28. Surface Integrals

Once we had parametric curves, we could integrate on those curves with line integrals. Now that we have parametric surfaces, we can integrate on surfaces.

If we have a surface $\large S$ given by $\large G(u,v)$ with parameter domain $\large R$, the surface integral of $\large S$ over $\large R$ is:

$$\large \iint_S f(x,y,z)\, dS = \iint_R f(x(u,v),y(u,v),z(u,v))\left|\vec{n}(u,v)\right|\, dA$$

Like all integrals, a surface integral is a limit of a Riemann sum:

surfaceInt

^{^{Image from: Briggs, Calculus Early Transcendentals, 3rd Edition}}

The parameter domain $\large R$ is broken up into rectangles, and these are mapped via $\large G(u,v)$ to the surface. The area element $\large dA$ in $\large R$ gets mapped to a parallelogram tangent to the surface, and the area of the parallelogram is $\large \left|\vec{T}_u\times\vec{T}_v \right|dudv=\left| \vec{n}\right| dA$.

Evaluating Surface Integrals

The steps to a surface integral problem are:

Find the parametric surface $\large G(u,v)$ along with the parameter domain.
Find the normal vector $\large \vec{n}(u,v)$.
Replace $\large x, y$, and $\large z$ in the integrand with their values on the surface in terms of the parameters $\large (u,v)$.
You now have a double integral!

The following video shows an example of a surface integral. It assumes that we know $\large \left| \vec{n}\right|=R^2\sin\phi$ for a sphere of radius $\large R$:

Some facts to remember

For a sphere of radius $\large R$, the magnitude of the normal vector is $\large R^2\sin\phi$.
$\large \iint_S 1 dS$ is the surface area of $\large S$
If $\large \rho(x,y,z)$ is a density on the surface, $\large \iint_S \rho(x,y,z) dS$ is the total amount (mass, charge, etc.).
The average value of a function $\large f(x,y,z)$ on a surface $\large S$ is $\large \displaystyle \frac{\iint_S f(x,y,z) dS}{\iint_S 1 dS}$