Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$S$ 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 $\nabla x \mathbf{F} \,d\,\mathbf{S}$

share|improve this question
    
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
add comment

1 Answer

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.