Surface Integrals

The Scalar Surface Integral

Let's start small. Imagine a thin curved sheet of metal - say, a dome - and suppose its mass is unevenly distributed: at each point $(x,y,z)$ the surface density (mass per unit area) is some function $H(x,y,z)$. What's the total mass of the dome?

If the surface were flat we'd just integrate $H\,dA$ over its footprint. But a curved sheet has more area than its shadow on the ground. So we cut $S$ into tiny patches, each with area $\Delta_i \sigma$, pick a sample point $(x_i^*, y_i^*, z_i^*)$ in each, and add up $H(x_i^*, y_i^*, z_i^*)\,\Delta_i \sigma$. Refining the partition gives the limit Kaplan writes as

That's the scalar surface integral (Kaplan eq. 5.74). Geometrically, $d\sigma$ is the area of an infinitesimal patch of $S$. Special case: if $H \equiv 1$, we get the surface area of $S$. Otherwise, we're computing a weighted total - mass, charge, heat content, anything distributed over the surface.

The whole game in the next two cards is: how do we actually compute $d\sigma$? We've two natural ways of describing a surface, and we'll need a recipe for each.

Graph Form: $z = g(x,y)$

The simplest description of a surface is "$z$ as a function of $x$ and $y$." Think of $S$ as a hilly landscape sitting above its shadow $R$ in the $xy$-plane. We want to express $d\sigma$ in terms of the easier $dA = dx\,dy$ on $R$.

Here's the picture. A small rectangle $dx \times dy$ in the plane lifts to a small parallelogram on $S$. Parametrize the surface as $\mathbf{r}(x,y) = (x,\,y,\,g(x,y))$. The two edge vectors of that parallelogram are

$$\mathbf{r}_x = (1, 0, g_x), \qquad \mathbf{r}_y = (0, 1, g_y).$$

Their cross product gives a normal vector whose length is the lift factor:

$$\mathbf{r}_x \times \mathbf{r}_y = (-g_x,\,-g_y,\,1), \qquad \|\mathbf{r}_x \times \mathbf{r}_y\| = \sqrt{1 + g_x^2 + g_y^2}.$$

This is Kaplan eq. 5.70 - and the same factor we met for surface area in §4.7. The square root is the "secant of the tilt angle": the steeper the surface, the more area it has compared to its shadow.

Worked example. Compute $\iint_S xy\,d\sigma$ where $S$ is the part of the plane $z = 2 - x - y$ in the first octant.

First, what's the footprint? "First octant" means $x \ge 0$, $y \ge 0$, $z \ge 0$. The third constraint $z \ge 0$ becomes $x + y \le 2$. So $R$ is the triangle with vertices $(0,0)$, $(2,0)$, $(0,2)$.

Now $g_x = -1$, $g_y = -1$, so

$$d\sigma = \sqrt{1 + 1 + 1}\,dx\,dy = \sqrt{3}\,dx\,dy.$$

The integral becomes

$$\iint_S xy\,d\sigma = \sqrt{3}\int_0^2 \int_0^{2-x} xy\,dy\,dx = \sqrt{3}\int_0^2 x \cdot \frac{(2-x)^2}{2}\,dx.$$

Expand $(2-x)^2 = 4 - 4x + x^2$ and integrate term by term:

$$\frac{\sqrt{3}}{2}\int_0^2 (4x - 4x^2 + x^3)\,dx = \frac{\sqrt{3}}{2}\left[2x^2 - \frac{4x^3}{3} + \frac{x^4}{4}\right]_0^2 = \frac{\sqrt{3}}{2}\!\left(8 - \frac{32}{3} + 4\right) = \frac{\sqrt{3}}{2}\cdot\frac{4}{3} = \boxed{\frac{2\sqrt{3}}{3}}.$$

Notice how the geometric factor $\sqrt{3}$ comes out front because the surface is flat - it has constant tilt. For curved surfaces, the square root sits inside the integral and we can rarely pull it out.

Parametric Form: $\mathbf{r}(u,v)$

Spheres, tori, and helicoids don't fit in graph form - they double back on themselves. The general description uses two parameters:

$$\mathbf{r}(u,v) = (x(u,v),\,y(u,v),\,z(u,v)),\qquad (u,v)\in R_{uv}.$$

The same parallelogram argument works. A small rectangle $du\times dv$ in parameter space lifts to a parallelogram with edges $\mathbf{r}_u\,du$ and $\mathbf{r}_v\,dv$. Its area is the magnitude of the cross product:

where $E$, $F$, $G$ are the entries of the first fundamental form:

$$E = \mathbf{r}_u \cdot \mathbf{r}_u,\qquad F = \mathbf{r}_u \cdot \mathbf{r}_v,\qquad G = \mathbf{r}_v \cdot \mathbf{r}_v.$$

Kaplan introduces this notation in §5.10 (eq. 5.71-5.72). One identity worth checking with bare hands once: $\|\mathbf{a} \times \mathbf{b}\|^2 = \|\mathbf{a}\|^2 \|\mathbf{b}\|^2 - (\mathbf{a}\cdot\mathbf{b})^2$, which is exactly $EG - F^2$.

A trap to name. At the poles of a sphere $\mathbf{r}(\phi,\theta) = (\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi)$, $\mathbf{r}_\phi \times \mathbf{r}_\theta$ vanishes - both lines of longitude collapse to a point. These are coordinate singularities, not geometric ones; the sphere itself is perfectly smooth at the poles. Such measure-zero defects don't affect the integral. Kaplan requires $EG - F^2 \neq 0$ for the parametrization to be useful, but a single bad point is fine.

Orientation: Picking a Side

For a scalar integral $\iint_S H\,d\sigma$, we never had to pick a side. Mass is mass; it doesn't matter which way we look at the dome. But for flux - "how much of a vector field crosses $S$" - we have to commit to a direction. Out of a closed box is positive; into it is negative. We need a notion of "out."

Kaplan's setup (§5.9): a smooth surface is oriented if we can choose a continuous unit-normal field $\mathbf{n}$ on it. The choice of $\mathbf{n}$ versus $-\mathbf{n}$ is the orientation. A surface that admits such a choice at all is called orientable or two-sided. For a graph $z = g(x,y)$ the upper normal has positive $z$-component:

$$\mathbf{n}_{\text{up}} = \frac{(-g_x,\,-g_y,\,1)}{\sqrt{1 + g_x^2 + g_y^2}}.$$

For the parametric form, $\mathbf{n} = \pm(\mathbf{r}_u \times \mathbf{r}_v)/\|\mathbf{r}_u \times \mathbf{r}_v\|$, and we pick the sign once and stick with it.

The classic counterexample. The Mobius strip is not orientable: walk around it once with $\mathbf{n}$ pointing "up" and you come back with it pointing "down." There is no continuous global choice. Such a surface has only one side, and flux through it is undefined - we simply don't try.

Piecewise-smooth surfaces (a cube, a cylinder with caps) are oriented face by face, with the normals on adjacent faces agreeing along their shared edge in the sense that, walking the edge, the surface lies on the same side (Kaplan Fig. 5.27).

Flux: $\iint_S \mathbf{v}\cdot\mathbf{n}\,d\sigma$

Now that we have an oriented surface and a normal $\mathbf{n}$, we define the flux of a vector field $\mathbf{v} = L\,\mathbf{i} + M\,\mathbf{j} + N\,\mathbf{k}$ across $S$:

Physically, if $\mathbf{v}$ is the velocity of a fluid, this measures the volume of fluid crossing $S$ per unit time in the $\mathbf{n}$ direction. The dot product picks out the component of $\mathbf{v}$ that's actually piercing the surface; tangential flow contributes nothing.

For a graph $z = g(x,y)$ with upper normal, something lovely happens. The unit normal is $\mathbf{n} = (-g_x, -g_y, 1)/\sqrt{1 + g_x^2 + g_y^2}$, and $d\sigma = \sqrt{1 + g_x^2 + g_y^2}\,dx\,dy$. The square roots cancel:

That's Kaplan eq. 5.80 with the upper-normal sign. With a lower normal we'd flip every term. The cancellation is exactly why the graph case is so popular in practice - no awkward square roots.

Worked example. Compute the upward flux of $\mathbf{v} = x\,\mathbf{i} + y\,\mathbf{j} + z\,\mathbf{k}$ through the paraboloid $z = 4 - x^2 - y^2$ over the disk $x^2 + y^2 \le 4$.

Take $g(x,y) = 4 - x^2 - y^2$, so $g_x = -2x$ and $g_y = -2y$. Then $L = x$, $M = y$, $N = z = 4 - x^2 - y^2$. Plug in:

$$-L\,g_x - M\,g_y + N = -x(-2x) - y(-2y) + (4 - x^2 - y^2) = 2x^2 + 2y^2 + 4 - x^2 - y^2 = x^2 + y^2 + 4.$$

Polar coordinates on the disk: $x^2 + y^2 = r^2$, $dx\,dy = r\,dr\,d\theta$.

$$\iint_S \mathbf{v}\cdot\mathbf{n}\,d\sigma = \int_0^{2\pi}\!\int_0^2 (r^2 + 4)\,r\,dr\,d\theta = 2\pi \int_0^2 (r^3 + 4r)\,dr = 2\pi\left[\frac{r^4}{4} + 2r^2\right]_0^2 = 2\pi\,(4 + 8) = \boxed{24\pi}.$$

Sanity check: $\mathbf{v} = (x,y,z)$ is the radial outward field, so upward flux through the paraboloid should be positive - and it is. We could later cross-check this against the Divergence Theorem on the closed region capped by the disk $z = 0$; that's a great exercise once we've covered §5.11.

Component Form: $L\,dy\,dz + M\,dz\,dx + N\,dx\,dy$

Kaplan writes the same flux integral in a third notation that connects directly to the Divergence Theorem and to differential forms. Define

$$\iint_S L\,dy\,dz \;=\; \iint_S L\cos\alpha\,d\sigma, \quad \iint_S M\,dz\,dx \;=\; \iint_S M\cos\beta\,d\sigma, \quad \iint_S N\,dx\,dy \;=\; \iint_S N\cos\gamma\,d\sigma,$$

where $(\cos\alpha, \cos\beta, \cos\gamma)$ are the direction cosines of $\mathbf{n}$. Adding gives Kaplan eq. 5.79:

Three separate-looking surface integrals are really one flux integral. The "$dy\,dz$" piece projects $S$ onto the $yz$-plane; the "$dz\,dx$" piece onto the $zx$-plane; the "$dx\,dy$" piece onto the $xy$-plane. Each projection comes with a sign determined by the orientation.

Why does Kaplan bother with this notation? Two reasons. First, the Divergence Theorem (§5.11) is stated most cleanly in component form: $\iint_S L\,dy\,dz + M\,dz\,dx + N\,dx\,dy = \iiint_R (\partial L/\partial x + \partial M/\partial y + \partial N/\partial z)\,dV$. Second, on coordinate-aligned faces (e.g. a face of a cube where $\mathbf{n} = \mathbf{k}$), only one of the three terms survives - which makes box-flux calculations very fast.