∂φ~S1 =

−2 sinφ2 cosφ

0

∂

∂z~S1 =

001

d~S1 =

2 cosφ2 sinφ

0

dφ dz

Does it point outside? Check e.g. φ = z = 0 in which case it should point towardsto +ve x− axis, and it does, so we got the sign right.

• Surface 2

∂

∂φ~S2 =

−r sinφr cosφ

0

∂

∂r~S2 =

cosφsinφ

0

d~S2 =

00−r

dφ dr

which points towards −z as we would want for the base.

• Surface 3: The lid will have the same d~S as the base, except we’ll have to choosethe sign such it points outwards which is upwards for the lid:

d~S3 =

00r

dφ dr

C. Fill it all in


D. Put the surface into the field

~A(x, y, z) =

4x−2y2

z2

~A(~S1) =

8 cosφ−8 sin2 φ

z2

~A(~S2) =

4r cosφ−2r sin2 φ

0

~A(~S3) =


9

E. Put everything into the integral

I = I1 + I2 + I3

I1 =

¨

S1

~A( ~S1) · d~S1

=

3ˆ

0

2πˆ

0

8 cosφ−8 sin2 φ

z2

·

2 cosφ2 sinφ

0

dφ dz

= 16

3ˆ

0

2πˆ

0

(cos2 φ− sin3 φ

)dφ dz

= 16

3ˆ

0

π dz

= 48π

I2 =

¨

S2

~A( ~S2) · d~S2

=

2ˆ

0

2πˆ

0


0

·

00−r

dφ dr

= 0


I3 =

¨

S3

~A( ~S3) · d~S3

=

2ˆ

0

2πˆ

0


9

·

00r

dφ dr

=

2ˆ

0

2πˆ

0

9rdφ dr

= 9 · 2π2ˆ

0

r dr

= 36π

The total integral:I = I1 + I2 + I3 = 84π

ii. by using the divergence theorem, i.e. calculating an appropriate volume integralThis should be less painful. Let’s first calculate the divergence

~∇ · ~A = 4− 4y + 2z

Now the volume integral. We’ll be using cylindrical co-ordinates again. Substitutingx = r cosφ, y = r sinφ, z = z:

~∇ · ~A = 4− 4r sinφ+ 2z

Our integration limits are 0 < r < 2, 0 < φ < 2π, 0 < z < 3. The volume element incylindrical co-ordinates is dV = r dr dφ dz. So,

I =

‹

∂V

~A · d~S

=

˚

V

~∇ · ~A dV

=

3ˆ

0

2φˆ

0

2ˆ

0

(4− 4r sinφ+ 2z) r dr dφ dz

= 2

3ˆ

0

2φˆ

0

2ˆ

0

(2r − 2r2 sinφ+ zr) dr dφ dz

= 2

3ˆ

0

2φˆ

0

(4− 16

3sinφ+ 2z

)dφ dz

= 2

3ˆ

0

(8π + 4πz) dz

= 8π

3ˆ

0

(2 + z) dz

= 8π

(6 +

9

2

)

= 84π


Which is the same result as previously.

(Note that this is quite different from a proof. All you have to show here is that it works forthis one example; in a proof you would have to show that it works in all cases.)

(e) For

~B =

−yx0

Evaluate the surface integral ¨

S

~B · d~S

over the surface depicted below:

Where the circle shown in black is the unit circle in the x− y plane, i.e. the circle is given by

x2 + y2 = 1, z = 0

Hint: If the divergence of a field ~F is zero, then it can always be written as ~F = ~∇× ~A, whichis useful information even if we don’t know ~A (and make no attempt at calculating it).

This is an open surface - so the only integral theorem the could work is Stokes’ Theorem.For this we need ~F = ~∇× ~A, which we can check (without calculating ~A) by evaluating ~∇ · ~F .

If ~∇ · ~F = 0, then ~F can be written as ~F = ~∇× ~A.

~∇ · ~F = ~∇ · ~F=

∂(−y)

∂x+∂x

∂y+∂0

∂z= 0 + 0 + 0 = 0

So we can use Stokes’ Theorem. Stokes’ theorem is¨S

(~∇× ~A

)· d~S =

˛∂S

~A · d~r

This states that the surface integral of ~F = ~∇× ~A only depends on the boundary of the surface~S. So even if we don’t know ~A we can simplify the surface integral of ~F over this surface bypicking another surface with the same boundary (we can do what as long as ~F can be written

as ~F = ~∇× ~A, i.e. ~∇ · ~F = 0). The obvious surface to pick is the unit disk in the x, y planewith the same boundary. So let’s go through the usual 3-step process



~S =

r cosφr sinφ

0


ii. Find d~S

∂

∂r~S =

cosφsinφ

0

∂

∂φ~S =

−r sinφr cosφ

0

d~S =∂

∂r~S × ∂

∂φ~S dr dφ

=

00r

dr dφ

=

001

r dr dφ

iii. Put it all in


~F (x, y, z) =

−yx0


Now replace all instances of x with the x-componentof ~S, all instances of y with the y-component of ~Sand all instances of z with the z-component of ~S:

~F (~S) = ~F (Sx, Sy, Sz)

= ~F (r cosφ, r sinφ, 0)

=

−r sinφr cosφ

0

B. Put everything into the integral:¨

complicated surface

~F (~S) · d~S =

¨

disk

~F (~S) · d~S

=

¨

disk

−r sinφr cosφ

0

·

001

r dr dφ

=

¨

disk

(−r sinφ · 0 + r cosφ · 0 + 0 · 1) r dr dφ

=

¨

disk

0 r dr dφ

= 0


(f) Repeat the calculation for the same surface, but now the field is

~C =

001

Same as above with different field:


~S =

r cosφr sinφ

0


ii. Find d~S (see above)

d~S =

001

r dr dφ

iii. Put it all in


~C(x, y, z) =

001


Now replace all instances of x with the x-componentof ~S, all instances of y with the y-component of ~S andall instances of z with the z-component of ~S (trivial,here):

~C(~S) =

001


B. Put everything into the integral:

¨

complicated surface

~C(~S) · d~S =

¨

disk

~C(~S) · d~S

=

¨

disk

001

·

001

r dr dφ

=

¨

disk

(0 · 0 + 0 · 0 + 1 · 1) r dr dφ

=

¨

disk

1 r dr dφ

So it’s just the area of the disk, which you probablyknow - we’ll calculate it anyway:

=

2πˆ

0

R̂

0

r dr dφ

= 2π

R̂

0

r dr

= 2π

[1

2r2

]R

0

= 2π1

2R2

= π R2

(unit circle, so R=1):

= π

(g) (Relevant but beyond exam level) Show that, for any simple path γ in the x − y plane, thearea enclosed by the path is given by:

A =1

2

˛γ

(−yx

)· d~r

Hint: A simple path is one where each point is passed-through only once as you go along thepath - a circle is simple, the figure 8 isn’t, because the path describing the figure 8 crossesitself. Usually, this is nothing to worry about, because you will normally only deal with asimple path.

• To use Stoke’s theorem, let’s go into 3-dimensions.

A =1

2

˛γ

(−yx

)· d~r

with d~r =

(dxdy

)is the same as

A =1

2

˛γ

−yx0

· d~r


with ~γ =

x(t)y(t)

0

and d~r =

dxdydz

.

• The enclosed area is given by

A =

¨S

dx dy

which can be written (somewhat over-complicatedly), as

A =

¨S

001

· d~S

where d~S is the vectorial surface element of the x− y plane:

~S =

xy0

,

d~S =∂

∂x~S × ∂

∂y~S dx dy

=

001

dx dy

• To use Stoke’s theorem, you will want to express

A =

¨S

001

· d~S

as

A =

¨S

~∇× ~V · d~S

So that you can put this into:

A =

˛∂S

~V · d~r

which will then give the result.

• So your task is now to find a vector ~V such that

001

= ~∇× ~V

and combining it all together to prove the claim.

Following the extensive hint, what remains to do is to find ~V . The method is guessing. Thereare several options. One of them (and the one that’s useful here), is

~V =1

2

−yx0


(Note that guessing vector potentials will not be part of the exam!) With this we get

Area =

¨

S

001

· d~S

=

¨

S

~∇× ~V · d~S

=

˛

∂S

~V · d~r

=1

2

˛

∂S

−yx0

· d~r

=1

2

˛

∂S

(−yx

)· d~r(2D)

(h) Use this result to calculate the area enclosed by the following path:

(a cosφb sinφ

)

which describes an ellipse in the x− y plane.

I =

˛1

2

(−yx

)· d~r

The usual 3-step process:

i. Parameterise the path - that’s been done for us

~γ =

(a cosφb sinφ

), φmin = 0, φmax = 2π

Feel free to re-name the parameter to t if you wish.

ii. Find d~r

d~r =d~γ

dφdφ

=

(−a sinφb cosφ

)dφ

iii. Put it all in

A. Put the path into the field. The field is ~F (x, y) = 12

(−yx

)

~F (γx(φ), γy(φ)) = ~F (a cosφ, b sinφ)

=

(−b sinφa cosφ

)


B. Put it all into the integral

I =

˛1

2

(−yx

)· d~r

=

2πˆ

0

1

2

(−b sinφa cosφ

)·(−a sinφb cosφ

)dφ

=

2πˆ

0

1

2

(ab sin2 φ+ ab cos2 φ

)dφ

=

2πˆ

0

1

2abdφ

=1

2ab

2πˆ

0

dφ

=1

2ab · 2π

= ab π

(i) Evaluate ˛~F · d~r

around the circle

(x− 2)2 + (y − 3)2 = 9, z = 0

where

~F =

x2 + yz2

2x− y3

0

Hint: Use Stoke’s theorem.The hint is to use Stokes’ theorem, so let’s integrate ~∇× ~F over a surface bound by the closed

path given - the simplest such surface is the disk in x, y centred at (x, y) = (2, 3). Start with

finding ~∇× ~F :

~G = ~∇× ~F =

∂0∂y −

∂2x−y3∂z

− ∂0∂x + ∂(x2+yz2)

∂z∂2x−y3∂x − ∂(x2+yz2)

∂y

=

02yz

2− z2

So:

i. Parameterise the disk (same as before, only now it’s not centred at (0, 0), but at (2, 3);the radius is 3.

~S =

r cosφ+ 2r sinφ+ 3

0



ii. Find d~S

∂

∂r~S =

cosφsinφ

0

∂

∂φ~S =

−r sinφr cosφ

0

d~S =∂

∂r~S × ∂

∂φ~S dr dφ

=

00r

dr dφ

=

001

r dr dφ

iii. Put it all in


~G(x, y, z) =

02yz

2− z2


Now replace all instances of x with the x-componentof ~S, all instances of y with the y-component of ~S andall instances of z with the z-component of ~S (since

d~S is

001

we know that only the z component

matters, but I’ll do them all, here):

~G(~S) = ~G(Sx, Sy, Sz)

= ~G(r cosφ+ 2, r sinφ+ 3, 0)

=

02 · (r sinφ+ 3) · 0

2− 02

=

002


B. Everything into integral:

¨

complicated surface

~G(~S) · d~S =

¨

disk

~G(~S) · d~S

=

¨

disk

002

·

001

r dr dφ

=

¨

disk

2 r dr dφ

= 2

3ˆ

0

2πˆ

0

r dr dφ

= 4π

3ˆ

0

r dr

= 2π32

= 18π

(j) Verify Stoke’s theorem for

~A =

2x− y−yz2

−y2z

Where ~S is the upper half surface of the sphere given by

x2 + y2 + z2 = 1

i.e. calculate the surface integral of ~∇ × ~A over the half sphere, and the line integral of ~Aover its boundary, and compare the result. (This is quite different from a proof. All you haveto show here is that it works for this one example; in a proof you would have to show that itworks in all cases.)

Surface Find ~F = ~∇× ~A

~F = ~∇× ~A

= ~∇×

2x− y−yz2

−y2z

=

∂(−y2z)∂y − ∂(−yz2)

∂z

−∂(−y2z)∂x + ∂(2x−y)

∂z∂(−yz2)∂x − ∂(2x−y)

∂y

=

−2yz + 2yz−0 + 00 + 1

=

001


Now, usually, we’d pick the simplest surface possible (the disk), but here we are explicitlyasked to verify Stokes’ theorem, so we are not supposed to use it. So we’ll go through thecalculation over the spherical surface. Not so bad, but to be avoided if you can use somethingeasier such as the disk (and usually you are allowed to use Stokes’ theorem, and then you canuse the disk):

i. Parameterise the half-sphere with r = 1

~S =


cos θ

θmin = 0, θmax =1

2π, φmin = 0, φmax = 2π

ii. Find d~S

∂

∂θ~S =


∂

∂φ~S =


0

d~S =∂

∂θ~S × ∂

∂φ~S dθ dφ

=

sin2 θ cosφsin2 θ sinφ

cos2 φ cos θ sin θ + sin2 φ cos θ sin θ

dθ dφ

=

sin2 θ cosφsin2 θ sinφcos θ sin θ

dθ dφ

=


cos θ

sin θ dθ dφ

btw, this happens to be

= ~S sin θ dθ dφ

iii. Put it all in

A. Put the surface into the field (trivial, since ~F constant):

~F (x, y, z) =

001


B. Put everything into the integral:¨

half−sphere

~F (~S) · d~S =

¨

half−sphere

~F (~S) · d~S

=

2πˆ

0

12πˆ

0

001

·


cos θ

sin θ dθ dφ

=

2πˆ

0

12πˆ

0

cos θ sin θ dθ dφ

=

2πˆ

0

12πˆ

0

1

2sin 2θ dθ dφ

= 2π

12πˆ

0

1

2sin 2θ dθ

= 2π

[−1

4cos 2θ

] 12π

0

= 2π

(1

4+

1

4

)

= π

And now the line integral

i. Parameterise path

~γ(t) =

cos tsin t

0

tmin = 0, tmax = 2π

ii. Find d~r

d~r =d~γ

dtdt

=

− sin tcos t

0

dt

iii. Put it all in

A. Put the path into the field. The field is

~A(x, y, z) =

2x− y−yz2

−y2z

~A(γx(t), γy(t), γz(t)) = ~A(cos t, sin t, 0)

=

2(cos t)− sin t−(sin t) · 02

−(sin t)2 · 0

=

2 cos t− sin t00


B. Put it all all into the integral

I =

˛

γ

~A · d~r

=

2πˆ

0

2 cos t− sin t00

·

− sin tcos t

0

dt

=

2πˆ

0

(−2 cos t sin t+ sin2 t

)dt

Integral of the odd function cos t · sin t from 0 to 2πis zero, while the integral of sin2 t from 0 to 2π is π

= π

(k) Curl (Supplementary)

i. If

~v = ~ω × ~r

where ~ω is a constant vector, prove that

~ω =1

2~∇× ~v

Hint: Use nabla-calculus techniques

~A×(~B × ~C

)=(~A · ~C

)~B −

(~A · ~B

)~C

Marks: 7 points for part a


~∇× ~v = ~∇× (~ω × ~r)

= ~∇×( ↓~ω ×~r

)+ ~∇×

(~ω×

↓~r

)

Discard 1st term since ~ω is constant, so any kind ofderivative of ~ω is zero.

= ~∇×(~ω×

↓~r

)

expand vector triple product

=

(~∇·↓~r

)~ω −

(~∇ · ~ω

) ↓~r

Re-arrange until arrows to the right, others to theleft of ~∇.

= ~ω

(~∇·↓~r

)−(~ω · ~∇

) ↓~r

Drop arrows:

= ~ω(~∇ · ~r

)−(~ω · ~∇

)~r

insert: ~r =

xyz

and expand

= ~ω

(∂x

∂x+∂y

∂y+∂z

∂z

)−(ωx

∂

∂x+ ωy

∂

∂y+ ωz

∂

∂z

)

xyz

= ~ω (3)−

(ωx

∂∂x + ωy

∂∂y + ωz

∂∂z

)x(

ωx∂∂x + ωy

∂∂y + ωz

∂∂z

)y(

ωx∂∂x + ωy

∂∂y + ωz

∂∂z

)z

= ~ω (3)−

ωx∂x∂x + ωy

∂x∂y + ωz

∂x∂z

ωx∂y∂x + ωy

∂y∂y + ωz

∂y∂z

ωx∂z∂x + ωy

∂z∂y + ωz

∂z∂z

= 3~ω −

ωx · 1 + 0 + 00 + ωy · 1 + 00 + 0 + ωz · 1

= 3~ω −

ωxωyωz

= 3~ω − ~ω~∇× ~v = 2~ω

and hence

~ω =1

2~∇× ~v

ii. Interpret the result physically. Marks: 3 points for part b~v is the velocity of a point on a body spinning with angular velocity |~ω| around the axisgiven by ~ω.

Chapter 8

Nabla Calculus

8.1 How it works (Supplementary)

• Here (and only here),~∇ (XY Z)

means any combination of vector or scalar fields, with any kind of “product” between them,be it cross product, dot product, or the normal product between two scalars, or scalar and avector. For example, the following applies to

~∇(~A ·(~B × ~C

)), ~∇×

(~A×

(~B × ~C

)), ~∇×

((φψ)

(~B × ~C

))

... etc.

(a) Apply what is in essence the product rule (and if you look at the components, this isexactly what it is), indicating with arrows which vector/scalar is to be differentiated ina given term (the others are treated like constants in that term):

~∇ (XY Z) = ~∇(↓X Y Z

)+ ~∇

(X↓Y Z

)+ ~∇

(XY

↓Z

)

(b) Use the usual rules of vector algebra (treat ~∇ as a normal vector during that process),until, in each term, only vectors with arrows (i.e. those being differentiated) are to the

right of the ~∇ operator, and all others are on its left.

(c) Once done, remove the arrows.

Of course, instead of the arrows, you can also use the notation favoured by Boas, where yougive the nabla operator at each term a subscript:

~∇ (XY Z) = ~∇X (XY Z) + ~∇Y (XY Z) + ~∇Z (XY Z)

Apart from the notation, the process is exactly the same. The one thing to be careful aboutin this notation is not to confuse this with directional derivatives that have a similar notation.

If you get stuck, you can always revert to doing this component by component. This isallowed, but will take more time (in some cases too much time to be a good idea as an examtactic), and is more error prone.

• For many nabla calculus questions, and in many other situations, you’ll need to know and usethe rules for scalar and vector triple products, which are:

123

124 CHAPTER 8. NABLA CALCULUS

– The scalar product is invariant under cyclic permutations:

~A ·(~B × ~C

)= ~C ·

(~A× ~B

)= ~B ·

(~C × ~A

)

– The scalar triple product swaps sign if you swap any two vectors:

~A ·(~B × ~C

)= − ~A ·

(~C × ~B

)

~A ·(~B × ~C

)= ~C ·

(~B × ~A

)

– The scalar triple product is evaluated as the determinant of the matrix you get whenyou write the vectors as the columns of a 3 × 3 matrix, (this fact is not needed here,but good to remember and added for completeness):

~A ·(~B × ~C

)= det( ~A, ~B, ~C)

Also not to be forgotten, although not needed in the questions below: The scalar tripleproduct corresponds to the volume of the parallelepiped spanned by the three vectors

– Triple vector product:

~A×(~B × ~C

)=(~A · ~C

)~B −

(~A · ~B

)~C

This is a difficult one to remember, but it’s got to be done.

8.2 Examples and Exercises (Supplementary)

Prove

(a)~∇ ·(φ ~A)

= ~A ·(~∇φ)

+ φ(~∇ · ~A

)

(b)~∇×

(φ ~A)

= φ(~∇× ~A

)− ~A×

(~∇φ)

Example solution:

i.

~∇×(φ ~A)

= ~∇×(↓φ ~A

)+ ~∇×

(φ↓~A

)

ii. At this stage (and only here) we treat ~∇ like any other vector. In order to manoeuvreall the vectors/scalars with arrows to its right, and the others to its left, we’ll use inthis question:

• ~A× ~B = − ~B × ~A

• You can swap the order in products of scalars with scalars, and scalars with vectors,at will (scalars are just numbers, after all), e.g. ~A×φ~B = φ ~A× ~B (this is of courseusually only true if none of the vectors/scalars is an operator, but that’s the point- at this stage we are allowed to treat everything like a normal vector, we alreadydid the operator-bit at step 1).

8.2. EXAMPLES AND EXERCISES (SUPPLEMENTARY) 125

~∇×(φ ~A)

= ~∇×(↓φ ~A

)+ ~∇×

(φ↓~A

)

= −(↓φ ~A

)× ~∇+ φ~∇×

↓~A

= − ~A× ~∇↓φ +φ ~∇×

↓~A

iii.~∇×

(φ ~A)

= − ~A× ~∇φ+ φ ~∇× ~A

(c)~∇ ·(~A× ~B

)= ~B ·

(~∇× ~A

)− ~A ·

(~∇× ~B

)

(d)~∇×

(~A× ~B

)=(~B · ~∇

)~A− ~B

(~∇ · ~A

)+ ~A

(~∇ · ~B

)−(~A · ~∇

)~B

Hint: You can find this in the lecture notes.

(e)~∇(~A · ~B

)=(B · ~∇

)~A−

(A · ~∇

)~B + ~B

(~∇ · ~A

)+ ~A

(~∇ · ~B

)

126 CHAPTER 8. NABLA CALCULUS

Chapter 9

Integration How-To

9.1 Line, Surface and Volume Integrals

Line integrals and surface integrals follow a pattern. Before you start, check if you can usean integral theorem to simplify the problem. Then:

(a) Parameterise

(b) Differentiate

(c) Calculate

9.1.1 Calculating Line Integrals

(a) Parameterise the path

~r(t) =

x(t)y(t)z(t)

The path is a 1-D object, so you’ll need exactly one parameter. Visualise this as the equationfor a particle’s position as a function of time. At this point, also identify the integration limitsin terms of t. Note: Sometimes, instead of ~r(t), we might use another letter, such as ~γ(t),which means the same thing.

(b) Differentiate the path, calculate vectorial path segment

d~r =d~r

dtdt

(c) Fill it all in

i. Put the path into the vector field, i.e. replace x, y, z in ~F (x, y, z) with the x, y, z com-ponent of ~r(t).

~F (x, y, z) =

Fx(x, y, z)Fy(x, y, z)Fz(x, y, z)

127

128 CHAPTER 9. INTEGRATION HOW-TO

becomes

~F (~r(t)) =

Fx(x(t), y(t), z(t))Fy(x(t), y(t), z(t))Fy(x(t), y(t), z(t))

For example, if ~F =

−yx0

and ~r(t) =

t2

t0

, then ~F (~r(t)) =

−tt2

0

. Some

prefer to use as your parameter x, y or z, as in ~r(x) =

xx2

0

(where x = −t translates

one to the other, make sure you keep track of what happens to the bounaries). That’sperfectly allowed, but it has two disadvantages. The main one is that it does not alwayswork (e.g. for a full circle). The second one is a matter of taste: I find it makes it easier

to get ~F (~r(t)) wrong if I have x, y, z in both, ~F and ~r. Therefore I tend to use t or u asmy parameter, or something else that does not look like a coordintate.

ii. Put it all into the integral

I =

tmaxˆ

tmin

~F (~r(t)) · d~rdt

(t) dt

=

tmaxˆ

tmin

(Fx(x(t), y(t), z(t))

dx

dt

+Fy(x(t), y(t), z(t))dy

dt

+Fz(x(t), y(t), z(t))dz

dt

)dt

which is now a normal one-dimensional integral. Note that this is only one path tosuccess - and one that will always work - but there are often short-cuts and there areother, equally valid approaches. Whatever you do, make sure you end up with anintegral over exactly one parameter.1.

Parameterising Paths and surfaces

There is an entire section at the end of this document (page 133) with examples on parameterisedpaths and surfaces. You can find more information and explanations of how to do this both in thelecture notes and in the answers to the problem sets.

9.2 Surface Integrals

9.2.1 Calculating the Flux of a vector field

I =

¨

S

~F (x, y, z) · d~S

1Sometimes you’ll find that´~F · d~r is written as

´(Fx dx + Fy dy + Fz dz). That is OK if you like it. Just do not forget

that dx, dy, dz are still related by the path! So you still have to parameterise the path and at the end write dx, dy, dz asdifferentials of one single parameters, e.g. dx

dtdt, etc.


(a) Parameterise the surface

~S(u, v) =

Sx(u, v)Sy(u, v)Sz(u, v)

What are the limits of the surface region you want to integrate over in terms of u, v?

(b) Calculate

d~S = ±∂~S

∂u× ∂~S

∂vdu dv

The sign depends on the orientation of the surface. If the orientation matters, check that yougot it right - if not, swap the sign of d~S. The orientation matters...

i. ... when the surface is close. d~S should point outwards.

ii. ... when you are using Stoke’s theorem. The right hand rule relates surface directionand line integral.

iii. ... if you split up your surface, d~S must be consistent between the different parts.

iv. ... whenever you care if something flowing in a given direction through the surfaceshould be a positive or negative flux.

v. ... not at all for open surfaces

To check if the orientation is correct, it is good enough to check it at one convenient pointwhere you can easily calculate d~S and know which way it should point. There are only twopossible choices. If the d~S you got initially points into the wrong direction, use −d~S instead.

(c) Put it all in:

I = ±umaxˆ

umin

vmax(u)ˆ

vmin(u)

~F (Sx(u, v), Sy(u, v), Sz(u, v)) ·(∂~S

∂u× ∂~S

∂v

)dv du

(which to use, + or −, is determined by the surface orientation - this can be arbitrary in somecases, then both answers are correct).

9.2.2 The surface integral of a scalar function

The procedure for calculating the surface integral of a scalar function (let’s call it ρ) is exactly the

same as the above, except that we replace ~F · d~S with ρ∣∣∣d~S∣∣∣.

(a) Parameterise the surface

~S(u, v) =

Sx(u, v)Sy(u, v)Sz(u, v)

What are the limits of the surface region you want to integrate over in terms of u, v?

(b) Calculate∣∣∣d~S∣∣∣ =

∣∣∣∣∣∂~S

∂u× ∂~S

∂v

∣∣∣∣∣ du dv

The surface orientation doesn’t matter, here, since we are only interested in∣∣∣d~S∣∣∣ anyway.


(c) Put it all in:

I =

umaxˆ

umin

vmax(u)ˆ

vmin(u)

ρ (Sx(u, v), Sy(u, v), Sz(u, v))

∣∣∣∣∣∂~S

∂u× ∂~S

∂v

∣∣∣∣∣ dv du

9.2.3 Volume Integrals

Volume integrals are easier than line or surface integrals since there is no need to “parameterisethe volume”. But you will still need to pick convenient co-ordinates and be able to express theintegration limits in terms of your co-ordinates.

˚

region

ρ(x, y, z)dV =

zmaxˆ

zmin

ymax(z)ˆ

ymin(z)

xmax(y,z)ˆ

xmin(y,z)

ρ(x, y, z) dx dy dz

You can change the order in which you perform the integration, which can be useful if it is, say,easier to parameterise the integration limits in z as a function of x, y than those in x as functionsof y, z:

˚

region

ρ(x, y, z)dV =

xmaxˆ

xmin

ymax(x)ˆ

ymin(x)

zmax(x,y)ˆ

zmin(x,y)

ρ(x, y, z) dz dy dx

If you are lucky, the integration limits are just constants. This can often be achieved by changingto co-ordinates that reflect the symmetry of the problem. See the next section.

9.3 Coordinate transformations

Choose coordinates that reflect the symmetry of the problem you want to tackle - this nearlyalways makes it easier. Spend a minute thinking about which coordinate system you will use foryour integration. Choosing the right coordinate system can mean the difference between 5 minutesand 20 minutes spent on a calculation.

(a)

dx =dx

dudu

(b)

dx dy =

∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ du dv

(c)

dx dy dz =

∣∣∣∣∂(x, y, z)

∂(u, v, w)

∣∣∣∣ du dv dw

9.4. INTEGRAL THEOREMS 131

Where∣∣∣ ∂(x,y,z)∂(u,v,w)

∣∣∣ is the Jacobi Determinant, given by:

!!!!!(x, y, z)

!(u, v, z)

!!!! !

!!!!!!

!x!u!y!u!z!u

!x!v!y!v!z!v

!x!w!y!w!z!w

!!!!!!

top (“numerator”): rows

bottom (“denominator”): columns

x-row

y-row

z-row

u-col v-col w-col

The Jacobi determinant has the convenient property

∣∣∣∣∂(x, y, z)

∂(u, v, z)

∣∣∣∣ =1∣∣∣∂(u,v,w)

∂(x,y,z)

∣∣∣

Try it out for polar, cylindrical and spherical co-ordinates and make sure you get the followingresults, which you should memorise:

• Polar

dx dy = r dr dφ

• Cylindrical

dx dy dz = r dr dφ dz

• Spherical

dx dy dz = r2 sin θ dr dθ dφ

A good exercise that connects the different sections discussed here, is to parameterise the x − y

plane in terms of polar coordinates: ~S(r, φ) =

r cosφr sinφ

0

, and then calculate

∣∣∣d~S∣∣∣ which should

also give you the area element in polar co-ordinates,∣∣∣d~S∣∣∣ = r dr dφ

9.4 Integral Theorems

You need to know the following integral theorems:

• Integrating a gradient field:

• Stokes’ Theorem

• Divergence Theorem

You should be able to use them to

• Make integration easier, for example


– Use the divergence theorem to transform a surface integral over a closed surface into avolume integral. ‹

∂V

~F · d~S =

˚

V

~∇ · ~F dV

This is a big time saver for nearly all integrals over closed surfaces. You will have tospot closed surfaces yourself - in the exam, it won’t usually be pointed out for you.

– Use Stokes’ theorem to convert a surface integral over an open surface of a field ~B =~∇× ~A over a complicated surface, into one over a simpler surface with the same boundary.¨

complicated surface

~B · d~Sc =

¨

simple surface

~B · d~Ss if ~∇ · ~B = 0

To apply this theorem, you don’t need to know ~A, and you should not expect the fieldin the question to be written as ~∇ × ~A. Simply check that ~∇ · ~B = 0, and you knowthat it can be written as ~B = ~∇× ~A, that’s all you need. ~B. Note that the field is notnecessarily called ~B, it can be called ~F or ~G or anything.

– Use Stokes’ theorem to convert a complicated line integral over a closed line, e.g.

˛∂S

12y − x2 + ecos(x2)

y2 − 12x cos(y)

z2 − 4Γ(|z|)

· d~r

into a simpler surface integral, in this example:

¨S

~∇×

y − x2 + ecos(x2)

y2 − cos(y)z2 − 4Γ(|z|)

· d~S =

¨S

00−1

· d~S

– Use Stoke’s theorem to convert a surface integral over an open surface into a line integral(less often useful, but sometimes it is, and you might get asked to do it anyhow. You’ll

need to know ~A).

– If a field is conservative (~∇× ~F = 0 ⇔ ~F = ~∇φ on simply connected region), find thefield’s potential φ(x, y, z) once, and use that to calculate line integrals trivially.

Notice that the names of the fields are completely arbitrary, so just because I used ~B for asolenoidal field in this summary, something called ~B in a question or an exam will not neces-sarily be a solenoidal field, and conversely, ~F or ~K might be solenoidal fields or conservative,or none of the above.

• Use them for simple proofs, such as

– Show that the surface integral over a closed surface for any field ~F with ~∇ · ~F = 0 iszero.

– Show that the surface integral over a closed surface for any field ~F = ~∇× ~A is zero (use

that ~∇ · (~∇× ( ~A)) ≡ 0).

– Using Stokes’ theorem, show that (on a simply connected region), the line integral over

any closed path of ~F = −~∇φ is zero.

– Show that the surface integral over any closed surface of ~F =

yzx10

xy

is zero.

– ... etc

9.5. PARAMETERISED PATHS AND SURFACES 133

9.5 Parameterised Paths and Surfaces

9.5.1 Examples for parameterised paths

• The straight line from ~r0 =

x0

y0

z0

to ~r1 =

x1

y1

z1

:

~r(t) = ~r0 + (~r1 − ~r0) t

=

x0 + (x1 − x0)ty0 + (y1 − y0)tz0 + (z1 − z0)t

with tmin = 0 and tmax = 1

• The path from (0, 0, 0) to ~r1 in three sections parallel to the coordinate axes. We take it firstparallel to the x axis, then parallel to the y axis, then parallel to the z axis:

~r1 =

x1t00

tmin = 0, tmax = 1

~r2 =

x1

y1t0

tmin = 0, tmax = 1

~r3 =

x1

y1

z1t

tmin = 0, tmax = 1

The total integral will be the sum of the path integrals of each of the paths, i.e. if we call thetotal path ~γ: ˆ

γ

~F · d~r =

ˆ

γ1

~F · d~r +

ˆ

γ2

~F · d~r +

ˆ

γ3

~F · d~r

• The closed path given byx2 + y2 = R2, z = 0

This is a circle in the x, y plane, centred at (0, 0):

~r =

R cos tR sin t

0

, tmin = 0, tmax = 2π

• The closed path given by

(x− x0)2 + (y − y0)2 = R2, z = 0

This is a circle in the x, y plane, centred at (x0, y0):

~r =

R cos t+ x0

R sin t+ y0

0

, tmin = 0, tmax = 2π

• The closed path that is a cicle of radius R, in the x, y plane, centred at (x0, y0): See above.


• The semi-circular path from (0,−1, 0) to (0,+1, 0), going through (1, 0, 0). This is the “x > 0”part of the unit cirlce centred at (0, 0, 0):

~r =

cos tsin t

0

, tmin = −π

2tmax =

π

2

• The path given by

y = x2, z = 0

from x = y = 0 to x = y = 1

~r =

tt2

0

tmin = 0, tmax = 1


y = x2, z = 705

from x = y = 0 to x = y = 1

~r =

tt2

705

tmin = 0, tmax = 1


z = x2, y = 0

from x = z = 1 to x = 5, z = 25

~r =

t0t2

tmin = 1, tmax = 5


z = x2, y = x

from x = y = z = 1 to x = 5, y = 5, z = 25

~r =

ttt2

tmin = 1, tmax = 5

• The closed path formed by the straight lines connecting (0,−1, 0), (1, 0, 0), (0, 1, 0), (−1, 0, 0)and back to (0,−1, 0). Split this up into 4 paths, your line-integral will be the sum of theline integrals over each of these paths. (By the way, here ~r(t) depends on which part of thesepaths we are on, ~γ1, ~γ2, ~γ3, ~γ4. You could call them ~r1(t) etc, but I find this this confusing,


since we tend to use the notation that ~r1 is for fixed ~r and not paths.)

~γ1 =

0−10

+

100

−

0−10

t

=

t−1 + t

0

, tmin = 0, tmax = 1

~γ2 =

100

+

010

−

100

t

=

1− tt0


~γ3 =

010

+

−100

−

010

t

=

−t

1− t0


~γ4 =

−100

+

0−10

−

−100

t

=

−1 + t−t0


9.5.2 Examples of parameterised surfaces

• x− y plane in cartesian coordinates (good for rectangles)

~S(x, y) =

xy0

Limits:

– For example the square with the corners (0, 0, 0), (1, 0, 0), (1, 2, 0), (0, 2, 0)

xmin = 0, xmax = 1, ymin = 0, ymax = 2

– A disk of radius R with center (0, 0, 0), in x− y plane:

xmin = −√

1− y2, xmax =√

1− y2, ymin = 0, ymax = 1

• x− y plane in polar coordinates (good for disks)

~S(r, φ) =

r cosφr sinφ

0

Limits example: disk of radius R with center (0, 0, 0), in x− y plane:

rmin = 0, rmax = R, φmin = 0, φmax = 2π


• A plane given by

ax+ by + cz = d

~S(x, y) =

xy

1c (d− ax− by)

Limits example: the section of the plane that is in the 1st octant, i.e. where x, y, z are allpositive (for this example we’ll assume a > 0, b > 0 and d > 0 - similar reasoning works fornegative parameters, too). That x, y, z are positive translates to the following conditions:

x > 0

y > 0

z = d− ax− by > 0

The first two lead to:

xmin = 0

ymin = 0

The third

ymax =1

b(d− ax)

Adding an obvious 4th condition

ymax > ymin

leads to

1

b(d− ax) > 0

x < d/a

and therefore xmax = d/a. This was a tricky one.

• A plane parallel to the x− y plane at z = m

~S(x, y) =

xym

or

~S(x, y) =

r cosφr sinφm

• A plane parallel to the x− z plane at y = m

~S(x, z) =

xmz

or

~S(r, φ) =

r cosφm

r sinφ


• A plane parallel to the y − z plane at x = m

~S(y, z) =

myz

or

~S(r, φ) =

mr cosφr sinφ

• Sphere, centred at origin, with radius R

~S(φ, θ) =

R sin θ cosφR sin θ sinφR cos θ

Limits examples:

– Full sphere:φmin = 0, φmax = 2π, θmin = 0, θmax = π

– The half-sphere with z > 0

φmin = 0, φmax = 2π, θmin = 0, θmax =π

2

– The half-sphere with z < 0

φmin = 0, φmax = 2π, θmin =π

2, θmax = π

– The section with x > 0, y > 0 and z > 0

φmin = 0, φmax =π

2, θmin =

π

2, θmax =

π

2

• Cylinder wall, centered at z − axis, with radius R

~S(φ, z) =

R cosφR sinφz

Limits examples:

– Full cylinder of height 5, with its base on x− y plane:

φmin = 0, φmax = 2π, zmin = 0, zmax = 5

– Section of the cylinder with x > 0 and y > 0, same height as above:

φmin = 0, φmax =π

2, zmin = 0, zmax = 5

• Paraboloid, given byz = x2 + y2

in cartesian parameters

~S(x, y) =

xy

x2 + y2


• The same paraboloid, cylindrical parameters

~S(r, φ) =

r cosφr sinφr2

Because they come up so often, I recommend that you learn the sphere and the cylinder by heart(you already know them - just spherical co-ordinates with r = const for the sphere, and cylindricalco-ordinates with r = const for the cylinder). But of course you will have to be able to deal withany surface. Make sure you understand how the others come about, so you could derive themyourself if necessary. Parametersing disks, squares, rectangles etc in any of the coordinate planes(e.g. x− y plane) should also become second nature.

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