Surface integral and what it computes

Question

Can anyone tell me what the surface integral of $f(x,y,z)$ actually calculates in a physical sense in three dimensions?

I know that the surface integral computes the surface area of some given surface:

$$\iint_D ||r_u\times r_v||dA$$

However what does the surface integral below calculate ($f(x,y,z)$ is a continuous function defined on the surface $S$): $$\iint_S f(x,y,z)dS=\iint_D f(r(u,v))||r_u\times r_v||dA$$

Does it calculate volume? Can anyone explain to me what it computes?

Edit: If a smooth parametric surface is given by the parametrization: $$r(u, v) = <x(u, v), y(u, v), z(u, v)>, (u,v)\in D$$ and $r$ is possibly injective on $D$ except possibly on the boundary of $D$, then the surface area of $S$ over $D$ is given by:

$$\iint_D ||r_u\times r_v||dA$$

Answer 1

Let the domains of integration $B,S,L,C$ and $SW$ be a body, surface, lamina, curve, and a straight wire, respectively, and $x,y,z$ be Cartesian variables.

• The following integrals give the integration domain's length/area/volume: \begin{align}\int_{SW} &\:\mathrm dx \\\int_C &\:\mathrm ds \\\iint_L &\:\mathrm dx\,\mathrm dy \\\int_S &\:\mathrm dS \\\iiint_B &\:\mathrm dx\,\mathrm dy\,\mathrm dz\end{align}
• If the integration domain's (mass) density function is $f,$ then the following integrals give its mass: \begin{align}\int_{SW} f(x) &\:\mathrm dx \\\int_C f(x,y,z) &\:\mathrm ds \\\iint_L f(x,y) &\:\mathrm dx\,\mathrm dy \\\int_S f(x,y,z) &\:\mathrm dS \\\iiint_B f(x,y,z) &\:\mathrm dx\,\mathrm dy\,\mathrm dz\end{align}
• The following integrals give the signed area/volume between the integration domain and the curve $y=f(x)$ or surface $z=f(x,y)$: \begin{align}\int_{SW} f(x) &\:\mathrm dx \\\int_C f(x,y) &\:\mathrm ds \\\iint_L f(x,y) &\:\mathrm dx\,\mathrm dy\end{align}