= Solution
This is a <scalar surface integral>, not an oriented <flux>. Split the whole closed <surface> into the bottom disc, curved side and tilted top; shared edge curves have zero <surface area>.
On the bottom, $z=0$ and the integrand is $y$. Its integral over the symmetric disc $x^2+y^2\leq3$ vanishes by oddness:
$$
I_{\mathrm{bottom}}=0.
$$
On the top, $z=4-y$, so $y+z=4$. The graph has $z_x=0$, $z_y=-1$, giving $dS=\sqrt{1+z_x^2+z_y^2}\,dx\,dy=\sqrt2\,dx\,dy$. Hence
$$
I_{\mathrm{top}}=4\sqrt2\,(3\pi)=12\pi\sqrt2.
$$
For the side, let $R=\sqrt3$ and parametrize by $(R\cos\phi,R\sin\phi,z)$, with $0\leq z\leq4-R\sin\phi$. The <surface area element> is $R\,dz\,d\phi$. Thus
$$
\begin{aligned}
I_{\mathrm{side}}
&=R\int_0^{2\pi}\int_0^{4-R\sin\phi}(R\sin\phi+z)\,dz\,d\phi\\
&=R\int_0^{2\pi}\left[8-\frac{R^2}2\sin^2\phi\right]d\phi\\
&=16\pi R-\frac{\pi R^3}{2}=\frac{29\pi\sqrt3}{2}.
\end{aligned}
$$
Adding all three pieces gives
$$
\boxed{\int_S(y+z)\,dS=\pi\left(12\sqrt2+\frac{29\sqrt3}{2}\right).}
$$
The tilted top remains above the bottom everywhere because $4-y\geq4-\sqrt3>0$. The calculation is an instance of a <scalar surface integral over a cylinder with one tilted cap>; reversing a surface orientation would not change any of its terms.
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