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 $$\dint{D} 1\cdot dS.$$ We have already seen that if $D$ is a region in the plane, the area of $D$ may be computed with $$\dint{D} 1\cdot dA,$$ 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 $$\dint{D} f(x,y)\, dA,$$ so we can compute integrals over surfaces in space, using $$\dint{D} f(x,y,z)\, dS.$$ 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 16.7.1 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.)

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 $${\bf r}_\theta\times{\bf r}_\phi = \langle -\cos\theta\sin^2\phi,-\sin\theta\sin^2\phi,-\cos\phi\sin\phi\rangle$$ and $$ |{\bf r}_\theta\times{\bf r}_\phi| = |\sin\phi| = \sin\phi,$$ 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 $$\int_0^{2\pi}\int_0^{\pi/2} \cos\phi\sin\phi\,d\phi\,d\theta=\pi.$$

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 $$\int_0^{2\pi}\int_0^{\pi/2} z\cos\phi\sin\phi\,d\phi\,d\theta =\int_0^{2\pi}\int_0^{\pi/2} \cos^2\phi\sin\phi\,d\phi\,d\theta ={2\pi\over 3},$$ so the center of mass is at $(0,0,2/3)$. $\square$

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
$$\dint{D} {\bf F}\cdot{\bf N}\,dS=\dint{D} {\bf F}\cdot\,d{\bf S},$$
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. 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 16.7.1. 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. Then cut along the line and see what happens.) 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.

Assuming that the quantities involved are well behaved, however, the flux of the vector field across the surface ${\bf r}(u,v)$ is $$\dint{D} {\bf F}\cdot{\bf N}\,dS =\dint{D}{\bf F}\cdot {{\bf r}_u\times{\bf r}_v\over|{\bf r}_u\times{\bf r}_v|} |{\bf r}_u\times{\bf r}_v|\,dA =\dint{D}{\bf F}\cdot ({\bf r}_u\times{\bf r}_v)\,dA.$$ 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 16.7.2 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.

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 ${\bf r}_u=\langle -v\sin u, v\cos u,0\rangle$, ${\bf r}_v=\langle \cos u, \sin u, 1\rangle$, 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 $$\eqalign{ \int_0^{2\pi}\int_0^1 \langle x,y,z^4\rangle\cdot\langle v\cos u,v\sin u,-v\rangle \,dv\,du &=\int_0^{2\pi}\int_0^1 xv\cos u+yv\sin u-z^4v\,dv\,du\cr &=\int_0^{2\pi}\int_0^1 v^2\cos^2 u+ v^2\sin^2 u-v^5\,dv\,du\cr &=\int_0^{2\pi}\int_0^1 v^2-v^5\,dv\,du={\pi\over3}.\cr }$$ $\square$

## Exercises 16.7

Here is a Sage cell if you'd like to use it, with an intial example.

**Ex 16.7.1**
Find the center of mass of an object that occupies the upper
hemisphere of $x^2+y^2+z^2=1$ and has density $x^2+y^2$.
(answer)

**Ex 16.7.2**
Find the center of mass of an object that occupies the
surface $z=xy$, $0\le x\le1$, $0\le y\le 1$ and has density $\sqrt{1+x^2+y^2}$.
(answer)

**Ex 16.7.3**
Find the center of mass of an object that occupies the
surface $\ds z=\sqrt{x^2+y^2}$, $1\le z\le4$ and has density $x^2z$.
(answer)

**Ex 16.7.4**
Find the centroid of the surface of a right circular cone of
height $h$ and base radius $r$, not including the base.
(answer)

**Ex 16.7.5**
Evaluate $\ds \dint{D} \langle 2,-3,4\rangle\cdot {\bf
N}\,dS$, where $D$ is given by $z=x^2+y^2$, $-1\le x\le 1$, $-1\le
y\le 1$, oriented up.
(answer)

**Ex 16.7.6**
Evaluate $\ds \dint{D} \langle x,y,3\rangle\cdot {\bf
N}\,dS$, where $D$ is given by $z=3x-5y$, $1\le x\le 2$, $0\le
y\le 2$, oriented up.
(answer)

**Ex 16.7.7**
Evaluate $\ds \dint{D} \langle x,y,-2\rangle\cdot {\bf
N}\,dS$, where $D$ is given by $z=1-x^2-y^2$, $x^2+y^2\le1$,
oriented up.
(answer)

**Ex 16.7.8**
Evaluate $\ds \dint{D} \langle xy, yz,zx\rangle\cdot {\bf
N}\,dS$, where $D$ is given by $z=x+y^2+2$, $0\le x\le 1$, $x\le
y\le 1$, oriented up.
(answer)

**Ex 16.7.9**
Evaluate $\ds \dint{D} \langle e^x, e^y,z\rangle\cdot {\bf
N}\,dS$, where $D$ is given by $z=xy$, $0\le x\le 1$, $-x\le
y\le x$, oriented up.
(answer)

**Ex 16.7.10**
Evaluate $\ds \dint{D} \langle xz,yz,z\rangle\cdot {\bf
N}\,dS$, where $D$ is given by $z=a^2-x^2-y^2$, $x^2+y^2\le b^2$,
oriented up.
(answer)

**Ex 16.7.11**
A fluid has density 870 kg/m$^3$ and flows with velocity ${\bf v} =
\langle z,y^2,x^2\rangle$, where distances are in meters and the
components of ${\bf v}$ are in meters per second. Find the rate of flow
outward through the portion of the cylinder $x^2+y^2 = 4$, $0\leq
z\leq 1$ for which $y\ge 0$.
(answer)

**Ex 16.7.12**
Gauss's Law says that the net charge, $Q$,
enclosed by a closed surface, $S$, is
$$Q=\epsilon_0 \dint{} {\bf E}\cdot {\bf N}\,dS$$
where ${\bf E}$ is an electric field and $\epsilon_0$ (the
permittivity of free space) is a known constant; $\bf N$ is oriented
outward.
Use Gauss's Law to find the charge contained in the cube with vertices
$(\pm 1, \pm 1, \pm 1)$ if the electric field is
${\bf E} = \langle x,y,z\rangle$.
(answer)