calculus 4 - mt202s...

CALCULUS 4 - MT202S Homework/Examples/Notes

Recommended Reading:

Lang, Loomis Calculus of several variables, Ch. 12-15Jacob, Evans A Course in Analysis Volume 2 Differentiation and Integration of Functions of SeveralVariables Vector Calculus, Part 4,5Edwards Advanced calculus of several variables, Ch. 4,5Marsden, Tromba Vector Calculus, Ch. 5,6,7,8Larsen, Hostetler, Edwards Essential Calculus, Ch. 5,12Adams, Essex Calculus Several Variables, Ch. 14

CALCULUS 4 - MT202S - ASSIGNMENT - 1

A1. Draw the sets Ωi in the plane R2 given below.

(a) Recall that the absolute value |x| of a real number x is

|x| =

x if x > 00 if x = 0−x if x < 0

.

DrawΩa = (x, y) |3y|+ |x| ≤ 6

Hint: Example E1

Solution: If (x, y) ∈ Ωa then also (±x,±y) ∈ Ωa. Thus we need only consider points withnonnegative coordinates.

x-6 -3 0 3 6

y

-2

-1

0

1

2

Ωa

(b) Ωb = (x, y) x2 ≤ 4 and (1 + x2) |y| ≤ 1.Solution: x2 ≤ 4 if and only if −2 ≤ x ≤ 2. Thus Ωb lies between the lines x = −2 and x = 2.Because of the second condition, “−1 ≤ (1 + x2)y ≤ 1 ≤ 1”, the set Ωb is bounded by the

graphs y = 11+x2

and y = − 11+x2

.

x

y

(c) Ωc =

(st, se−t) 1 ≤ s ≤ 2, 12≤ t ≤ 2

Hint: Start with the graph of the exponential function on the interval [1

2, 2]:

x0 0.5 1 2 3 4

y

0

1

2

For s ∈ [1, 2], where does the point (st, se−t) lie with respect to the point (t, e−t)? Look atexample E2

Solution:

x0 0.5 1 2 3 4

y

0

1

2

Ωc

A2. Compute the areas of the first two regions in problem A1, i.e. the areas of ΩA1a and ΩA1b.

Try the third, ΩA1c, without help from the MSC.

Solution: The first, (x, y) |3y|+ |x| ≤ 6 consists of four rectangular triangles with sides 6 and2. Hence area (x, y) |3y|+ |x| ≤ 6 = 24.

The second, (x, y) x2 ≤ 4 and (1 + x2) |y| ≤ 1, is twice the area under the graph of the functionf(x) = 1

1+x2on the interval x ∈ [−2, 2]. Hence

area

(x, y) x2 ≤ 4 and (1 + x2) |y| ≤ 1

= 2

∫ 2

−2

1

1 + x2dx = [arctan(x)]x=2

x=−2 = 4 arctan(2)

The third region,

(st, se−t) 1 ≤ s ≤ 2, 12≤ t ≤ 2

, lies between the lines x = 1/2 and x = 4. If

l(S(x0)) denotes the length of the intersection S(x0) of this region with (x0, y) y ∈ R, then

area

(st, se−t) 1 ≤ s ≤ 2,

1

2≤ t ≤ 2

=

∫ 4

1/2

l(S(x0)) dx0 .

In order to compute l(S(x0)), we need to distinguish 3 cases:

x

0 0.5 1 2 3 4

y

0

1

2case I case II case III

The point (x0, y) is in the set if

I: 1/2 ≤ x0 ≤ 1: e−x0 ≤ y ≤ 2x0e−1/2

II: 1 ≤ x0 ≤ 2: e−x0 ≤ y ≤ 2e−x0/2

III: 2 ≤ x0 ≤ 4: x0e−2

2≤ y ≤ 2e−x0/2

and the area becomes∫ 1

1/2

2xe−12 − e−x dx+

∫ 2

1

2e−x2 − e−x dx+

∫ 4

2

2e−x2 − e−2

2x dx

=15

4√e− 6

e2

A3. Compute the volume of

U =

(x, y, z) ∈ R3 1 ≤ x ≤ yz ≤ z ≤ 2.

Try to draw the set!

Solution: If (x, y, z) ∈ U then 1 ≤ x ≤ z ≤ 2, and for given x, z satisfying this condition the point(x, y, z) is in U if and only if x

z≤ y ≤ 1, i.e. y ∈ [x

z, 1], an interval of length (1− x

z). Hence

volU =

∫ 2

1

∫ 2

x

(1− x

z

)dz dx

=

∫ 2

1

[z − x ln(z)]z=2z=x dx

=

∫ 2

1

2− x− x ln(2) + x ln(x) dx

=

[2x− x2

2(1 + ln(2)) +

1

2x2 ln(x)− 1

4x2

]x=2

x=1

= 2− 3(3 + 2 ln(2))

4+ 2 ln(2) = −1

4+

1

2ln(2) .

A4. Compute ∫ 1

−1

∫ x

−x

∫ x−y

0

∫ xz

0

∫ u

0

v + x dv du dz dy dx .

Hint: Simply compute the integrals starting with the innermost. Use symmetry where you can,example E8.

Solution: ∫ 1

−1

∫ x

−x

∫ x−y

0

∫ xz

0

∫ u

0

v + x dv du dz dy dx

=

∫ 1

−1

∫ x

−x

∫ x−y

0

∫ xz

0

1

2u2 + xu du dz dy dx

=

∫ 1

−1

∫ x

−x

∫ x−y

0

1

6(xz)3 +

1

2x(xz)2 dz dy dx

=

∫ 1

−1

∫ x

−x

∫ x−y

0

1

6x3z3 +

1

2x3z2 dz dy dx

=

∫ 1

−1

∫ x

−x

1

24x3(x− y)4 +

1

6x3(x− y)3 dy dx

=

∫ 1

−1

∫ 2x

0

1

24x3w4 +

1

6x3w3 dw dx

=

∫ 1

−1

25

120x3x5 +

24

24x3x4 dx

=

∫ 1

−1

25

120x8 +

24

24x7 dx

=

∫ 1

0

26

120x8 dx =

26

3 ∗ 5 ∗ 23 ∗ 32=

8

33 ∗ 5=

8

135

Please hand up your solutions to A1-A4 on Friday, 14/2 before 4pm.


A5. Sketch the region between the curves y = 20− x2 and y = x4 and compute its area.

Solution:

x

y = 20− x2

y = x4

-3 -2 -1 0 1 2 3

y

(x, y) x4 ≤ y ≤ 20− x2

In order to compute the intersection points of the given boundary curves we need to find x so that

20− x2 = x4 i.e. x = ±2 .

Slicing along the x-axis gives for the area that

area

(x, y) x4 ≤ y ≤ 20− x2

=

∫ 2

−2

20− x2 − x4 dx =

[20x− x3

3− x5

5

]x=2

x=−2

= 80− 16

3− 64

5=

15× 80− 5× 16− 3× 64

15=

58× 16

15=

928

15.

A6. Sketch the region W in R3 given by

W =

(x, y, z) 0 ≤ x ≤ y2 ≤ z ≤ 1

and compute its volume.

Hint: W consists of two pieces, one with y ≥ 0 and the other with y ≤ 0. What are the possible y-coordinates of points (x, y, z) in W? If y0 is such a value, then the slice S(y0), i.e. the the intersectionof W with the plane y = y0 := x, y0, z x, z ∈ R is a rectangle.

Solution: There are x, z ∈ R with (x, y, z) ∈ W if and only if 0 ≤ y2 ≤ 1, i.e. if −1 ≤ y ≤ 1. Thusthe slice S(y0) is a rectangle,

S(y0) = W ∩ x, y0, z x, z ∈ R =x, y0, z 0 ≤ x ≤ y2

0, y20 ≤ z ≤ 1

x

-1 0 1

z

-1

0

1

y

-1

0

1

z = 1, y = 1

z = 1, x = 0

z = 1, y = −1

z = y2, x = 0z = y2, x = y2

z = 1, x = y2

Since S(y0) is a rectangle with side lengths 1− y20 and y2

0, its area is

areaS(y0) = (1− y20)y2

0 = y20 − y4

0

and the volume of W is

volW =

∫ 1

−1

areaS(y0) dy0 =

∫ 1

−1

y20 − y4

0 dy0 =2

3− 2

5=

4

15.

A7. Sketch the region

W =

(r cos(α), r sin(α)) 1 ≤ r ≤ 2, 0 ≤ α ≤ 1

r

.

in the plane and compute its area.

Hint: Start drawing the region (r cos(α), r sin(α)) 1 ≤ r ≤ 2, any αSolution: The arcs Cr =

(r cos(α), r sin(α)) 0 ≤ α ≤ 1

r

all have the same length 1.

x

0 1 2

y

0

1

2

C1C1.2

C1.4C1.6C1.8 C2

The region can be parametrized by

φ : Q :=

(r, α) r ∈ [1, 2], α ∈ [0,

1

r]

→ W , φ(r, α) = (r cos(α), r sin(α)) .

The Jacobian of φ is

d(r,α)φ =

(cos(α) −r sin(α)sin(α) r cos(α)

),

det d(r,α)φ = r(cos(α)2 + sin(α)2) = r .

The area of W is therefore

areaW =

∫Q

|det dφ| =

A8. Sketch the regionsU = (2x+ 2y, 2x) 0 ≤ x, y ≤ 1 and

W =

(2x+ 2y, 2x) x2 + y2 ≤ 1.

and compute their area.

Hint: You do not need to compute any integral here! Start with the sets Q = (x, y) 0 ≤ x, y ≤ 1and D = (x, y) x2 + y2 ≤ 1. The same map φ takes Q → U and D → V . What is this map?How does it transform areas?

Solution: If B denotes the ball of radius 1 in the plane, then W = AB where A is the matrix(2 20 2

).

x-2 -1 0 1 2

y

-2

-1

0

1

2

B 2B

Q

(1 10 1

)Q

(2 20 2

)Q

(2 20 2

)B

The area of W isareaW = areaAB = |detA| areaB = 4π .



A9. Consider the curveC = ((1 + t) cos(t), (1 + t) sin(t)) t ∈ [0, π]

oriented so that it starts in p = (1, 0) and ends in q = (−1− π, 0).

(a) Plot the curve C.

(b) Consider the parametrization

c : [0, π]→ C with c(t) = ((1 + t) cos(t), (1 + t) sin(t)) .

Compute the velocity c′(t) and its norm ‖c′(t)‖.(c) Express the length of the curve C as an integral.

(d) Evaluate this integral in elementary terms.

Hint: You might take

arcsinh(u) = 2

∫ u

0

√1 + t2 dt− u

√1 + u2

as the definition of arcsinh.

A10. Let F be the vector field in R2 given by

F (x, y) = (x, y)

and let C be the curve of problem A9

(a) Compute the line integral∫C〈F ds〉.

Hint: See ?? and ??.

(b) Find a function f on R2 so that F (x, y) = grad(x,y)f for all x, y.

Hint: Guess and try, a polynomial for instance. Recall that the gradient of a differentiablefunction f on Rn is the vector field gradf so that

grad(x1,...xn)f := gradf(x1, . . . xn) =

∂∂x1f(x1, . . . xn)

∂∂x2f(x1, . . . xn)

...∂∂xn

f(x1, . . . xn)

For a method to find a function with a given gradient look at example ??.

A11. Let F be the vector field on R4 given by

F (x, y, z, w) = (−y, x,−w, z)

and let c : [0, 2π] be the curve with

c(t) = (cos(t), sin(t), 1, t2) .

(a) Compute

∫c

〈F ds〉.

(b) Is F conservative?

Hint: Recall that a conservative vector field is a vector field whose line integral over everyclosed curve is zero.


Examples

E1. The setΩ =

(x, y) ∈ R2 |2x+ 3y|+ |x+ y| ≤ 3

is the region bounded by four lines: By the triangle inequality, the condition “|2x+ 3y|+|x+ y| ≤ 3”implies the four inequalities

−3 ≤ (2x+ 3y)± x+ y ≤ 3

i.e.−3 ≤ 3x+ 4y ≤ 3 and − 3 ≤ x+ 2y ≤ 3 .

Thus Ω is bounded by the four lines

−3 = 3x+ 4y , 3 = 3x+ 4y , − 3 = x+ 2y , 3 = x+ 2y ,

as shown in the diagram.

x-9 -6 -3 -1 0 1 3 6 9

y

-6

-3

-1

0

1

3

6

Ω

3 = x+ 2y

3 = 3x+ 4y

−3 = x+ 2y

−3 = 3x+ 4y

The area of Ω is twice the area of the blue/pink region which consists of a parallelogram and atriangle whose areas are 2× 3 and 1

2× 2× 3 respectively is. Thus the area of Ω is

areaΩ = 2×(

2× 3 +1

2× 2× 3

)= 18

E2. In order to draw the set

Ω =

(u+ cos(t), u+ sin(t)) 0 ≤ u ≤ 1,π

6≤ t ≤ π

3

,

we start with the curve

(cos(t), sin(t)) π6≤ t ≤ π

3

and shift it.

x

-1 0 1 2

y

-1

0

1

2

Ω

(√3

2, 1

2

)(

12,√

32

)

The slices of Ω in direction of the red arrow each have length√

2, hence the area of Ω is√

2× thewidth of the region, which is

w =

√√√√2

(√3

2− 1

2

)2

=

√(√

3− 1)2

2,

areaΩ =√

2× w =√

3− 1 .

E3. Transformation Formula: If φ : [a1, b1]×· · ·× [an, bn]→ Ω is a diffeomorphism and f a continuousfunction on Ω ⊂ Rn, then∫

Ω

f =

∫ bn

an

· · ·∫ b1

a1

f(φ(x1, . . . , xn))∣∣det d(x1,...,xn)φ

∣∣ dx1 . . . dxn .

The map φ is called a parametrization of Ω.

To compute the area ofΩ =

(uv, u− v2) ∈ R2 1 ≤ u, v ≤ 2

we use the parametrization

φ : [1, 2]× [1, 2]→ Ω , φ(u, v) = (uv, u− v2) .

x0 1 2 3 4

y

-5

-4

-3

-2

-1

0

1

2v = 1

v = 1.5

v = 2

u = 1

u = 1.5

u = 2

The derivative (Jacobian)

d(u,v)φ =

(v u1 −2v

)has determinant

det d(u,v)φ = det

(v u1 −2v

)= −2v2 − u .

SAGE: var(’u,v’) ; det(jacobian((u*v,u-v^2),(u,v)))

Hence, for 1 ≤ u, v ≤ 2, ∣∣det d(u,v)φ∣∣ = 2v2 + u

and

areaΩ =

∫ 2

1

∫ 2

1

∣∣det d(u,v)φ∣∣ du dv

=

∫ 2

1

∫ 2

1

2v2 + u du dv

=

∫ 2

1

2v2 +4− 1

2dv

=

∫ 2

1

2v2 +3

2dv

=2

3(8− 1) +

3

2=

37

6

E4. We compute the area of ΩA1c of problem A1,

ΩA1c =

(st, se−t) 1 ≤ s ≤ 2,

1

2≤ t ≤ 2

φ←−−−− [1, 2]× [

1

2, 2]

where the map φ is given byφ(s, t) = (st, se−t) .

The differential of this has determinant

det d(s,t)φ = det

(t se−t −se−t

)= se−t(−t− 1) .

For 1 ≤ s ≤ 2, 12≤ t ≤ 2 we therefore have∣∣det d(s,t)φ

∣∣ = (1 + t)se−t ,

hence

areaΩA1c =

∫ 2

12

∫ 2

1

(1+t)se−t ds dt =3

2

∫ 2

12

(1+t)e−t dt =3

2

[−2e−t − te−t

]t=2

t= 12

=3

2

(5

2e−1/2 − 4e−2

)

=15

4√e− 6

e2

E5. To compute the area ofU = (xy, x− xy) 1 ≤ x, y ≤ 2

we use the parametrization φ : [1, 2]× [1, 2]→ U , φ(x, y) = (xy, x− xy). Its Jacobian is

d(x,y)φ =

(y 1− yx −x

),

det d(x,y)φ = −xy − x(1− y) = −x∣∣det d(x,y)φ∣∣ = x for x ∈ [1, 2] .

Thus the area is

areaU =

∫ 2

1

∫ 2

1

x dy dx =

∫ 2

1

x dx =3

2.

E6. We compute the volume of

V =

(eu+v, tu, tv) 0 ≤ u, v ≤ 1, 1 ≤ t ≤ 2.

The parametrization

φ : [0, 1]× [0, 1]× [1, 2]→ V , φ(u, v, t) = (eu+v, tu, tv) .

has Jacobian

d(u,v,t)φ =

eu+v eu+v 0t 0 u0 t v

with determinant

det

eu+v eu+v 0t 0 u0 t v

= det

eu+v 0 0t −t u0 t v

(subtract the first row from the second)

= eu+v det

(−t ut v

)= −t(u+ v)eu+v .

Hence for 0 ≤ u, v ≤ 1, 1 ≤ t ≤ 2, ∣∣det d(u,v,t)φ∣∣ = t(u+ v)eu+v .

By the transformation formula, the volume of V is the (triple) integral

volV =

∫ 1

0

∫ 1

0

∫ 2

1

t(u+ v)eu+v dt du dv

3

2

∫ 1

0

∫ 1

0

(u+ v)eu+v du dv

=3

2

∫ 1

0

∫ 1

0

ueuev + veuev du dv

E7. Polar Coordinates: Every point (x, y) ∈ R2 can be written in the form

(x, y) = (r cos(α), r sin(α)) = r(cos(α), sin(α)) with r ≥ 0, α ∈ R .

r is uniquely determined by (x, y),

r =√x2 + y2 ,

but α only up to integer multiples of 2π.

For instance, consider the region

S =

(x, y) 1 ≤ x2 + y2 ≤ 9 and |x| ≤ y.

x-3 -2 -1 0 1 2 3

y

r =√x2 + y2 = 1

r =√x2 + y2 = 2

r =√x2 + y2 = 3

α = 1/4π

α = 3/8π

α = 1/2π

α = 5/8π

α = 3/4π

This can be parametrized by the function φ : [1, 3]× [π4, 3π

4]→ S,

φ(r, α) = (r cos(α), r sin(α)) .

The derivative of this parametrization is given by

d(r,α)φ =

(cos(α) −r sin(α)sin(α) r cos(α)

),

det d(r,α)φ = r(cos(α)2 + sin(α)2) = r .

SAGE: var(’r,a’) ; det(jacobian((r*cos(a),r*sin(a)),(r,a)))

or

SAGE: var(’r,a’) ; (det(jacobian((r*cos(a),r*sin(a)),(r,a)))).simplify_trig()

to simplify trigonometric expressions, here sin2 + cos2 to 1.

For the area this gives

areaS =

∫ 3π4

π4

∫ 3

1

∣∣det d(r,α)φ∣∣ dr dα =

∫ 3π4

π4

∫ 3

1

r dr dα

=π

2

[r2

2

]r=3

r=1

= 2π

E8. In order to compute the integral∫ 2

−2

∫ a

−a

∫ b

−b

∫ abc

−abcv + a+ b+ c dv dc db da

we first use that the function f with f(v) = v is odd, hence its integral over a symmetric interval

vanishes,∫ h−h v dv = 0. Thus∫ 2

−2

∫ a

−a

∫ b

−b

∫ abc

−abcv + a+ b+ c dv dc db da

=

∫ 2

−2

∫ a

−a

∫ b

−b

∫ abc

−abca+ b+ c dv dc db da

=

∫ 2

−2

∫ a

−a

∫ b

−b[(a+ b+ c)v]v=abc

v=−abc dc db da

=

∫ 2

−2

∫ a

−a

∫ b

−b2(a+ b+ c)abc dc db da

=

∫ 2

−2

∫ a

−a

∫ b

−b2abc2 dc db da

=

∫ 2

−2

∫ a

−a2ab

2

3b3 db da

=4

3

∫ 2

−2

∫ a

−aab4 db da

=8

3× 5

∫ 2

−2

a6 da =16× 27

3× 5× 7=

2014

105

We compute the area of

W =

(r cos(t), r sin(t)) 0 ≤ t ≤ 2π, 1 + t ≤ r ≤ 3

2+ t

.

We use the parametrization φ : Q → W by polar coordinates, φ(r, t) = (r cos(t), r sin(t)). Thedomain

Q =

(r, t) 0 ≤ t ≤ 2π, 1 + t ≤ r ≤ 3

2+ t

of the parametrization is the parallelogram bounded by the lines

t = 0 , t = 2π , 1 + t = r ,3

2+ t = r

t-1 0 1 2 3 4 5 6 7

r

-1

0

1

2

3

4

5

6

7

8

9

t = 0 t = π

r = t+ 1

r = t+ 32

To sketch W , we sketch the curves

Cλ = (r cos(t), r sin(t)) 0 ≤ t ≤ 2π, λ+ t = r

for λ ∈ [0, 12],

x-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

y

-7

-6

-5

-4

-3

-2

-1

0

1

2

3

4

λ = 1

λ = 1.25

λ = 1.5

Since the Jacobian determinant of the parametrization is

d(r,t)φ = r

the area of W is

areaW =

∫Q

∣∣det d(r,t)φ∣∣ =

∫ 2π

0

∫ 32

+t

1+t

r dr dt

=

∫ 2π

0

(32

+ t)2 − (1 + t)2

2dt =

∫ 2π

0

(12

+ 1 + t)2 − (1 + t)2

2dt

=

∫ 2π

0

1 + t+ 14

2dt =

5π

4+ π2 .

E9. Surface Area Let φ : Q → S ⊂ RN be a parametrization of the surface S, Q ⊂ Rn, n ≤ N . Let fbe a continuous function on S. Then∫

S

f =

∫Q

f(φ(x))√

det(dxφtdxφ) dnx .

in particular, the volume/area of S is

areaS =

∫Q

√det(dxφtdxφ) dnx .

For example, we compute the area of the cone

C =

(x, y, z) 1 ≤ x ≤ 2,√y2 + z2 = x

= (x, x cos(α), x sin(α)) 1 ≤ x ≤ 2, 0 ≤ α ≤ 2π

parametrized by

φ : Q = (x, α) 1 ≤ x ≤ 2, 0 ≤ α ≤ 2π = [1, 2]× [0, 2π]︸︷︷︸⊂R2

→ C ⊂ R3 ,

φ(x, α) = (x, x cos(α), x sin(α)) .

x

0

1

2

y

-2-1

01

23

4

z

-1

0

1

2

The Jacobian of the parametrization is

d(x,α)φ =

1 0cos(α) −x sin(α)sin(α) x cos(α)

with transpose d(x,α)φt =

(1 cos(α) sin(α)0 −x sin(α) x cos(α)

),

d(x,α)φtd(x,α)φ =

(1 cos(α) sin(α)0 −x sin(α) x cos(α)

) 1 0cos(α) −x sin(α)sin(α) x cos(α)

=

(2 00 x2

),

det d(x,α)φtd(x,α)φ = (x+ 1)x = 2x2 ,√

det(dxφtdxφ) =√

2x . (E9)

Thus the area of S is

areaC =

∫Q

√det(dxφtdxφ) d2x =

∫ 2

1

∫ 2π

0

√2 x dα dx =

∫ 2

1

2π√

2 x dx

= 3√

2π .

E10. Area of a Graph of a Function: If h : U ⊂ Rk → S ⊂ R is a continuously differentiable function,then the graph of h is the surface

S = (x1, . . . , xk, h(x1, . . . , xk)) (x1, . . . , xk) ∈ U .

The parametrization φ : U → S,

φ(x1, . . . , xk) = (x1, . . . , xk, h(x1, . . . , xk))

of S has derivative at x = (x1, . . . , xk) given by the (k + 1)× k-matrix

dxφ =

1 0 · · · 00 1 · · · 0...

. . ....

0 0 · · · 1∂h∂x1

(x) ∂h∂x2

(x) · · · ∂h∂xk

(x)

=

(Ik

gradxh

)

Its transpose is the k × (k + 1)-matrix

dxφt =

(Ik gradxh

).

In order to compute the determinant of the k × k-matrix dxφtdxφ, we choose coordinates so that

gradxh =

(∂h

∂x1

(x),∂h

∂x2

(x), · · · , ∂h∂xk

(x)

)=

(∂h

∂x1

(x), 0, · · · , 0).

Then

dxφtdxφ =

1 0 · · · 0 ∂h

∂x1(x)

0 1 · · · 0 0...

. . ....

...0 0 · · · 1 0

1 0 · · · 00 1 · · · 0...

. . ....

0 0 · · · 1∂h∂x1

(x) 0 · · · 0

=

1 +

(∂h∂x1

(x))2

0 · · · 0

0 1 · · · 0...

.... . .

...0 0 · · · 1

.

The determinant of this isdet dxφ

tdxφ = 1 + ‖gradxh‖2 .

Thus the area of the graph S of h is

areaS =

∫U

√1 + ‖gradxh‖

2 dkx . (E10)

Example: We compute the area of the graph of the function

h : D =

(x, y) x2 + y2 ≤ 1→ R , h(x, y) = x2 − y2 .

The gradient of h is given by

grad(x,y)h =

(2x−2y

)

has norm ∥∥grad(x,y)h∥∥2

=

∥∥∥∥( 2x−2y

)∥∥∥∥2

= 4(x2 + y2) ,

hence

areaS =

∫D

√1 + 4(x2 + y2)︸︷︷︸

=:g(x,y)

dx dy .

SAGE: You can type the surface area formula (E10) for a graph almost directly into sage, you needto put in the correct bound for the integrals. What needs to be put in below in place of the xxxxx ?

h(x,y)=x^2-y^2 ;

integral(integral(

sqrt(1+norm(derivative(h))^2)(x,y),

y,xxxx,xxxx),

x,-1,1)

We compute this in polar coordinates,

D = φ(Q) , Q = [0, 2π)× [0, 2π) , φ(r, α) = (r cos(α), r sin(α)) .

Since det d(r,α)φ = r,

areaS =

∫Q

g(φ(r, α)) det d(r,α)φ dr dα

=

∫ 2π

0

∫ 1

0

√1 + 4r2 r dr dα

= 2π

∫ 1

0

√1 + 4r2 r dr

= 2π

[(1 + 4r2)

3/2

12

]r=1

r=0

=π

6(5√

5− 1) .

E11. We compute the area of the surface

S =(t,(t+

α

2π

)cos(α),

(t+

α

2π

)sin(α)

)0 ≤ α ≤ 3π, 1 ≤ t ≤ 2

⊂ R3

x

0

1

2

y

-2 -1 0 1 2 3 4

z

-1

0

1

2

We parametrize S by

φ : W = (t, α) 0 ≤ α ≤ 3π, 1 ≤ t ≤ 2 = [0, 3π]× [1, 2]→ S

φ(t, α) =(t,(t+

α

2π

)cos(α),

(t+

α

2π

)sin(α)

)d(t,α)φ =

1 0

cos(α) −(t+ α

2π

)sin(α) + cos(α)

2π

sin(α)(t+ α

2π

)cos(α) + sin(α)

2π

.

det(d(t,α)φ

td(t,α)φ)

= det

(2 1

2π1

2π

(t+ α

2π

)2+(

12π

)2

)Hence

areaS =

∫ 3π

0

∫ 2

1

√2t2 +

2tα

π+α2 + 1

2π2− 1

4π2dt dα .

E12. A (parametrized) curve in Rn is a piecewise continuously differentiable function c : [a, b] → Rn.“piecewise” here means that the curve need not be differentable at finitely many points of the interval[a, b]. This is to allow some corners. The following are parametrized curves.

(a) Parabola “y = x2”: c : [−1, 1]→ R2, c(t) = (t, t2) x-1 0 1

y = x2

y

(b) Square: c : [0, 4]→ R2, c(t) =

(t, 0) if 0 ≤ t ≤ 1(1, t− 1) if 1 ≤ t ≤ 2(1− (t− 2), 1) if 2 ≤ t ≤ 3(0, 1− (t− 3)) if 3 ≤ t ≤ 4

x0 1

y

(c) Neil’s Parabola “y2 = x3”: c : [−1, 1]→ R2, c(t) = (t2, t3)

x-1 0 1

y

-1

0

1 y2 = x3

y2 = x3

(d) Helix: c : [0, 2π]→ R3, c(t) =

(cos(t), sin(t),

t

2π

)

x

01

2

y

-2-1

01

2

z

-1

0

1

2

E13. Length of a parametrized curve: The integral

length (c) =

∫ b

a

‖c′(t)‖ dt .

is called the (arc)length of the parametrized curve c : [a, b]→ Rn.

We compute the lengths of the curves in E12:

(a) Parabola: c : [−1, 1]→ R2, c(t) = (t, t2), has derivative

c′(t) = (1, 2t)

with norm‖c′(t)‖ =

√1 + (2t)2 =

√1 + 4t2 .

Hence the length integral is

length (c) =

∫ 1

−1

‖c′(t)‖ dt =

∫ 1

−1

‖c′(t)‖ dt

=

∫ 1

−1

√1 + 4t2 dt =

arcsinh(√

2)√2

+√

3 .

(b) Square: c : [0, 4]→ R2,

c(t) =

(t, 0) if 0 ≤ t ≤ 1(1, t− 1) if 1 ≤ t ≤ 2(1− (t− 2), 1) if 2 ≤ t ≤ 3(0, 1− (t− 3)) if 3 ≤ t ≤ 4

This has derivative

c′(t) =

(1, 0) if 0 ≤ t ≤ 1(0, 1) if 1 ≤ t ≤ 2(−1, 0) if 2 ≤ t ≤ 3(0,−1) if 3 ≤ t ≤ 4

and its norm is always 1, hence

length (c) =

∫ 4

0

‖c′(t)‖ dt = 4 .

(c) Neil’s Parabola: c : [−1, 1]→ R2, c(t) = (t2, t3) has derivative

c′(t) = (2t, 3t2) with norm ‖c′(t)‖ =√

(2t)2 + (3t2)2 =√

4t2 + 9t4 .

Hence the length integral becomes

length (c) =

∫ 1

−1

‖c′(t)‖ dt =

∫ 1

−1

√4t2 + 9t4 dt

= 2

∫ 1

0

√4 + 9t2 t dt︸︷︷︸

d t2

2

= 2

∫ 1/2

0

√4 + 18u du =

4

3× 18

[(4 + 18u)

32

]u=1/2

u=0=

2

27

(133/2 − 8

)

(d) Helix: c : [0, 2π]→ R3, c(t) =

(cos(t), sin(t),

t

2π

)has speed

c′(t) =

(− sin(t), cos(t),

1

2π

)with norm

‖c′(t)‖ =

√(− sin(t))2 + cos(t)2 +

1

4π2=

√1 +

1

4π2.

Hence the length integral becomes

length (c) =

∫ 2π

0

‖c′(t)‖ dt =

∫ 2π

0

√1 +

1

4π2dt

= 2π

√1 +

1

4π2=√

4π2 + 1 .

E14. In many cases, you will not be able to express the length integral in elementary terms. For instance,the curve

c : [0, 2]→ R2 , c(t) = (t, t3)

has length∫ 2

0

‖c′(t)‖ dt =

∫ 2

0

∥∥∥∥∥ d

du

∣∣∣∣∣u=t

(uu3

)∥∥∥∥∥ dt =

∫ 2

0

∥∥∥∥( 13t2

)∥∥∥∥ dt =

∫ 2

0

√1 + (3t2)2 dt =

∫ 2

0

√1 + 9t4 dt

E15. Or the integral is difficult: The graph of the exponential function x 7→ ex, 0 ≤ x ≤ 2, is parametrizedby

c : [0, 2]→ R2 , c(t) = (t, et)

and has length∫ 2

0

‖c′(t)‖ dt =

∫ 2

0

∥∥∥∥∥ d

dv

∣∣∣∣∣v=t

(vev

)∥∥∥∥∥ dt =

∫ 2

0

∥∥∥∥(1et

)∥∥∥∥ dt =

∫ 2

0

√1 + (et)2 dt =

∫ 2

0

√1 + e2t dt

= −√

2 +√e4 + 1 +

1

2log(√

2 + 1)− 1

2log(√

2− 1)− 1

2log(√

e4 + 1 + 1)

+1

2log(√

e4 + 1− 1)

sage: integrate(sqrt(1+e^(2*x)),x,0,2)

-sqrt(2) + sqrt(e^4 + 1) + 1/2*log(sqrt(2) + 1) - 1/2*log(sqrt(2) - 1)

- 1/2*log(sqrt(e^4 + 1) + 1) + 1/2*log(sqrt(e^4 + 1) - 1)

E16. Length of a Curve: Let C ⊂ Rn be a curve and c : [a, b]→ C a parametrization.

Then the length of the curve C is

lengthC =

∫ b

a

‖c′(t)‖ dt .

is called the (arc)length of the curve c.

Example: Consider the curve C = (x, y) ∈ R2 x2 + y2 = 1. Since we can write every pair of realnumbers whose squares add up to 1 in the form (cos(t), sin(t)) with some t ∈ [0, 2π] (the “angle”),we can parametrize C,

C =

(x, y) ∈ R2 x2 + y2 = 1

=

(cos(t), sin(t))︸︷︷︸=c(t)

t ∈ [0, 2π]

= c(t) t ∈ [0, 2π]

wherec : [0, 2π]→ C , c(t) = (cos(t), sin(t)) .

The parametrization c is injective on (0, 2π) = [0, 2π] \ 0, 2π. The speed c′ of c is

c′(t) =d

dt′

∣∣∣∣∣t′=t

(cos(t′)sin(t′)

)=

(− sin(t)cos(t)

),

‖c′(t)‖ =√

(− sin(t))2 + (cos(t))2 = 1 .

We compute the lenght,

length (C) = length (c) =

∫ 2π

0

‖c′(t)‖ dt =

∫ 2π

0

1 dt = 2π .

E17. This is easy if the curve C ⊂ Rn is given explicitly, i.e. in the form

C = c(t) t ∈ [a, b]

with some parametrized simple curve c : [a, b]→ Rn.

Example: Consider the logarithmic spiral

C =

(t cos(ln(t)), t sin(ln(t)) t ∈ [0, e4π].

We can use the parametrization

c : [0, e4π]→ C , c(t) = (t cos(ln(t)), t sin(ln(t))

to compute the length of C,

length (C) = length (c) =

∫ e4π

0

∥∥∥∥∥ d

dt′

∣∣∣∣∣t′=t

(t′ cos(ln(t′))t′ sin(ln(t′))

)∥∥∥∥∥ dt

=

∫ e4π

0

∥∥∥∥(cos(ln(t))− sin(ln(t))sin(ln(t)) + cos(ln(t))

)∥∥∥∥ dt

=

∫ e4π

0

(cos(ln(t))− sin(ln(t)))2 + (sin(ln(t)) + cos(ln(t)))2 dt

=

∫ e4π

0

2 cos(ln(t))2 + 2 sin(ln(t))2 dt = 2e4π .

calculus 4 - mt202s...

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