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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 $x^2+y^2\le 3$ vanishes by oddness:On the top, $z=4-y$, so $y+z=4$. The graph has $z_x=0$, $z_y=-1$, giving $dS=\sqrt{1+z_x^2+z_y^2}dxdy=\sqrt{2}dxdy$. HenceFor the side, let $R=\sqrt{3}$ and parametrize by $(R\cos \varphi ,R\sin \varphi ,z)$, with $0\le z\le 4-R\sin \varphi$. The surface area element is $Rdzd\varphi$. ThusAdding all three pieces givesThe tilted top remains above the bottom everywhere because $4-y\ge 4-\sqrt{3}>0$. 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.

For the scalar surface integral over an obliquely cut cylinder, split the scalar surface integral into the curved cylinder and its two planar caps. Write $s=z-x$, so the solid has $x^2+y^2\le 4$ and $-3\le s\le 2$. On the curved cylinder, use the parametrizationThe surface area element follows fromConsequently the curved contribution isOn either cap $z=x+s$, the surface area element is $\sqrt{1+z_x^2+z_y^2}dxdy=\sqrt{2}dxdy$ over a radius-two disk. The lower cap has $x-z=3$ and the upper cap has $x-z=-2$. Thusand the complete scalar surface integral isThe surface area element is unsigned: opposite outward normals on the caps do not introduce extra minus signs.

A scalar surface integral integrates a scalar field against the unsigned surface area element. For a regular parametrization $\mathbf{r}(u,v)$, it is $\int f(\mathbf{r}(u,v))\Vert {\mathbf{r}}_u\times {\mathbf{r}}_v\Vert dudv$. Reversing orientation does not change it. For example, the scalar surface integral of the constant one is the surface area; a signed flux instead integrates the normal component of a vector field.

Let $S$ be the full boundary of $x^2+y^2\le R^2$, $0\le z\le H-\alpha y$, with $H>|\alpha |R$. The bottom contribution is zero by symmetry. On the top, the integrand is the constant $H$ and the surface area element is $\sqrt{1+{\alpha }^2}dxdy$. On the side, set $y=R\sin \varphi$ and integrate in $z$ first: the result is $R[H^2-{\alpha }^2{R}^2{\sin }^2\varphi ]/2$. Integrating from zero to $2\pi$ gives the displayed scalar surface integral. No orientation signs enter; a scalar area integral is different from a flux.

Let $S$ be the whole boundary of $x^2+y^2\le R^2$, $a\le z-\alpha x\le b$, where $R>0$ and $a<b$. On the curved cylinder, the parametrization $(R\cos \theta ,R\sin \theta ,\alpha R\cos \theta +s)$ has surface area element $Rd\theta ds$ and $\alpha x-z=-s$. Each planar cap has area $\pi R^2\sqrt{1+{\alpha }^2}$. Summing the side and both caps gives the scalar surface integralThe surface area element is unsigned; cap orientation does not change the signs of the scalar integrand. This separates the geometric area factor from the field values and avoids mistaking a scalar surface integral for a flux.