Green's Theorem

Section11.5Green's Theorem

Objectives

In this section you will...

See applications of double integrals: Circulation Density and Flux Density
Learn to use Green's Theorem to compute circulation/work and flux

Now that we have double integrals, it's time to make some of our circulation and flux exercises from the line integral section get extremely simple. We'll start by defining the circulation density and flux density for a vector field \(\vec F(x,y)=\left\lt M,N\right>\) in the plane.

Definition11.5.1Circulation Density and Flux Density (Divergence)

Let \(\vec F(x,y)=\left\lt M,N\right>\) be a continuously differentiable vector field. At the point \((x,y)\) in the plane, create a circle \(C_a\) of radius \(a\) centered at \((x,y)\text{,}\) where the area inside of \(C_a\) is \(A_a=\pi a^2\text{.}\) The quotient \(\ds \frac{1}{A_a}\oint_{C_a} \vec F \cdot \vec T ds\) is a circulation per area. The quotient \(\ds \frac{1}{A_a}\oint_{C_a} \vec F \cdot \vec n ds\) is a flux per area.

The circulation density of \(\vec F\) at \((x,y)\) we define to be
\begin{equation*} \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=N_x-M_y = \lim_{a\to 0} \frac{1}{A_a}\oint_{C_a} \vec F \cdot d\vec r = \lim_{a\to 0} \frac{1}{A_a}\oint_{C_a} Mdx+Ndy. \end{equation*}
We will not prove that the partial derivative expressions \(N_x-M_y\) and \(M_x+N_y\) are actually equal to the limits given here. That is best left to an advanced course.
The divergence, or flux density, of \(\vec F\) at \((x,y)\) we define to be
\begin{equation*} \frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}=M_x+N_y=\lim_{a\to 0} \frac{1}{A_a}\oint_{C_a} \vec F \cdot \vec n ds = \lim_{a\to 0} \frac{1}{A_a}\oint_{C_a} Mdy-Ndx. \end{equation*}

In the definitions above, we could have replaced the circle \(C_a\) with a square of side lengths \(a\) centered at \((x,y)\) with interior area \(A_a\text{.}\) Alternately, we could have chosen any collection of curves \(C_a\) which “shrink nicely” to \((x,y)\) and have area \(A_a\) inside. Regardless of which curves you chose, it can be shown that

\begin{equation*} N_x-M_y=\lim_{a\to 0} \frac{1}{A_a}\oint_{C_a} \vec F \cdot \vec T ds \text{ and } M_x+N_y=\lim_{a\to 0} \frac{1}{A_a}\oint_{C_a} \vec F \cdot \vec n ds. \end{equation*}

To understand what the circulation and flux density mean in a physical sense, think of \(\vec F\) as the velocity field of some gas.

The circulation density tells us the rate at which the vector field \(\vec F\) causes objects to rotate around points. If circulation density is positive, then particles near \((x,y)\) would tend to circulate around the point in a counterclockwise direction. The larger the circulation density, the faster the rotation. The velocity field of a gas could have some regions where the gas is swirling clockwise, and some regions where the gas is swirling counterclockwise.
The divergence, or flux density, tells us the rate at which the vector field causes object to either flee from \((x,y)\) or come towards \((x,y)\text{.}\) For the velocity field of a gas, the gas is expanding at points where the divergence is positive, and contracting at points where the divergence is negative.

We are now ready to state Green's Theorem. Ask me in class to give an informal proof as to why this theorem is valid.

Theorem11.5.2Green's Theorem

Let \(\vec F(x,y)=(M,N)\) be a continuously differentiable vector field, which is defined on an open region in the plane that contains a simple closed curve \(C\) and the region \(R\) inside the curve \(C\text{.}\) Then we can compute the counterclockwise circulation of \(\vec F\) along \(C\text{,}\) and the outward flux of \(\vec F\) across \(C\) by using the double integrals

\begin{equation*} \oint_{C} \vec F \cdot \vec T ds=\iint_R N_x-M_y dA \text{ and } \oint_{C} \vec F \cdot \vec n ds=\iint_R M_x+N_y dA. \end{equation*}

Subsection11.5.1Practice with Green's Theorem

Let's now use this theorem to rapidly find circulation (work) and flux.

Exercise11.5.1

Consider a rectangle \(C\) where the boundary of the rectangle are \(2\leq x\leq 7\) and \(0\leq y\leq 3\text{.}\) The rectangle is subject to a vector field \(\vec F=(2x+3y,4x+5y)\text{.}\)

(a)

Compute \(N_x-M_y\)

(b)

Compute \(M_x+N_y\text{.}\)

(c)

Find the circulation of \(\vec F\) along \(C\)

Hint

This should just reduce to a statement about area.

(d)

Find the flux of \(\vec F\) along \(C\)

Hint

This should just reduce to a statement about area.

Note: If you tried doing this without Green's theorem, you would have to parametrize 4 line segments, compute 4 integrals, and then sum the results.

Exercise11.5.2

Consider the vector field \(\vec F=(x^2+y^2,3x+5y)\) and \(C\text{,}\) the circle \((x-3)^2+(y+1)^2=4\) (oriented counterclockwise).

(a)

Start by computing \(N_x-M_y\) and \(M_x+N_y\text{.}\)

(b)

Now find both the circulation and flux of \(\vec F\) along \(C\text{.}\)

Hint

You should be able to reduce the integrals to facts about the area and centroid.

Exercise11.5.3Understanding Question

Explain why you need or can use a centroid to reduce the integrals. This will help you in the next exercise.

Exercise11.5.4

Repeat the previous exercise, but change the curve \(C\) to the boundary of the triangular region \(R\) with vertexes at \((0,0)\text{,}\) \((3,0)\text{,}\) and \((3,6)\text{.}\) You can complete this exercise without having to set up the bounds on any integrals, if you reduce the integrals to facts about area and centroids. You are welcome to look up the centroid of a triangular region without computing it.

Subsection11.5.2Computational Practice

These are provided to help you achieve better skills in basic computational answers.