# Stokes Theorem

1. Nov 28, 2006

### Borat321

1. The problem statement, all variables and given/known data
Let M be the capped cylindrical surface which is the union of two surfaces, a cylinder given by x^2 + y^2 = 9, 0 ≤ z ≤ 1, and a hemispherical cap defined by x^2 + y^2 + (z−1)^2 = 9, z ≥ 1. For the vector field F = (zx + z^2y + 2 y, z^3yx+ 8 x, z^4x^2), compute doubleintM (∇×F) ·dS in any way you like.

doubleintM (∇×F) ·dS = ?

2. Relevant equations

I thought the line integral would make the most sense in solving this - I wanted to take the line integral with respect to the bottom circle. parameterized, it is (3costheta, 3sintheta, 1).

Line int = F(r) (rprime)

3. The attempt at a solution

I have no idea what's going on here - I tried to take the line integral by saying r=(3sint,3cost,0) Since z=0 for circle the only parts that matter in the F is the 2y and the 8x, but I don't think that's right.

Integral w/ limits 0 to 2pi

(2(3sintheta),8(3costheta),0) dot product (-3sintheta, 3costheta, 0)

Can anyone help me with this one?

2. Nov 29, 2006

### HallsofIvy

Staff Emeritus
Not quite. That is the bottom of the hemispherical cap. The bottom of M is the circle in the xy-plane: $(3 cos\theta , 3 sin\theta ,0)$

Oh! Okay, now you moved down to the correct circle!

$F= (zx+ z^2y+ 2y, z^3yx+8x, z^4x)$ which, on z= 0 becomes
$F= (2y, 8x, 0)$. Good. That's what you have. But $s= (3 cos(\theta), 3 sin(\theta), 0)$ so $ds= (-3 sin(\theta), 3 cos(\theta), 0)d\theta$. Your integral should be:
$$\int_{\theta= 0}^{2\pi}(6sin(\theta ), 24cos(\theta ),0)\cdot (-3sin(\theta ),3cos(\theta ),0)d\theta$$.
Never forget the "d" in an integral!