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.
Applications
I still am not happy with this. I think I want to have them
draw more regions from inequalities, before I have them give
inequalities. They struggled with 1. I need more like #1. With
the inclass practice I'm giving them now, they are doing great,
but I still think this could be improved. I might just give
them a double integral, and tell them that the bounds describe
a region. Draw the region. This might help them put a
disconnect between the integrand and the bounds. I'll examine
this again next time.
Before we introduce integration, let's practice using
inequalities to describe regions in the plane. In first semester
calculus, we often use the inequalities $a\leq x\leq b$ and
$g(x)\leq y\leq f(x)$ to describe the region above $g$ below $f$
for $x$ between $a$ and $b$. We trapped $x$ between two
constants, and $y$ between two functions. Sometimes we wrote
$c\leq y\leq d$ where $g(y)\leq x\leq f(y)$ to describe the
region to the right of $g$ and left of $f$ for $y$ between $c$
and $d$. We need to practice writing inequalities in this form,
as these inequalities provide us the bounds of integration for
double integrals.
Consider the region $R$ in the $xy$plane that is below the
line $y=x+2$, above the line $y=2$, and left of the line $x=5$.
We can describe this region by saying for each $x$ with $0\leq
x\leq 5$, we want $y$ to satisfy $2\leq y\leq x+2$. In set
builder notation, we would write $$R=\{(x,y)\mid 0\leq x\leq
5, 2\leq y\leq x+2\}.$$ The symbols $\{$ and $\}$ are used to
enclose sets, and the symbol $\mid$ stands for “such
that”. We read the above line as “$R$ equals the
set of $(x,y)$ such that zero is less than $x$ which is less
than 5, and 2 is less than $y$ which is less than $x+2$.”
 Describe the region $R$ by saying for each $y$ with
$c\leq y\leq d$, we want $x$ to satisfy $a(y)\leq x\leq
b(y)$. In other words, find constants $c$ and $d$, and
functions $a(y)$ and $b(y)$, so that for each $y$ between $c$
and $d$, the $x$ values must be between the functions $a(y)$
and $b(y)$.
 Write your last answer in the set builder notation
$$R=\{(x,y)\mid c\leq y\leq d, a(y)\leq x\leq b(y)\}.$$
[Hint: If you're struggling, then draw the 4 curves given by
$0=x$, $x= 5$, $2=y$ and $y= x+2$. Then shade either above,
below, left, or right of the line (as appropriate).]
For each region $R$ below, draw the region and give a set of
inequalities of the form $a\leq x\leq b, c(x)\leq y\leq d(x)$
or in the form $c<y<d, a(y)\leq x\leq b(y)$. In class,
we'll give whichever one you did not.
I like to let 3 people present this.
 The region $R$ is above the line $x+y=1$ and inside the
circle $x^2+y^2=1$.
 The region $R$ is below the line $y=8$, above the curve
$y=x^2$, and to the right of the $y$axis.
 The region $R$ is bounded by $2x+y=3$, $y=x$, and
$x=0$.
We're now ready to discuss double integrals. Just as single
integrals gave us the area under a function over an interval,
double integrals will give us the volume under a function, above
a region in the plane. We'll introduce double integrals by
looking at cross sections of a solid. You did this in first
semester calculus, but you always used geometric shapes, for
which we know the area, to create the cross sections. Let's start
with a review of this idea, and then jump into double integrals.
Consider the parabola $z=9x^2$ for $0\leq x\leq 3$. Revolve
this parabola half way around the $z$axis, and compute the
volume of the solid that is inside the paraboloid and above the
$xy$plane. Do so by considering horizontal semicircular cross
sections.
The area of each cross section is the
area of a semicircle, which is $A=\frac{1}{2}\pi r^2 = \frac12
\pi x^2$ as the radius is $x$. We thicken each cross section up
by multiplying by $dz$, the height of each circular disc. Since
the height has a $dz$, we replace $x$ with $x=\sqrt{9z}$ in
the area formula to get a little bit of volume as $dV =
\frac{1}{2}\pi (\sqrt{9z})^2 dz$. The $z$ values range from
$0$ to $9$, which means the volume of the solid is the integral
$$V =\int_0^9 \frac{1}{2}\pi (\sqrt{9z})^2 dz
=\frac{\pi}{2}\int_0^9 9z dz =\frac{\pi}{2}
\left(9z\frac{z^2}{2}\right)\bigg_0^9 =\frac{81\pi}{4} $$
Let's now consider the exact same solid, but instead of
cutting horizontally, let's cut the object vertically. The first
problem has you cut the solid up using slices that are parallel
to the $x$axis (so keeping $y$ constant). The next has you
repeat the problem, but this time using vertical slices that are
parallel to the $y$axis (so keeping $x$ constant). The idea is
exactly the same. Just find the area of a slice, thicken it up
(using $dx$ or $dy$), and then integrate.
Consider the solid domain $D$ in space that is beneath
the surface $f(x,y)=9x^2y^2$ and above the $xy$plane, where
the $x$ values satisfy $x\geq 0$. The region is half of a
parabolic solid. Our goal in this problem is to find the volume
of the solid $D$.
 Please
click on this sage link. In this picture, you'll see the
solid $D$ drawn. You'll also see several cross sections of
the surface (half parabolas), each one obtained by letting
$y$ equal a constant ($y=2, 1,0,1,2$).
 When $y_i=0$, the half parabola has area $\int_0^3
9x^2dx$. When $y_i=2$, the half parabola has area $\int_0^?
5x^2dx$. When $y=y_i$, explain why the area of each of the
cross sections from the first part is $$A_{y=y_i}=\ds
\int_0^{\sqrt{9y_i^2}} (9x^2y_i^2) dx.$$
 In the first Sage picture above, we cut the solid into 6
pieces. We could cut the solid into more pieces.
Click on this Sage link to see what happens if we cut
the solid into 12 pieces, and then fatten up each cross
section by $dy=1/2$ units, obtaining 12 tiny bits of volume
$dV$. Explain why the total volume of the solid $D$ equals
$$\ds\int_{3}^{3} \left(\int_0^{\sqrt{9y^2}}(9x^2y^2)
dx\right) dy.$$
The integral above is called an iterated integral because
you first compute the inside integral and then you compute the
outside integral (you iteratively integrate). Often the
parenthesis are not written because we know that the inside
integral should be performed first without writing the
parenthesis. We could also explicitly emphasize which variables
go with each bound by writing $$\ds\int_{y=3}^{y=3}
\left(\int_{x=0}^{x=\sqrt{9y^2}}9x^2y^2 dx\right) dy.$$ This
latter approach is not commonly used, but can save a beginner
from making simple errors.
The bounds of the integral $$\ds\int_{3}^{3}
\left(\int_0^{\sqrt{9y^2}}(9x^2y^2) dx\right) dy$$ describe
a region $R$ in the plane, namely $$3\leq y\leq 3
\quad\text{and}\quad 0\leq x\leq \sqrt{9y^2}.$$ Draw this
region $R$ in the $xy$plane. Then give bounds to describe the
region alternately by first stating constants which trap $x$
(so $a\leq x\leq b$) and then functions which trap $y$ (so
$c(x)\leq y \leq d(x)$). Use these new bounds to write an
iterated integral $$\ds\int_{x=a}^{x=b}
\left(\int_{y=c(x)}^{y=d(x)}9x^2y^2 dy\right) dx$$ that gives
the exact same volume of the solid $D$ from the previous
problem.
I have not formally defined a definite integral. I will
probably do that next semester, when I have more time. Right
now, I'll lecture that bit in class. It will get added at some
point, provided it is needed. I'm not really sure I want to
formally define it in the problem set. It's never formally USED
in textbooks (though formally define). Perhaps the best spot
for the formal definition is in an analysis course.
In the two problems above, we computed the volume of solid
by considering cross sections of the solid. We could also cut the
solid up in both the $x$ and $y$ dimensions. This would result in
tiny rectangles in the $xy$ plane with area $dA=dxdy=dydx$, and
the solid would have height $f(x,y)$ above these rectangles. This
means we would have a little bit of volume written as
$$dV=fdA=fdxdy=fdydx.$$ Adding up these little bits of volume
gives us the double integral $$V = \iint_R fdA=\iint_R
fdydx=\iint_R fdxdy.$$ We can either set up the bounds with $x$
on the inside, or $y$ on the inside. We'll get the same answer.
When we set up the integral with bounds, we call it an iterated
integral, and write. $$\ds \int_a^b \int_{c(x)}^{d(x)}f(x,y)dydx
\quad \text{or} \quad \ds \int_c^d
\int_{a(y)}^{b(y)}f(x,y)dxdxy.$$
A double integral is written $\ds \iint_R f(x,y)dA$. We just
have to state what the region $R$ is to talk about a double
integral. The formal definition of a double integrals involves
slicing the region $R$ up into tiny rectangles of area $dxdy$,
multiplying each rectangle by a height $f$, and then summing
over all rectangles. This process is repeated as the length and
width of the rectangles shrinks to zero at similar rates, with
the double integral being the limit of this process. An
iterated integral is a double integral where we have actually
set up the bounds as either $$\ds \int_a^b
\int_{c(x)}^{d(x)}f(x,y)dydx \quad \text{or} \quad \ds \int_c^d
\int_{a(y)}^{b(y)}f(x,y)dxdxy.$$ We'll focus mostly on setting
up iterated integrals in this course.
Consider the region $R$ in the plane that is bounded by the
line $y=x+2$ and the parabola $y=x^24$. Distances are measured
in $cm$.
 Draw the region $R$, and give bounds of the form $a\leq
x\leq b$, $c(x)\leq y\leq d(x)$ to describe the region.
 A metal plate occupies the region $R$. The metal plate
was constructed to have a density of $\delta (x,y)=(y+4)$
g/$cm^2$. Explain why the mass of the plate is the double
integral $\ds\iint_R \delta dA$.

Compute the double integral $\ds\iint_R (y+4) dA$ by
setting up an iterated integral (use the bounds from part
1) and then performing each integral. Start with the inside
integral, and then compute the outside integral. Check your
work with the link in the margin. You can use this Sage
link to check any double integral. If you think you have
the bounds right, use this Sage link to draw the region
your bounds describe. If it doesn't the draw the region you
thought, then your bounds are off. Trial and error is a
powerful tool here. You've got to try, and fail, and then
make adjustments. This is the key to mastering double
integrals.
Consider the iterated integral $\ds \int_0^3\int_x^3
e^{y^2}dydx$.
 Write the bounds as two inequalities ($0\leq x\leq 3$ and
$?\leq y\leq ?$). Then draw and shade the region $R$
described by these two inequalities.
 Swap the order of integration from $dydx$ to $dxdy$. This
forces you to describe the region using two inequalities of
the form $c\leq y\leq d$ and $a(y)\leq x\leq b(y)$. This is
the key.
 Use your new bounds to compute the integral by hand
(you'll need a $u$substitution $u=y^2$ on the outer
integral).
 Now use
Sage to check your work. Then also use Sage to compute
the original the original integral $\ds \int_0^3\int_x^3
e^{y^2}dydx$, and tell us what the inner integral equals (if
you see $i$, $\sqrt{\pi}$, and erf, then you did this
correctly).
Consider the region $R$ in the plane that is trapped between
the curves $x=2y$ and $x=y^2$. We would like to compute
$\iint_R (y) dA$ over this region $R$. Set up both iterated
integrals. Then compute one of them. Explain why your answer is
negative.
In the line integral chapter, we introduce the ideas of
average value, centroid, and center of mass. We now extend those
ideas to regions in the plane, in exactly the same way. For
example, the average value formula in the line integral section
was $\bar f = \dfrac{\int_C fdx}{\int_C ds}$. For double
integrals, we just change $ds$ to $dA$, and add an integral.
Average value formula
This gives the formula $\bar f = \dfrac{\iint_R
fdA}{\iint_R dA}.$ The same substitution works on all the
integrals from before. We now have $dm = \delta dA$ instead of
$dm=\delta ds$, as now density is a mass per area, instead of a
mass per length. We obtained the arc length of a curve $C$ by
computing $s=\int_C ds$, as we just add up little bits of arc
length. We can obtain the area of a region $R$ by computing
$A=\iint_R dA$, as we just add up little bits of area.
Centroid Formula
The centroid of a region $R$ in the plane is $$ \left(\bar
x = \frac{\iint_R x dA}{\iint_R dA}, \bar y = \frac{\iint_R y
dA}{\iint_R dA}\right) $$
Center of Mass Formula
and the center of mass is $$ \left(\bar x = \frac{\iint_R x
dm}{\iint_R dm}, \bar y = \frac{\iint_R y dm}{\iint_R dm}\right),
\text{ where $dm=\delta dA$}. $$ %I talk about
inertial and radii of gyration in my course. Jason, I think you
leave it out. That's why I commented out this portion. I decided
to more this from the line integral to the double integral
chapter. Here it is.
I used the variable $d$ to stand for radius
of rotation. I DO NOT use $r$ because too many students replace
it with the polar coordinate $r$, especially when we get to
double integrals. I tried $d$, but now students were thinking
it was a differential. I need a different variable. I've used
$(rad)^2$ before, and it works, it's just awkward. Jason, if
you have a good idea, email me. I would like to discuss this.
We could use $(\text{dist})^2$ or $(\text{radius of
rotation})^2$. Maybe the last is the best. However, I would
really like to write $I=\int ?^2 dm$ without it taking up half
a board. A variable would be good. Capital $R$ doesn't
work.
One of the main reasons we are studying
mass, center of mass, centroids, etc., is so that we can
understand energy. The transfer of energy (for example from
kinetic to electrical and then back from electrical to kinetic)
is one of the most important ideas in modern innovations. Some of
you may have already had a physics class, in which you learned
that the kinetic energy of an object with mass $m$ moving at
speed $v$ is $$KE = \frac{1}{2}mv^2.$$ If an object has a large
mass, it takes a lot of work (transfer of energy) to get the
object moving. Mass is an object's resistance to straight line
motion. When something rotates, we need a handy way to compute
its kinetic energy. We'll show that the kinetic energy of an
object that is rotating about a line $L$, and has an angular
velocity of $\omega$ radians per second about the line, is
precisely
Compare the two formulas $KE =
\frac{1}{2}mv^2$ and $KE = \frac{1}{2}I\omega^2$. If we replace
speed with angular speed, then we replace mass with inertia.
Heavy objects are hard to push. Objects with large inertia are
hard to rotate. Just as mass is an objects resistance to being
moved, inertia is an objects resistance to being rotated.
Engineers build Ibeams so that a large portion of the mass is
far from the axis of rotation. This causes a large inertia,
which prevents Ibeams from rotating.
$$KE = \frac{1}{2}I \omega^2,$$ where $I$
is the (second) moment of inertia. If an object has a large
inertia, it takes a lot of work (transfer of energy) to get the
object rotating. Inertia is like an object's resistance to
rotational motion.
We can obtain the moment of inertia by
integrating $I=\iint_R (d)^2 dm$ where $d$ is the radius of
rotation about the axis $L$ of rotation, which means that $d$
is distance from a point $(x,y,z)$ to the axis of rotation $L$.
If the line $L$ is one of the coordinate axes, then we obtain
the key formulas $$ I_x = \iint_R (y^2)dm,\quad I_y = \iint_R
(x^2)dm,\quad I_z = \iint_R (x^2+y^2)dm .$$
If you have never worked with kinetic
energy before, you may skip the next problem and then just
practice using these formulas.
% \instructor{One student asked if this
had to do with figure skating. OF course. I spun around in a
circle with my arms out, and then quickly brought them in. I
also like to pick up a table/desk, and show them how easy it is
to rotate the object if I use an axis near the center. Then I
try to grab the edge of the desk and rotate it, it doesn't
work. The only real thing I want them to master is that the
inertia gets really large (grows quadratically) with distance
to an axis. So I do something memorable to help them remember
that. We just did the normal acceleration problem in the motion
unit, so I try to connect it to that.}% Suppose that an object,
whose mass is $m$, is attached to a string (whose mass is so
small we'll ignore it). We rotate the object about a point,
where the angular velocity is $\omega$ radians per second. The
length of the string (distance from the point to the center of
rotation) is $d$.
 Explain why the speed of the object
$v=\omega d$.

We know an object's kinetic energy
$KE=\frac{1}{2}mv^2$.
The quantity $I=d^2m$ is called the
moment of inertia. However, this formula assumes that all
the mass is located at a single point.
Explain why the kinetic energy of
the rotating object is $KE =
\frac{1}{2}d^2m\omega^2$.
 Let's take our object, located at the
point $P(x,y)$, and rotate it about the $x$axis, still with
angular velocity $\omega$. Find the kinetic energy of this
point. [All you need is the distance $d$ from the point
$P(x,y)$ to the $x$axis.]

We can think of a region $R$ in the plane
as thousands of points $P(x,y)$, each with mass $dm=\delta
dA$. As we rotate an entire object about the $x$axis with
angular velocity $\omega$, each little piece contributes a
small amount of kinetic energy. Explain why the total
kinetic energy of the region $R$, when rotated about the
$x$ axis at angular speed $\omega$, is
The inertia about the $x$ axis is $I_x
= \ds \iint_R (y^2)dm = \iint_R (y^2)\delta dA$.
$$KE= \frac{1}{2}\left(\iint_R
(y^2)dm\right)\omega^2$$.

How does the formula above change if we
instead rotate about the $y$axis? What if we rotate about
the origin?
Feel free to ask in class about how
this connects to figure skating.
Consider the triangular region $R$ in the
first quadrant, bounded by the line $\ds
\frac{x}{5}+\frac{y}{7}=1$. Assume that the density of the
object is a constant $\delta = c$.
 Draw the region $R$, and give bounds
for performing double integrals over this region. Check your
answer with Sage
(use any $f$ you want for the integrand, it doesn't matter as
you just want to make sure you got the bounds
right).

Set up an integral formula to compute the
center of mass $\bar x$ of the region $R$. Compute any
integrals by hand to show that $\bar x =
\frac{5}{3}$.
Remember you can check your work with
Sage.
Then state a guess for $\bar
y$.
 Set up an integral formula to compute
the moments of inertia $I_x$, $I_y$, and $I_z$. Use
technology to compute $I_x$.

If the triangular region had its
corners at $(0,0)$, $(b,0)$, and $(0,h)$, state the center
of mass $\bar x$ and the moment of inertia $I_x$. You
should be able to complete this part by replacing the 5's
and 7's in your formula with $b$'s and $h$'s, provided you
first factor any really large numbers, like $1715 = 5\cdot
343=5\cdot 7\cdot ?...$ (I'll let you finish
factoring).
When we found average value, we wanted a
height $\bar f$ such that the area under $f$ and the area under
$\bar f$ were the same. As an equation, we wrote $$\iint_R \bar f
dA = \iint_R f dA,$$ and then since $\bar f$ is constant, we
pulled it out of the integral, and then solved for $\bar f$ to
get $$ \bar f \iint_R dA = \iint_R f dA \quad \text{or}\quad \bar
f = \frac{\iint_R f dA }{\iint_R dA} .$$ This same process gives
the center of mass. We could replace the variable distance $x$ in
$\int_C x dm$ with the constant distance $\bar x$, and then solve
for $\bar x$ in the equation $\int_C \bar xdm = \int_C x dm$. If
all the mass were located at one spot, what would that spot have
to be for the first moments of mass to be the same. The radii of
gyration are obtained in the exact same manner. We'll think of a
radius of gyration as a rotational center of mass.
%
% Suppose a planar region $R$ has
density $\delta(x,y)$. The inertia about a line $L$ we know is
$I_x=\iint_R d^2 \delta(x,y)dA$, where $d$ is the radius of
rotation (distance to the line $L$). What constant radius $R$
should we replace the variable radius $d$ with so that $\int_C
R^2 dm = \int_C d^2 dm$. Explain why the radius of rotation
about the $z$axis is $$R_z = \sqrt{\frac{\iint_R
(x^2+y^2)\delta(x,y)dA}{\iint_R \delta(x,y)dA}}.$$ [Hint:
Remember that $R$ is constant. Read the paragraph
above.]
You only needed to show how to obtain the
radius of gyration about the $z$ axis. All three radii of
gyration are found using the formulas $$ R_x =
\sqrt{\frac{\iint_R (y^2+z^2)dm}{\iint_R dm}},\quad R_y =
\sqrt{\frac{\iint_R (x^2+z^2)dm}{\iint_R dm}}, \text{ and } R_z =
\sqrt{\frac{\iint_R (x^2+y^2)dm}{\iint_R dm}}. $$ where
$dm=\delta(x,y) dA$. Note that in 2 dimensions, we have $z=0$, so
the formulas for $R_x$ and $R_y$ are simpler. %ends the
part that involves the radius of gyration.
Consider the rectangular region $R$ in the $xy$plane described
by $\{(x,y)\ \ 2\leq x\leq 11, 3\leq y\leq 7\}$.
 Set up an integral formula which would give $\bar y$ for
the centroid of $R$. Then evaluate the integral.
 State $\bar x$ from geometric reasoning.
 Set up an integral to give the moment
of inertia about the $y$axis if $\delta=5$. Note that $z=0$
in the $xy$plane.
 Set up an integral to give the $R_x$ if
the density is $\delta(x,y)=xy^2$.
Consider the region in the plane that is bounded by the curves
$x=y^23$ and $x=y1$. A metal plate occupies this region in
space, and its temperature function on the plate is give by the
function $T(x,y)=2x+y$. Find the average temperature of the
metal plate.
\label{centroid trick} Consider the region $R$ that is the
circular disc which is inside the circle $(x2)^2+(y+1)^2=9$.
The centroid is clearly $(2,1)$, and the area is
$A=\pi(3)^2=9\pi$. We can use these fact to simplify many
integrals that require integrating over the region $R$.
 Compute $\iint_R 3dA = 3\iint_RdA$. [How can area help
you?]
 Explain why $\iint_R x dA = \bar x A$ for any region $R$,
and then compute $\iint_R x dA$ for the circular disc. [You
don't need to set up any integrals at all.]
 Compute the integral $\iint_R 5x+2y dA$ by using centroid
and area facts.
Consider the region $R$ in the $xy$plane that is formed from
two rectangular regions. The first region $R_1$ satisfies
$x\in[2,2]$ and $y\in[0,7]$. The second region $R_2$ satisfies
$x\in[5,5]$ and $y\in[7,10]$. Find the centroids of $R_1$,
$R_2$ and then finally $R$.
Let $R$ be the region in the plane with $a\leq x\leq b$ and
$g(x)\leq y\leq f(x)$. Let $A$ be the area of $R$.
When you use double integrals to find centroids, the formulas
for the centroid are the same for both $\bar x$ and $\bar y$.
In other courses, you may see the formulas on the left,
because the ideas will be presented without requiring
knowledge of double integrals. Integrating the inside
integral from the double integral formula gives the single
variable formulas.
 Set up an iterated integral to compute the area of $R$.
Then compute the inside integral. You should obtain a
familiar formula from firstsemester calculus.
 Set up an iterated integral formula to compute $\bar x$
for the centroid. By computing the inside integral, show why
$\ds\bar x = \frac{1}{A}\int_a^b x (fg)dx$.
 If the density depends only on $x$, so $\delta = \delta
(x)$, set up an iterated integral formula to compute $\bar y$
for the center of mass. Explain why $$\ds\bar y =
\frac{1}{mmahtmath}\int_a^b
\frac{1}{2}(f^2g^2)\delta(x)dx.$$
Switching Coordinates: The Jacobian
We now want to
explore how to perform $u$substitution in high dimensions. Let's
start with a review from first semester calculus.
Consider the integral $\ds\int_{1}^4 e^{3x} dx$.
 Let $u=3x$. Solve for $x$ and then compute $dx$.
 Explain why $\ds\int_{1}^4 e^{3x} dx=\int_{3}^{12}e^u
\left(\frac{1}{3}\right)du$.
 Explain why $\ds\int_{1}^4 e^{3x} dx=\int_{12}^{3}e^u
\left\frac{1}{3}\right du$.
 If the $u$values are between $3$ and $2$, what would
the $x$values be between? How does the length of the $u$
interval $[3,2]$ relate to the length of the corresponding
$x$ interval?
In the problem above, we used a change of coordinates
$u=3x$, or $x=1/3 u$. By taking derivatives, we found that
$dx=\frac{1}{3}du$. The negative means that the orientation of
the interval was reversed. The fraction $\frac13$ tells us that
lengths $dx$ using $x$ coordinates will be $1/3$rd as long as
lengths $du$ using $u$ coordinates. When we write $dx =
\frac{dx}{du}du$, the number $\frac{dx}{du}$ is called the
Jacobian of $x$ with respect to $u$. The Jacobian tells us how
lengths are altered when we change coordinate systems. We now
generalize this to polar coordinates. Before we're done with this
section, we'll generalize the Jacobian to any change of
coordinates.
Consider the polar change of coordinates $x=r\cos\theta$ and
$y=r\sin\theta$, which we could just write as $$\vec
T(r,\theta)=(r\cos\theta,r\sin\theta).$$
 Compute the derivative $D\vec T(r,\theta)$. You should
have a 2 by 2 matrix.
 We need a single number from this matrix that tells us
something about area. Determinants are connected to area.
Compute the determinant of $D\vec T(r,\theta)$ and
simplify.
The determinant you found above is called the Jacobian of
the polar coordinate transformation. Let's summarize these
results in a theorem.
Ask me in class to give you an informal picture approach that
explains why $dxdy=rdrd\theta$.
% If we use the polar coordinate transformation
$x=r\cos\theta, y=r\sin\theta$, then we can switch from $(x,y)$
coordinates to $(r,\theta)$ coordinates if we use
$$dxdy=rdrd\theta.$$ The number $r$ is called the Jacobian
of $x$ and $y$ with respect to $r$ and $\theta$. If we require
all bounds for $r$ to be nonnegative, we can ignore the
absolute value. If $R_{xy}$ is a region in the $xy$ plane that
corresponds to the region $R_{r\theta}$ in the $r\theta$ plane
(where $r\geq 0$), then we can write $$\iint_{R_{xy}} f(x,y)
dxdy = \iint_{R_{r\theta}} f(r\cos\theta,r\sin\theta) r\
drd\theta.$$
We need some practice using this idea. We'll start by
describing regions using inequalities on $r$ and $\theta$.
For each region $R$ below, draw the region in the $xy$plane.
Then give a set of inequalities of the form $a\leq r\leq b,
\alpha(r)\leq \theta \leq \beta(r)$ or
$\alpha<\theta<\beta, a(\theta)\leq r\leq b(\theta)$. For
example, if the region is the inside of the circle $x^2+y^2=9$,
then we could write $0\leq \theta\leq 2\pi$, $0\leq r\leq 3$.
 The region $R$ is the quarter circle in the first
quadrant inside the circle $x^2+y^2=25$.
 The region $R$ is below $y=\sqrt{9x^2}$, above $y=x$,
and to the right of $x=0$.
 The region $R$ is the triangular region below $y=\sqrt 3
x$, above the $x$axis, and to the left of $x=1$.
% Consider the opening problem for this unit. We want to
find the volume under $f(x,y)=9x^2y^2$ where $x\geq0$ and
$z\geq 0$. We obtained the integral formula $$\iint_R f dA =
\ds\int_{y=3}^{y=3} \int_{x=0}^{x=\sqrt{9y^2}}9x^2y^2 dx
dy.$$
 Write bounds for the region $R$ by giving bounds for $r$
and $\theta$.
 Rewrite the double integral as an iterated integral with
bounds for $r$ and $\theta$. Don't forget the Jacobian (as
$dxdy=rdrd\theta$).
 Compute the integral in the previous part by hand.
[Suggestion: you'll want to simplify $9x^2y^2$ to $9r^2$
before integrating.]
Find the centroid of a semicircular disc of radius $a$ ($y\geq
0$). Actually compute any integrals.
After
doing this, in class we'll set up the integral formulas needed
to find $R_y$, the radius of gyration about the $y$axis,
assuming the density is $\delta(x,y)=x^2+y^2$.
This is a great place to comment in class about the ability
to do the integrals separately.
Compute the integral $\ds
\int_{0}^{1}\int_{\sqrt{1x^2}}^{\sqrt{1x^2}}
\frac{2}{(1+x^2+y^2)^2}dydx$. [Hint: try switching coordinate
systems to polar coordinates. This will require you to first
draw the region of integration, and then then obtain bounds for
the region in polar coordinates.]
Optional problem:
The problem about integrating $e^{x^2}$ from 0 to infinity.
Make it optional, and then give them some hints. I'll work on
this next semester. I want to prepare them for the normal
distribution. Any student who want to tackle this problem
should be asked if they want to become a math major. :)
We're now ready to define the Jacobian of any
transformation.
Suppose $\vec T(u,v)=(x(u,v),y(u,v))$ is a differentiable
coordinate transformation. To find the Jacobian of this
transformation, we first find the derivative of $\vec T$. This
is a square matrix, so it has a determinant, which should give
us information about area. As the determinant may be positive
or negative, we then take the absolute value to obtain the
Jacobian. So the Jacobian of the transformation $\vec T$ is the
absolute value of the determinant of the derivative.
For a tongue twister, say “the absolute value of the
determinant of the derivative” ten times really fast.
Notationally we write $$J(u,v) = \frac{\partial
(x,y)}{\partial (u,v)} = \det(D\vec T(u,v)).$$
Consider the transformation $u=x+2y$ and $v=2xy$.
 Solve for $x$ and $y$ in terms of $u$ and $v$. Then
compute the Jacobian $\frac{\partial (x,y)}{\partial
(u,v)}$.
 We were give $u$ and $v$ in terms of $x$ and $y$, so we
could have directly computed $\frac{\partial (u,v)}{\partial
(x,y)}$. Do so now.
 Make a conjecture about the relationship between
$\frac{\partial (x,y)}{\partial (u,v)}$ and $\frac{\partial
(u,v)}{\partial (x,y)}$.
There are a lot of different uses of the word Jacobian. It
sometimes includes the absolute value, sometimes does not.
Sometimes you stop at the derivative, sometimes you take the
determinant. Which one is correct? I'm going with the one
that includes the absolute values as well. That way I can say
$r$ is the Jacobian for polar, and for spherical we have
$\rho^2\phi$ (not $\rho^2\phi$). If we want to address this
in general, it should be in an appendix, with maybe a
marginpar or footnote.
Suppose that $f$ is integrable over a region $R_{xy}$ in the
$xy$ plane. Suppose that $\vec T(u,v)=(x(u,v),y(u,v))$ is a
coordinate transformation that has the Jacobian $\ds
\frac{\partial (x,y)}{\partial (u,v)} $. Suppose the region
$R_{uv}$ in the $uv$plane corresponds to the region $R_{xy}$
in the $xy$plane. Provided the Jacobian is nonzero except
possibly on regions with zero area, we can then write
$$\iint_{R_{xy}} f(x,y) dxdy = \iint_{R_{uv}} f(x(u,v),y(u,v))
\frac{\partial (x,y)}{\partial (u,v)} dudv.$$ We can remember
this in differential form as $$dxdy = \frac{\partial
(x,y)}{\partial (u,v)} dudv.$$
Let's use this to rapidly find the area inside of an
ellipse.
Consider the region $R$ inside the ellipse
$\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2=1$.
We'll consider the change of coordinates given by $u=(x/a)$ and
$v=(y/b)$.
 Draw the region $R$ in the $xy$plane. After substituting
$u=x/a$ and $v=y/b$, draw the region $R_{uv}$ in the
$uv$plane. You should have a circle. What is the area inside
this circle in the $uv$plane?
 Solve for $x$ and $y$, and then compute the Jacobian
$\dfrac{\partial (x,y)}{\partial (u,v)}$. Show how to get the
same result from directly computing $\dfrac{\partial
(u,v)}{\partial (x,y)}$.
 We know the area in the $xy$plane of the ellipse is
$\iint_{R_{xy}} dxdy$. Use the previous theorem to switch to
an integral over the region $R_{uv}$. Then evaluate this
integral by using facts about area so prove that the area in
the $xy$ plane is $\pi a b$. [Hint: you don't actually have
to set up any bounds, rather just reduce this to an area
integral over the region $R_{uv}$.]
Let $R$ be the region in the plane bounded by the curves
$x+2y=1$, $x+2y=4$, $2xy=0$, and $2xy=8$. We want to compute
the integral $\iint_R xdxdy$. Draw the region $R$ in the
$xy$plane. Use the change of coordinates $u=x+2y$ and $v=2xy$
to evaluate this integral. Make sure you provide a sketch of
the region $R_{uv}$ in the $uv$plane (it should be a
rectangle). [Hints: what are the bounds for $u$ and $v$? You'll
want to solve for $x$ and $y$ in terms of $u$ and $v$, and then
you'll need a Jacobian.]
This is problem 7 in section 15.8.
Use the transformation $u=3x+2y$ and $v=x+4y$ to evaluate
the integral $$\iint_R (3x^2+14xy+8y^2)dxdy =\iint_R
(3x+2y)(x+4y)dxdy $$ for the region $R$ that is bounded by the
lines $y=(3/2)x+1$, $y=(3/2)x+3$, $y=(1/4)x$, and
$y=(1/4)x+1$.
Green's Theorem
Now that we have double integrals, it's
time to make some of our circulation and flux problems from the
line integral section get extremely simple. We'll start by
defining the circulation density and flux density for a vector
field $\vec F(x,y)=\left<M,N\right>$ in the plane.
\label{definition of flux density in 2D} Let $\vec
F(x,y)=\left<M,N\right>$ be a continuously differentiable
vector field. At the point $(x,y)$ in the plane, create a
circle $C_a$ of radius $a$ centered at $(x,y)$, where the area
inside of $C_a$ is $A_a=\pi a^2$. The quotient $\ds
\frac{1}{A_a}\oint_{C_a} \vec F \cdot \vec T ds$ is a
circulation per area. The quotient $\ds
\frac{1}{A_a}\oint_{C_a} \vec F \cdot \vec n ds$ is a flux per
area.
In the definitions above, we could have replaced the circle
$C_a$ with a square of side lengths $a$ centered at $(x,y)$ with
interior area $A_a$. Alternately, we could have chosen any
collection of curves $C_a$ which “shrink nicely” to
$(x,y)$ and have area $A_a$ inside. Regardless of which curves
you chose, it can be shown that $$N_xM_y=\lim_{a\to 0}
\frac{1}{A_a}\oint_{C_a} \vec F \cdot \vec T ds \quad \text{ and
} \quad M_x+N_y=\lim_{a\to 0} \frac{1}{A_a}\oint_{C_a} \vec F
\cdot \vec n ds.$$ To understand what the circulation and flux
density mean in a physical sense, think of $\vec F$ as the
velocity field of some gas.
 The circulation density tells us the rate at which the
vector field $\vec F$ causes objects to rotate around points.
If circulation density is positive, then particles near $(x,y)$
would tend to circulate around the point in a counterclockwise
direction. The larger the circulation density, the faster the
rotation. The velocity field of a gas could have some regions
where the gas is swirling clockwise, and some regions where the
gas is swirling counterclockwise.
 The divergence, or flux density, tells us the rate at which
the vector field causes object to either flee from $(x,y)$ or
come towards $(x,y)$. For the velocity field of a gas, the gas
is expanding at points where the divergence is positive, and
contracting at points where the divergence is negative.
We are now ready to state Green's Theorem. Ask me in class
to give an informal proof as to why this theorem is valid.
I draw a curve. I then cut the interior up into little
rectangular pieces, and ask them to consider the sum of the
flux along every single little rectangular piece inside. I show
them how the circulation (or flux) along any interior edge is
computed twice but with opposite signs. This means that the
integrals along every interior edge vanish. We then have the
circulation along the entire edge equal to the sum of a bunch
of tiny circulations. Multipy and divide by the area inside
each rectangle. Taking a limit as the size of the rectangles
shrinks to zero gives us a double integral of the circulation
per area. This is Green's theorem
Let $\vec F(x,y)=(M,N)$ be a continuously differentiable vector
field, which is defined on an open region in the plane that
contains a simple closed curve $C$ and the region $R$ inside
the curve $C$. Then we can compute the counterclockwise
circulation of $\vec F$ along $C$, and the outward flux of
$\vec F$ across $C$ by using the double integrals $$ \oint_{C}
\vec F \cdot \vec T ds=\iint_R N_xM_y dA \quad \text{ and }
\quad \oint_{C} \vec F \cdot \vec n ds=\iint_R M_x+N_y dA.$$
Let's now use this theorem to rapidly find circulation
(work) and flux.
See 16.4 for more practice. Try doing a bunch of these, as
they get really fast.
Consider the vector field $\vec F=(2x+3y,4x+5y)$. Start
by computing $N_xM_y$ and $M_x+N_y$. If $C$ is the boundary of
the rectangle $2\leq x\leq 7$ and $0\leq y\leq 3$, find both
the circulation and flux of $\vec F$ along $C$. You should be
able to reduce the integrals to facts about area. [If you tried
doing this without Green's theorem, you would have to
parametrize 4 line segments, compute 4 integrals, and then sum
the results.]
Consider the vector field $\vec F=(x^2+y^2,3x+5y)$. Start by
computing $N_xM_y$ and $M_x+N_y$. If $C$ is the circle
$(x3)^2+(y+1)^2=4$ (oriented counterclockwise), then find both
the circulation and flux of $\vec F$ along $C$. You should be
able to reduce the integrals to facts about the area and
centroid.
Repeat the previous problem, but change the curve $C$ to the
boundary of the triangular region $R$ with vertexes at $(0,0)$,
$(3,0)$, and $(3,6)$. You can complete this problem without
having to set up the bounds on any integrals, if you reduce the
integrals to facts about area and centroids. You are welcome to
look up the centroid of a triangular region without computing
it.