
# 16.7: Surface Integrals


In the integral for surface area,

$\int_a^b\int_c^d |{\bf r}_u\times{\bf r}_v|\,du\,dv,$

the integrand $$|{\bf r}_u\times{\bf r}_v|\,du\,dv$$ is the area of a tiny parallelogram, that is, a very small surface area, so it is reasonable to abbreviate it $$dS$$; then a shortened version of the integral is

$\iint\limits_{D} 1\cdot dS. \nonumber$

We have already seen that if $$D$$ is a region in the plane, the area of $$D$$ may be computed with

$\iint\limits_{D} 1\cdot dA,\nonumber$

so this is really quite familiar, but the $$dS$$ hides a little more detail than does $$dA$$. Just as we can integrate functions $$f(x,y)$$ over regions in the plane, using

$\iint\limits_{D} f(x,y)\, dA,\nonumber$

so we can compute integrals over surfaces in space, using

$\iint\limits_{D} f(x,y,z)\, dS. \nonumber$

In practice this means that we have a vector function $${\bf r}(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle$$ for the surface, and the integral we compute is

$\int_a^b\int_c^d f(x(u,v),y(u,v),z(u,v))|{\bf r}_u\times{\bf r}_v|\,du\,dv.$

That is, we express everything in terms of $$u$$ and $$v$$, and then we can do an ordinary double integral.

Example $$\PageIndex{1}$$:

Suppose a thin object occupies the upper hemisphere of $$x^2+y^2+z^2=1$$ and has density $$\sigma(x,y,z)=z$$. Find the mass and center of mass of the object. (Note that the object is just a thin shell; it does not occupy the interior of the hemisphere.)

Solution

We write the hemisphere as $${\bf r}(\phi,\theta)= \langle \cos\theta\sin\phi, \sin\theta\sin\phi, \cos\phi\rangle$$, $$0\le\phi\le \pi/2$$ and $$0\le\theta\le 2\pi$$. So $${\bf r}_\theta = \langle -\sin\theta\sin\phi, \cos\theta\sin\phi, 0\rangle$$ and $${\bf r}_\phi =\langle \cos\theta\cos\phi, \sin\theta\cos\phi, -\sin\phi\rangle$$. Then

${\bf r}_\theta\times{\bf r}_\phi =\langle -\cos\theta\sin^2\phi,-\sin\theta\sin^2\phi,-\cos\phi\sin\phi\rangle \nonumber$

and

$|{\bf r}_\theta\times{\bf r}_\phi| = |\sin\phi| = \sin\phi,\nonumber$

since we are interested only in $$0\le\phi\le \pi/2$$. Finally, the density is $$z=\cos\phi$$ and the integral for mass is

$\int_0^{2\pi}\int_0^{\pi/2} \cos\phi\sin\phi\,d\phi\,d\theta=\pi.\nonumber$

By symmetry, the center of mass is clearly on the $$z$$-axis, so we only need to find the $$z$$-coordinate of the center of mass. The moment around the $$x$$-$$y$$ plane is

$\int_0^{2\pi}\int_0^{\pi/2} z\cos\phi\sin\phi\,d\phi\,d\theta =\int_0^{2\pi}\int_0^{\pi/2} \cos^2\phi\sin\phi\,d\phi\,d\theta ={2\pi\over 3},\nonumber$

so the center of mass is at $$(0,0,2/3)$$.

Now suppose that $${\bf F}$$ is a vector field; imagine that it represents the velocity of some fluid at each point in space. We would like to measure how much fluid is passing through a surface $$D$$, the flux across $$D$$. As usual, we imagine computing the flux across a very small section of the surface, with area $$dS$$, and then adding up all such small fluxes over $$D$$ with an integral. Suppose that vector $$\bf N$$ is a unit normal to the surface at a point; $${\bf F}\cdot{\bf N}$$ is the scalar projection of $$\bf F$$ onto the direction of $$\bf N$$, so it measures how fast the fluid is moving across the surface. In one unit of time the fluid moving across the surface will fill a volume of $${\bf F}\cdot{\bf N}\,dS$$, which is therefore the rate at which the fluid is moving across a small patch of the surface. Thus, the total flux across $$D$$ is

$\iint\limits_{D} {\bf F}\cdot{\bf N}\,dS=\iint\limits_{D} {\bf F}\cdot\,d{\bf S},$

defining $$d{\bf S}={\bf N}\,dS$$. As usual, certain conditions must be met for this to work out; chief among them is the nature of the surface. As we integrate over the surface, we must choose the normal vectors $$\bf N$$ in such a way that they point "the same way'' through the surface. For example, if the surface is roughly horizontal in orientation, we might want to measure the flux in the "upwards'' direction, or if the surface is closed, like a sphere, we might want to measure the flux "outwards'' across the surface. In the first case we would choose $$\bf N$$ to have positive $$z$$ component, in the second we would make sure that $$\bf N$$ points away from the origin.

NOn-orientable surfaces: Möbius strips

Unfortunately, there are surfaces that are not orientable: they have only one side, so that it is not possible to choose the normal vectors to point in the "same way'' through the surface. The most famous such surface is the Möbius strip shown in Figure $$\PageIndex{1}$$. It is quite easy to make such a strip with a piece of paper and some tape. If you have never done this, it is quite instructive; in particular, you should draw a line down the center of the strip until you return to your starting point. No matter how unit normal vectors are assigned to the points of the Möbius strip, there will be normal vectors very close to each other pointing in opposite directions.

Figure $$\PageIndex{1}$$: A Möbius strip.

Assuming that the quantities involved are well behaved, however, the flux of the vector field across the surface $${\bf r}(u,v)$$ is

\begin{align} \iint\limits_{D} {\bf F}\cdot{\bf N}\,dS &=\iint\limits_{D}{\bf F}\cdot {{\bf r}_u\times{\bf r}_v\over|{\bf r}_u\times{\bf r}_v|} |{\bf r}_u\times{\bf r}_v|\,dA \\[4pt] &=\iint\limits_{D}{\bf F}\cdot ({\bf r}_u\times{\bf r}_v)\,dA. \end{align}

In practice, we may have to use $${\bf r}_v\times{\bf r}_u$$ or even something a bit more complicated to make sure that the normal vector points in the desired direction.

Example $$\PageIndex{2}$$:

Compute the flux of $${\bf F}=\langle x,y,z^4\rangle$$ across the cone $$z=\sqrt{x^2+y^2}$$, $$0\le z\le 1$$, in the downward direction.

Solution

We write the cone as a vector function:

$${\bf r}=\langle v\cos u, v\sin u, v\rangle$$, $$0\le u\le 2\pi$$ and $$0\le v\le 1$$.

Then $${\bf r}_u=\langle -v\sin u, v\cos u,0\rangle$$ and $${\bf r}_v=\langle \cos u, \sin u, 1\rangle$$ and $${\bf r}_u\times{\bf r}_v=\langle v\cos u,v\sin u,-v\rangle$$.

The third coordinate $$-v$$ is negative, which is exactly what we desire, that is, the normal vector points down through the surface. Then

\begin{align*} \int_0^{2\pi} \int_0^1 \langle x,y,z^4 \rangle\cdot\langle v\cos u,v\sin u,-v\rangle \,dv\,du &=\int_0^{2\pi} \int_0^1 xv \cos u+yv\sin u-z^4v\,dv\,du \\[4pt] &=\int_0^{2\pi} \int_0^1 v^2\cos^2 u+ v^2\sin^2 u-v^5\,dv\,du \\[4pt] &=\int_0^{2\pi}\int_0^1 v^2-v^5\,dv\,du \\[4pt] &={\pi\over3}. \end{align*}

## Contributors

• Integrated by Justin Marshall.