vector iii 2015

Surface Integral

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Sk/EUM114/ Vector 3 Lecture/2015

Surface Integral

Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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Stokes Theorem

Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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Surface Integral

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Triple Integrals - Divergence Theorem of Gauss (1777-1855)

The triple integral is a generalization of the double integral introduced above. For defining this integral,

one considers a function , ,f x y z defined in a bounded closed region T of space. One subdivides T by

planes parallel to the three coordinate planes. Then, one numbers the parallelepipeds inside T from 1 to n.

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In each such parallelepiped, one chooses an arbitrary point, say , ,k k kx y z in the kth

parallelepiped, and form the sum

1

, ,n

n k k k k

k

J f x y z v

(1)

wherekv is the volume of the kth parallelepiped. This limit is called the triple integral of

, ,f x y z over the region T, and is denoted by

, , or , ,T T

f x y z dxdydz f x y z dv (2)

S T T

F nd div Fdxdydz div Fdv (3)

Example

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Compute ,S

F n d where 2 2 , , 2 4F x y z x i yj z k and S is the surface of the cylinder

2 2 4, 0 2x y z .

Solution

div 2 2 8F x z , so

2 2

2 2 2 2

2

04

2

4 4

2 2 8 2 2 8

z=2 2 2 4 4 20

z=0

S T x y

x y x y

F n x z dzdxdy x z dzdxdy

xz z z dxdy x dxdy

Let cosx r and siny r , then 2 2 2 2 2 2 24 cos sin 2x y r r r r .

The differential area in the polar coordinate system becomes dA dxdy rdrd .)

2 2 2 2

2

0 0 0 0

2

0

4 cos 20) 4 cos 5

8 4 cos 10 80

3

r rdrd r r drd

d

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Example

By transforming to a triple integral, evaluate

3 2 2S

I x dydz x y dxdz x z dxdy

where S is the closed surface consisting of the cylinder 2 2 2 0 z bx y a and the dicks z=0

and z=b. From (6), it is evident that

3 2 2, ,P x Q x y R x z

Hence, by taking advantage of the symmetry of the circle (evaluate only 1/4 of the circle), the

corresponding triple integral takes the from by (6)

2 2

3/22 2 2 2 2 2

0 0 0 0 0

13 4 5 20

3

b a a y b a

T

I x x x dxdydz x dxdydz a y dydz

Let sin , cosy a dy a d , then

3/23/2 3/2

2 2 2 2 2 2 3 31 sin cos cosa y a a a

and

43/2 /2

2 2 4 4 4

0 0

1 1 1 3cos

3 3 3 8 2 16

a aa y dy a d a

Hence, the integral I becomes

Surface Integral

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44

0

520

16 4

b aI dz a b

Example

If F and n are expressed in terms of their Cartesian components such that

cos , cos , cos ,x y z

F n x n i y n j z n k i j kr r r

where

2 2 2r x y z and the region T is the sphere 2 2 2 2x y z a

Evaluate the triple integral :

Solution

Assuming that 0r

2 2 2 2 2 2 2 2 2

2 2 2 32 2 2

/x y z x x y zP x r x

x x x y z rx y z

Likewise,

2 2

3

Q r y

y r

and

2 2

3

R r z

z r

Hence,

2 2

3

3 20

r rdiv F if r

r r

and

2

T

dxdydzr

Since the triple integral is discontinuous when r=0, care must be exercised. Returning back to

the equivalent surface integral reveals that

2 2 22

2 2 2 1 4

S S S

x y zF nd d d a

r r r

The integral is simply the surface area of a sphere of radius a, thus avoiding an integral

containing a discontinuity at r=0.

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The triple integral may be evaluated by transforming the Cartesian coordinates into the spherical

coordinates. From the sketch, it is clear that

In this case, r . 2 sindv dxdydz r drd d

2 22 2

0 0 0 0 0

22 2 2

0

2 12 sin sin

cos 2 1 1 40

a

T

dxdydz r drd d a d dr r

a d a a

As expected, it yields the same result.

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Hints on Surface Itegrals

Case 1. Parametrize and Integrate

Write the parametric equation r(u,v) of the surface. Find dS = (ru x rv) du dv. Take the dot

product with F and integrate over u and v.

Case 2. Use the Divergence (Gausss) Theorem a) The Divergence Theorem has exactly the same flavor as Stokes Theorem.

You go up one dimension

You take some kind of derivative of the integrand

You integrate over the region bounded by the closed shape

b) In the xz plane, ( ) ( ) ( )dS dx i dz k dxdz j

c) In the yz plane, ( ) ( ) ( )dS dy j dz k dydz i

Case 3.

In Stokes theorem, Curl (F ) =0, implies that F = Grad (f.) for some f.

Then it is easy to find the potential f and evaluate at the end points. Another way of saying this is that the

integral depends only on the boundary of the curve, namely the end points. Thus, one could replace the

curve by the simplest curve with the same boundary, namely a line joining the end points. In practice one

never does this since it is simpler to evaluate f at the end points.

Is

No

Is S closed?

No

Parametrize and Integrate

Yes

Yes

Is S closed?

No

Choose simplest surface S with the

same boundary

Yes

=0

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In Gausss theorem, if Div(F)=0, implies that F=Curl(A) for some A.

Now it is not easy to find the vector potential A. But the situation is similar. The integral depends only on

the boundary, so one can replace the original surface S with the simplest surface that has the same

boundary. So, for example, if the surface is the upper hemisphere z=sqrt(a^2-x^2-y^2), where the

computation of the differential of surface is nasty, one could use the surface z=0, (where dS = dx dy k)

and integrate over a circle of radius a.

Think of this as the butterfly net catcher theorem. If Div (F) =0 then there are no sources or sinks inside

the surface, the flux of butterflies coming into the surface net depends only on the Rim of the net.

Case 4.

If Div(F)=0 and the surface is closed, then there are no sources or sinks inside the surface, so the net Flux

is zero! Whatever goo comes in, must come out..

So Why is this Important?

Gausss and Stokes theorem are central to Physics. Without Gradients, Curls and Divergences, one could not give a full quantitative description of Gravitation, Electrodynamics, Fluid Dynamics, Aerospace

Engineering, Atmospheric Sciencesyou get the point.

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Summary of Surface Integrals

Type

of Int. Notation

Definition and

Interpretations/Applications Standard/Direct Way to Compute

Major Theorems that

Indicate How to Compute

Indirectly

surf

ace

in

teg

ral

of

a

real-

valu

ed f

un

ctio

n

( , , )S

f x y z dS

*

,1 1

lim ( )m n

ij ijm n

i j

f P S

(mass/moments of a surface)

22

( , , ) 1D

dz dzf x y z dA

dx dy

, where

( , )z g x y is the function whose graph is the

surface S of integration and D is the

projection of S; OR ( ( , )) u vD

f u v dA r r r where D is the pre-image (in the uv-plane) of

S

Su

rface

(or

flu

x)

inte

gra

l of

a v

ecto

r

fiel

d

S dSF , OR

S dSnF (this notation infers

outward orientation

if S is closed and

upward orientation

if S is not closed)

m

i

n

j

ijijijnm

SPP1 1

**

,lim nF

(rate of flow through a surface)

( , , ) ( , , ) ( , , )D

g gP x y z Q x y z R x y z dA

x y

where z = g(x,y) is the function whose graph

is the surface S of integration and D is the

projection of S; OR

D vu dA)( rrF , where D is the pre-image (in the uv-plane) of S

Stokes Theorem:

SC dd SFrF curl where C is the boundary of S

Divergence Theorem:

ES dVd FSF div where S is the boundary of E

Note: There are analogous notations, definitions, and formulas for situations in which the roles of x, y, and/or z are swapped.

Based on an upward

orientation of S

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