I've been working on this problem:
Evaluate the surface integral: $\iint_{S}\ y\ z\ dS$, $S$ is the portion of $x^2+y^2=1$ with $x\geq0$ and $z$ between $z=1$ and $z=4-y$.
This is my work:
$$x=\cos{\left(\theta\right)} \text{ and } y=\sin{\left(\theta\right)}\\\vec{r}\left(z,\theta\right)=\cos{\left(\theta\right)}\vec{i}+\sin{\left(\theta\right)}\vec{j}+z\vec{k}; 1\le z\le4-\sin{\left(\theta\right)} \text{ and } 0\le\theta\le2\pi\\\left|\begin{matrix}i&j&k\\0&0&1\\-\sin{\left(\theta\right)}&\cos{\left(\theta\right)}&0\\\end{matrix}\right|=-\cos{\left(\theta\right)}\vec{i}+-\sin{\left(\theta\right)}\vec{j}+0\vec{k}\\\left|\left|-\cos{\left(\theta\right)}\vec{i}+\ -\sin{\left(\theta\right)}\vec{j}+0\vec{k}\right|\right|=1\\\int_{0}^{2\pi}\int_{1}^{4-\sin{\left(\theta\right)}}{\sin{\left(\theta\right)}z\ dz\ d\theta}\\\int_{0}^{2\pi}{\frac{\sin{\left(\theta\right)}\left(4-\sin{\left(\theta\right)}\right)^2}{2}-\frac{\sin{\left(\theta\right)}}{2}d\theta}=-4\pi$$
I don't see any obvious error in my work (although that doesn't mean there is none :) ), but I saw another answer giving $-2\pi$. I don't know that the other answer is correct, but I want to know - is mine? I did not use the same method as they did. I did check it with $x=\sin(\theta)$ and $y=\cos(\theta)$, and I got the same answer (which is good). Any insights about where I went wrong?
EDIT: Is it because of $x\geq0$? That cuts the cylinder in half which would give $-2\pi$. Is there a way to incorporate this into my integral?
You can include the restriction for $x\geq 0$ by requiring that $\theta\in(-\frac{\pi}{2},\frac{\pi}{2} )$. That yields the correct expression for the integral since
$$\frac{1}{2}\int_{-\pi/2}^{\pi/2}\sin\theta[(4-\sin\theta)^2-1]d\theta=-4\int_{-\pi/2}^{\pi/2}d\theta~\sin^2\theta=-2\pi$$
as requested.