A line integral of a function adds $f$ times small lengths along a curve. A surface integral of a function adds $f$ times small areas over a surface. It gives the mass of a thin curved sheet, the average temperature over a dome, or the charge on a curved plate. The area factor comes from surface area.
Key idea
For a smooth surface $S$ given by $\mathbf{r}(u, v)$, $(u, v)$ in $D$, and a continuous function $f$:
$$\iint_S f\,dS = \iint_D f(\mathbf{r}(u, v))\,|\mathbf{r}_u \times \mathbf{r}_v|\,dA.$$
For a graph $z = g(x, y)$ over $D$: $dS = \sqrt{1 + g_x^2 + g_y^2}\,dA$. On a sphere of radius $a$: $dS = a^2\sin\varphi\,d\varphi\,d\theta$. On a cylinder of radius $a$: $dS = a\,d\theta\,dz$.
Why it works
A small parameter rectangle of area $\Delta u\,\Delta v$ becomes a patch of area about $|\mathbf{r}_u \times \mathbf{r}_v|\,\Delta u\,\Delta v$ (see the previous lesson). On that patch $f$ is nearly constant, so the patch contributes about $f \cdot |\mathbf{r}_u \times \mathbf{r}_v|\,\Delta u\,\Delta v$. Adding gives a Riemann sum for the double integral over $D$. With $f = 1$ the formula gives the area. Like $\int_C f\,ds$, the value does not depend on the parametrization or on which side of the surface is chosen.
Worked examples
Worked example: A tilted triangle
Evaluate $\iint_S z\,dS$, where $S$ is the part of the plane $x + y + z = 1$ in the first octant.
- $z = 1 - x - y$ over the triangle $T$: $x, y \ge 0$, $x + y \le 1$. $dS = \sqrt{1 + 1 + 1}\,dA = \sqrt{3}\,dA$.
- $$\iint_T (1 - x - y)\,dA = \int_0^1 \int_0^{1 - x} (1 - x - y)\,dy\,dx = \int_0^1 \frac{(1 - x)^2}{2}\,dx = \frac{1}{6}.$$
- Answer: $\frac{\sqrt{3}}{6}$.
Answer: $\frac{\sqrt{3}}{6}$
Worked example: On a sphere
Evaluate $\iint_S z^2\,dS$ over the unit sphere.
- $z = \cos\varphi$ and $dS = \sin\varphi\,d\varphi\,d\theta$.
- $$\int_0^{2\pi} \int_0^\pi \cos^2\varphi\sin\varphi\,d\varphi\,d\theta = 2\pi \cdot \left[-\frac{\cos^3\varphi}{3}\right]_0^\pi = 2\pi \cdot \frac{2}{3} = \frac{4\pi}{3}.$$
- Check by symmetry: $x^2$, $y^2$, $z^2$ have equal integrals and add up to $\iint 1\,dS = 4\pi$, so each is $\frac{4\pi}{3}$.
Answer: $\frac{4\pi}{3}$
Worked example: On a cylinder
Evaluate $\iint_S (x^2 + z)\,dS$ over the side of $x^2 + y^2 = 4$, ${}0 \le z \le 3$.
- $x = 2\cos\theta$, $dS = 2\,d\theta\,dz$.
- $$\int_0^{2\pi} \int_0^3 (4\cos^2\theta + z) \cdot 2\,dz\,d\theta = 2\int_0^{2\pi} \left(12\cos^2\theta + \frac{9}{2}\right) d\theta = 2(12\pi + 9\pi) = 42\pi.$$
Answer: ${}42\pi$
Common mistake
Integrating over the shadow. $\iint_S f\,dS$ is not $\iint_D f\,dA$: the surface is usually bigger than the region below it. In the first example, leaving out $\sqrt{3}$ gives $\frac{1}{6}$, the answer for a flat triangle on the floor. Always find $|\mathbf{r}_u \times \mathbf{r}_v|$ (or $\sqrt{1 + g_x^2 + g_y^2}$) first.