Scalar surface integral (${\int }_{S}fdS$)

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.