Green's theorem examples

Background

• Green's theorem

Remembering the formula

"Was it ${\displaystyle \frac{\partial Q}{\partial x}}$‍ or ${\displaystyle \frac{\partial Q}{\partial y}}$‍?"

"Which term is subtracted again?"

• The line integral of a vector field $\mathbf{\text{F}}(x,y)$‍ around a closed curve $C$‍ measures the fluid rotation around that boundary $C$‍.
• The double integral of the curl of $\mathbf{\text{F}}$‍ adds up all the tiny little bits of fluid rotation within the region $R$‍ enclosed by $C$‍.
• Intuitively, it makes sense that these should be related. And in fact, they are equal.

Warning: Green's theorem only applies to curves that are oriented counterclockwise. If you are integrating clockwise around a curve and wish to apply Green's theorem, you must flip the sign of your result at some point.

How do you know when to use Green's theorem?

"Mathematics is not a spectator sport" - George Polya

Example 1: Line integral $\to$‍ Area

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  Clockwise

  A

  Clockwise
• (Choice B)  

  Counterclockwise

  B

  Counterclockwise

Check

$P(x,y)=$‍

$Q(x,y)=$‍

Check

Answer

The line integral in Green's theorem looks like this:

${\displaystyle {\oint }_{C}Pdx+Qdy}$‍

And here's the integral we need to compute:

${\displaystyle {\oint }_{C}3ydx+4xdy}$‍

Just matching terms, we should apply

$P(x,y)=3y$‍

and

$Q(x,y)=4x$‍

${\displaystyle \frac{\partial Q}{\partial x}}=$‍

${\displaystyle \frac{\partial P}{\partial y}}=$‍

Check

Answer

${\displaystyle \frac{\partial Q}{\partial x}}={\displaystyle \frac{\partial }{\partial x}}(4x)=4$‍

${\displaystyle \frac{\partial P}{\partial y}}={\displaystyle \frac{\partial }{\partial y}}(3y)=3$‍

${\displaystyle {\iint }_R({\displaystyle \frac{\partial Q}{\partial x}}-{\displaystyle \frac{\partial P}{\partial y}})dA=}$‍

Check

Answer

Start by plugging in the expressions for ${\displaystyle \frac{\partial Q}{\partial x}}$‍ and ${\displaystyle \frac{\partial P}{\partial y}}$‍ that you found in the previous problem.

$\begin{array}{rl}{\iint }_R({\displaystyle \frac{\partial Q}{\partial x}}-{\displaystyle \frac{\partial P}{\partial y}})dA& ={\iint }_R(4-3)dA\\ \\ & ={\iint }_{R}dA\end{array}$‍

Now before you go crazy describing the region $R$‍ with the appropriate bounds for $x$‍ and $y$‍, notice what this integral represents. Since there is no function inside it is just adding up tiny bits of area of $R$‍, so it will equal the area of $R$‍ itself!

Since $R$‍ is a circle with radius $2$‍, that area is $\pi r^2=\pi (2{)}^2=4\pi$‍. Therefore our line integral evaluates to $4\pi$‍.

Example 1 debrief

Example 2: Two function graphs

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  Clockwise

  A

  Clockwise
• (Choice B)  

  Counterclockwise

  B

  Counterclockwise

Check

$P(x,y)=$‍

$Q(x,y)=$‍

Check

Answer

The line integral in Green's theorem looks like this:

${\displaystyle {\oint }_{D}Pdx+Qdy}$‍

And here's the integral we need to compute:

${\displaystyle {\oint }_D{x}^{2}ydx-y^{2}dy}$‍

Just matching terms, we should apply

$P(x,y)=x^{2}y$‍

and

$Q(x,y)=-y^2$‍

${\displaystyle \frac{\partial Q}{\partial x}}=$‍

${\displaystyle \frac{\partial P}{\partial y}}=$‍

Check

Answer

${\displaystyle \frac{\partial Q}{\partial x}}={\displaystyle \frac{\partial }{\partial x}}(-y^2)=0$‍

${\displaystyle \frac{\partial P}{\partial y}}={\displaystyle \frac{\partial }{\partial y}}(x^{2}y)=x^2$‍

$x_1=$‍

$x_2=$‍

$y_1(x)=$‍

$y_2(x)=$‍

Check

Answer

Let's start with the easier ones. The bounds for $y$‍ are explicitly given in the problem: The region is defined to be above the graph $y=(x^2-4)(x^2-1)$‍ and below the graph $y=4-x^2$‍. This means the inner integral will look like this:

${\displaystyle {\int }_{(x^2-4)(x^2-1)}^{4-x^2}\text{<yadda yadda yadda>}dy}$‍

The trickier part is knowing the bounds of $x$‍. The definition of our region doesn't even mention $x$‍. Let's take a look at the graphs of the $y$‍-bounds:

The bounds on the $x$‍-values are determined by where these two graphs intersect on the left and right. Just looking at the graph, you can probably guess that they intersect at $x=-2$‍ and $x=2$‍. You can also check this more precisely by plugging in $2$‍ and $-2$‍ to each equation.

If you did not have the graph to look at to guess-and-check, you would solve the equation $(x^2-4)(x^2-1)=4-x^2$‍.

This means the double integral as a whole will have the following setup:

${\displaystyle {\int }_{-2}^2{\int }_{(x^2-4)(x^2-1)}^{4-x^2}\stackrel{\text{We’ll put something here in just a sec}}{\overbrace{\text{<yadda yadda yadda>}}}dydx}$‍

${\displaystyle {\oint }_D{x}^{2}ydx-y^{2}dy=}$‍

Check

Answer

$\begin{array}{rl}& {\int }_{-2}^2\left({\int }_{(x^2-4)(x^2-1)}^{4-x^2}\underset{\text{Replace with values from step }3}{\underbrace{({\displaystyle \frac{\partial P}{\partial y}}-{\displaystyle \frac{\partial Q}{\partial x}})}}dy\right)dx\\ \\ & {\int }_{-2}^2\left({\int }_{(x^2-4)(x^2-1)}^{4-x^2}\underset{\text{Factor out from }y\text{ integral}}{\underbrace{(x^2-0)}}dy\right)dx\\ \\ & {\int }_{-2}^2\left(x^2\underset{\text{Simply the difference of bounds}}{\underbrace{{\int }_{(x^2-4)(x^2-1)}^{4-x^2}dy}}\right)dx\\ \\ & {\int }_{-2}^2{x}^2(\underset{\text{Factor out }{x}^2-4}{\underbrace{(4-x^2)-(x^2-4)(x^2-1)}})dx\\ \\ & {\int }_{-2}^2{x}^2(x^2-4)\underset{-x^2}{\underbrace{(-1-(x^2-1))}}dx\\ \\ & {\int }_{-2}^2-x^4(x^2-4)dx\\ \\ & {\int }_{-2}^2-(x^6-4x^4)dx\\ \\ & {\int }_{-2}^{2}4x^4-x^{6}dx\\ \\ & {[4{\displaystyle \frac{x^5}{5}}-{\displaystyle \frac{x^7}{7}}]}_{x=-2}^{x=2}\\ \\ & (4{\displaystyle \frac{2^5}{5}}-{\displaystyle \frac{2^7}{7}})-(4{\displaystyle \frac{(-2{)}^5}{5}}-{\displaystyle \frac{(-2{)}^7}{7}})\\ \\ & (4{\displaystyle \frac{32}{5}}-{\displaystyle \frac{128}{7}})\underset{\text{This is just adding the same amount again}}{\underbrace{-(4{\displaystyle \frac{-32}{5}}-{\displaystyle \frac{-128}{7}})}}\\ \\ & 2(4{\displaystyle \frac{32}{5}}-{\displaystyle \frac{128}{7}})\\ \\ & 2({\displaystyle \frac{128}{5}}-{\displaystyle \frac{128}{7}})\\ \\ & 256({\displaystyle \frac{1}{5}}-{\displaystyle \frac{1}{7}})\\ \\ & 256({\displaystyle \frac{7}{35}}-{\displaystyle \frac{5}{35}})\\ \\ & 256\left({\displaystyle \frac{2}{35}}\right)\\ \\ & \boxed{{\displaystyle {\displaystyle \frac{512}{35}}}}\end{array}$‍

Example 2 debrief

Sneaky area calculations

${\displaystyle \frac{\partial Q}{\partial x}}-{\displaystyle \frac{\partial P}{\partial y}}=1$‍,

Example 3: Area of a seashell

Choose 1 answer:

Choose 1 answer:

• (Choice A)  

  Clockwise

  A

  Clockwise
• (Choice B)  

  Counterclockwise

  B

  Counterclockwise

Check

Choose all answers that apply:

Choose all answers that apply:

• (Choice A)  

  $\begin{array}{rl}P(x,y)& =x\\ Q(x,y)& =y\end{array}$‍

  A

  $\begin{array}{rl}P(x,y)& =x\\ Q(x,y)& =y\end{array}$‍
• (Choice B)  

  $\begin{array}{rl}P(x,y)& =-y\\ Q(x,y)& =0\end{array}$‍

  B

  $\begin{array}{rl}P(x,y)& =-y\\ Q(x,y)& =0\end{array}$‍
• (Choice C)  

  $\begin{array}{rl}P(x,y)& =0\\ Q(x,y)& =x\end{array}$‍

  C

  $\begin{array}{rl}P(x,y)& =0\\ Q(x,y)& =x\end{array}$‍
• (Choice D)  

  $\begin{array}{rl}P(x,y)& =-y/2\\ Q(x,y)& =x/2\end{array}$‍

  D

  $\begin{array}{rl}P(x,y)& =-y/2\\ Q(x,y)& =x/2\end{array}$‍

Check

Answer

All but the first choice are correct. When you take the definitions of $P$‍ and $Q$‍ in the first choice, both partial derivatives are $0$‍.

$xdy-ydx=$‍

$dt$‍

Check

Answer

$\begin{array}{rl}& \underset{\text{Plug in functions }x(t)\text{, }y(t)}{\underbrace{xdy-ydx}}\\ \\ & \underset{\text{Factor out }dt}{\underbrace{x(t)\left({\displaystyle \frac{dy}{dt}}dt\right)-y(t)\left({\displaystyle \frac{dx}{dt}}dt\right)}}\\ \\ & \underset{\text{Plug in definitions of }x(t)\text{ and }y(t)}{\underbrace{(x(t){\displaystyle \frac{dy}{dt}}-y(t){\displaystyle \frac{dx}{dt}})}}dt\\ \\ & (t\cos (t)\underset{\text{Product rule}}{\underbrace{{\displaystyle \frac{d}{dt}}(t\sin (t))}}-t\sin (t)\underset{\text{Product rule}}{\underbrace{{\displaystyle \frac{d}{dt}}(t\cos (t))}})dt\\ \\ & (t\cos (t)(t\cos (t)+\sin (t))-t\sin (t)(-t\sin (t)+\cos (t)))dt\\ \\ & (t^2{\cos }^2(t)+\cancel{t\cos (t)\sin (t)}+t^2{\sin }^2(t)-\cancel{t\sin (t)\cos (t)})dt\\ \\ & (t^2\underset{1}{\underbrace{({\cos }^2(t)+{\sin }^2(t))}})dt\\ \\ & t^{2}dt\end{array}$‍

${\displaystyle {\int }_{\text{Spiral}}{\displaystyle \frac{1}{2}}(xdy-ydx)=}$‍

Check

Answer

Start by plugging in the value for $xdy-ydx$‍ that you found in the last problem, and put in the bounds $t=0$‍ and $t=2\pi$‍:

$\begin{array}{rl}{\int }_{\text{Spiral}}{\displaystyle \frac{1}{2}}\underset{t^{2}dt}{\underbrace{(xdy-ydx)}}& ={\int }_0^{2\pi }{\displaystyle \frac{1}{2}}{t}^{2}dt\\ \\ & ={\left[{\displaystyle \frac{1}{6}}{t}^3\right]}_{t=0}^{t=2\pi }\\ \\ & ={\displaystyle \frac{(2\pi {)}^3}{6}}-{\displaystyle \frac{0^3}{6}}\\ \\ & ={\displaystyle \frac{8{\pi }^3}{6}}\end{array}$‍

Summary

• Green's theorem can turn tricky line integrals into more straight-forward double integrals.
• To know if Green's theorem will actually make a line integral simpler, ask the following two questions:

• Also, consider if the region encompassed by the curve $C$‍ will be easy to describe with a double integral, or if it has a known area.
• You can compute the area of a region with the following line integral around its counterclockwise-oriented boundary:

  ${\displaystyle {\oint }_C{\displaystyle \frac{1}{2}}(xdy-ydx)}$‍