Math Core

A curve is described by one parameter, $\mathbf{r}(t)$ . A surface needs two, $\mathbf{r}(u, v)$ . Once you can parametrize a surface, you can find its area, integrate a function over it (a surface integral), and measure how much of a vector field flows through it (a flux integral). These are the two-dimensional cousins of arc length and line integrals, and they set up Stokes' theorem and the divergence theorem.

Parametric surfaces

A parametric surface is the set of points

\mathbf{r}(u, v) = \langle x(u, v),\ y(u, v),\ z(u, v)\rangle, \qquad (u, v) \text{ in a region } D.

Some standard parametrizations:

surface	parametrization	parameter domain
graph $z = g(x, y)$	$\langle x,\ y,\ g(x, y)\rangle$	the region below the surface
cylinder $x^2 + y^2 = a^2$	$\langle a\cos\theta,\ a\sin\theta,\ z\rangle$	$0 \le \theta \le 2\pi$ , $z$ in some interval
sphere $x^2 + y^2 + z^2 = a^2$	$\langle a\sin\phi\cos\theta,\ a\sin\phi\sin\theta,\ a\cos\phi\rangle$	$0 \le \phi \le \pi$ , $0 \le \theta \le 2\pi$
plane through $P_0$ containing $\mathbf{a}$ , $\mathbf{b}$	$P_0 + u\mathbf{a} + v\mathbf{b}$	all $u, v$

The sphere uses the spherical angles: $\phi$ is measured down from the positive $z$ -axis and $\theta$ is the usual polar angle.

Tangent vectors, normals and surface area

Holding $v$ fixed and varying $u$ traces a curve on the surface with tangent vector $\mathbf{r}_u = \partial\mathbf{r}/\partial u$ ; similarly $\mathbf{r}_v$ . These two tangent vectors span the tangent plane, so their cross product $\mathbf{r}_u\times\mathbf{r}_v$ is a normal vector to the surface.

A small rectangle $\Delta u$ by $\Delta v$ in the parameter domain maps to a small patch on the surface that is nearly a parallelogram with sides $\mathbf{r}_u\,\Delta u$ and $\mathbf{r}_v\,\Delta v$ . Its area is $\lvert\mathbf{r}_u\times\mathbf{r}_v\rvert\,\Delta u\,\Delta v$ .

Surface area and the surface element

The area element on the surface is $dS = \lvert\mathbf{r}_u\times\mathbf{r}_v\rvert\,dA$ , and

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

For a graph $z = g(x, y)$ : $\mathbf{r}_x\times\mathbf{r}_y = \langle -g_x,\ -g_y,\ 1\rangle$ , so $dS = \sqrt{1 + g_x^2 + g_y^2}\,dA$ .

For a sphere of radius $a$ : $dS = a^2\sin\phi\,d\phi\,d\theta$ .

Worked example: A cylinder

Find the area of the cylinder $x^2 + y^2 = 4$ for $0 \le z \le 3$ .

With $\mathbf{r}(\theta, z) = \langle 2\cos\theta,\ 2\sin\theta,\ z\rangle$ : $\mathbf{r}_\theta = \langle -2\sin\theta,\ 2\cos\theta,\ 0\rangle$ and $\mathbf{r}_z = \langle 0, 0, 1\rangle$ . Then $\mathbf{r}_\theta\times\mathbf{r}_z = \langle 2\cos\theta,\ 2\sin\theta,\ 0\rangle$ , which has length $2$ . So

\text{Area} = \int_0^3\!\!\int_0^{2\pi} 2\,d\theta\,dz = 12\pi,

which matches circumference times height, $4\pi\cdot 3$ .

Worked example: A paraboloid

Find the area of the part of $z = x^2 + y^2$ that lies below the plane $z = 2$ .

Here $g_x = 2x$ , $g_y = 2y$ , so $dS = \sqrt{1 + 4x^2 + 4y^2}\,dA$ . The region below is the disk $x^2 + y^2 \le 2$ . In polar coordinates, with $u = 1 + 4r^2$ :

\text{Area} = \int_0^{2\pi}\!\!\int_0^{\sqrt{2}} \sqrt{1 + 4r^2}\,r\,dr\,d\theta = 2\pi\cdot\frac{1}{12}\Big[(1 + 4r^2)^{3/2}\Big]_0^{\sqrt{2}} = \frac{\pi}{6}(27 - 1) = \frac{13\pi}{3}.

Surface integrals of functions

Just as $\int_C f\,ds$ weights a curve by $f$ , a surface integral weights a surface by $f$ :

\iint_S f(x, y, z)\,dS = \iint_D f\big(\mathbf{r}(u, v)\big)\,\lvert\mathbf{r}_u\times\mathbf{r}_v\rvert\,dA.

If $f$ is a density (mass per unit area) of a thin sheet, this is the sheet's mass. With $f = 1$ it is the surface area.

Worked example: Integrating over a hemisphere

Evaluate $\displaystyle\iint_S z\,dS$ , where $S$ is the upper half of the unit sphere.

On the unit sphere $z = \cos\phi$ and $dS = \sin\phi\,d\phi\,d\theta$ , with $0 \le \phi \le \tfrac{\pi}{2}$ :

\iint_S z\,dS = \int_0^{2\pi}\!\!\int_0^{\pi/2}\cos\phi\sin\phi\,d\phi\,d\theta = 2\pi\cdot\frac{1}{2} = \pi.

Oriented surfaces and flux

To measure flow through a surface you need to say which side is which. A surface is oriented when you choose a continuous unit normal $\mathbf{n}$ at every point. For a closed surface such as a sphere, the standard choice is the outward normal. For a graph $z = g(x, y)$ , "upward" means the normal $\langle -g_x, -g_y, 1\rangle$ , whose $z$ -component is positive. (A Möbius strip has only one side and can't be oriented.)

Definition

Flux integral

The flux of $\mathbf{F}$ across the oriented surface $S$ is

\iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_S \mathbf{F}\cdot\mathbf{n}\,dS = \iint_D \mathbf{F}\big(\mathbf{r}(u, v)\big)\cdot(\mathbf{r}_u\times\mathbf{r}_v)\,dA,

where the parametrization is chosen so that $\mathbf{r}_u\times\mathbf{r}_v$ points in the direction of $\mathbf{n}$ .

For a graph $z = g(x, y)$ with upward orientation and $\mathbf{F} = \langle P, Q, R\rangle$ :

\iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_D \left(-P\,g_x - Q\,g_y + R\right) dA.

If $\mathbf{F}$ is the velocity of a fluid, the flux is the volume of fluid crossing $S$ per unit time, in the direction of $\mathbf{n}$ . Notice that the $\lvert\mathbf{r}_u\times\mathbf{r}_v\rvert$ factors cancel: $\mathbf{n}\,dS = (\mathbf{r}_u\times\mathbf{r}_v)\,dA$ .

Worked example: Flux through a paraboloid

Find the flux of $\mathbf{F} = \langle y,\ x,\ z\rangle$ upward through the part of $z = 4 - x^2 - y^2$ above the $xy$ -plane.

Here $g_x = -2x$ and $g_y = -2y$ , so

-P\,g_x - Q\,g_y + R = 2xy + 2xy + (4 - x^2 - y^2) = 4xy + 4 - r^2.

Over the disk $r \le 2$ , the term $4xy$ integrates to $0$ by symmetry. The rest is

\int_0^{2\pi}\!\!\int_0^2 (4 - r^2)\,r\,dr\,d\theta = 2\pi\left(8 - 4\right) = 8\pi.

Common mistake

Scalar surface integrals use $\lvert\mathbf{r}_u\times\mathbf{r}_v\rvert$ (a length); flux integrals use $\mathbf{r}_u\times\mathbf{r}_v$ itself (a vector, dotted with $\mathbf{F}$ ). And for flux, check that your normal points the way the problem asks. If it points the wrong way, the answer changes sign.

Tip

On a sphere of radius $a$ centered at the origin, the outward unit normal is simply $\mathbf{n} = \dfrac{\langle x, y, z\rangle}{a}$ . So for $\mathbf{F} = \langle x, y, z\rangle$ , $\mathbf{F}\cdot\mathbf{n} = a$ everywhere and the flux is $a\cdot 4\pi a^2 = 4\pi a^3$ .

Practice

Practice 1

For the plane $\mathbf{r}(u, v) = \langle u,\ v,\ 3 - u - 2v\rangle$ , compute the normal vector $\mathbf{r}_u\times\mathbf{r}_v$ .

Enter a point like (2, -3)

Practice 2

Find the area of the part of the plane $z = 1 + 2x + 2y$ that lies above the rectangle $0 \le x \le 1$ , $0 \le y \le 2$ .