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$S$ is parametrized by $X(s,t) = (s\cos(t), s\sin(t), t)$, $0 \leq s \leq 1$ and $0 \leq t \leq \frac{\pi}{2}$

$$\mathbf{F} = z \mathbf{i} + x \mathbf{j} + z \mathbf{k}$$

I have two things preventing me from solving this: first, I do not know how to find $d\,\mathbf{s}$ for $X(s,t) = (s\cos(t), s\sin(t), t)$ and I am not sure how to take out the parameterization of $X$ and find a surface to work with for the double integral over $s$ of $\nabla x \mathbf{F} \,d\,\mathbf{S}$

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Note: The name is "Stokes", not "Stoke", so it's "Stokes's Theorem", not "Stoke's Theorem". – Arturo Magidin May 4 '11 at 4:27
I apologize for the mistake! Thanks for correcting me =) – Sir Winford May 4 '11 at 5:05
up vote 1 down vote accepted

How comfortable are you with surface integrals? $X(s,t)$, as defined in your post, is a parametrization of $S$. Computing $\int_S \nabla \times \mathbf{F} d\mathbf{S}$ seems like a standard pre-Stokes' homework problem about surface integrals.

To compute the line integral side of Stokes' theorem, you'll need to parametrize the boundary of $S$. Notice that the domain of $X$ in the $st$-plane is a rectangle (with sides 1 and $\pi/2$). $X$ sends each point of the rectangle to a point on $S$, and it sends the boundary of the rectangle to the boundary of $S$. Can you see how to parametrize the boundary now? (Make sure it's oriented correctly!) Now all you have to do is compute an ordinary line integral.

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Alright, I was able to figure out the surface integral...which leaves me with the line integral. I've never been very good at parameterizing surfaces, but I'm trying to figure that out now. – Sir Winford May 4 '11 at 5:05

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