Stokes' Theorem Let $C$ be the following, let $C$ be the curve of intersection of the cylinder $x^2 + y^2 = 1$ and the given surface $z = f(x,y)$, oriented counterclockwise around the cylinder. Use Stokes' theorem to compute the line integral by first converting it to a surface integral.
(a) $\int_C (y \, \mathrm{d}x + z \, \mathrm{d}y + x \, \mathrm{d}z),\quad z=x \cdot y$. 
I'm having a problem setting up the problem. I appreciate any assistance.
 A: Just to get you started, here's the details for (a). Start by finding the curl of the vector field $\mathbf{F}=\langle y,z,x\rangle$. You get
$$\nabla\times\mathbf{F}=\det\begin{pmatrix} \mathbf{i} &\mathbf{j}&\mathbf{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\y&z&x\end{pmatrix}=\langle0,-1,-1 \rangle$$
Stokes's Theorem says the line integral of $\mathbf{F}$ around $C$ is equal to the surface integral of $\nabla\times\mathbf{F}$ over any surface $S$ having $C$ as a boundary (written $C=\partial S$). The easiest such surface would just be the part of the surface $z=xy$ lying inside the cylinder $x^2+y^2=1$. Call $S$ this surface, and parametrize it in polar coordinates by $\mathbf{S}(r,\theta)=\langle r\cos\theta, r\sin\theta, r^2\cos\theta\sin\theta\rangle$ for $r\in[0,1]$ and $\theta\in[0,2\pi)$. The $z$ component here comes from the fact that $z=xy$, and in polar coordinates $x=r\cos\theta$ and $y=r\sin\theta$. 
Then we just need to compute $$\int\int_S\langle 0,-1,-1\rangle\cdot \mathbf{dS}=\int_0^{2\pi}\int_0^1\langle 0,-1,-1\rangle\cdot\left(\frac{\partial\mathbf{S}}{\partial r}\times\frac{\partial\mathbf{S}}{\partial\theta}\right)\,dr\,d\theta$$
But we have
$$\frac{\partial\mathbf{S}}{\partial r}=\langle\cos\theta,\sin\theta,2r\cos\theta\sin\theta \rangle=\langle \cos\theta,\sin\theta,r\sin(2\theta)\rangle$$
and
$$\frac{\partial\mathbf{S}}{\partial \theta}=\langle -r\sin\theta, r\cos\theta, r^2(\cos^2\theta-\sin^2\theta) \rangle=r\langle-\cos\theta,\sin\theta,r\cos(2\theta)\rangle$$
Then
$$\frac{\partial\mathbf{S}}{\partial r}\times\frac{\partial\mathbf{S}}{\partial\theta}=\det\begin{pmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\\cos\theta&\sin\theta&r\sin(2\theta)\\-r\sin\theta&r\cos\theta&r^2\cos(2\theta)\end{pmatrix}$$
The first component doesn't matter, because we're dotting it with $\langle0,-1,-1\rangle$ anyway. The second component is $-r^2(\sin(2\theta)\sin\theta+\cos(2\theta)\cos\theta)$ and the third component is just $r$. You can tell this is the right orientation because the third component is positive and hence pointing up rather than down.
Taking the dot product, our integral becomes
$$\int_0^{2\pi}\int_0^1\left(r^2\sin(2\theta)\sin(\theta)+r^2\cos(2\theta)\cos\theta-r\right)\,dr\,d\theta$$
Doing the integration with respect to $r$ we get
$$\int_0^{2\pi}\left(\frac{1}{3}(\sin(2\theta)\sin\theta+\cos(2\theta)\cos\theta)-\frac{1}{2}\right)\,d\theta$$
I'll leave it to you to finish -- with the hint that $\sin(2\theta)\sin\theta+\cos(2\theta)\cos\theta$ simplifies very nicely if you remember your sum/difference formulas.
