# Problem with stokes' theorem

1. Nov 8, 2009

### s_gunn

1. The problem statement, all variables and given/known data

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

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

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

2. Relevant equations

$$\int\int_{S}(curl \vec{F})\cdot\vec{n} dA = \oint_{C}\vec{F}\cdot d\vec{r}$$

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!

2. Nov 8, 2009

### lanedance

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

3. Nov 8, 2009

### lanedance

$\vec{dr}$ 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 $\vec{dr} = (dx, 0, 0)$

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

4. Nov 8, 2009

### s_gunn

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!!

5. Nov 8, 2009

### s_gunn

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!!!

6. Nov 8, 2009

### lanedance

$$\vec{F} \bullet \vec{dr} = (2z^2,6x,0)\bullet (0,1,0)dy = 6x.dy |_{x=0}$$