Surface Area
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.5: “Triple Integrals” (surface area is in sec 5.4) |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-4-triple-integrals |
| Textbook used in class | Stewart, Calculus, Section 15.6: “Surface Area” (Examples 1, 2) |
Quick Reference
Surface area of $z = f(x,y)$ over region $D$ in the $xy$-plane: $$A = \iint_D \sqrt{1 + \left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2}\,dA.$$
Notation: sometimes written $A = \iint_D \sqrt{1 + f_x^2 + f_y^2}\,dA$.
Motivation
In single-variable calculus, arc length integrates $\sqrt{1+(y')^2}$ to account for the fact that a curve is longer than its projection onto the $x$-axis. The surface area formula is the two-dimensional analogue: the surface $z = f(x,y)$ may be tilted relative to the $xy$-plane, so a tiny patch of area $dA$ in the $xy$-plane corresponds to a tilted patch on the surface. The factor $\sqrt{1+f_x^2+f_y^2}$ is the “stretching factor” that corrects for the tilt.
Key Concept
Derivation outline. At a point $(a,b,f(a,b))$, the surface is approximated by its tangent plane. A small parallelogram in the $xy$-plane with sides $\Delta x$ and $\Delta y$ projects to a parallelogram on the tangent plane with sides $\langle 1,0,f_x\rangle\Delta x$ and $\langle 0,1,f_y\rangle\Delta y$. The area of this tilted parallelogram is the magnitude of their cross product: $$\langle 1,0,f_x\rangle\times\langle 0,1,f_y\rangle = \langle -f_x,-f_y,1\rangle,$$ $$\text{magnitude} = \sqrt{f_x^2+f_y^2+1}.$$
Integrating over $D$ gives the surface area formula.
Worked Examples
Example 1. Find the surface area of the part of the paraboloid $z = x^2+y^2$ that lies below $z = 4$. (Stewart 15.6, Example 2.)
$f_x = 2x$, $f_y = 2y$. Projection $D$: $x^2+y^2 \leq 4$ (disk of radius 2).
$$A = \iint_D \sqrt{1+4x^2+4y^2}\,dA.$$
In polar: $\sqrt{1+4r^2}$, $D$: $0 \leq r \leq 2$, $0 \leq \theta \leq 2\pi$.
$$A = \int_0^{2\pi}\int_0^2\sqrt{1+4r^2}\,r\,dr\,d\theta = 2\pi\int_0^2 r\sqrt{1+4r^2}\,dr.$$
Let $u = 1+4r^2$, $du = 8r\,dr$:
$$= 2\pi\cdot\frac{1}{8}\int_1^{17}\sqrt{u}\,du = \frac{\pi}{4}\cdot\left[\frac{2u^{3/2}}{3}\right]_1^{17} = \frac{\pi}{6}(17^{3/2}-1).$$
the surface area of $z = f(x,y)$ over $D$ equals the area of $D$. The area of $D$ is $\iint_D 1\,dA$. The surface area is $\iint_D\sqrt{1+f_x^2+f_y^2}\,dA$, which is larger whenever the surface is not horizontal ($f_x \neq 0$ or $f_y \neq 0$). The flat case $f = c$ (constant) gives $\sqrt{1+0+0} = 1$ and the surface area equals the area of $D$, but any tilt increases it.
Leveled Practice
Problem 1. Find the surface area of the portion of the plane $z = 1 - x - y$ over the triangle $D$ with vertices $(0,0)$, $(1,0)$, $(0,1)$.
Show answer
$f_x = -1$, $f_y = -1$. $\sqrt{1+1+1} = \sqrt{3}$.
$A = \sqrt{3}\cdot\text{Area}(D) = \sqrt{3}\cdot\frac{1}{2} = \frac{\sqrt{3}}{2}$.