Surface Area of Parametric Surfaces
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 6.6: “Parametric Surfaces and Their Areas” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/6-6-parametric-surfaces-and-their-areas |
| Textbook used in class | Stewart, Calculus, Section 16.6: “Parametric Surfaces and Their Areas” (Examples 5, 6) |
Quick Reference
Surface area of parametric surface $\mathbf{r}(u,v)$ over parameter domain $D$: $$A = \iint_D |\mathbf{r}_u\times\mathbf{r}_v|\,dA.$$
For a graph $z = f(x,y)$: $|\mathbf{r}_x\times\mathbf{r}_y| = \sqrt{1+f_x^2+f_y^2}$ (recovers the formula from Section 15.6).
Motivation
The surface area formula for parametric surfaces generalizes the formula $\iint\sqrt{1+f_x^2+f_y^2}\,dA$ for graphs. The cross product $\mathbf{r}_u\times\mathbf{r}_v$ plays the role of $\langle -f_x,-f_y,1\rangle$: it is the vector normal to the surface, and its magnitude $|\mathbf{r}_u\times\mathbf{r}_v|$ is the area scaling factor from the parameter domain $D$ to the actual surface.
Key Concept
Think of the parameter domain $D$ as a flat square, and the surface as what you get by bending, stretching, and warping that square. The factor $|\mathbf{r}_u\times\mathbf{r}_v|$ at each point measures how much a tiny rectangle in $D$ has been stretched to become the corresponding patch on the surface.
Worked Example
Find the surface area of the sphere $x^2+y^2+z^2 = a^2$. (Stewart 16.6, Example 5.)
Parametrize: $\mathbf{r}(\phi,\theta) = \langle a\sin\phi\cos\theta, a\sin\phi\sin\theta, a\cos\phi\rangle$, $0 \leq \phi \leq \pi$, $0 \leq \theta \leq 2\pi$.
$\mathbf{r}_\phi = \langle a\cos\phi\cos\theta, a\cos\phi\sin\theta, -a\sin\phi\rangle$.
$\mathbf{r}_\theta = \langle -a\sin\phi\sin\theta, a\sin\phi\cos\theta, 0\rangle$.
$\mathbf{r}_\phi\times\mathbf{r}_\theta = a^2\langle \sin^2\phi\cos\theta, \sin^2\phi\sin\theta, \sin\phi\cos\phi\rangle$.
$|\mathbf{r}_\phi\times\mathbf{r}_\theta| = a^2\sin\phi\sqrt{\sin^2\phi(\cos^2\theta+\sin^2\theta) + \cos^2\phi} = a^2\sin\phi$.
$$A = \int_0^{2\pi}\int_0^\pi a^2\sin\phi\,d\phi\,d\theta = 2\pi\cdot a^2\cdot[-\cos\phi]_0^\pi = 2\pi\cdot a^2\cdot 2 = 4\pi a^2.$$
This confirms the standard formula for the surface area of a sphere.
for a sphere of radius $a$, the parameter domain area is $4\pi a^2$. The parameter domain is the rectangle $[0,\pi]\times[0,2\pi]$, which has area $2\pi^2$. The surface area of the sphere is $4\pi a^2$. These are different: the parameter domain is NOT the sphere’s surface. The factor $|\mathbf{r}_\phi\times\mathbf{r}_\theta| = a^2\sin\phi$ converts from parameter area to surface area.
Common Misconceptions
the surface area of a parametric surface equals the area of its parameter domain $D$.
This is the concept-image-conflicts-definition error. The parameter domain $D$ is a flat region in the $uv$-plane; the surface $S$ is what results after bending and stretching that region according to $\mathbf{r}(u,v)$. The area of $D$ is $\iint_D dA$, whereas the area of $S$ is $\iint_D |\mathbf{r}_u \times \mathbf{r}_v|\,dA$. The factor $|\mathbf{r}_u \times \mathbf{r}_v|$ measures the local stretching and is generally not equal to $1$. For a sphere of radius $a$, the parameter domain has area $2\pi^2$ while the sphere has area $4\pi a^2$.
Leveled Practice
Problem 1. Find the surface area of the part of the cylinder $x^2+y^2 = 4$ between $z = 0$ and $z = 3$.
Show answer
Parametrize: $\mathbf{r}(\theta,z) = \langle 2\cos\theta, 2\sin\theta, z\rangle$, $0 \leq \theta \leq 2\pi$, $0 \leq z \leq 3$.
$\mathbf{r}_\theta = \langle -2\sin\theta, 2\cos\theta, 0\rangle$, $\mathbf{r}_z = \langle 0, 0, 1\rangle$.
$\mathbf{r}_\theta\times\mathbf{r}_z = \langle 2\cos\theta, 2\sin\theta, 0\rangle$.
$|\mathbf{r}_\theta\times\mathbf{r}_z| = 2$.
$A = \int_0^{2\pi}\int_0^3 2\,dz\,d\theta = 2\pi\cdot 6 = 12\pi$.
(This matches the formula $2\pi rh = 2\pi(2)(3) = 12\pi$.)