Surface Integrals

In this lesson you’ll learn how to integrate over surfaces in 3D space. There are two types: scalar surface integrals (accumulating a function over a surface, weighted by area) and vector surface integrals (computing the flux of a vector field through a surface). These are the 3D analogs of line integrals.

In Lesson 14, line integrals accumulated values along curves. Surface integrals do the same thing, but over 2D surfaces in 3D space. Instead of parametrizing a curve with one parameter t, you parametrize a surface with two parameters u and v.

A scalar surface integral accumulates a function f over a surface S. If S is parametrized by $\mathbf{r}(u, v)$ over a parameter domain D, the formula is

$$\iint_S f \, dS = \iint_D f(\mathbf{r}(u,v)) \, \|\mathbf{r}_u \times \mathbf{r}_v\| \, du \, dv$$

The cross product $\mathbf{r}_u \times \mathbf{r}_v$ gives a vector perpendicular to the surface at each point. Its magnitude $\|\mathbf{r}_u \times \mathbf{r}_v\|$ is the area scaling factor, analogous to the speed $\|\mathbf{r}'(t)\|$ in line integrals. When f = 1, the scalar surface integral gives the surface area of S.

A vector surface integral computes the flux of a vector field F through an oriented surface S:

$$\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v) \, du \, dv$$

The dot product $\mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v)$ measures how much of F passes through the surface at each point. Positive flux means the field flows in the direction of the normal. Negative flux means it flows against the normal.

For a surface given as z = g(x, y) (a graph), the formulas simplify. The surface element becomes

$$dS = \sqrt{1 + \left(\frac{\partial g}{\partial x}\right)^2 + \left(\frac{\partial g}{\partial y}\right)^2} \, dx \, dy$$

This is often easier to work with than a full parametrization.

Example 1: Surface area of the unit sphere

Compute $\iint_S 1 \, dS$ over the unit sphere (f = 1 gives surface area).

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The unit sphere is the surface of all points at distance 1 from the origin. We’re integrating f = 1 over this surface, which just adds up all the tiny area elements to get the total surface area.

Parametrize the sphere using spherical coordinates with $\mathbf{r}(\varphi, \theta) = \langle \sin\varphi\cos\theta,\; \sin\varphi\sin\theta,\; \cos\varphi \rangle$ where phi goes from 0 to pi and theta goes from 0 to $2\pi$

The cross product magnitude works out to $\|\mathbf{r}_\varphi \times \mathbf{r}_\theta\| = \sin\varphi$ (this is a standard result for the unit sphere).

$$\iint_S 1 \, dS = \int_0^{2\pi} \int_0^{\pi} \sin\varphi \, d\varphi \, d\theta$$ $$= \int_0^{2\pi} \left[-\cos\varphi\right]_0^{\pi} d\theta = \int_0^{2\pi} 2 \, d\theta = 4\pi$$

The surface area of the unit sphere is $4\pi,$ which matches the formula $4\pi r^2$ with r = 1.

Example 2: Flux through the upper hemisphere

Compute the flux of $\mathbf{F} = \langle 0,\; 0,\; z \rangle$ through the upper hemisphere of the unit sphere, with outward-pointing normal.

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The orange arrows show the outward normal vectors at sample points on the hemisphere. The field F = $\langle 0, 0, z \rangle$ points straight up, and z is positive on the upper hemisphere, so the field flows outward through the surface everywhere.

Parametrize the upper hemisphere: $\mathbf{r}(\varphi, \theta) = \langle \sin\varphi\cos\theta,\; \sin\varphi\sin\theta,\; \cos\varphi \rangle$ with phi from 0 to pi/2 and theta from 0 to $2\pi$

The outward normal (cross product) for the unit sphere is $\mathbf{r}_\varphi \times \mathbf{r}_\theta = \sin\varphi \cdot \langle \sin\varphi\cos\theta,\; \sin\varphi\sin\theta,\; \cos\varphi \rangle$

Compute $\mathbf{F} \cdot (\mathbf{r}_\varphi \times \mathbf{r}_\theta)$ where F at the surface point is $\langle 0, 0, \cos\varphi \rangle$:

$$\mathbf{F} \cdot (\mathbf{r}_\varphi \times \mathbf{r}_\theta) = \cos\varphi \cdot \sin\varphi \cdot \cos\varphi = \sin\varphi\cos^2\varphi$$ $$\iint_S \mathbf{F} \cdot d\mathbf{S} = \int_0^{2\pi} \int_0^{\pi/2} \sin\varphi\cos^2\varphi \, d\varphi \, d\theta$$

The inner integral uses the substitution u = cos(phi):

$$\int_0^{\pi/2} \sin\varphi\cos^2\varphi \, d\varphi = \left[-\frac{\cos^3\varphi}{3}\right]_0^{\pi/2} = 0 - \left(-\frac{1}{3}\right) = \frac{1}{3}$$ $$\text{Flux} = \int_0^{2\pi} \frac{1}{3} \, d\theta = \frac{2\pi}{3}$$

The flux is $\frac{2\pi}{3}$. The field F = $\langle 0, 0, z \rangle$ pushes upward, and the hemisphere captures that upward flow.

Example 3: Surface integral over a graph

Compute $\iint_S z \, dS$ where S is the surface z = x + y over the unit square $0 \leq x \leq 1,\; 0 \leq y \leq 1$

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The tilted plane z = x + y rises from 0 at the origin to 2 at the corner (1, 1). The green outline shows the unit square domain in the xy-plane below.

For a graph z = g(x, y) = x + y, the partial derivatives are $g_x = 1$ and $g_y = 1,$ so

$$dS = \sqrt{1 + 1 + 1} \, dx \, dy = \sqrt{3} \, dx \, dy$$

On the surface, z = x + y, so

$$\iint_S z \, dS = \int_0^1 \int_0^1 (x + y)\sqrt{3} \, dx \, dy = \sqrt{3} \int_0^1 \int_0^1 (x + y) \, dx \, dy$$ $$= \sqrt{3} \int_0^1 \left[\frac{x^2}{2} + xy\right]_0^1 dy = \sqrt{3} \int_0^1 \left(\frac{1}{2} + y\right) dy$$ $$= \sqrt{3} \left[\frac{y}{2} + \frac{y^2}{2}\right]_0^1 = \sqrt{3} \cdot 1 = \sqrt{3}$$

Surface integrals connect math to physical measurements on surfaces:

• Gauss’s Law in electromagnetism says the flux of the electric field through a closed surface equals the enclosed charge divided by a constant. This is a surface integral.
• Fluid dynamics uses flux integrals to compute how much fluid passes through a membrane or pipe cross-section per unit time.
• Game engines use surface integrals (or discrete approximations) for lighting calculations: how much light energy hits each polygon of a 3D model.
• Heat transfer engineering computes the total heat flow through a surface using flux integrals of the temperature gradient.