Surface integral and Stokes theorem I read the following in a book and I have trouble understanding it. Help would be appreciated.
Let $Π$ be a plane in $\Bbb R^3$ that is parallel to the $x$-axis but not perpendicular to the $xy$-plane and $S$ is the intersection of $Π$ with the cylinder $x^2+y^2=a^2$. Then the book says that: 
$\int_S(yz-y)dx+(xz+x)dy=2πa^2$, which I have trouble showing.
I tried using Stokes but didn't get far with it. Could someone please explain it to me?
 A: All you need is a straightforward calculation. If the plane is $cy+dz=1$, with $d\ne0$ (to avoid being perpendicular to $xy$), we can parametrize the curve as 
$$
r(t)=\left(a\cos t,a\sin t,\frac{1-c\sin t}d\right).
$$
Then
$$
r'(t)=\left(-a\sin t,a\cos t,-\frac{c\cos t}d\right).
$$
Then, writing the field as $(y(z-1),x(z+1),0)$, we have (saving the substitution of $x,y,z$ for later in the computation)
\begin{align}
\int_S(yz-y)dx+(xz+x)dy
&=\int_0^{2\pi}(y(z-1),x(z+1),0)\cdot\left(-a\sin t,a\cos t,-\frac{c\cos t}d\right)\,dt\\ \ \\
&=\int_0^{2\pi}\left[-a\sin t\, y(z-1)+a\cos t\,x(z-1)\right]\,dt\\ \ \\
&=\int_0^{2\pi}\left[-a^2\sin^2 t\, (z-1)+a^2\cos^2 t\,(z-1)\right]\,dt\\ \ \\
&=\int_0^{2\pi}\left[a^2+a^2\,z\,(\cos^2 t-\sin^2 t)\right]\,dt\\ \ \\
&=2\pi a^2+\int_0^{2\pi}\left[a^2\left(\frac{1-c\sin t}d\right)\,(\cos^2 t-\sin^2 t)\right]\,dt\\ \ \\
&=2\pi a^2+\frac{a^2}d\,\int_0^{2\pi}(\cos^2t-\sin^2t)\,dt-\frac {a^2c}d\left[\int_0^{2\pi}(\sin t\,\cos^2t+\sin^3t)\,\,dt\right]\\ \ \\
&=2\pi a^2.
\end{align}
