Surface Area of a Parametric Surface

To find the area of a curved surface, cut the parameter region into small rectangles and look at what each one becomes on the surface: a small, nearly flat parallelogram. Its area comes from the cross product, and adding them up gives a double integral. For graphs this is the formula from surface area by double integrals; the parametric version covers every surface.

Key idea

For a smooth parametric surface $\mathbf{r}(u, v)$, $(u, v)$ in $D$, covered once: $$\text{Area} = \iint_D |\mathbf{r}_u \times \mathbf{r}_v|\,dA.$$ Useful values: sphere of radius $a$, $|\mathbf{r}_\varphi \times \mathbf{r}_\theta| = a^2\sin\varphi$; cylinder of radius $a$, $|\mathbf{r}_\theta \times \mathbf{r}_z| = a$; graph $z = f(x, y)$, $\sqrt{1 + f_x^2 + f_y^2}$.

Why it works

The parameter rectangle $[u, u + \Delta u] \times [v, v + \Delta v]$ maps to a small patch with corners near $\mathbf{r}(u, v)$, $\mathbf{r}(u, v) + \mathbf{r}_u\Delta u$ and $\mathbf{r}(u, v) + \mathbf{r}_v\Delta v$. That patch is almost the parallelogram with sides $\mathbf{r}_u\Delta u$ and $\mathbf{r}_v\Delta v$, whose area is $|\mathbf{r}_u \times \mathbf{r}_v|\,\Delta u\,\Delta v$. Adding over all cells and letting them shrink gives the integral. (The same idea, in the plane, gave the Jacobian.)

The cone $\langle v\cos u, v\sin u, v \rangle$, ${}0 \le v \le 2$. The grid cells get wider as $v$ grows: their area factor $|\mathbf{r}_u \times \mathbf{r}_v| = \sqrt{2}\,v$ grows with the distance from the tip.

For a graph $z = f(x, y)$, use $\mathbf{r}(x, y) = \langle x, y, f(x, y) \rangle$: then $\mathbf{r}_x \times \mathbf{r}_y = \langle -f_x, -f_y, 1 \rangle$ and the formula becomes $\iint_D \sqrt{1 + f_x^2 + f_y^2}\,dA$, the one from Calculus 3. The parametric version is needed when the surface is not a graph over a region of the $xy$-plane: a whole sphere, a cylinder standing on the $xy$-plane, a surface that folds over itself. One caution: the formula counts area each time the parametrization covers it, so choose parameter ranges that cover the surface exactly once. On a sphere, ${}0 \le \varphi \le \pi$ and ${}0 \le \theta \le 2\pi$ do; letting $\varphi$ run to ${}2\pi$ would count the sphere twice.

Worked examples

Worked example: A sphere

Find the area of the sphere of radius $a$.

$$\mathbf{r}(\varphi, \theta) = \langle a\sin\varphi\cos\theta, a\sin\varphi\sin\theta, a\cos\varphi \rangle,$$ and $|\mathbf{r}_\varphi \times \mathbf{r}_\theta| = a^2\sin\varphi$.
$$\int_0^{2\pi} \int_0^\pi a^2\sin\varphi\,d\varphi\,d\theta = 2\pi a^2 \cdot 2 = 4\pi a^2.$$

Answer: ${}4\pi a^2$

Worked example: A cone

Find the area of the cone $z = \sqrt{x^2 + y^2}$ for ${}0 \le z \le 2$.

From the previous lesson, $$|\mathbf{r}_u \times \mathbf{r}_v| = |\langle v\cos u, v\sin u, -v \rangle| = \sqrt{2}\,v.$$
$$\int_0^{2\pi} \int_0^2 \sqrt{2}\,v\,dv\,du = 2\pi \cdot \sqrt{2} \cdot 2 = 4\sqrt{2}\,\pi.$$

Answer: ${}4\sqrt{2}\,\pi$

Worked example: A helicoid

Find the area of the spiral ramp $\mathbf{r}(u, v) = \langle u\cos v, u\sin v, v \rangle$, ${}0 \le u \le 1$, ${}0 \le v \le 2\pi$.

$\mathbf{r}_u = \langle \cos v, \sin v, 0 \rangle$, $\mathbf{r}_v = \langle -u\sin v, u\cos v, 1 \rangle$, so $\mathbf{r}_u \times \mathbf{r}_v = \langle \sin v, -\cos v, u \rangle$ with length $\sqrt{1 + u^2}$.
Area $$= 2\pi\int_0^1 \sqrt{1 + u^2}\,du = 2\pi \cdot \frac{1}{2}\left[\sqrt{2} + \ln(1 + \sqrt{2})\right] = \pi\left[\sqrt{2} + \ln(1 + \sqrt{2})\right] \approx 7.21.$$

Answer: $\pi\left[\sqrt{2} + \ln(1 + \sqrt{2})\right]$

Surface Area of a Parametric Surface

Worked examples

Practice

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