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.
 Describe how to integrate a function along a curve. Use
line integrals to find the area of a sheet of metal with height
$z=f(x,y)$ above a curve $\vec r(t)=\left(x,y\right)$ and the
average value of a function along a curve.
 Find the following geometric properties of a curve:
centroid, mass, center of mass, inertia, and radii of
gyration.
 Compute the work (flow, circulation) and flux of a vector
field along and across piecewise smooth curves.
 Determine if a field is a gradient field (hence
conservative), and use the fundamental theorem of line
integrals to simplify work calculations.
You'll have a chance to teach your examples to your peers
prior to the exam. This table contains a summary of the
key ideas for this chapter. Here are some extra
homework problems which line up with the text. {\noindent
%\footnotesize
{lcllll}\hline Topic (11th ed.)
&Sec &Basic Practice &Good Problems &Thy/App
&Comp \\\hline Line integrals & 16.1&18, 922,
2332 & & &3336 \\\hline Work, Flow, Circulation,
Flux & 16.2&716, 2528, 3740 &1724, 2930, 4144
&4546 &4752 \\\hline Gradient Fields &
16.2&16 & & & \\\hline Gradient Fields &
14.5&18 & & & \\\hline Potentials &
16.3&112,1324 & 2533 & 3438&
\\\hline
} {\noindent %\footnotesize
{lcllll}\hline Topic (12th ed.)
&Sec &Basic Practice &Good Problems &Thy/App
&Comp \\\hline Line integrals & 16.1&18, 926,
3342 &2732 & &4346 \\\hline Work, Flow,
Circulation, Flux & 16.2&712, 1924, 3136, 4750
&1318, 2530, 3738, & 5154 &5560 \\\hline
Gradient Fields & 16.2&16 & & & \\\hline
Gradient Fields & 14.5&110 & & & \\\hline
Potentials & 16.3&112,1324 & 2533 &
3438& \\\hline
}
Arclength  $s = \int_C ds= \int_a^b
\left\frac{d\vec r}{dt}\rightdt$ 
Surface
Area  $\sigma = \int_C d\sigma=\int_C f ds = \int_a^b f
\left\frac{d\vec r}{dt}\rightdt$ 
Average
Value  $\bar f = \frac{\int f ds}{\int ds}$ 
Work, Flow, Circulation  $W=\int_C d\text{Work} =
\int_C (\vec F\cdot \vec T) ds = \int_C \vec F\cdot d\vec r
= \int_C Mdx+Ndy$ 
Flux  $\text{Flux} = \int_C
d\text{Flux} = \int_C \vec F\cdot \vec n ds = \oint_C
MdyNdx$ 
Mass  $m=\int_C dm = \int_C \delta ds
$ 
Centroid  $\left(\bar x,\bar y,\bar z\right)
=\left(\frac{\int x ds}{\int_C ds},\frac{\int y ds}{\int_C
ds},\frac{\int z ds}{\int_C ds}\right)$ 
Center of
Mass  $\left(\bar x,\bar y,\bar z\right)
=\left(\frac{\int x dm}{\int_C dm},\frac{\int y dm}{\int_C
dm},\frac{\int z dm}{\int_C dm}\right)$ 
First moment of mass  $M_{yz}=\int x dm$,
$M_{xz} = \int x dm$, $M_{xy}=\int x dm$ 
(Second) Moment of Inertia  $I_x = \int (y^2+z^2)
dm$, $I_y = \int (x^2+z^2) dm$, $I_z = \int (x^2+y^2) dm$

Radius of Gyration  $R =
\sqrt{I/m}$ 
Fund. Thm. Calculus 
$f(b)f(a)=\int df = \int \frac{df}{dx}dx = \int_a^b
f^\prime(x)dx$ 
Fund. Thm of Line Int.  $f(B)f(A)=\int_C
\vec \nabla f \cdot d\vec r$
Outline of proof: $\int_C df = \int_C
\frac{df}{ds}ds = \int_a^b \frac{df/dt}{ds/dt}ds =
\int_a^b \vec \nabla f \cdot \frac{d\vec r}{dt}dt =
\int_C \vec F \cdot d \vec r$ 
There is a YouTube playlist to go along with this
chapter. Each video is about 46 minutes long.
You'll also find the following links to Sage can help you
speed up your time spent on homework.
If you would like homework problems from
the text that line up with the ideas we are studying, please use
the following tables. {\noindent \footnotesize
{lcllll}\hline Topic (11th ed.)
&Sec &Basic Practice &Good Problems &Thy/App
&Comp \\\hline Line integrals & 16.1&18, 922,
2332 & & &3336 \\\hline Work, Flow, Circulation,
Flux & 16.2&716, 2528, 3740 &1724, 2930, 4144
&4546 &4752 \\\hline Gradient Fields &
16.2&16 & & & \\\hline Gradient Fields &
14.5&18 & & & \\\hline Potentials &
16.3&112,1324 & 2533 & 3438&
\\\hline
} {\noindent \footnotesize
{lcllll}\hline Topic (12th ed.)
&Sec &Basic Practice &Good Problems &Thy/App
&Comp \\\hline Line integrals & 16.1&18, 926,
3342 &2732 & &4346 \\\hline Work, Flow,
Circulation, Flux & 16.2&712, 1924, 3136, 4750
&1318, 2530, 3738, & 5154 &5560 \\\hline
Gradient Fields & 16.2&16 & & & \\\hline
Gradient Fields & 14.5&110 & & & \\\hline
Potentials & 16.3&112,1324 & 2533 &
3438& \\\hline
}
Area
I draw and cut out the area between a
curve and the axis on a sheet of paper and draw a representative
rectangle on it. We talk about the singlevariable way to find
area and label the rectangle and label it with $f(x)$ and $dx$
and $dA=f(x)dx$. Then I draw a curve on the whiteboard and put
the $x$axis of my sheet on the curve to show how we now are
computing an area above a curve. Students see how the $dx$
becomes $ds$, and how $d\sigma=f(x,y)ds$. Then we do the problem
below and show the Sage graph.
In firstsemester calculus, we
learned that the area under a function $f(x)$ above the $x$axis
is given by $A = \int_a^b f(x) dx$. The quantity $dA= f(x) dx$
represents a small bit of area whose length is $dx$ and whose
height is $f(x)$. To get the total area, we just added up the
little bits of area, which is why $$A=\int dA = \int_a^b f(x)
dx.$$ Now we want to generalize this
See
Sage for a picture of this sheet.
Consider the surface in space that is below the
function $f(x,y)=9x^2y^2$ and above the curve $C$
parametrized by $\vec r(t)=(2\cos t, 3\sin t)$ for
$t\in[0,2\pi]$. Think of this region as a metal plate that has
been stood up with its base on $C$ where the height above each
spot is given by $z=f(x,y)$.

Draw the curve $C$ in the $xy$plane. If we cut the
curve up into lots of tiny segments, each having length
$ds$. Explain why the length of each segment is
approximately $$ds=\sqrt{(2\sin t)^2+(3\cos t)^2}dt.$$
[Hint: think of the particle moving around the curve, and a
tiny segment being the path of the particle over a small
time interval $dt$. Distance traveled in a small segment is
rate (i.e., speed) multiplied by time.]

We'll use $\sigma$ (a lowercase Greek letter
“sigma”) to stand for surface area.
We let $\sigma$ be the area of the metal sheet that
lies above $C$ and under $f$. Explain why $\sigma$ is given
by the integral $$\sigma = \int_C f ds =
\int_0^{2\pi}(9(2\cos t)^2(3\sin t)^2)\sqrt{(2\sin
t)^2+(3\cos t)^2}dt.$$ You'll need to explain why a little
bit of surface area $d\sigma$ is $d\sigma =fds$.

Find the surface area $\sigma$ of the metal sheet.
[Use technology to do this integral.]
Our results from the problem above suggest the following
definition.
The line integral is also called the path integral, contour
integral, or curve integral.
Let $f$ be a function and let $C$ be a piecewise smooth
curve whose parametrization is $\vec r(t)$ for $t\in[a,b]$.
We'll require that the composition $f(\vec r(t))$ be continuous
for all $t\in [a,b]$. Then we define the line integral of $f$
over $C$ to be the integral $$\int_C f ds = \int_a^b f(\vec
r(t))\frac{ds}{dt}dt = \int_a^b f(\vec r(t))\left\frac{d\vec
r}{dt}\rightdt.$$
When we ask you to set up a line integral, it means that you
should do steps 1‐3, so that you get an integral with a single
variable and with bounds that you could plug into a computer or
complete by hand.
Notice that this definition suggests the following four
steps. These four steps are the key to computing any line
integral.
 Start by getting a parametrization $\vec r(t)$ for $a\leq
t\leq b$ of the curve $C$.
 Find the speed by computing the velocity $\dfrac{d\vec
r}{dt}$ and then the speed $\left\dfrac{d\vec
r}{dt}\right$.
 Multiply $f$ by the speed, and replace each $x$, $y$,
and/or $z$ with what it equals in terms of $t$.

Integrate the product from the previous step.
Practice doing this by hand on every problem, unless it
specifically says to use technology. Some of the integrals
are impossible to do by hand.
These quick problems help the students get
the hang of line integrals and practice parametrizing curves.
\marginparbmw{See 16.1: 932. Some problems give you a
parametrization, some expect you to come up with one on your
own.}
To practice matching parameterizations to
curves, try 16.1:18.}
Consider the function $f(x,y)=3xy+2$.
Let $C$ be a circle of radius 4 centered at the origin.
Compute $\int_C fds$. [You'll have to come up with your own
parameterization.]
The big idea behind integrals is that we are adding up a
bunch of small quantities (above, it was $d\sigma=fds$) to get a
total quantity (surface area). In line integrals, we add up the
value of $f$ weighted by a length $ds=\abs{\frac{d\vec
r}{dt}}dt$. This concept can be extended to curves that are more
than just plane curves.
If you've forgotten how to parametrize line segments, see
this problem.
See Larson 15.2, especially examples 1–5.
Do the following integrals
 The principles of line integrals apply even if the curve
is not a 2d curve. Let $f(x,y,z)=x^2+y^22z$ and let $C$ be
two coils of the helix $\vec r(t)=(3\cos t, 3\sin t, 4t)$,
starting at $t=0$. Remember that the parameterization means
$x=3\cos t$, $y=3\sin t$, and $z=4t$. Compute $\int_Cf ds$.
[You will have to find the end bound yourself. How much time
passes to go around two coils?]

Let $f(x,y,z)=x^2+3yz$. Let $C$ be the straight
line segment from $(1,0,0)$ to $(0,4,5)$. Compute $\int_C f
ds$.
See
this problem if
you forgot how to parametrize plane curves.
See Larson 15.2: 1‐20, 63‐70.
Check your answer with
Sage.
Let $f(x,y)=x^2+y^225$. Let $C$ be the portion of the
parabola $y^2=x$ between $(1,1)$ and $(4,2)$. We want to
compute $\int_C fds$.
 Draw the curve $C$ and the function $f(x,y)$ on the same
3D $xyz$ axes.
 Without computing the line integral $\int_C fds$,
determine if the integral should be positive or negative.
Explain why this is so by looking at the values of $f(x,y)$
at points along the curve $C$. Is $f(x,y)$ positive,
negative, or zero, at points along $C$?
 Parametrize the curve and set up the line integral
$\int_C f ds$. [Hint: if you let $y=t$, then $x=$? What
bounds do you put on $t$?]
 Use technology to compute $\int_C fds$ to get a numeric
answer. Was your answer the sign that you determined
above?
See Larson 12.5.
Compute $\int_C 1 ds$ for the following curves $C$ and
interpret your answers.
 $C$ is the curve $\vec r(t)=(2\cos t, 2\sin t)$, $t\in
[0,2\pi]$.
 $C$ is the curve $\vec r(t)=(\cos t, t, \sin t)$,
$t\in[0,\pi]$.
See 11.5: 2128.
(Challenge) Set up (do not evaluate) an integral formula to compute
the length of the polar coordinate rose $r=2\cos 3\theta$. [Hint: how can you use the ideas from
the previous problem along with the parametric equation for the polar curve?]
Work, Flow, Circulation, and Flux
Let's turn our
attention to the work done by a vector field representing a force
as we move through the field. Work is a transfer of energy.
 A tornado picks up a couch and applies forces to the couch
as the couch swirls around the center. Work measures the
transfer of energy from the tornado to the couch, giving the
cough its kinetic energy.
 When an object falls, gravity does work on the object. The
work done by gravity converts potential energy to kinetic
energy.
 If we consider the flow of water down a river, it's gravity
that gives the water its kinetic energy. We can place a
hydroelectric dam next to river to capture a lot of this
kinetic energy. Work transfers the kinetic energy of the river
to rotational energy of the turbine, which eventually ends up
as electrical energy available in our homes.
When we study work, we are really studying how energy is
transferred. This is one of the key components of modern life.
Let's start with simple review. Recall from Section~TODO that the work done by a vector field
$\vec F$ through a displacement $\vec d$ is the dot product $\vec
F\cdot \vec d$.
An object moves from $A=(6,0)$ to $B=(0,3)$. Along the way, it
encounters the constant force $\vec F = (2,5)$. How much work
is done by $\vec F$ as the object moves from $A$ to $B$? See
\footnote{The displacement is $BA=(6,3)$. The work is $\vec
F\cdot \vec d = (2,5)\cdot(6,3) = 12+15=3$.}.
An object moves from $A=(6,0)$ to $B=(0,3)$. A parametrization
of the objects path is $\vec r(t) = (6,3)t+(6,0)$ for $0\leq
t\leq 1$.
 For $0\leq t\leq .5$, the force encountered is $\vec F =
(2,5)$. For $.5\leq t\leq 1$, the force encountered is
$(2,6)$. How much work is done in the first half second? How
much work is done in the last half second? How much total
work is done?
 If we encounter a constant force $\vec F$ over a small
displacement $d\vec r$, explain why the work done is \ $\ds
dW = \vec F\cdot d\vec r =F\cdot \frac{d\vec r}{dt}dt $.

You can visualize what's happening in this problem as
follows. A sailboat sails in a straight line between two
points. As the sailboat moves through the water and the
wind, the wind blows at the sailboat from different
directions.
%
% Suppose that the force constantly changes as we
move along the curve. At time $t$, we encounter the force
$F(t) = (2,5+3t)$, which we could think of as the wind
blowing stronger and stronger to the north. Explain why the
total work done by this force along the path is $$\ds
W=\int \vec F\cdot d\vec r = \int_0^1 (2,5+3t)\cdot
(6,3)dt.$$ Then compute this integral. It should be
slightly larger than the first part.
 (Optional) If you are familiar with the units of energy,
complete the following. What are the units of $\vec F$,
$d\vec r$, and $dW$.
We know how to compute work when we move along a straight
line. Prior to problem TODO
on page \pageref{first work problem}, we made the following
statements.
If a force $F$ acts through a displacement $d$, then the most
basic definition of work is $W=Fd$, the product of the force
and the displacement. This basic definition has a few
assumptions.
 The force $F$ must act in the same direction as the
displacement.
 The force $F$ must be constant throughout the
displacement.
 The displacement must be in a straight line.
We used the dot product to remove the first assumption, and
we showed in problem TODO
that the work is simply the dot product $$W=\vec F\cdot \vec r,$$
where $\vec F$ is a force acting through a displacement $\vec r$.
The previous problem showed that we can remove the assumption
that $\vec F$ is constant, by integrating to obtain $$W=\int \vec
F \cdot d\vec r = \int_a^b F\cdot \frac{d\vec r}{dt}dt, $$
provided we have a parametrization of $\vec r$ with $a\leq t\leq
b$. The next problem gets rid of the assumption that $\vec r$ is
a straight line.
%
Please use this
Sage link to check your work.
% Suppose that we move along the circle $\vec r(t) =
(3\cos t,3\sin t)$. As we move along this circle, we encounter
a rotation force $\vec F(x,y) = (2y,2x)$.
 Draw the curve $\vec r(t)$. Then at several points on the
curve, draw the vector field $\vec F(x,y)$. For example, at
the point $(3,0)$ you should have the vector $\vec
F(3,0)=(2(0),2(3))=(0,6)$, a vector sticking straight up 6
units. Are we moving with the vector field, or against the
vector field?
 Explain why we can state that a little bit of work done
over a small displacement is $dW = \vec F\cdot d\vec r$. Why
does it not matter that $\vec r$ moves in a straight
line?
 Explain why the work done by $\vec F$ along the circle
$C$
We put the $C$ under the integral $\int_C$ to remind us
that we are integrating along the curve $C$. This means
we need to get a parametrization of the curve $C$, and
give bounds before we can integrate with respect to $t$.
is $$W = \int_C\left(2y,2x\right)\cdot d\vec r =
\int_0^{2\pi}\left(2(3\sin t),2(3\cos
t)\right)\cdot(3\sin t, 3\cos t)dt.$$\ Then integrate to
show that the work done by $\vec F$ along this circle is
$36\pi$.
It's time for a definition.
The work done by a vector field $\vec F$, along a curve $C$
with parametrization $\vec r(t)$ for $a\leq t\leq b$ is $$W =
\int_C \vec F\cdot d\vec r= \int_a^b \vec F\cdot \frac{d\vec
r}{dt}dt.$$ If we let $\vec F = (M,N)$ and we let $\vec
r(t)=(x,y)$, so that $d\vec r = (dx,dy)$, then we can write
work in the differential form $$W = \int_C \vec F\cdot d\vec r=
\int_C (M,N)\cdot (dx,dy) = \int_C Mdx+Ndy.$$
Consider the curve $y=3x^25x$ for $2\leq x\leq 1$. Give a
parametrization of this curve. See \footnote{Whenever you have
a function of the form $y=f(x)$, you can always use $x=t$ and
$y=f(t)$ to parametrize the curve. So we can use $\vec r(t) =
(t, 3t^25t)$ for $2\leq t\leq 1$ as a parametrization.}.
Please use this
Sage link to check
your work.
% Consider the parabolic curve $y=4x^2$ for $1\leq
x\leq 2$, and the vector field $\vec F(x,y) = (2x+y,x)$.
 Give a parametrization $\vec r(t)$ of the parabolic curve
that starts at $(1,3)$ and ends at $(2,0)$. See the review
problem above if you need a hint.
 Compute $d\vec r$ and state $dx$ and $dy$. What are $M$
and $N$ in terms of $t$?
 Compute the work done by $\vec F$ to move an object along
the parabola from $(1,3)$ to $(2,0)$ (i.e. compute $\int _C
Mdx+Ndy$). Check your answer with Sage.
 How much work is done by $\vec F$ to move an object along
the parabola from $(2,0)$ to $(1,3)$. In general, if you
traverse along a path backwards, how much work is done?
Again consider the vector field $\vec F(x,y) = (2x+y,x)$. In
the previous problem we considered how much work was done by
$\vec F$ as an object moved along the the parabolic curve
$y=4x^2$ for $1\leq x\leq 2$. We now want to know how much
work is done to move an object along a straight line from
$(1,3)$ to $(2,0)$.
 Give a parametrization $\vec r(t)$ of the straight line
segment that starts at $(1,3)$ and ends at $(2,0)$. Make
sure you give bounds for $t$.
 Compute $d\vec r$ and state $dx$ and $dy$. What are $M$
and $N$ in terms of $t$?
 Compute the work done by $\vec F$ to move an object along
the straight line path from $(1,3)$ to $(2,0)$. Check your
answer with Sage.
When you enter your curve in Sage, remember to type the
times symbol in “(3*t1, ...)”. Otherwise,
you'll get an error.
 Optional (we'll discuss this in class if you don't have
it). How much work does it take to go along the closed path
that starts at $(2,0)$, follows the parabola $y=4x^2$ to
$(3,1)$, and then returns to $(2,0)$ along a straight line.
Show that this total work is $W=9$.
The examples above showed us that we can compute work along
any closed curve. All we have to do is parametrize the curve,
take a derivative, and then compute $dW = \vec F \cdot d\vec r$.
This gives us a little bit of work along a curve, and we sum up
the little bits of work (integrate) to find the total work. In
the examples above, the vector fields represented forces.
However, vector fields can represent much more than just forces.
The vector field might represent the flow of water down a river,
or the flow of air across an airplane wing. When we think of the
vector field as a velocity field, a natural question is: how much
of the fluid flows along our curve (the \emph{flow})?
Alternately, we might ask how much of the fluid goes across our
curve (the \emph{flux}). Flow along a curve is directly related
to the lift of an airplane wing (which occurs when the flow along
the top of the wing is different than the flow below the wing).
The flux across a curve will quickly take us to powering a wind
mill as wind flows across the surface of a blade (once we hit 3D
integrals). Flux is also extremely important in studying electric
and magnetic fields.
If the unit tangent vector is $\vec T = \dfrac{(3,4)}{5}$, give
two unit vectors that are orthogonal to $\vec T$. See
\footnote{We just reverse the order and change a sign to get
$\vec N_1 = \dfrac{(4,3)}{5}$ and $\vec N_1 =
\dfrac{(4,3)}{5}$ as the orthogonal vectors.}.
We used the formulas $$W = \int_C \vec F\cdot d\vec r = \int_C
(M,N)\cdot(dx,dy)$$ to compute the work done by $\vec F$ along
a curve $C$ parametrized by $\vec r = (x,y)$.
 Explain why $W = \int_C \vec F\cdot \vec T ds$. [Why does
$\vec T ds = d\vec r$? Look up $\vec T$ in the last chapter.]
%
 Show that $\vec F\cdot \vec T$ is the projection of $\vec
F$ onto the vector $T$ (the amount of work in the tangential
direction). [Just write down the formula for a projection.
How long is $\vec T$? This should be really fast.]

% We know that $\vec T = \dfrac{(dx,
dy)}{\sqrt{(dx)^2+(dy)^2}}$. Suppose $\vec n$ is a unit
normal vector to the curve. Give two options for $\vec n$.
[Hint: Look at the review problem.]
 We know $\vec T ds = (dx,dy)$. Why does $\vec n ds$ equal
$(dy,dx)$ or $(dy,dx)$?
 The integral $W = \int_C \vec F\cdot \vec T ds$ measures
how much of the vector field flows along the curve. What does
the integral $\Phi = \int_C \vec F\cdot \vec n ds$
measure?
Consider the curve $\vec r(t) = (5\cos t, 5\sin t)$, and the
vector field $\vec F(x,y) = (3x, 3y)$. This is a radial field
that pushes things straight outwards (away from the origin).
 Compute the work $\ds W= \int_C (M,N)\cdot (dx,dy)$ and
show that is equals zero. (Can you give a reason why it
should be zero?)
See
Sage for the work
calculation.
 To get a normal vector, we could change $(dx,dy)$ to
$(dy,dx)$ or to $(dy,dx)$. Compute both $\ds \int_C
(M,N)\cdot (dy,dx)$ and $\ds \int_C (M,N)\cdot (dy,dx)$.
(They should differ by a sign.) Both integrals measure the
flow of the field across the curve, instead of along the
curve.
 If we want flux to measure the flow of a vector field
outwards across a curve, then the flux of this vector field
should be positive. Which vector, $(dy,dx)$ or $(dy,dx)$,
should we choose above for $n$.
See
Sage for the flux
computation
 (Challenge, we'll discuss in class.) Suppose $\vec r$ is
a counterclockwise parametrization of a closed curve. The
outward normal vector would always point to the right as you
move along the curve. Prove that $(dy,dx)$ always points to
the right of the curve. [Hint: If you want a right pointing
vector, what should $\vec B=\vec T\times \vec n$ always equal
(either $(0,0,1)$ or $(0,0,1)$). Use the fact that $\vec
B\times \vec T = \vec n$ to get $\vec n$.]
% Suppose $C$ is a smooth curve with parametrization
$\vec r(t)=(x,y)$. Suppose that $\vec F(x,y)$ is a vector field
that represents the velocity of some fluid (like water or air).
 We say that $C$ is closed curve if $C$ begins and ends at
the same point.
 We say that $C$ is a simple curve if $C$ does not cross
itself.
 The flow of $\vec F$ along $C$ is the integral
$$\text{Flow} = \int_C (M,N)\cdot (dx,dy) = \int_C
Mdx+Ndy.$$
 If $C$ is a simple closed curve parametrized counter
clockwise, then the flow of $\vec F$ along $C$ is called
circulation, and we write $\text{Circulation} = \oint_C
Mdx+Ndy$
 The flux of $\vec F$ across $C$ is the flow of the fluid
across the curve (an area/second). If $C$ is a simple closed
curve parametrized counter clockwise, then the outward flux
is the integral $$\text{Flux} = \Phi = \int_C(M,N)\cdot
(dy,dx) =\int_C MdyNdx .$$
If you haven't yet, please watch the YouTube videos for
% Consider the vector field $\vec F(x,y) = (2x+y,x+2y)$.
When you construct a plot of this vector field, you'll notice
that it causes objects to spin outwards in the clockwise
direction. Suppose an object moves counterclockwise around a
circle $C$ of radius 3 that is centered at the origin. (You'll
need to parameterize the curve.)
 Should the circulation of $\vec F$ along $C$ be positive
or negative? Make a guess, and then compute the circulation
$\oint_C Mdx+Ndy$. Whether your guess was right or wrong,
explain why you made the guess.
 Should the flux of $\vec F$ across $C$ be positive or
negative? Make a guess, and then actually compute the flux
$\oint_C MdyNdx$. Whether your guess was right or wrong,
explain why you made the guess.
 Please use this Sage link
to check both computations.
We'll tackle more work, flow, circulation, and flux
problems, as we proceed through this chapter. Work, flow,
circulation, and flux are all examples of line integrals.
Remember that work, flow, and circulation are $$W=\int_C (\vec
F\cdot \vec T)ds =\int_C (M,N)\cdot(dx,dy) = \int_C Mdx+Ndy,$$
while the formula for flux is $$\Phi=\int_C (\vec F\cdot \vec
n)ds =\int_C (M,N)\cdot(dy,dx) = \int_C Mdy+Ndx.$$ Do you see
how these are both the line integral of a function $f = \vec
F\cdot \vec T$ or $f=\vec F\cdot \vec n$ along a curve $C$. The
function $f$ inside the integrand does not have to represent the
height of a sheet. We'll use it to represent lots of things.
Let's practice two more work/flux problems, to sharpen our skills
with these concepts.
% \instructor{Answers—
Sage.
}% Let $\vec F=(y,x+y)$ and $C$ be the triangle with vertices
$(2,0)$, $(0,2)$, and $(0,0)$.
 Look at a drawing of $C$ and the vector field (see margin
for the Sage link). We'll move along the triangle in a
counter clockwise manner. Without doing any computations, for
each side of the triangle make a guess to determine if the
flow along that edge is positive, negative, or zero.
Similarly, guess the sign of the flux along each edge.
Explain.
 Obtain three parameterizations for the edges of the
triangle. One of the parameterizations is $\vec r(t) =
(0,2)t+(0,2)$.
 Now find the counterclockwise circulation (work) done by
$\vec F$ along $C$. You'll have three separate calculations,
one for each side. We'll do the flux computation in class.
Check your work on each piece with the Sage calculator.
See
Sage. Think of an airplane wing as you solve this
problem.
% \instructor{Answers—
Sage.
Again, we'll discuss why some are positive, and some are
negative. I want to emphasize flow in, and flow out. One is
clearly positive, the other clearly negative. The overall sum
is positive. It should be obvious with a picture.}% Consider
the vector field $\vec F=(2xy,x)$. Let $C$ be the curve that
starts at $(2,0)$, follows a straight line to $(1,3)$, and
then back to $(2,0)$ along the parabola $y=4x^2$.
 Look at a drawing of $C$ and the vector field (see margin
for the Sage link). If we go counterclockwise around $C$, for
each part of $C$, guess the signs of the counterclockwise
circulation and the flux (positive, negative, zero).
 Find the flux of $\vec F$ across $C$. There are two
curves to parametrize. Make sure you traverse along the
curves in the correct direction. [Hint: You should get
integer values along both parts. Check your work with
Sage, but make sure you show
us how to do the integrals by hand.]
Ask me in class to change the vector fields above, and
examine what happens with different vectors fields. In
particular, it's possible to have any combination of values for
circulation and flux. We'll be able to use technology to rapidly
compute many values.
In both of the preceding problems, both the circulation and
flux are positive. Everything is always positive. This would
give a student a false impression that work and flux are always
positive. We need to change some computations. Make sure to
look into this. It needs to change.
Average Value
The concept of averaging values together
has many applications. In firstsemester calculus, we saw how to
generalize the concept of averaging numbers together to get an
average value of a function. We'll review both of these concepts.
Later, we'll generalize average value to calculate centroids and
center of mass.
Throughout the
section, I point out how each formula is a variation on one of
the patterns below.
Suppose a class takes a test and there are
three scores of 70, five scores of 85, one score of 90, and two
scores of 95. We will calculate the average class score, $\bar
s$, four different ways to emphasize four ways of thinking
about the averages. We are emphasizing the pattern of the
calculations in this problem, rather than the final answer, so
it is important to write out each calculation completely in the
form $\bar s = \blank{1cm}$ before calculating the number $\bar
s$.

$\bar s=\frac{\sum \text{values}}{\text{number of
values}}$
Compute the average by adding 11 numbers together
and dividing by the number of scores. Write down the whole
computation before doing any arithmetic.

$\bar s=\frac{\sum (\text{value}\cdot\text{weight})}{\sum
\text{weight}}$
Compute the numerator of the fraction in the
previous part by multiplying each score by how many times
it occurs, rather than adding it in the sum that many
times. Again, write down the calculation for $\bar s$
before doing any arithmetic.

$\bar s=\sum (\text{value}\cdot\text{(% of stuff)})$
Compute $\bar s$ by splitting up the fraction in
the previous part into the sum of four numbers. This is
called a “weighted average” because we are
multiplying each score value by a weight.

$\text{(number of values)}\bar s = \sum \text{values}$
$(\sum \text{weight})\bar s = \sum
(\text{value}\cdot\text{weight})$
Another way of thinking about the average $\bar s$
is that $\bar s$ is the number so that if all 11 scores
were the same value $\bar s$, you'd have the same sum of
scores. Write this way of thinking about these computations
by taking the formulas for $\bar s$ in the first two parts
and multiplying both sides by the denominator.
In the next problem, we generalize the above ways of
thinking about averages from a discrete situation to a continuous
situation. You did this in firstsemester calculus when you did
average value using integrals.
Suppose the price of a stock is \$10 for one day. Then the
price of the stock jumps to $20 for two days. Our goal is to
determine the average price of the stock over the three days.
 Why is the average stock price not \$15?
 Let $f(t) = \begin{cases}10
&0<t<1\\20&1<t<3\end{cases}$, the price
of the stock for the threeday period. Draw the function $f$,
and find the area under $f$ where $t\in[0,3]$.
 Now let $y=\bar f$. The area under $\bar f$ over $[0,3]$
is simply width times height, or $(ba)\bar f$. What should
$\bar f$ equal so that the area under $\bar f$ over $[0,3]$
matches the area under $f$ over $[0,3]$.
 We found a constant $\bar f$ so that the area under $\bar
f$ matched the area under $f$. In other words, we solved the
equation below for $\bar f$: $$\int_a^b \bar f dx = \int_a^b
f dx$$ Solve for $\bar f$ symbolically (without doing any of
the integrals). This quantity is called the average value of
$f$ over $[a,b]$.
I also write $\bar f=\int_a^b f
\frac{dx}{\int_a^b dx}$ to emphasize the weighted average
approach
The formula for $\bar f$ in the previous part
resembles at least one of the ways of calculating averages
from the previous problem.
Which ones and why?
I talk about ants building a mound. Then after
removing the ants, so none get hurt, you shake their tank. The
average value is the height of the dirt. The mountains filled in
the valleys.
Ask me in class about the “ant farm”
approach to average value.
Consider the elliptical curve $C$ given by the
parametrization $\vec r(t) = (2\cos t, 3\sin t)$. Let $f$ be
the function $f(x,y)=9x^2y^2$.
 Draw the surface $f$ in 3D. Add to your drawing the curve
$C$ in the $xy$ plane. Then draw the sheet whose area is
given by the integral $\int_C f ds$.
 What's the maximum height and minimum height of the
sheet?
 We'd like to find a constant height $\bar f$ so that the
area under $f$, above $C$, is the same as area under $\bar
f$, above $C$. This height $\bar f$ is called the average
value of $f$ along $C$.
%
\marginparbmw{Please read
Isaiah
40:4 and
Luke
3:5. These scriptures should help you remember how to
find average value. }%
Again, I emphasize how
this relates to the ways of computing averages from
this problem.
Explain why the average value of $f$ along $C$ is $$\bar f
= \frac{\int_C f ds}{\int_C ds}.$$ Connect this formula
with the ways of thinking about averages from
this problem.
[Hint: The area under $\bar f$ above $C$ is $\int_C \bar f
ds$. The area under $f$ above $C$ is $\int_C f ds$. Set
them equal and solve for $\bar f$. ]

Use a computer to evaluate the integrals $\int_C f
ds$ and $\int_C ds$, and then give an approximation to the
average value of $f$ along $C$. Is your average value
between the maximum and minimum of $f$ along $C$? Why
should it be?
After this problem, I like to emphasize that they
should have noticed a linear growth rate, and then I show them
how I would have guessed the answer.
The temperature
$T(x,y,z)$ at points on a wire helix $C$ given by $\vec r(t) =
(\sin t, 2t, \cos t)$ is known to be $T(x,y,z)=x^2+y+z^2$. What
are the temperatures at $t=0$, $t=\pi/2$, $t=\pi$, $t=3\pi/2$
and $t=2\pi$? You should notice the temperature is constantly
changing. Make a guess as to what the average temperature is
(share with the class why you made the guess you made—it's OK
if you're wrong). Then compute the average temperature of the
wire using the integral formula from the previous problem. You
can do all these computations by hand.
Physical Properties
It feels like this section has too many problems—some of them
seem redundant. I should go back through and trim down these
problems, or trim down parts of problems.
A number of physical properties of realworld objects can
be calculated using the concepts of averages and line integrals.
We explore some of these in this section. Additionally, many of
these concepts and calculations are used in statistics.
Centroids
Let $C$ be a curve. If we look at all of the $x$coordinates of
the points on $C$, the “center” $x$coordinate,
$\bar x$, is the average of all these $x$coordinates.
Likewise, we can talk about the averages of all of the $y$
coordinates or $z$ coordinates of points on the function ($\bar
y$ or $\bar z$, respectively). The centroid of an object
is the geometric center $(\bar x, \bar y, \bar z)$, the point
with coordinates that are the average $x$, $y$, and $z$
coordinates.
Notice the word “average” in the definition
of the centroid. Use the concept of average value to explain
why the coordinates of the centroid are the formulas below.
[Hint: If we have a curve $C$ with parametrization $\vec r(t)$
and function $f$ so that $f(\vec r(t))$ is continuous, then
we've developed in
this problem a formula for the average
value of $f$ along $C$. What function $f(x,y,z)$ gives the
$x$coordinate of a point?] $$ \bar x = \frac{\int_C x
ds}{\int_C ds},\quad \bar y = \frac{\int_C y ds}{\int_C
ds},\quad \text{and}\quad \bar z = \frac{\int_C z ds}{\int_C
ds}. $$ Notice that the denominator in each case is just the
arc length $s=\int_C ds$.
Let $C$ be the semicircular arc
$\vec r(t)=(a\cos t, a\sin t)$ for $t\in[0,\pi]$. Without doing
any computations, make an educated guess for the centroid
$(\bar x, \bar y)$ of this arc. Then compute the integrals
given in
the centroid problem above to
find the actual centroid. Share with the class your guess, even
if it was incorrect.
Mass
Jason made a comment that aluminum and copper won't combine. So
this should be changed eventually.
Density is generally a mass per unit volume. However, when
talking about a curve or wire, as in this chapter, it's simpler
to let density be the mass per unit length. Sometimes an object
is made out of a composite material, and the density of the
object is different at different places in the object. For
example, we might have a straight wire where one end is aluminum
and the other end is copper. In the middle, the wire slowly
transitions from being all aluminum to all copper. The centroid
is the midpoint of the wire. However, since copper has a higher
density than aluminum, the balance point (the center of mass)
would not be at the midpoint of the wire, but would be closer to
the denser and heavier copper end. In this section, we'll develop
formulas for the mass and center of mass of such a wire. In future mechanical
engineering courses, you would learn how to determine the density
$\delta$ (mass per unit length) at each point on such a composite
wire.
Suppose a wire $C$ has the parameterization $\vec r(t)$
for $t\in[a,b]$. Suppose the wire's density at a point
$(x,y,z)$ on the wire is given by the function $\delta(x,y,z)$.
You'll learn to calculate this function in a
future class. For the purposes of our class, we'll just assume
we know what $\delta(x,y,z)$ is.
 Consider a small portion of the curve at $t=t_0$ of
length $ds$. Explain why the mass of the small portion of the
curve is $dm=\delta(\vec r(t_0)) ds$.
 Explain why the mass $m$ of an object is given by the
formulas below (explain why each equals sign is true):
$$m=\int_C dm = \int_C \delta ds = \int_a^b \delta(\vec r(t))
\left\frac{d\vec r}{dt}\rightdt.$$
See Larson 15.2:23‐26 for more
practice.
A wire lies along the straight segment from $(0,2,0)$
to $(1,1,3)$. The wire's density (mass per unit length) at a
point $(x,y,z)$ is $\delta(x,y,z)=x+y+z$.
 Is the wire heavier at $(0,2,0)$ or at $(1,1,3)$?
 What is the total mass of the wire? [You'll need to
parameterize the line as your first step—see
this problem if you need
a refresher.]
Center of mass
Here I introduce the center of mass of an
object, talk about moments, and talk about how the center of mass
formula is also an averaging of the coordinates, where the weight
is the amount of mass at a particular coordinate value (i.e.,
$xdm$).
The center of mass of an object is the point where the
object balances. In order to calculate the $x$coordinate of the
center of mass, we average the $x$coordinates, but we weight
each $x$coordinate with its mass. Similarly, we can calculate
the $y$ and $z$ coordinates of the center of mass. The next
problem helps us reason about the center of mass of a collection
of objects. Calculating the center of mass of a collection of
objects is important, for example, in astronomy when you want to
calculate how two bodies orbit each other.
After a
student presents, this is a great time to connect the averages
back to
this problem.
I point out that we can think about the object as 2 points of
the same mass at $P_1$ and 3 points of the same mass at $P_2$.
This suggests averaging 5 things with method 1. Alternately, I
suggest the approach $\frac{2}{5}P_1+\frac{3}{5}P_2$,
suggesting a weighted average.
Suppose two objects are
positioned at the points $P_1=(x_1,y_1,z_1)$ and
$P_2=(x_2,y_2,z_2)$. Our goal in this problem is to understand
the difference between the centroid and the center of mass.
 Find the centroid of two objects.
I found many students struggled with setting
up a really simple sum. In class, after they present this
one, I would suggest actually taking time to show them how to
write the problem in summation notation with 2 points. It
will prepare them for the proof of center of mass coming
up.
Suppose both objects have the same mass of 2 kg. Find
the center of mass.
 If the mass of the object at point $P_1$ is 2 kg, and the
mass of the object at point $P_2$ is 5 kg, will the center of
mass be closer to $P_1$ or $P_2$? Give a physical reason for
your answer before doing any computations. Then find the
center of mass $(\bar x, \bar y, \bar z)$ of the two points.
[Hint: You should get $\bar x= \frac{2x_1+5x_2}{2+5}$.]
 If we consider a system with 3 points, what formula gives
the center of mass in the $x$ direction? What if there are 4
points, 5 points, or $n$ points?
This problem reinforces what you just did with two points in
the previous problem. However, it now involves two people on a
seesaw.
 My daughter and her friend are sitting on a seesaw. Both
girls have the same mass of 30 kg. My wife stands about 1 m
behind my daughter. We'll measure distance in this problem
from my wife's perspective. We can think of my daughter as a
point mass located at $(1\text{m},0)$ whose mass is $30$ kg.
Suppose her friend is located at $(5\text{m},0)$. Suppose the
kids are sitting just right so that the seesaw is perfectly
balanced. That means the the center of mass of the girls is
precisely at the pivot point of the seesaw. Find the distance
from my wife to the pivot point by finding the center of mass
of the two girls.
 My daughter's friend has to leave, so I plan to take her
place on the seesaw. My mass is 100 kg. Her friend was
sitting at the point $(5,0)$. I would like to sit at the
point $(a,0)$ so that the seesaw is perfectly balanced.
Without doing any computations, is $a>5$ or $a<5$?
Explain.
 Suppose I sit at the spot $(x,0)$ (perhaps causing my
daughter or I to have a highly unbalanced ride). Find the
center of mass of the two points $(1,0)$ and $(x,0)$ whose
masses are $30$ and $100$, respectively (units are meters and
kilograms).
 Where should I sit so that the seesaw is perfectly
balanced (what is $a$)?
In
this problem, we
focused on a system of points with mass. In this problem, we look at a continuous wire.
 Suppose now that we have a wire located along a curve
$C$. The density of the wire is known to be $\delta(x,y,z)$
(which could be different at different points on the curve).
Imagine cutting the wire into a thousand or more tiny chunks.
Each chunk would be centered at some point $(x_i,y_i,z_i)$
and have length $ds_i$. Explain why the mass of each little
chunk is $dm_i\approx\delta ds_i$.
 Give a formula for the center of mass in the $y$
direction of the thousands of points $(x_i,y_i,z_i)$, each
with mass $dm_i$. [This should almost be an exact copy of the
first part.] Then explain why $$\bar y = \frac{\int_C y
dm}{\int_C dm}=\frac{\int_C y \delta ds}{\int_C \delta
ds}.$$
Ask me in class to show you another way to obtain the
formula for center of mass. It involves looking at masses
weighted by their distance (called a moment of mass). You may
have already seen an idea similar to this in other classes, like
statics (using moments of force) or statistics (in computing
expected values). For quick reference, the formulas for the
centroid of a wire along $C$ are $$ \bar x = \frac{\int_C x
ds}{\int_C ds},\quad \bar y = \frac{\int_C y ds}{\int_C ds},\quad
\text{and}\quad \bar z = \frac{\int_C z ds}{\int_C ds}.
\quad\text{(Centroid)} $$ If the wire has density $\delta$, then
the formulas for the center of mass are
The quantity $\int_C x dm$ is sometimes called the first moment
of mass about the $yz$plane (so $x=0$). Notationally, some
people write $M_{yz} =\int_C x ds$. Similarly, we could write
$M_{xz}=\int_C y dm$ and $M_{xy}=\int_C zdm$. With this
notation, we could write the center of mass formulas as $$(\bar
x,\bar y,\bar z) = \left( \frac{M_{yz}}{m}, \frac{M_{xz}}{m},
\frac{M_{xy}}{m} \right) . $$
$$ \bar x = \frac{\int_C x dm}{\int_C dm},\quad \bar y =
\frac{\int_C y dm}{\int_C dm},\quad \text{and}\quad \bar z =
\frac{\int_C z dm}{\int_C dm}, \quad\text{(Center of mass)} $$
where $dm=\delta ds$. Notice that the denominator in each case is
just the mass $m=\int_C dm$. We'll often use the notation $(\bar
x, \bar y,\bar z)$ to talk about both the centroid and the center
of mass. If no density is given in a problem, then $(\bar x, \bar
y,\bar z)$ is the centroid. If a density is provided, then $(\bar
x, \bar y,\bar z)$ refers to the center of mass. If the density
is constant, it doesn't matter (the centroid and center of mass
are the same, which is what the seesaw problem showed).
I purposefully put this problem in to show students
how to generalize a formula from prime numbers to any number. I
mention how I use primes (5, 7, 11, 13) when I'm looking for a
pattern. I want to help them develop this skill a little.
However, if you are pressed for time, then skip 5.
Check your work with
Sage
Suppose a wire with density $\delta(x,y)=x^2+y$ lies
along the curve $C$ which is the upper half of a circle around
the origin with radius $7$.
 Parametrize $C$ (find $\vec r(t)$ and the domain for
$t$).
 Where is the wire heavier, at $(7,0)$ or $(0,7)$?
[Compute $\delta$ at both.]
 In this problem, we
showed that the centroid of the wire is $(\bar x, \bar y) =
\left(0,\frac{2(7)}{\pi}\right)$. We now seek the center of
mass. Before computing, will $\bar x$ change? Will $\bar y$
change? How will each change? Explain.
 Set up the integrals needed to find the center of mass.
Then use technology to compute the integrals. Give an exact
answer (involving fractions), rather than a numerical
approximation.
 Change the radius from 7 to $a$, and guess what the
center of mass will be. (This is why you need the exact
answer above, not a numerical answer).
The Fundamental Theorem of Line Integrals
In this final
section, we'll return to the concept of work. Many vector fields
are actually the derivative of a function, called the
\emph{potential} function of the vector field. In this section,
we'll first discuss finding the potential of such a vector field.
Then we'll show that the work done by such a vector field is the
difference in the potential at the end points. This is much
simpler than calculating a line integral.
% Let $\vec F$ be a vector field. A potential for the
vector field is a function $f$ whose derivative equals $\vec
F$. So if $Df=\vec F$, then we say that $f$ is a potential for
$\vec F$. When we want to emphasize that the derivative of $f$
is a vector field, we call $Df$ the gradient of $f$ and write
$Df = \vec \nabla f$.
The symbol $\vec \nabla f$ is read “the gradient of
$f$” or “del f.”
If $\vec F$ has a potential, then we say that $\vec F$ is
a gradient field.
We'll quickly see that if a vector field has a potential,
then the work done by the vector field is the difference in the
potential. If you've ever dealt with kinetic and potential
energy, then you hopefully recall that the change in kinetic
energy is precisely the difference in potential energy. This is
the reason we use the word “potential.”
% Let's practice finding gradients and potentials.
 Let $f(x,y) = x^2+3xy+2y^2$. Find the gradient of $f$,
i.e., find $Df(x,y)$. Then compute $D^2f(x,y)$ (you should
get a square matrix). What are $f_{xy}$ and $f_{yx}$?
 Consider the vector field $\vec F(x,y)=(2x+y,x+4y)$. Find
the derivative of $\vec F(x,y)$ (it should be a square
matrix). Then find a function $f(x,y)$ whose gradient is
$\vec F$ (i.e., $Df=\vec F$). What are $f_{xy}$ and
$f_{yx}$?

% Consider the vector field $\vec
F(x,y)=(2x+y,3x+4y)$. Find the derivative of $\vec F$. Why
is there no function $f(x,y)$ so that $Df(x,y)=\vec
F(x,y)$? [Hint: what would $f_{xy}$ and $f_{yx}$ have to
equal?]
Based on your observations in the previous problem, we have
the following key theorem.
Let $\vec F$ be a vector field that is everywhere continuously
differentiable. Then $\vec F$ has a potential if and only if
the derivative $D\vec F$ is a symmetric matrix. We say that a
matrix is symmetric if interchanging the rows and columns
results in the same matrix (so if you replace row 1 with column
1, and row 2 with column 2, etc., then you obtain the same
matrix).
% For each of the following vector fields, find a
potential, or explain why none exists.
 $\vec F(x,y)=(2xy, 3x+2y)$
 $\vec F(x,y)=(2x+4y, 4x+3y)$
 $\vec F(x,y)=(2x+4xy, 2x^2+y)$
 $\vec F(x,y,z)=(x+2y+3z,2x+3y+4z,2x+3y+4z)$
 $\vec F(x,y,z)=(x+2y+3z,2x+3y+4z,3x+4y+5z)$
 $\vec F(x,y,z)=(x+yz,xz+z,xy+y)$
 $\ds \vec F(x,y) = \left(\frac{x}{1+x^2}+\arctan
(y),\frac{x}{1+y^2}\right)$
If a vector field has a potential, then there is an
extremely simple way to compute work. To see this, we must first
review the fundamental theorem of calculus. The second half of
the fundamental theorem of calculus states
If $f$ is continuous on $[a,b]$ and $F$ is an antiderivative
of $f$, then $F(b)F(a) = \int_a^b f(x) dx$.
If we replace $f$ with $f'$, then an antiderivative of
$f'$ is $f$, and we can write,
If $f$ is continuously differentiable on $[a,b]$, then
$$f(b)f(a)=\int_a^b f'(x) dx.$$
This last version is the version we now generalize.
% Suppose $f$ is a continuously differentiable function,
defined along some open region containing the smooth curve $C$.
Let $\vec r(t)$ be a parametrization of the curve $C$ for
$t\in[a,b]$. Then we have $$f(\vec r(b))f(\vec r(a))=\int_a^b
Df(\vec r(t))D\vec r(t)\ dt.$$
Notice that if $\vec F$ is a vector field, and has a
potential $f$, which means $\vec F = Df$, then we could rephrase
this theorem as follows.
Suppose $\vec F$ is a vector field that is continuous along
some open region containing the curve $C$. Suppose $\vec F$ has
a potential $f$. Let $A$ and $B$ be the start and end points of
the smooth curve $C$. Then the work done by $\vec F$ along $C$
depends only on the start and end points, and is precisely
$$f(B)f(A)=\int_C \vec F\cdot d\vec r = \int_C Mdx+Ndy.$$ The
work done by $\vec F$ is the difference in a potential,
$f(B)f(A)$.
If you are familiar with kinetic energy, then you should
notice a key idea here. Work is a transfer of energy. As an
object falls, energy is transferred from potential energy to
kinetic energy. The total kinetic energy at the end of a fall is
precisely equal to the difference between the potential energy at
the top of the fall and the potential energy at the bottom of the
fall (neglecting air resistance). So work (the transfer of
energy) is exactly the difference in potential energy.
The proof of the fundamental theorem of line integrals is
quite short. All you need is the fundamental theorem of
calculus, together with the chain rule (
TODO).
Suppose $f(x,y)$ is continuously differentiable, and
suppose that $\vec r(t)$ for $t\in[a,b]$ is a parametrization
of a smooth curve $C$. Prove that $f(\vec r(b))f(\vec r(a)) =
\int_a^b Df(\vec r(t))D\vec r(t)\ dt$. [Let $g(t) = f(\vec
r(t))$. Why does $g(b)g(a) = \int_a^b g'(t)dt$? Use the chain
rule (matrix form) to compute $g'(t)$. Then just substitute
things back in.]
% For each vector field and curve below, find the work
done by $\vec F$ along $C$. In other words, compute the
integral $\int_C Mdx+Ndy$ or $\int_C Mdx+Ndy+Pdz$.

% Let $\vec F(x,y) = (2x+y,x+4y)$ and $C$ be the
parabolic path $y=9x^2$ for $x$ from $3$ to $2$.

% Let $\vec F(x,y,z) = (2x+yz,2z+xz,2y+xy)$ and $C$
be the straight segment from $(2,5,0)$ to $(1,2,3)$.
[Hint: If you parametrize the curve, then you've done the
problem the HARD way. Did you try to find a potential function
for the vector field?]
See
Sage—$C_1$ and $C_2$ are in blue, and several possible
$C_3$ are shown in red.
% Let $\vec F = (x,z,y)$. Let $C_1$ be the curve which
starts at $(1,0,0)$ and follows a helical path $(\cos t, \sin
t, t)$ to $(1,0,2\pi)$. Let $C_2$ be the curve which starts at
$(1,0, 2\pi)$ and follows a straight line path to $(2,4,3)$.
Let $C_3$ be any smooth curve that starts at $(2,4,3)$ and ends
at $(0,1,2)$.
In the problem above, the path we took to get from one
point to another did not matter. The vector field had a
potential, which meant that the work done did not depend on the
path traveled.
We say that a vector field is \emph{conservative} if the
integral $\int_C \vec F\cdot d\vec r$ does not depend on the
path $C$. We say that a curve $C$ is piecewise smooth if it can
be broken up into finitely many smooth curves.
Compute $\ds \int \frac{x}{\sqrt{x^2+4}}dx$. See \footnote{ Let
$u=x^2+4$, which means $du=2xdx$ or $dx=\frac{du}{2x}$. This
means $$\ds \int \frac{x}{\sqrt{x^2+4}}dx = \int
\frac{x}{\sqrt{u}}\frac{du}{2x} = \frac{1}{2}\int u^{1/2}du =
\frac{1}{2}\frac{u^{1/2}}{1/2} = \sqrt{u} = \sqrt{x^2+4}. $$ }.
\instructor{While $\vec F$ is not continuously differentiable
everywhere (it's not at the origin), we can still use the
fundamental theorem of line integrals if we stay away from the
origin. This shows that $\vec F$ is a conservative vector
field, provided we dodge the origin.}% The gravitational vector
field is directly related to the radial field $\ds\vec F =
\frac{\left(x,y,z\right)}{(x^2+y^2+z^2)^{3/2}}$. Show that
this vector field is conservative by finding a potential for
$\vec F$. Then compute the work done by an object that moves
from $(1,2,2)$ to $(0,3,4)$ along ANY path that avoids the
origin. [See the review problem just before this if you're
struggling with the integral.]
Suppose $\vec F$ is a gradient field. Let $C$ be a piecewise
smooth closed curve (closed means the starting and ending
points are the same). Compute $\int_C \vec F\cdot d\vec r$ (you
should get a number). Explain how you know your answer is
correct.
“Nice” sets (Optional)
\label{sec:nicesets}
The above results connecting path independence and gradient
fields assume that the function is defined on a
“nice” domain set. Here we will discuss in (brief)
technical detail what a “nice” set is and what goes
wrong when you have a set that is not so nice. This would all be
explored in much more detail in an advanced calculus, real
analysis, or algebraic topology class.
Let $\vec F=\langle M,N,P\rangle$ be continuous on an open
set $D$. In the diagram in Figure~TODO, the arrows represent implications
(i.e., if there is an arrow from one statement pointing to
another, then the first statement being true implies that the
second statement is true). A label on the arrow means that the
condition has to be satisfied for the implication to be hold. The
definitions of the terms are in the book (or you can ask me).
Using terms from the book, a “nice” set is a set that
is open, connected, and simply connected. Roughly, this means
that $D$ does not include its boundary and is a single region
without any holes going through it. For such a “nice”
set, all four statements in Figure~TODO are equivalent (i.e., either they
are all true or they are all false for a given vector field $\vec
F$). If the set $D$ is not connected, then pathindependence does
not imply the vector field is a gradient field. In this case,
discrepancies can occur between the two definitions for
“conservative vector field”. There is a homework
problem that shows that the bottom arrow does not hold when $D$
is not simply connected. When $D$ is not connected and simply
connected, we lose the nice, easy test in the lower left of the
figure for determining exactly when a vector field is a gradient
field.