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Problem with stokes' theorem

by s_gunn
Tags: stokes, theorem
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s_gunn
#1
Nov8-09, 10:34 AM
P: 34
1. The problem statement, all variables and given/known data

Let: [tex]\vec{F}(x,y,z) = (2z^{2},6x,0),[/tex] and S be the square: [tex]0\leq x\leq1, 0\leq y\leq1, z=1. [/tex]

a) Evaluate the surface integral (directly):
[tex]\int\int_{S}(curl \vec{F})\cdot\vec{n} dA[/tex]

b) Apply Stokes' Theorem and determine the integral by evaluating the corresponding contour integral.

2. Relevant equations

[tex]\int\int_{S}(curl \vec{F})\cdot\vec{n} dA = \oint_{C}\vec{F}\cdot d\vec{r}[/tex]

3. The attempt at a solution

a) Basically I took the curl of F and got (0, 4z, 6) and dotted it with the normal vector which is either (0,0,+1) or (0,0,-1) as no orientation was given in the question! which gave me a value of +6 or -6.

b) now this is the part which has confused me!! How on earth do I find the corresponding contour integral?? I have no idea how to find dr and I'm not sure what to do about parametizing the square.

Any help would be greatly appreciated!!!

P.S. I have a similar problem involving Gauss' Law so am hoping to kill two birds with one stone!
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lanedance
#2
Nov8-09, 10:46 AM
HW Helper
P: 3,307
the normal to use for the first part is determeined by a right hand rule, to line up with the direction you go around the contour

now to do the contour integral, see it as the sum of 4 line intergals, one for each side of the square
lanedance
#3
Nov8-09, 10:49 AM
HW Helper
P: 3,307
[itex] \vec{dr} [/itex] will be an infinitesimal in the direction of the line, be careful with direction. eg if the direction is the +ve x, it essentially reduces to [itex] \vec{dr} = (dx, 0, 0) [/itex]

Pick a way to go around say counter clockwise form above and stick to it, setting the [itex] \vec{dr} [/itex] direction & limits accordingly

s_gunn
#4
Nov8-09, 11:20 AM
P: 34
Problem with stokes' theorem

I still don't get it!!

if i name the corners of the square a,b,c and d for (0,0,1), (0,1,1), (1,1,1) and (1,0,1) respectively, then will dr be:
a->b: (0,dy,0)
b->c: (dx,0,0)
c->d: (0,-dy,0)
d->a: (-dx,0,0)?

But then when I dot product with f=(2z^2,6x,0) and integrate each seperately, I get:
6x + 2z^2 - 6x - 2z^2 = 0

Forgive me if I'm being stupid but we've only ever done stokes' law the other way round so I've never had any experience this way!!
s_gunn
#5
Nov8-09, 02:29 PM
P: 34
Can anyone find a way to explain this to me or point me in the right direction? I've been searching online and my books for hours and the only examples that I find are the opposite way round!!!
lanedance
#6
Nov8-09, 06:32 PM
HW Helper
P: 3,307
Quote Quote by s_gunn View Post
I still don't get it!!

if i name the corners of the square a,b,c and d for (0,0,1), (0,1,1), (1,1,1) and (1,0,1) respectively, then will dr be:
a->b: (0,dy,0)
b->c: (dx,0,0)
c->d: (0,-dy,0)
d->a: (-dx,0,0)?

But then when I dot product with f=(2z^2,6x,0) and integrate each seperately, I get:
6x + 2z^2 - 6x - 2z^2 = 0

Forgive me if I'm being stupid but we've only ever done stokes' law the other way round so I've never had any experience this way!!
show your working!

these are definte integrals, so you should end up with a number for each integral, not a function...

so for a->b:(0,0,1)->(0,1,1) along the y axis
notice x = 0, z=1, only y is changing
[tex] \vec{F} \bullet \vec{dr} = (2z^2,6x,0)\bullet (0,1,0)dy = 6x.dy |_{x=0}[/tex]

so there is actually no field component parallel to dy in the integral...

note the dot product method is a slight shortcut in notation, really should parameteris the line in terms of a variable & integrate, though as teh intergals are along axes in this question we can take the shortcut


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