(508696274) vector iii 2015

Surface Integral

Surface Integral

1Sk/EUM114/ Vector 3 Lecture/2015

Surface Integral


Surface Integral

Stoke’s Theorem


Surface Integral


Surface Integral

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

The triple integral is a generalization of the double integral introduced above.

For defining this integral,

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

planes parallel to the three coordinate planes. Then, one numbers the parallelepipeds inside T from 1 to n.16

Sk/EUM114/ Vector 3 Lecture/2015

Surface Integral

xk , yk , zk In each such parallelepiped, one chooses an arbitrary point, say in the kth

parallelepiped, and form the sum

n

f xk , yk , zk vk

k 1

(1)Jn

where vk 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

f x, y, z dxdydz f x, y, z dvT

or (2)T

Fˆ nˆd div Fˆdxdydz div

Fˆdv

(3)S T T

Example


Surface Integral

Compute Fˆ nˆ d , whereS

Fˆ x, y, z x2iˆ 2 yˆj 4z2kˆ and S is the surface of the cylinder

x2 y2 4, 0 z 2 .

Solution

Fˆdiv 2x 2 8z , so

2

Fˆ nˆ 2x 2 8z dzdxdy

2x 2 8z dzdxdy

0x2 y2 4S T

2 xz 2 z 4 z 2 z=2 dxdy 4 x 20 dxdy

x2 y2 4

x2

y2 4

z=0

Let x r cos and y r sin , then x2 y2 4 r 2 cos2 r 2 sin2 r 2 r 2 .

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

2 2 2 2 r cos 5r drd4r cos 20)rdrd 4 2 0 0 0 0

2 8 4 cos 10 d 800 3


Surface Integral

Example

By transforming to a triple integral, evaluate

I x3dydz x2 y dxdz x2 z dxdy S

x2 y2 a2 0 z bwhere S is the closed surface consisting of the cylinder and the dicks z=0

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

P x3 , Q x2 y, R x2 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)

a 1b a a2 y23/ 2 b

I 3x2 x2 x2 dxdydz 4 5 a2 y2 dydzx2 dxdydz 20 0 0 30 0 0 T

Let y a sin , dy a cos d , then

3/ 2

3/ 2

a2 3/ 2

a2

cos2

a3

cos3

a2 y2 1 sin2

and

1 3 a41 3/ 2 1 /2 aa2 y2 dy a4 cos4 d a4 3 3 3 8 2 160 0

Hence, the integral I becomes


Surface Integral

4b a 5I 20 dz

a4b16 40

Example

F

ˆ

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

x iˆ

y ˆj z

kˆ nˆ cos x, n iˆ cos y, n ˆj cos z, n kˆ Fˆr r r

where

r x2 y2 z 2 and the region T is the sphere x2 y2 z2 a2

Evaluate the triple integral :

Solution

Assuming that r 0

x2 y2 z 2 x2 / x2 y2 z 2 r 2 x2P x

x x x2 y2 z 2 r3

x2 y2 z 2

Likewise,

Q r 2 y2 R r 2 z 2

and r 3 r 3y z

Hence,

3r 2 r 2 2 2 rdiv F if r 0 and dxdydzr3

r T

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

the equivalent surface integral reveals that

2 2 2 x

y

z Fˆ nˆd d 1d 4 a2S

r 2 Sr 2 r 2

S

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

containing a discontinuity at r=0.


Surface Integral

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 . dv dxdydz r 2 sin drdd

2 2 a 1 2

dxdydz 2 r 2 sin drd d a2 sin d dr r0 0 0 0 0

T

2 cos 1 1 4 a2 2 2 d 2 a a0 0

As expected, it yields the same result.


S (dx)i (dz)k (dxdz) j

S (dy) j (dz)k (dydz)i

Surface Integral

Hints on Surface Itegrals

surface S’ with the

Case 1. Parametrize and IntegrateWrite 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 (Gauss’s) Theorema) The Divergence Theorem has exactly the same flavor as Stoke’s 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)

c)

In the xz plane, d

In the yz plane, d

Case 3.In Stoke’s 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 thecurve by the simplest curve with the same boundary, namely a line joining the end points. In practice onenever does this since it is simpler to evaluate f at the end points.22Sk/EUM114/ Vector 3 Lecture/2015

Is

No Yes

Is S closed? Is S closed?

No Yes No Yes

Parametrize andIntegrate

Choose simplest

same boundary=0

Surface Integral

In Gauss’s theorem, ifDiv(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 insidethe 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 Fluxis zero! Whatever goo comes in, must come out..So Why is this Important?

Gauss’s and Stoke’s theorem are central to Physics. Without Gradients, Curls and Divergences, one couldnot give a full quantitative description of Gravitation, Electrodynamics, Fluid Dynamics, AerospaceEngineering, Atmospheric Sciences…you get the point.


Surface Integral

Summary of Surface Integrals

Indicate How to Compute

dz dz

f ( x, y, z)

1 dA , where dx dy D

f (P* )Slimm,n

1 1S

Du v

S

S

dS , OR

ndS

C S

* *F (r r )dA , where D is the pre-image

m,nS E

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.


inte

gral

of

a ve

ctor

Type of Int.

Notation Definition andInterpretations/Applications

Standard/Direct Way to ComputeMajor Theorems that

Indirectly

surf

ace

inte

gral

of

a re

al-v

alu

ed f

un

ctio

n

f ( x, y, z)dS

m n

ij ij

i j

(mass/moments of a surface)

2 2

z g(x, y) is the function whose graph is the surface S of integration and D is the

projection of S; OR f (r(u, v)) r r dA

where D is the pre-image (in the uv-plane) ofS

Su

rfac

e (o

r fl

ux)

fiel

d

F

F

(this notation infers outward orientation if S is closed and upward orientation if S is not closed)

Based on an upward orientation of S

m n

lim FPij nPij S ij

i1 j 1

(rate of flow through a surface)

g gD

P(x, y, z) x

Q(x, y, z) y

R(x, y, z) dA

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 u v

(in the uv-plane) of S

Stokes’ Theorem:

F dr curlF dS where

C

is the boundary of SDivergence Theorem:

F dS divFdV where

S

(508696274) vector iii 2015

Documents

x2 x2 y2 z

x r cos

x2px x x x2 y2 z 2r3x2

dicks z

fr r rwherer x2 y2 z

z dxdydz f x

r x2 zhence

cos2 r