Parametrization for the following surface Evaluate $\int\int_{S}\ y^2\ dS$, where S is that part of the cylinder $x^2 + z^2 = 1$ between the planes $y=0$ and $y=3-x$.
I tried parametrizing the the above surface $S$ using the following
$r(x,z)=<x,3-x,z>$ where $(x, z)\in D$, $D := x^2 +z^2 \leq 1$
From there I proceed with the double integration using polar coordinates. But the answer I yield is wrong. Hence I speculate that my perimetrization is wrong. 
What should be the correct perimetrization ?
Any help or insight is deeply appreciated.
 A: A possible parametrization is the following one :
\begin{align}
X\left(t,\theta\right)&=\left(x(t,\theta),y(t,\theta),z(t,\theta)\right)\\
&=\left(\cos\theta,t,\sin\theta\right) \\
&=u_{\theta}+tN
\end{align}
where $u_\theta:=(\cos\theta,0,\sin\theta)$ and $N:=(0,1,0)$.
At any point $X(t,\theta)$ of the surface, a basis of the tangent space (plane here) is then constituted by the tangent vectors
\begin{align}
X_t(t,\theta)=N,
\end{align}
\begin{align}
X_\theta(t,\theta)=u'_\theta:=(-\sin\theta,0,\cos\theta).
\end{align}
The surface element is then
\begin{align}
\mathrm{d}S&=|X_t\wedge X_\theta |\mathrm{d}t\mathrm{d}\theta\\
&=|N\wedge u'_\theta |\mathrm{d}t\mathrm{d}\theta\\
&=|-u_\theta| \mathrm{d}t\mathrm{d}\theta\\
&= \mathrm{d}t\mathrm{d}\theta
\end{align}
so between the planes $\{y=0\}$ and $\{y=3-x\}$, the integral of $y^2$ is
\begin{align}
\int_{0\leq y\leq3-x}y^2\mathrm{d}S&=\int_{0\leq t\leq3-\cos\theta}\int_{0<\theta<2\pi}t^2\mathrm{d}t\mathrm{d}\theta\\
&=\int_{0<\theta<2\pi}\frac{\left(3-\cos\theta\right)^3}{3}\mathrm{d}\theta\\
&=\frac{1}{3}\int_{0<\theta<2\pi}\left(27-27\cos\theta+9\cos^2\theta-\cos^3\theta\right)\mathrm{d}\theta\\
&=21\pi.
\end{align}
