Surface Integrals of Scalar Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.7: “Surface Integrals” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-7-surface-integrals |
| Textbook used in class | Stewart, Calculus, Section 16.7: “Surface Integrals” (Examples 1, 2) |
Quick Reference
Surface integral of $f$ over $S$: $$\iint_S f\,dS = \iint_D f(\mathbf{r}(u,v))\,|\mathbf{r}_u\times\mathbf{r}_v|\,dA.$$
For a graph $z = g(x,y)$: $$\iint_S f\,dS = \iint_D f(x,y,g(x,y))\,\sqrt{1+g_x^2+g_y^2}\,dA.$$
When $f = 1$: $\iint_S 1\,dS = $ surface area.
Motivation
Surface integrals of scalar functions extend line integrals of scalar functions from curves to surfaces. If $f(x,y,z)$ is the density of a thin shell with shape $S$, then $\iint_S f\,dS$ gives the total mass of the shell. If $f = 1$, the integral returns the surface area.
Key Concept
The surface element $dS = |\mathbf{r}_u\times\mathbf{r}_v|\,dA_{uv}$ plays the role of the arc-length element $ds = |\mathbf{r}'(t)|\,dt$ in line integrals. It converts from the parameter domain to actual surface area, accounting for stretching.
Worked Example
Evaluate $\iint_S\sqrt{1+x^2+y^2}\,dS$ where $S$ is the helicoid $\mathbf{r}(u,v) = \langle u\cos v, u\sin v, v\rangle$, $0 \leq u \leq 1$, $0 \leq v \leq \pi$. (Stewart 16.7, Example 1.)
$\mathbf{r}_u = \langle \cos v, \sin v, 0\rangle$, $\mathbf{r}_v = \langle -u\sin v, u\cos v, 1\rangle$.
$\mathbf{r}_u\times\mathbf{r}_v = \langle \sin v\cdot 1 - 0, 0 - \cos v, u\cos^2 v + u\sin^2 v\rangle = \langle \sin v, -\cos v, u\rangle$.
$|\mathbf{r}_u\times\mathbf{r}_v| = \sqrt{\sin^2 v+\cos^2 v+u^2} = \sqrt{1+u^2}$.
$f(\mathbf{r}(u,v)) = \sqrt{1+u^2\cos^2 v + u^2\sin^2 v} = \sqrt{1+u^2}$.
$$\iint_S\sqrt{1+x^2+y^2}\,dS = \int_0^\pi\int_0^1\sqrt{1+u^2}\cdot\sqrt{1+u^2}\,du\,dv = \pi\int_0^1(1+u^2)\,du = \pi\left[u+\frac{u^3}{3}\right]_0^1 = \pi\cdot\frac{4}{3} = \frac{4\pi}{3}.$$
surface integrals of scalar functions are orientation-dependent. Surface integrals of scalar functions ($\iint_S f\,dS$) are NOT orientation-dependent, because $dS = |\mathbf{r}_u\times\mathbf{r}_v|\,dA \geq 0$ is always positive. Only surface integrals of vector fields ($\iint_S\mathbf{F}\cdot d\mathbf{S}$, flux integrals) depend on orientation, because they use $\mathbf{r}_u\times\mathbf{r}_v$ (with sign) rather than $|\mathbf{r}_u\times\mathbf{r}_v|$.
Common Misconceptions
$\iint_S f\,dS$ and $\iint_S \mathbf{F}\cdot d\mathbf{S}$ are computed by the same formula.
This is the input-output-confusion error. The scalar surface integral uses $dS = |\mathbf{r}_u \times \mathbf{r}_v|\,dA$, the magnitude of the cross product, and is independent of orientation. The vector surface integral (flux) uses $d\mathbf{S} = (\mathbf{r}_u \times \mathbf{r}_v)\,dA$, the cross product itself, and depends on the chosen orientation. Using the magnitude where the signed vector is needed -- or vice versa -- produces an answer that ignores direction and orientation entirely.
Leveled Practice
Problem 1. Find $\iint_S(x+y)\,dS$ where $S$ is the part of the plane $x+y+z = 1$ in the first octant.
Show answer
Graph $z = 1-x-y$ over the triangle $D$: $x\geq 0$, $y\geq 0$, $x+y\leq 1$.
$g_x = -1$, $g_y = -1$. $dS = \sqrt{1+1+1}\,dA = \sqrt{3}\,dA$.
$\iint_S(x+y)\,dS = \sqrt{3}\iint_D(x+y)\,dA = \sqrt{3}\int_0^1\int_0^{1-x}(x+y)\,dy\,dx$.
Inner: $\int_0^{1-x}(x+y)\,dy = [xy+y^2/2]_0^{1-x} = x(1-x)+(1-x)^2/2$.
$= (1-x)[x + (1-x)/2] = (1-x)(x+1)/2 = (1-x^2)/2$.
Outer: $\int_0^1(1-x^2)/2\,dx = (1/2)[x-x^3/3]_0^1 = (1/2)(2/3) = 1/3$.
$\iint_S(x+y)\,dS = \sqrt{3}/3 = \sqrt{3}/3$.