Area of the part of the cylinder $x^2+y^2=2ay$ outside the cone $z^2=x^2+y^2$ Problem: Find the area of the part of the cylinder $x^2+y^2=2ay$ that lies outside the cone $z^2=x^2+y^2$. 
My attempt:
So I thought we could do this by projecting the surface onto the $yz$-plane and taking the surface integral of the function $x=g(y,z)=\sqrt{z^2-y^2}$. I.e letting $S$ be the surface and $E$ be the projection onto the $yz$-plane where we have a $2$ before the integral over $E$ since we have both $x<0$ and $0\leq x$:
\begin{align*}\iint_{\mathcal{S}}x \ \mathrm{d}S &=2\iint_{E}x\underbrace{\sqrt{1+\left(\frac{\partial x}{\partial y}\right)^2+\left(\frac{\partial x}{\partial z}\right)^2} \ \mathrm{d}z\mathrm{d}y}_{\mathrm{d}S} \\
&=2\iint_{E}x\sqrt{1+\frac{z^2}{x^2}+\frac{y^2}{x^2}} \ \mathrm{d}z\mathrm{d}y\\
&=2\iint_{E}\sqrt{x^2+z^2+y^2}\  \mathrm{d}z\mathrm{d}y\\
&=2\iint_{E}\sqrt{2}z\  \mathrm{d}z\mathrm{d}y
\end{align*}
Now in the projection it seems to me that we have the following bounds on  $z$ and $y$ since the cylinder has radius $a$ and the cone and the surface intersect at $z=\sqrt{2ay}$
$$0\leq z \leq \sqrt{2ay} \quad \text{and} \quad 0\leq y \leq 2a$$
so:
\begin{align*}2\iint_{E}\sqrt{2}z\  \mathrm{d}z\mathrm{d}y &= \sqrt{2}\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}2z \ \mathrm{d}z\mathrm{d}y \\
&=\sqrt{2}\int_{0}^{2a} 2ay \ \mathrm{d}y\\
&=4\sqrt{2}a^{2}\end{align*}
However my book says its $16a^2$ so what is my mistake(s)?
PS. I think this is also possible with polar coordinates but I would like to use the surface integral with projection onto the $yz$-plane.
PSDS. Picture is not totally acurate as $a=4$
Edit:
As Ninad Munshi pointed out I was projecting  the wrong surface and I used the wrong formula for the surface area. My thoughts are
Would it be correct to say that $\iint\mathrm{d}S$ is the surface area, and would $\mathrm{d}S$ be $\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2} dzdy$?
If so I still seem to be off by a factor of $2$ as 
\begin{align*}\iint_{\mathcal{S}} \mathrm{d}S &= 2 \iint_{E}\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2}dzdy \\
&=2\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}\sqrt{1+\left( \frac{a-y}{\sqrt{2ay-y^2}} \right)^2}dzdy=8a^{2}\end{align*}

 A: The issue was, as pointed out in the comments by Ninad Munshi, that the wrong surface was used to begin with aswell as the wrong integral was used.
The correct way to solve this question is to start with the cylinder $x^2+y^2=2ay$ that we wish to project onto the $yz$ plane. This is done by first calculating $\mathrm{d}S$ in
$$\iint_{\mathcal{S}}dS$$
which gives the surface area
We have that 
$$dS = \sqrt{1+\left(\frac{\partial x}{\partial y}\right)^{2}+\left(\frac{\partial x}{\partial z}\right)^{2}} \ \mathrm{d}z\mathrm{d}y=\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$
Now the problem I had is that when I projected the cylinder: I only considered the symmetric areas of $x<0$ and $0\leq x$ while we infact have two more symmetries: namely $z<0$ and $0\leq z$. 
In summary: We have four areas that are equal (and not two) so letting $E$ represent the area of the projection of the cylinder onto the $yz$-plane in the first octant we get:
$$\iint_{\mathcal{S}}\mathrm{d}S=4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y$$
The particular limits of $z$ and $y$ are still correct that is
$$0\leq z \leq \sqrt{2ay} \quad \text{and} \quad 0\leq y \leq 2a$$
So:
\begin{align*}4\iint_{E}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y & = 4\int_{0}^{2a}\int_{0}^{\sqrt{2ay}}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}z\mathrm{d}y \\
&= 4 \int_{0}^{2a}\sqrt{2ay}\sqrt{1+\frac{(a-y)^{2}}{2ay-y^2}}\ \mathrm{d}y\\
&= 4(4a^2)=16a^2\end{align*}
Which is the correct answer
