# Stokes' Theorem Question

1. Nov 27, 2009

### goliath11

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

Note: the bullets in the equations are dot products, the X are cross products

Evaluate: [over curve c]$$\oint$$( F $$\bullet$$ dr ) where F = < exp(x^2), x + sin(y^2) , z> and C is the curve formed by the intersection of the cone: z = $$\sqrt{(x^2 + y^2)}$$ and the cylinder: x^2 + (y-1)^2 = 1 oriented CCW looking down from + z axis

2. Relevant equations

I'm assuming this is a Stokes' theorem question, so:

[over curve C]$$\oint$$ F $$\bullet$$ dr = [over surface S]$$\int$$$$\int$$( curl(F) $$\bullet$$ dS)

3. The attempt at a solution

First, the surface looks like it's on the cone, so paramaterizing the cone in polar coordinates:

(S) r(t,r) = < rcost , rsint , r >

now for finding curl(F) i did: gradient X F = <0,0,1>

So according to the equation mentioned above,

[over curve C]$$\oint$$ F $$\bullet$$ dr = [over surface S]$$\int$$$$\int$$( curl(F) $$\bullet$$ dS)

and: [over surface S]$$\int$$$$\int$$( curl(F) $$\bullet$$ dS) = [over domain D]$$\int$$$$\int$$ curl(F) $$\bullet$$ < r(partial derivative with t) X r(partial derivative with r) > dA

so i found curl(F) already, < r(partial with t) X r(partial with r) > = <rcost, rsint, -r> , and dA = rdrdt
the domain is the domain of the parameters, (t,r), so plugging in the cylinder equation i get:

r = 2sint, so: 0 < r < 2sint and 0 < t < pi , and plugging all that in i get:

$$\int$$[0<t<pi]$$\int$$[0<r<2sint] <0,0,1> $$\bullet$$ <rcost,rsint,-t> r dr dt

i get this down to (-8/2)$$\int$$[0 to pi] (sint)^3 dt which is = -32/9.

I know the answer is supposed to = pi , but I have no idea what I'm doing wrong. Any help would be greatly appreciated, thanks for your time!

Last edited: Nov 27, 2009
2. Nov 27, 2009

### Dick

You don't really need to go to polar coordinates. Since the curl(F)=(0,0,1) you only need to find the z component of the vector dS. If you parameterize the cone as r=(x,y,sqrt(x^2+y^2)), just find the z component of dr/dx X dr/dy (both derivatives partial, of course). I think you'll find the integrand is constant over your domain.

3. Nov 27, 2009

### goliath11

Thanks for the quick reply. I see your point, and I can solve it with your method. However, I can't quite see what's wrong with using polar coords, it feels like I'm doing the same thing.

4. Nov 28, 2009

### Dick

It's mostly right. What's wrong is the 'r' in dA=r*dr*dt. The cross product of the two tangent vectors already gives you that 'r'. If Tr and Tt are the two tangents, you should just have dS=Tr X Tt dr dt, not Tr X Tt r*dr*dt.

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