Parametric Surfaces and Surface Area, multivariable and vector calculus lesson

A map of the world is a flat rectangle of latitude and longitude, yet every point on it names a place on a curved globe. A parametric surface works the same way. A flat region of parameters is bent into space, and a grid drawn on the flat region becomes a net of curves on the surface. Measuring how much each tiny grid square gets stretched is exactly how surface area is computed.

Key terms

Surface parametrization

Definition Let $D$ be a region in the $uv$ -plane. A parametrization is a continuous function

\begin{gathered} \mathbf{r}(u, v) = \langle x(u, v), y(u, v), z(u, v) \rangle, \\[0.4em] (u, v) \in D, \end{gathered}

and the set $S$ of all points $\mathbf{r}(u, v)$ with $(u, v)$ in $D$ is a parametric surface. The variables $u$ and $v$ are the parameters and $D$ is the parameter domain.

In plain words Each pair of numbers in $D$ tells you one point on the surface. Different formulas can describe the same surface.

Example The part of the paraboloid $z = x^2 + y^2$ above the unit disk is described by

\mathbf{r}(r, \theta) = \langle r\cos\theta,\ r\sin\theta,\ r^2 \rangle,

with $0 \le r \le 1$ and $0 \le \theta \le 2\pi$ . The edge $r = 1$ of the parameter rectangle goes to the rim $\langle \cos\theta, \sin\theta, 1 \rangle$ , the circle of radius $1$ at height $1$ , and the edge $r = 0$ collapses to the single point at the origin.

Grid curves

Definition Fixing $v = v_0$ and letting $u$ vary gives the curve $u \mapsto \mathbf{r}(u, v_0)$ on $S$ , and fixing $u = u_0$ gives $v \mapsto \mathbf{r}(u_0, v)$ . These are the grid curves of the parametrization. They are the images of the horizontal and vertical lines of the parameter domain.

In plain words Hold one parameter still and slide the other. You trace a curve on the surface, and the two families of such curves form a net.

Example On the paraboloid above, fixing $\theta = \theta_0$ gives

r \mapsto \langle r\cos\theta_0,\ r\sin\theta_0,\ r^2 \rangle,

an arc of a parabola that rises from the origin to the rim in the vertical half-plane at angle $\theta_0$ . Fixing $r = \frac{1}{2}$ gives the circle of radius $\frac{1}{2}$ at height $\frac{1}{4}$ .

Partial derivatives and their cross product

Definition If $\mathbf{r}$ has first partial derivatives at $(u_0, v_0)$ , then

\mathbf{r}_u = \left\langle \frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}, \frac{\partial z}{\partial u} \right\rangle,

and $\mathbf{r}_v$ is defined the same way. The vector $\mathbf{r}_u$ is the velocity of the grid curve with $v = v_0$ , and $\mathbf{r}_v$ that of the grid curve with $u = u_0$ , so both are tangent to $S$ wherever they are nonzero. Their cross product $\mathbf{r}_u \times \mathbf{r}_v$ is perpendicular to both, and its length $\lvert \mathbf{r}_u \times \mathbf{r}_v \rvert$ is the area of the parallelogram they span.

In plain words A tiny parameter rectangle of sides $\Delta u$ and $\Delta v$ lands on the surface as roughly a parallelogram with sides $\mathbf{r}_u \Delta u$ and $\mathbf{r}_v \Delta v$ . The cross product gives its normal direction and its area.

Example For the paraboloid,

\begin{aligned} \mathbf{r}_r &= \langle \cos\theta,\ \sin\theta,\ 2r \rangle, \\[0.4em] \mathbf{r}_\theta &= \langle -r\sin\theta,\ r\cos\theta,\ 0 \rangle. \end{aligned}

Their cross product and its length for $r \ge 0$ are

\begin{gathered} \mathbf{r}_r \times \mathbf{r}_\theta \\ = \langle -2r^2\cos\theta,\ -2r^2\sin\theta,\ r \rangle, \\[0.4em] \lvert \mathbf{r}_r \times \mathbf{r}_\theta \rvert = r\sqrt{4r^2 + 1}. \end{gathered}

The $z$ component $r$ is positive for $r > 0$ , so this normal points up into the bowl.

Regular surface

Definition Let $D$ be a closed bounded region in the $uv$ -plane with piecewise smooth boundary, and let $\mathbf{r}$ be $C^1$ on an open set containing $D$ . The parametrization $\mathbf{r}$ of $S = \mathbf{r}(D)$ is regular if $\mathbf{r}$ is one-to-one on the interior of $D$ and, at every interior point of $D$ ,

\mathbf{r}_u \times \mathbf{r}_v \neq \mathbf{0}.

A surface with such a parametrization is a regular, or smooth, surface patch. Both conditions may fail at boundary points of $D$ , which lets seams and edges that collapse to a point through. At a point where the cross product vanishes, $\mathbf{r}_u$ and $\mathbf{r}_v$ are parallel or one of them is zero, so they do not span a plane. Some books ask for a nonzero cross product at every point of $D$ , which is stricter.

In plain words Regular means the grid never collapses and never folds back onto itself away from the edge of $D$ . At every interior point the two grid directions stay independent, so they really span a plane.

Example The cone $\mathbf{r}(u, v) = \langle u\cos v, u\sin v, u \rangle$ has

\mathbf{r}_u \times \mathbf{r}_v = \langle -u\cos v,\ -u\sin v,\ u \rangle,

with length $\sqrt{2}\, \lvert u \rvert$ . Over $0 \le u \le 1$ , $0 \le v \le 2\pi$ it is nonzero at every interior point and the map is one-to-one there, so the parametrization is regular. The cross product vanishes on the edge $u = 0$ , which goes to the vertex, and there the cone has a sharp point and no tangent plane. The paraboloid parametrization is regular for the same reason, with its cross product vanishing only on the edge $r = 0$ . Yet the paraboloid does have a tangent plane at the origin, the plane $z = 0$ . So a zero cross product on the edge of $D$ may come from the surface or only from the choice of parameters, and regularity promises a tangent plane only at images of points where the cross product is nonzero.

Tangent plane

Definition Let $\mathbf{r}$ be regular and let $(u_0, v_0)$ be a point where $\mathbf{n} = \mathbf{r}_u \times \mathbf{r}_v \neq \mathbf{0}$ . The tangent plane to $S$ at $P_0 = \mathbf{r}(u_0, v_0)$ is the plane through $P_0$ with normal vector $\mathbf{n}$ , that is, the points $\mathbf{x}$ with

\mathbf{n} \cdot (\mathbf{x} - P_0) = 0.

In plain words The two grid directions at a point span a flat plane that just touches the surface there. Their cross product is its normal.

Example On the paraboloid at $r = \frac{1}{2}$ , $\theta = 0$ , the point is $(\frac{1}{2}, 0, \frac{1}{4})$ and the normal is $\langle -\frac{1}{2}, 0, \frac{1}{2} \rangle$ . Doubling it to $\langle -1, 0, 1 \rangle$ gives the tangent plane

-\left(x - \tfrac{1}{2}\right) + \left(z - \tfrac{1}{4}\right) = 0,

which simplifies to $z = x - \frac{1}{4}$ . The graph formula agrees. For $z = x^2 + y^2$ at $(\frac{1}{2}, 0)$ the tangent plane is $z = \frac{1}{4} + 1 \cdot (x - \frac{1}{2}) + 0 \cdot y$ , which is the same plane.

Surface area

Definition Let $S = \mathbf{r}(D)$ be given by a regular parametrization $\mathbf{r}$ over a closed bounded region $D$ with piecewise smooth boundary. The surface area of $S$ is

A(S) = \iint_D \lvert \mathbf{r}_u \times \mathbf{r}_v \rvert \, dA.

In plain words Add up the areas of all the tiny parallelograms. Each parameter square of area $\Delta u \, \Delta v$ is stretched by the factor $\lvert \mathbf{r}_u \times \mathbf{r}_v \rvert$ .

Example For the paraboloid over the unit disk,

\begin{aligned} A &= \int_0^{2\pi} \!\! \int_0^1 r\sqrt{4r^2 + 1} \, dr \, d\theta \\[0.4em] &= 2\pi \cdot \frac{5\sqrt{5} - 1}{12} = \frac{\pi (5\sqrt{5} - 1)}{6}, \end{aligned}

about $5.33$ . The flat unit disk below has area $\pi \approx 3.14$ , so the bowl is noticeably larger, as it should be.

Surface area of a graph

Definition Let $g$ be $C^1$ on an open set containing a closed bounded region $D$ of the $xy$ -plane with piecewise smooth boundary. The graph $z = g(x, y)$ over $D$ has the regular parametrization $\mathbf{r}(x, y) = \langle x, y, g(x, y) \rangle$ , with $\mathbf{r}_x \times \mathbf{r}_y = \langle -g_x, -g_y, 1 \rangle$ , which is never zero. Its area is

A = \iint_D \sqrt{1 + g_x^2 + g_y^2} \, dA.

In plain words A graph is the simplest surface. Steeper parts get a larger stretch factor, and a flat graph gets factor $1$ .

Example The part of the plane $x + y + z = 1$ in the first octant is the graph of $g = 1 - x - y$ over the triangle with vertices $(0, 0)$ , $(1, 0)$ , $(0, 1)$ , which has area $\frac{1}{2}$ . Since $g_x = g_y = -1$ , the area is

\sqrt{1 + 1 + 1} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}.

Overlap and coverage

Definition The area formula counts each point of $S$ as many times as the parametrization reaches it. It gives the area of $S$ when $\mathbf{r}$ covers all of $S$ and is one-to-one on the interior of $D$ . Overlap along the boundary of $D$ , such as a seam where $\theta = 0$ and $\theta = 2\pi$ meet or an edge that collapses to a point, does not matter, because the boundary of $D$ has zero area.

In plain words Every point of the surface should be reached, and apart from seams and edges, reached once. Running a parameter too far counts part of the surface twice.

Example The sphere $\mathbf{r}(\phi, \theta) = \langle 2\sin\phi\cos\theta, 2\sin\phi\sin\theta, 2\cos\phi \rangle$ with $0 \le \phi \le \pi$ and $0 \le \theta \le 2\pi$ has $\lvert \mathbf{r}_\phi \times \mathbf{r}_\theta \rvert = 4\sin\phi$ , and

\int_0^{2\pi} \!\! \int_0^\pi 4\sin\phi \, d\phi \, d\theta = 16\pi,

which is $4\pi \cdot 2^2$ . The seam $\theta = 0$ and the two poles are reached more than once, but they lie on the boundary of the parameter rectangle. Letting $\theta$ run from $0$ to $4\pi$ would wrap the sphere twice and give $32\pi$ , which is wrong.

Check yourself

Find the tangent plane to the surface $\mathbf{r}(u, v) = \langle u, v, uv \rangle$ at $u = 1$ , $v = 2$ .

Show the answer

The point is $(1, 2, 2)$ , with $\mathbf{r}_u = \langle 1, 0, 2 \rangle$ and $\mathbf{r}_v = \langle 0, 1, 1 \rangle$ there. Their cross product is $\langle -2, -1, 1 \rangle$ , so the plane is

-2(x - 1) - (y - 2) + (z - 2) = 0,

which simplifies to $z = 2x + y - 2$ .

Find the area of the cylinder $x^2 + y^2 = 4$ between $z = 0$ and $z = 3$ , using a parametrization that covers it once.

Show the answer

Use $\mathbf{r}(\theta, z) = \langle 2\cos\theta, 2\sin\theta, z \rangle$ with $0 \le \theta \le 2\pi$ and $0 \le z \le 3$ , which is one-to-one except on the seam. Here $\lvert \mathbf{r}_\theta \times \mathbf{r}_z \rvert = 2$ , so the area is $2 \cdot 2\pi \cdot 3 = 12\pi$ .

Someone computes the area of the paraboloid piece above using $0 \le r \le 1$ and $0 \le \theta \le 4\pi$ . What goes wrong?

Show the answer

Each point with $0 < r \le 1$ is reached at least twice, so the parametrization is not one-to-one on the interior of the parameter domain. The integral doubles to $\frac{\pi(5\sqrt{5} - 1)}{3}$ , twice the true area.

Parametric Surfaces and Surface Area

Key terms

Surface parametrization

Grid curves

Partial derivatives and their cross product

Regular surface

Tangent plane

Surface area

Surface area of a graph

Overlap and coverage

Check yourself

How this lesson connects