This unit covers the following ideas. In preparation
for the quiz and exam, make sure you have a lesson plan
containing examples that explain and illustrate the following
concepts.
Surface Area and Surface Integrals
In firstsemester
calculus, we learned how to compute integrals $\int_a^b f dx$
along straight (flat) segments $[a,b]$. This semester, in the
line integral unit, we learned how to change the segment to a
curve, which allowed us to compute integrals $\int_C fds$ along
any curve $C$, instead of just along curves (segments) on the
$x$axis. The integral $\int_a^b dx=ba$ gives the length of the
segment $[a,b]$. The integral $\int_C ds$ gives the length $s$ of
the curve $C$. In the double integral unit we learned how to
compute double integrals $\iint_R fdA$ along flat regions $R$ in
the plane. We'll now learn how to change the flat region $R$ into
a curved surface $S$, and then compute integrals of the form
$\iint_S fd\sigma$ along curved surfaces. The differential
$d\sigma$ stands for a little bit of surface area. We already
know that $\iint_R dA$ gives the area of $R$. We'll define
$\iint_S d\sigma$ so that it gives the surface area of $S$.
Consider the surface $S$ given by $z=9x^2y^2$ (we've seen
this surface many times). A parametrization of this surface is
$$\vec r(x,y) = (x,y,9x^2y^2).$$
 \instructor{See
Sage.}% Draw the surface $S$. Add to your surface plot
the parabolas given by $\vec r(x,0)$, $\vec r(x,1)$, and
$\vec r(x,2)$, as well as the parabolas given by $\vec
r(0,y)$, $\vec r(1,y)$, and $\vec r(2,y)$. You should have an
upside down paraboloid, with at least 6 different parabolas
drawn on the surface. These parabolas should divide the
surface up into a bunch of different patches. Our goal is to
find the area of each patch, where each patch is almost like
a parallelogram.
 Find $\ds \frac{\partial \vec r}{\partial x}$ and
$\ds\frac{\partial \vec r}{\partial y}$. At the point
$(2,1)$, draw both vectors. These vectors form the edges of a
parallelogram. Add that parallelogram to your picture.
 Show that the area of a parallelogram whose edges are the
vectors $\ds\frac{\partial \vec r}{\partial x}(x,y)$ and
$\ds\frac{\partial \vec r}{\partial y}$ is
$\sqrt{1+4x^2+4y^2}$. [Hint: think about the cross
product.]
 Find the area of the parallelogram whose edges are the
vectors $\ds\frac{\partial \vec r}{\partial x}dx$ and
$\ds\frac{\partial \vec r}{\partial y}dy$, where $dx$ and
$dy$ are to be determined.
In the previous problem, you showed that the area of the
parallelogram with edges given by $\frac{\partial \vec
r}{\partial x}dx$ and $\frac{\partial \vec r}{\partial y}dy$ is
$$d\sigma =\left \frac{\partial \vec r}{\partial x} \times
\frac{\partial \vec r}{\partial y}\right dxdy.$$ This little bit
of area approximates the area of a tiny patch on the surface. If
we add all these areas up, we should obtain the surface area.
Let $S$ be a surface. Let $\vec r(u,v)=(x,y,z)$ be a
parametrization of the surface, where the bounds on $u$ and $v$
form a region $R$ in the $uv$ plane. Then the surface area
element (representing a little bit of surface) is $$d\sigma
=\left \frac{\partial \vec r}{\partial u} \times
\frac{\partial \vec r}{\partial v}\right dudv = \left\vec
r_u\times\vec r_v\rightdudv.$$ The surface integral of a
continuous function $f(x,y,z)$ along the surface $S$ is
$$\iint_S f(x,y,z) d\sigma = \iint_R f(\vec r(u,v)) \left
\frac{\partial \vec r}{\partial u} \times \frac{\partial \vec
r}{\partial v}\right dudv.$$ If we let $f=1$, then the surface
area of $S$ is simply $$\sigma = \iint_S d\sigma = \iint_R
\left \frac{\partial \vec r}{\partial u} \times \frac{\partial
\vec r}{\partial v}\right dudv.$$
This definition tells us how to compute any surface
integral. The steps are almost identical to the line integral
steps.
 Start by getting a parametrization $\vec r$ of the surface
$S$ where the bounds form a region $R$.
 Find a little bit of surface area by computing $d\sigma
=\left \frac{\partial \vec r}{\partial u} \times
\frac{\partial \vec r}{\partial v}\right dudv.$
 Multiply $f$ by $d\sigma$, and replace each $x$, $y$, $z$
with what they equals from the parametrization.
 Integrate the previous function along $R$, your
parameterization's bounds.
Consider the surface $S$ given by $z=9x^2y^2$, for $z\geq 0$.
A parametrization of this surface is $$\vec r(x,y) =
(x,y,9x^2y^2),\quad \text{where } 9x^2y^2\geq 0.$$
 Give a set of inequalities for $x$ and $y$ that describe
the region $R$ over which we need to integrate. The
inequalities you give should be in a form that you can use
them as the bounds of a double integral.
 Find $d\sigma = \left\vec r_x\times \vec
r_y\rightdxdy$.
 Set up the surface integral $\iint_S d\sigma$ as an
iterated double integral over $R$.
 Convert the integral above to an integral in polar
coordinates (don't forget the Jacobian).
Consider the surface $S$ given by $z=9x^2y^2$, for $z\geq 0$.
A different parametrization of this surface is $$\vec
r(r,\theta) = (r\cos\theta,r\sin\theta,9r^2),\quad \text{where
} 9r^2\geq 0.$$
 Give a set of inequalities for $r$ and $\theta$ that
describe the region $R_{r\theta}$ over which we need to
integrate. The inequalities you give should be in a form that
you can use them as the bounds of a double integral.
 Find $d\sigma = \left\vec r_r\times \vec r_\theta
\rightdrd\theta$.
 Set up the surface integral $\iint_S d\sigma$ as an
iterated double integral over $R_{r\theta}$.
Find, actually compute, the surface area of the surface $S$
given by $z=9x^2y^2$, for $z\geq 0$. Do this by computing any
of the integrals from the previous two problems.
If a surface $S$ is parametrized by $\vec r(x,y) =
(x,y,f(x,y))$, show that $d\sigma = \sqrt{1+f_x^2+f_y^2}\ dxdy$
(compute a cross product). If $\vec r(x,z) = (x,f(x,z),z)$,
what does $d\sigma$ equal (compute a cross product  you should
see a pattern)?
Use the pattern you've discovered to quickly compute
$d\sigma$ for the surface $x=4y^2z^2$, and then set up an
iterated double integral that would give the surface area of
$S$ for $x\geq 0$.
\label{sphere surface area element} Consider the sphere
$x^2+y^2+z^2=a^2$. We'll find $d\sigma$ using two different
parameterizations.
 If you use the rectangular parametrization $\vec r(x,y) =
(x,y,\sqrt{a^2x^2y^2})$, what is $d\sigma$? [Hint, use the
previous problem.] Why can this parametrization only be use
if the surface has positive $z$values?

You'll want to memorize this result.
If you use the spherical parametrization $$\vec
r(\phi,\theta) =
(a\sin\phi\cos\theta,a\sin\phi\sin\theta,a\cos\phi),$$ show
that $$d\sigma = (a^2\sin\phi)d\phi d\theta=
(a^2\sin\phi) d\phi d\theta,$$ where we can ignore the
absolute values if we require $0\leq \phi\leq \pi$. Along
the way, you'll show that $$ \vec r_\phi\times \vec
r_\theta = a^2\sin \phi
(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi).$$
We can compute average value, centroids, center of mass,
moments of inertia, and radii of gyration as before. We just
replace $dA$ with $d\sigma$, and all the formulas are the same.
Consider the hemisphere $x^2+y^2+z^2=a^2$ for $z\geq 0$.
 Set up a formula that would give $\bar z$ for the
centroid of the hemisphere. I suggest you use a spherical
parametrization, as then the bounds are fairly simple, and we
know $d\sigma = (a^2\sin\phi) d\phi d\theta$ from the
previous problem.
 Compute the two integrals in your formula. %By doubling
the bottom integral, you'll have also shown that the surface
area of a sphere of radius $a$ is $\sigma = 4\pi a^2$.
 Set up an integral formula for $R_z$, the radius of
gyration about the $z$ axis, provided the density is
constant.
Flux across a surface
We now want to look at the flux of
a vector field across a surface $S$. In the line integral
section, we defined the outward flux of a vector field $F$ across
a curve $C$ to be the line integral $\ds \int_C \vec F\cdot \vec
n ds$, where $\vec n$ is a normal vector point out of region
enclosed by a curve $C$. When we want to find the flux of a
vector field across a surface, we must state in which direction
we want to compute the flux. We then must make sure that normal
vector $\vec n$ we choose to use actually points in the desired
direction. The flux of a vector field $\vec F$ across a surface
$S$ is the surface integral $$\text{Flux}=\Phi = \iint_S \vec
F\cdot \vec n d\sigma .$$ The next problem will help us simplify
the computation of $\vec nd\sigma$.
Consider again the surface $z=9x^2y^2$.
 Using the parametrization $\vec r(x,y) =(x,y,9x^2y^2)$,
find a unit normal vector $\vec n$ to the surface so that
$\vec n$ points upwards away from the $z$axis. State what
$d\sigma$ equals, as well as $\vec n d\sigma$. Make sure you
explain how you know the normal vector you give is pointing
upwards away from the $z$ axis.
 Using the parametrization $\vec r(r,\theta) =(r\cos
\theta,r\sin\theta ,9r^2)$, find a unit normal vector $\vec
n$ to the surface so that $\vec n$ points downwards towards
the $z$axis. State what $d\sigma$ equals, as well as $\vec n
d\sigma$. Make sure you explain how you know the normal
vector you give is pointing downwards towards the $z$
axis.
[For both parts above, the computations involved were
actually done in previous problems. You just need to compile
the information here.]
In the problem above, we showed that $\vec n d\sigma =
\pm(\vec r_x\times\vec r_y)dxdy$ and that $\vec n d\sigma =
\pm(\vec r_r\times\vec r_\theta)drd\theta$. We no longer need to
find the magnitude of the cross product, but we must determine
the correct sign to put on our cross product. This shows us that
we can write flux as $$\text{Flux}=\Phi = \iint_S \vec F\cdot
\vec n d\sigma = \iint_{R_{uv}} \vec F\cdot (\pm \vec r_u\times
\vec r_v) dudv .$$
Consider the cone $z^2=x^2+y^2$ and vector field $\vec F =
(2x+3y, x2y, yz)$. Set up an iterated integral that would give
the flux of $\vec F$ outwards (away from the $z$axis) for the
portion of the cone between $z=1$ and $z=3$. [Hint: Start by
parameterizing the cone by using a polar parametrization
$$x=r\cos\theta, y=r\sin\theta, z=?.$$ You should obtain bounds
for $r$ and $\theta$ that are constants. Compute the normal
vector and look at the third component to determine if it
points up or down. Then just plug everything into the formula.]
When the surface is flat, often you can determine the
normal vector without having to perform any cross products. We'll
now compute a flux of a vector field outwards across the 6 faces
of a cube.
Find the flux of $ \vec F=(x+y,y,z) $ outward across the
surface of the cube in the first quadrant bounded by {$
x=2,y=3,z=5 $}. The cube has 6 surfaces, so we have to compute
the flux across all 6 surfaces. Fill in the table below to
complete the flux across each surface, and then compute each
integral to find the total flux.
{cccccc} \hline Surface&$\vec r(u,v)$ &
$\vec n$ & $\vec F(\vec r(u,v))$ & $\vec F\cdot
\vec n$ & Flux\\\hline Back $x=0$&$
\left<0,y,z\right>$ & $
\left<1,0,0\right>$ & $\vec F(0,y,z) =
\left<y,y,z\right>$ & $y$& $\iint_{\text
Back} y d\sigma = \bar y
\sigma=(\frac{3}{2})(15)$\\\hline Front $x=2$& $
\left<2,y,z\right>$ & & $\vec F(2,y,z) =
\left<2+y,y,z\right>$ & & \\\hline Left
$y=0$& & & & & $0$ (Why?)\\\hline Right
$y=3$& $ \left<x,3,z\right>$ & $
\left<0,1,0\right>$ & $\vec F(x,3,z) =
\left<x+3,3,z\right>$ & 3 & 30 (Why?)
\\\hline Bottom $z=0$& & & & & \\\hline
Top $z=3$& & & & & \\\hline
You should be able to complete each integral by
considering centroids and surface area of each of the 6
different flat surfaces. Show that the total flux is 90.
In the double integral chapter, we learned a way to greatly
simplify flux computations when working with simple closed
curves. Green's theorem stated that $\int_C \vec F\cdot \vec n\
ds = \iint_R M_x+N_y dA$. The divergence of $\vec F$ is the
quantity $\text{div}(\vec F) = M_x+N_y$. This generalizes to
higher dimensions, and is called the divergence theorem. The next
problem illustrates how. We'll study this more in the triple
integral unit.
Consider the exact same vector field and box as the previous
problem. So we have the vector field $\vec F=(x+y,y,z) $ and
$S$ is the surface of the cube in the first quadrant bounded by
{$ x=2,y=3,z=5 $}.
 Compute the divergence of $\vec F$, which is
$\text{div}(\vec F) = M_x+N_y+P_z$.
 The divergence theorem states that if $S$ is a closed
surface (has an inside and an outside), and the inside of the
surface is the solid domain $D$, then the flux of $\vec F$
outward across $S$ equals the triple integral $$\iint_S\vec
F\cdot \vec n\ d\sigma = \iint\int_D \text{div}(\vec F)dV.$$
Use the divergence theorem to compute the flux of $\vec F$
across $S$. [Hint: Just as the area is found by adding up
little bits of area, which is what we mean by $A=\iint_R dA$,
the volume is found by adding up little bits of volume.]
In problem
TODO, we found $$\vec
n d\sigma = \vec r_\phi\times \vec r_\theta d\phi d\theta =
a^2\sin \phi
(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)d\phi d\theta$$
for a sphere of radius $a$. Use this to compute the outward
flux of $$\ds \vec
F=\frac{\left<x,y,z\right>}{(x^2+y^2+z^2)^{3/2}} $$
across a sphere of radius $a$. You should get a negative number
since the vector field has all arrows pointing in. [Hint:
Remember that for a sphere of radius $a$ we have
$a^2=x^2+y^2+z^2$. When you perform the dot product of $\vec F$
and $\vec n$, you'll save yourself a lot of time if you
remember that $\vec u\cdot \vec u = \vec u^2$; the dot
product of a vector with itself is the length squared.]
Repeat the previous problem, but this time don't use the
formula from problem
TODO. In fact, you
don't even need to parametrize the surface. Instead, if you are
at the point $(x,y,z)$ on a sphere of radius $a$, give a
formula for the outward pointing unit normal vector $\vec n$.
Give this formula by only using a geometric argument. Then find
the outward flux of {$\ds \vec
F=\frac{\left<x,y,z\right>}{(x^2+y^2+z^2)^{3/2}} $}
across a sphere of radius $a$. You should find that $\vec
F\cdot \vec n$ simplifies to a constant, so that you never
actually have to compute $d\sigma$. Then you can use known
facts about the surface area of a sphere.
A sizeable amount of text for Stokes' Theorem is commented out
in the source. It is not polished up yet, as I have not yet had
time to visit the topic in our course. It is an optional topic,
that I hope to get to soon.