A curve is traced by one parameter; a surface needs two. Just as $\mathbf{r}(t)$ describes a curve in space, a function $\mathbf{r}(u, v)$ of two parameters sweeps out a surface. This lets us describe the quadric surfaces and many others in a form that is ready for integration.
Key idea
A parametric surface is given by $\mathbf{r}(u, v) = \langle x(u, v), y(u, v), z(u, v) \rangle$ for $(u, v)$ in a region $D$ of the $uv$-plane. Its tangent vectors are $\mathbf{r}_u$ and $\mathbf{r}_v$, and
$$\mathbf{N} = \mathbf{r}_u \times \mathbf{r}_v$$
is a normal vector. The surface is smooth where $\mathbf{N} \ne \mathbf{0}$.
| Surface | Parametrization | Parameters |
|---|---|---|
| graph $z = f(x, y)$ | $\langle u, v, f(u, v) \rangle$ | $(u, v)$ in the domain |
| cylinder $x^2 + y^2 = a^2$ | $\langle a\cos u, a\sin u, v \rangle$ | ${}0 \le u \le 2\pi$ |
| sphere $x^2 + y^2 + z^2 = a^2$ | $\langle a\sin u\cos v, a\sin u\sin v, a\cos u \rangle$ | ${}0 \le u \le \pi$, ${}0 \le v \le 2\pi$ |
| cone $z = \sqrt{x^2 + y^2}$ | $\langle v\cos u, v\sin u, v \rangle$ | $v \ge 0$ |
| plane through $P_0$ spanned by $\mathbf{a}$, $\mathbf{b}$ | $P_0 + u\mathbf{a} + v\mathbf{b}$ | all $u$, $v$ |
Why it works
Why $\mathbf{r}_u \times \mathbf{r}_v$ is normal: holding $v$ fixed and letting $u$ vary traces a curve on the surface (a "$u$-curve"), and $\mathbf{r}_u$ is its tangent vector. Likewise $\mathbf{r}_v$ is tangent to a $v$-curve. Both lie in the tangent plane, so their cross product is perpendicular to it. If they are parallel (or one is zero), the cross product vanishes and the grid of curves collapses there, as at the tip of a cone.
Worked examples
Worked example: A plane
Parametrize the plane through $(1, 0, 2)$ spanned by $\langle 1, 1, 0 \rangle$ and $\langle 0, 1, -1 \rangle$, and find a normal vector.
- $\mathbf{r}(u, v) = \langle 1 + u, u + v, 2 - v \rangle$.
- $\mathbf{r}_u = \langle 1, 1, 0 \rangle$, $\mathbf{r}_v = \langle 0, 1, -1 \rangle$, and $\mathbf{r}_u \times \mathbf{r}_v = \langle -1, 1, 1 \rangle$.
- So the plane is $-(x - 1) + y + (z - 2) = 0$, that is, $-x + y + z = 1$.
Answer: $\langle 1 + u, u + v, 2 - v \rangle$, normal $\langle -1, 1, 1 \rangle$
Worked example: A cone
Find $\mathbf{r}_u \times \mathbf{r}_v$ for the cone $\langle v\cos u, v\sin u, v \rangle$. Where is it not smooth?
- $\mathbf{r}_u = \langle -v\sin u, v\cos u, 0 \rangle$, $\mathbf{r}_v = \langle \cos u, \sin u, 1 \rangle$.
- $\mathbf{r}_u \times \mathbf{r}_v = \langle v\cos u, v\sin u, -v \rangle$.
- This is $\mathbf{0}$ only when $v = 0$: the vertex, where the cone comes to a point.
Answer: $\langle v\cos u, v\sin u, -v \rangle$; not smooth at the vertex
Worked example: Recognizing a surface
What surface is $\mathbf{r}(u, v) = \langle 2\cos u, 2\sin u, v \rangle$?
- $x^2 + y^2 = 4\cos^2 u + 4\sin^2 u = 4$ for every $u$, while $z = v$ can be anything.
- It is the cylinder of radius ${}2$ around the $z$-axis.
Answer: the cylinder $x^2 + y^2 = 4$