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.
1. 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.
2. Find the following geometric properties of a curve: centroid, mass, center of mass, inertia, and radii of gyration.
3. Compute the work (flow, circulation) and flux of a vector field along and across piecewise smooth curves.
4. 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

{|l|c|l|l|l|l|}\hline Topic (11th ed.) &Sec &Basic Practice &Good Problems &Thy/App &Comp \\\hline Line integrals & 16.1&1-8, 9-22, 23-32 & & &33-36 \\\hline Work, Flow, Circulation, Flux & 16.2&7-16, 25-28, 37-40 &17-24, 29-30, 41-44 &45-46 &47-52 \\\hline Gradient Fields & 16.2&1-6 & & & \\\hline Gradient Fields & 14.5&1-8 & & & \\\hline Potentials & 16.3&1-12,13-24 & 25-33 & 34-38& \\\hline
} {\noindent %\footnotesize
{|l|c|l|l|l|l|}\hline Topic (12th ed.) &Sec &Basic Practice &Good Problems &Thy/App &Comp \\\hline Line integrals & 16.1&1-8, 9-26, 33-42 &27-32 & &43-46 \\\hline Work, Flow, Circulation, Flux & 16.2&7-12, 19-24, 31-36, 47-50 &13-18, 25-30, 37-38, & 51-54 &55-60 \\\hline Gradient Fields & 16.2&1-6 & & & \\\hline Gradient Fields & 14.5&1-10 & & & \\\hline Potentials & 16.3&1-12,13-24 & 25-33 & 34-38& \\\hline
}
 Arclength $s = \int_C ds= \int_a^b \left|\frac{d\vec r}{dt}\right|dt$ Surface Area $\sigma = \int_C d\sigma=\int_C f ds = \int_a^b f \left|\frac{d\vec r}{dt}\right|dt$ 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 $\text{Flux} = \int_C d\text{Flux} = \int_C \vec F\cdot \vec n ds = \oint_C Mdy-Ndx$ 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 4-6 minutes long.

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
{|l|c|l|l|l|l|}\hline Topic (11th ed.) &Sec &Basic Practice &Good Problems &Thy/App &Comp \\\hline Line integrals & 16.1&1-8, 9-22, 23-32 & & &33-36 \\\hline Work, Flow, Circulation, Flux & 16.2&7-16, 25-28, 37-40 &17-24, 29-30, 41-44 &45-46 &47-52 \\\hline Gradient Fields & 16.2&1-6 & & & \\\hline Gradient Fields & 14.5&1-8 & & & \\\hline Potentials & 16.3&1-12,13-24 & 25-33 & 34-38& \\\hline
} {\noindent \footnotesize
{|l|c|l|l|l|l|}\hline Topic (12th ed.) &Sec &Basic Practice &Good Problems &Thy/App &Comp \\\hline Line integrals & 16.1&1-8, 9-26, 33-42 &27-32 & &43-46 \\\hline Work, Flow, Circulation, Flux & 16.2&7-12, 19-24, 31-36, 47-50 &13-18, 25-30, 37-38, & 51-54 &55-60 \\\hline Gradient Fields & 16.2&1-6 & & & \\\hline Gradient Fields & 14.5&1-10 & & & \\\hline Potentials & 16.3&1-12,13-24 & 25-33 & 34-38& \\\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 single-variable 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 first-semester 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 $f(x,y)=9-x^2-y^2$ 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 $z=f(x,y)$.
1. 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.]
The height of the surface above each little arc is given by $f(x,y)$. Explain why the surface area of a little part of the surface is $$d\sigma = (9-4\cos^2t-9\sin^2t)\sqrt{(-2\sin t)^2+(3\cos t)^2}dt.$$
2. We'll use $\sigma$ (a lower-case 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 $d\sigma =fds$.
3. 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 $$\int_C f ds = \int_a^b f(\vec r(t))\frac{ds}{dt}dt = \int_a^b f(\vec r(t))\left|\frac{d\vec r}{dt}\right|dt.$$
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.

1. Start by getting a parametrization $\vec r(t)$ for $a\leq t\leq b$ of the curve $C$.
2. Find the speed by computing the velocity $\dfrac{d\vec r}{dt}$ and then the speed $\left|\dfrac{d\vec r}{dt}\right|$.
3. Multiply $f$ by the speed, and replace each $x$, $y$, and/or $z$ with what it equals in terms of $t$.
4. You should use the Sage line integral calculator to check all your answers.
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: 9-32. Some problems give you a parametrization, some expect you to come up with one on your own.}
To practice matching parameterizations to curves, try 16.1:1-8.}
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
1. The principles of line integrals apply even if the curve is not a 2d curve. Let $f(x,y,z)=x^2+y^2-2z$ 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?]
2. 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.
Let $f(x,y)=x^2+y^2-25$. Let $C$ be the portion of the parabola $y^2=x$ between $(1,-1)$ and $(4,2)$. We want to compute $\int_C fds$.
1. Draw the curve $C$ and the function $f(x,y)$ on the same 3D $xyz$ axes.
2. 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 $C$?
3. 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 $t$?]
4. 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.
1. $C$ is the curve $\vec r(t)=(2\cos t, 2\sin t)$, $t\in [0,2\pi]$.
2. $C$ is the curve $\vec r(t)=(\cos t, t, \sin t)$, $t\in[0,\pi]$.
See 11.5: 21-28.
See Sage for a plot.
(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 $B-A=(-6,3)$. 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 $0\leq t\leq 1$.
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?
2. If we encounter a constant force $\vec F$ over a small displacement $d\vec r$, explain why the work done is \ $\ds dW = \vec F\cdot d\vec r =F\cdot \frac{d\vec r}{dt}dt$.
3. 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.
%
See Sage
% 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.
4. (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.
%
% 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 $\vec F(x,y) = (-2y,2x)$.
1. 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?
2. 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?
3. 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 $y=3x^2-5x$ 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 $\vec r(t) = (t, 3t^2-5t)$ for $-2\leq t\leq 1$ as a parametrization.}.
% Consider the parabolic curve $y=4-x^2$ for $-1\leq x\leq 2$, and the vector field $\vec F(x,y) = (2x+y,-x)$.
1. 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.
2. Compute $d\vec r$ and state $dx$ and $dy$. What are $M$ and $N$ in terms of $t$?
3. 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.
4. 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 $y=4-x^2$ 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 $(2,0)$.
1. 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 $t$.
2. Compute $d\vec r$ and state $dx$ and $dy$. What are $M$ and $N$ in terms of $t$?
3. 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*t-1, ...)”. Otherwise, you'll get an error.
4. 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 $y=4-x^2$ 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 $\vec r = (x,y)$.
1. 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.] %
2. 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.]
3. % 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.]
4. We know $\vec T ds = (dx,dy)$. Why does $\vec n ds$ equal $(dy,-dx)$ or $(-dy,dx)$?
5. 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).
1. 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.
2. 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.
3. 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
4. (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 $$\text{Flux} = \Phi = \int_C(M,N)\cdot (dy,-dx) =\int_C Mdy-Ndx .$$
% 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.)
1. 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.
2. Should the flux of $\vec F$ across $C$ be positive or negative? Make a guess, and then actually compute the flux $\oint_C Mdy-Ndx$. Whether your guess was right or wrong, explain why you made the guess.
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.
Watch a YouTube video. Also, see Sage for a picture.
% \instructor{Answers—Sage. }% Let $\vec F=(-y,x+y)$ and $C$ be the triangle with vertices $(2,0)$, $(0,2)$, and $(0,0)$.
1. 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.
2. Obtain three parameterizations for the edges of the triangle. One of the parameterizations is $\vec r(t) = (0,-2)t+(0,2)$.
3. 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 $\vec F=(2x-y,x)$. 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 $y=4-x^2$.
1. 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).
2. 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 first-semester 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 $\bar s$.
1. $\bar s=\frac{\sum \text{values}}{\text{number of values}}$
Compute the average by adding 11 numbers together and dividing by the number of scores. Write down the whole computation before doing any arithmetic.
2. $\bar s=\frac{\sum (\text{value}\cdot\text{weight})}{\sum \text{weight}}$
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.
3. $\bar s=\sum (\text{value}\cdot\text{(% of stuff)})$
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.
4. $\text{(number of values)}\bar s = \sum \text{values}$
$(\sum \text{weight})\bar s = \sum (\text{value}\cdot\text{weight})$
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 first-semester 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.