Math 344: Calculus III

16.4 Green's Theorem

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16.4.1 The Fundamental Theorems

In an integral calculus course one studies the fundamental theorem of calculus for univariant functions.

Fundamental Theorem for Functions of One Variable

If $f:[a,b]\rightarrow\mathbb{R}$ is continuously differentiable on the interval $[a,b]$, then

$$ \require{color} \definecolor{brightblue}{rgb}{.267, .298, .812} \definecolor{darkblue}{rgb}{.08, .18, .28} \definecolor{palepink}{rgb}{1, .73, .8} \definecolor{softmagenta}{rgb}{.99,.34,.86} \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pypx}[2][x]{\dfrac{\partial #2}{\partial #1}} \newcommand{\dydx}[2][x]{\dfrac{d #2}{d #1}} \newcommand{\deltax}[2][x]{\frac{\Delta #2}{\Delta #1}} \newcommand{\rb}[1][r]{\mathbf{#1}} \newcommand{\rt}[1][t]{\rb(#1)} \int_a^b f'(x)\,dx = f(b) - f(a) $$

In order to extend this theorem to two dimensions we must first make some observations.

1. The interval $[a,b]$ is simply connected . One obtains a different result if the integral is computed over disjoint closed intervals.
2. The integrand contains the derivative of our function evaluated at the end points.
3. The end points are the boundary of our interval.
4. The derivative in the integrand must be integrable . We learned in integral calculus that any continuous function is integrable. There are weaker conditions on the integrand that will satisfy integrability and render the fundamental theorem true.

In two dimensions we have a more complicated situation. As in previous discussions we need a simply connected, closed region , and we need one with a boundary that is a simple closed curve whose parameterization is continuously differentiable. As we noticed above it is enough the derivative of the parameterization to be integrable.

Figure 1

The region in figure 1 is

• simply connected , that is it is made up of one connected region with no holes inside of it, and the boundary does not touch or cross itself.

• closed , that is it includes its boundary

• smooth , that is the boundary is differentiable; it has no discontinuities or corners.

• $C^1$, that is any parameterization of the boundary will be a differentiable function of one variable and its derivative will be a continuous function; one calls this continuously differentiable or smooth.

However there are weaker conditions that we can still use.

Figure 2

The region in figure 2

• is simply connected
• is closed
• has a piecewise smooth boundary

A piecewise smooth boundary has a parameterization that is defined by a finite number of smooth sections. An example of such a parameterization would be

$$ f(t) = \left\{\begin{array}{lcl} \langle t, 0 \rangle & \ & 0\le t\lt 4 \\ \langle 4, t-4 \rangle & \ & 4\le t \le 6 \\ \langle 10-t, \sqrt{10-t} \rangle & \ & 6\le t\lt 10 \end{array} \right. $$

What about these regions?

Figure 3

We will extend our fundamental theorem to include these regions at the end of this section. Though these regions are not simply connected, they still have a piecewise smooth boundary.

16.4.2 Hypotheses for Fundamental Theorems

Recall that in our first fundamental theorem we integrate $f'$. This will continue to be true for fundamental theorems. However functions of several variables have partial derivatives. We will consistently require our function of of several variables to have integrable first order partial derivatives .

The statement of this requirement takes many forms

• a function must have continuous first order partial derivatives for every variable
• a function must have a continuous gradient
• a vector function must be conservative
• the potential function of a vector field must have all continuous partials

The hypotheses for the region of integration usually requires the region to be

• simply connected or be made up of a finite number of simply connected disjoint regions or components

As we discussed above, the boundary of our region(s) must be

• $C^1$ or piecewise continuously differentiable
• positively oriented

Positive orientation relates the direction a particle travels along the boundary of a region to the sign of the area of the region. In three-dimensional space there are two possibilities and we chose one, the right hand rule . After all the Latin word for left is sinister .

The right-hand rule determines our choice of orientation . We use it to determine the direction of our cross product. It also determines our orientation for the path of a particle along the boundary.

Start by making the "thumbs up" gesture with your right hand. Next, extend your index (1st) finger and point it in the direction of the tangent vector of the boundary. Now, extend your middle (2nd) finger so that it makes a right angle with your index finger and point it toward the interior of the region. Your right thumb now points in the direction of the cross product. If you used your left hand here, your thumb points in the opposite direction.

Figure 4

We say that the region is on the left of the boundary. Positive orientation is more complicated for regions that are not simply connected. Here we will use a common notation for the boundary of a region. The partial derivative symbol $\partial$ in the figure means boundary . Hence $\partial D$ indicates the boundary of region $D$.

Why does orientation matter?

We learned in the first section that the path integral with respect to arc length

$$ \displaystyle\int_{C} f\,ds = \displaystyle\int_{-C} f\,ds $$
is not oriented. However the integral of a path in a vector field

$$ \displaystyle\int_{\partial D} \mathbf{F}\cdot d\rb $$
does have an orientation. This is due to the fact that the integrand $\mathbf{F}\cdot d\rb$, has a sign that depends on whether the direction of travel $\mathbf{T}$ corresponds to the direction of the vector field, or if the tangent vector points in a direction away from the vector field.

Figure 5

In figure 5 a path through a vector field with several tangent vectors to the path show that

1. when the tangent vector to the path points generally the same direction as the vector field at a point, the dot product is positive
2. when the tangent vector to the path is orthogonal to the vector field at a point, the dot product is zero
3. when the tangent vector to the path points generally in the opposite direction as the vector field a a point, the dot product is negative

The magnitude of the dot product will be affected by the length of the tangent vector, the length of the vector field vector, and the cosine of the angle between them

$$ \mathbf{F}\cdot d\rb = \|\mathbf{F}\|\,\|\rb\|\,\cos(\theta) $$
Remember this is a work integral that computes the work done by the vector field on a particle traveling along the path according to the parameterization.

16.4.3 Green's Theorem

Green's Theorem

Let $C$ be a positively oriented, piecewise smooth, simple closed curve in the plane and $D$ the simply connected region bounded by $C$. If vector field

$$ \rb[F](x,y) = \langle P(x,y), Q(x,y) \rangle $$
such that $P$ and $Q$ have continuous partial derivatives on an open region that contains $D$, then

$$ \displaystyle\int_C \rb[F]\cdot\rb = \displaystyle\int_C \langle P, Q\rangle\,d\rb = \displaystyle\int_C P\,dx + Q\,dy = \displaystyle\iint_D \left(\pypx{Q} - \pypx[y]{P} \right)\,dA $$

If $\rb[F]$ is a conservative vector field and $C$ is any simple closed curve in an open region in which all of the first order partial derivatives are continuous, we have

$$ \pypx[y]{P} = \pypx{Q} $$
This yields

$$ \displaystyle\int_C \rb[F]\cdot\rb = \displaystyle\int_C \langle P, Q\rangle\,d\rb = \displaystyle\int_C P\,dx + Q\,dy = \displaystyle\iint_D \left(\pypx{Q} - \pypx[y]{P} \right)\,dA = 0 $$

16.4.4 Using Green's Theorem

Example 1

Evaluate $\displaystyle\int_C x^6\, + xy^2\,dy$ where $C$ is the rectangle with endpoints $(-1,0)$, $(1,0)$, $(1,1)$ and $(-1,1)$.

Figure 6

As this curve is piecewise differentiable, we would need to compute four (4) separate path integrals and add them. However with Green's theorem we can replace the piecewise boundary with

$$ \begin{align*} \displaystyle\int_C x^6\, + xy^2\,dy &= \displaystyle\iint_D \left(\pypx{Q} - \pypx[y]{P} \right)\,dA \\ \\ &= \displaystyle\int_{-1}^1\displaystyle\int_0^1 \left(y^2 - 0\right)\,dydx = \displaystyle\int_{-1}^1 1\,dx \cdot \displaystyle\int_0^1 y^2\,dy \\ \\ &= \Big[ x \Big]_{-1}^1 \cdot \left[ \frac{y^3}{3} \right]_0^1 = \left[1 - (-1)\right]\cdot\left[ \frac{1}{3} - \frac{0}{3} \right] = \frac{2}{3} \end{align*} $$

Example 1 shows that the double integral was easier to evaluate than the path integral. This is not always the case.

Example 2

Let $D$ be a region bounded by a simple closed path $\partial D = C$ in the $xy$-plane. The coordinates of the centroid are given by

$$ \overline{x} = \frac{1}{A}\displaystyle\iint_D x\,dA,\qquad\overline{y} = \frac{1}{A}\displaystyle\iint_D y\,dA $$
where $A$ is the area of $D$. To compute $\overline{x}$ using Green's Theorem we require

$$ \begin{align*} \overline{x} &= \frac{1}{A}\displaystyle\iint_D x\,dA = \displaystyle\iint_{D} \left( \pypx{Q} - \pypx[y]{P} \right)\,dA \end{align*} $$
If we want to use Green's Theorem we need functions $P$ and $Q$ so that

$$ \begin{align*} \pypx{Q} - \pypx[y]{P} &= x \\ P &= 0 \end{align*} $$
We consider
$$ Q = \displaystyle\int \pypx{Q}\,dx = \displaystyle\int x\,dx = \frac{x^2}{2} + C $$
Setting $C = 0$ one obtains

$$ \overline{x} = \frac{1}{A}\displaystyle\iint_D x\,dA = \frac{1}{A}\displaystyle\int_{\partial D} 0\,dx + \frac{x^2}{2}\,dy = \frac{1}{2A}\displaystyle\int_C x^2\,dy $$
In order to compute $\overline{y}$ using Green's theorem we need

$$ \begin{align*} \overline{y} = \frac{1}{A}\displaystyle\iint_D y\,dA = \displaystyle\iint_D \left( \pypx{Q} - \pypx[y]{P} \right)\,dA \end{align*} $$
That is

$$ \begin{align*} \pypx{Q} - \pypx[y]{P} &= y \\ Q &= 0 \end{align*} $$
Hence $P = \displaystyle\int \pypx[y]{P}\,dy = \displaystyle\int -y\,dy = -\frac{y^2}{2} + C$. Letting $C = 0$ yields

$$ \overline{y} = \frac{1}{A}\displaystyle\iint_D y\,dA = \displaystyle\int_{\partial D} -\frac{y^2}{2}\,dx + 0\,dx = -\frac{1}{2A}\displaystyle\int_C = y^2\,dx $$
Thus

$$ \overline{x} = \frac{1}{2A}\displaystyle\int_C x^2\,dy,\qquad\overline{y} = -\frac{1}{2A}\displaystyle\int_C = y^2\,dx $$

Exercise 1

Find the centroid of a semi-circular region with radius $a$.

Figure 7

Check your work

Computing the integrals using Green's Theorem yields
$$ \begin{align*} A &= \frac{1}{2} \pi\,a^2 = \frac{a^2}{2}\pi \\ \\ \overline{x} &= \frac{1}{2A}\displaystyle\int_C x^2\,dy \\ \\ &= \frac{1}{a^2\pi}\left(\,\displaystyle\int_{\text{half circle}} x^2\,dy + \displaystyle\int_{-a}^a x^2\,dy\,\right) \\ \end{align*} $$
In the first integral we can use the parameterization
$$ \langle x(t), y(t) \rangle = \langle a\cos(t), a\sin(t) \rangle,\quad 0\le t\le\pi $$
Thus $dy = y'(t)\,dt = a\cos(t)\,dt$ In the second integral we can use the parameterization
$$ \langle x(t), y(t) \rangle = \langle -a,0\rangle(1-t) + \langle a,0 \rangle t = a\langle 2t-1, 0 \rangle,\qquad 0\le t\le1 $$
In this case $dy = y'(t)\,dt = 0\,dt$. Hence
$$ \begin{align*} \overline{x} &= \frac{1}{a^2\pi}\left(\,\displaystyle\int_0^{\pi} a^2\cos^2(t)\,a\cos(t)\,dt + \displaystyle\int_{-a}^a a^2(2t-1)^2\,0\,dt\,\right) \\ \\ &= \frac{a}{\pi}\displaystyle\int_0^{\pi}\left(1-\sin^2(t)\right)\,\cos(t)\,dt \\ \\ &= \frac{a}{\pi}\displaystyle\int \left(1-u^2\right)\,du = \frac{a}{\pi}\,\left[ 1 - \frac{\sin^3(t)}{3}\,\right]_0^{\pi} = 0\\ \\ \overline{y} &= \frac{1}{a^2\pi}\left(\,\displaystyle\int_0^{\pi} a^2\sin^2(t)\,(-a\sin(t))\,dt + \displaystyle\int_{-1}^a a^2(2t-1)\,0\,dt\,\right) \\ \\ &= \frac{a}{\pi}\displaystyle\int_0^{\pi} \left(1-\cos^2(t)\right)\,(-\sin(t))\,dt \\ \\ &= \frac{a}{\pi}\displaystyle\int \left(1 - u^2\right)\,du = \frac{a}{\pi}\,\left[\,1 - \frac{\cos^3(t)}{3}\,\right]_0^{\pi} = \frac{a}{\pi}\,\left[ 1 - \frac{-1}{3} - \left(\,1 - \frac{1}{3}\,\right)\,\right] = \frac{2a}{3\pi} \\ \end{align*} $$
The centroid of the semi-circular region is given by $\left(0,\frac{2a}{3\pi}\right)$.

16.4.6 Extending Green's Theorem

Figure 8

We can extend Green's Theorem to apply to regions with holes by dividing the region into simply connected regions.

Figure 9

Notice that the boundary curves are oriented so that we uphold the right-hand rule; the region $D$ is always on the left as the curve is traversed. This means that while positive direction for curve $C_1$ on the outer boundary of region $D$ is counter-clockwise, the positive direction for curve $C_2$ on the inner boundary of region $D$ is clockwise. Since the boundaries for simply connected regions $D'$ and $D''$ are piecewise differentiable and the path traverses the boundary in opposite directions on the lines we introduced we obtain

$$ \begin{align*} \displaystyle\iint_D \left(\pypx{Q} - \pypx[y]{P}\right)\,dA &= \displaystyle\iint_{D'} \left(\pypx{Q} - \pypx[y]{P}\right)\,dA + \displaystyle\iint_{D''} \left(\pypx{Q} - \pypx[y]{P}\right)\,dA \\ \\ &= \displaystyle\int_{\partial D'} P\,dx + Q\,dy + \displaystyle\int_{\partial D''} P\,dx + Q\,dy \\ \\ &= \displaystyle\int_{C_1} P\,dx + Q\,dy + \displaystyle\int_{C_2} P\,dx + Q\,dy \\ \\ &= \displaystyle\int_{C_1\cup C_2} P\,dx + Q\,dy \\ \\ &= \displaystyle\int_{C} P\,dx + Q\,dy \end{align*} $$

16.4.7 Using Green's Theorem to Evaluate General Regions

One of the values of Green's theorem is that is allows us to evaluate the flux (we will discuss this in more detail in a later section) of a vector field through a boundary. In the next example, we consider a general region that encloses the origin and want to compute the integral

$$ \int_C \mathbf{F}\cdot d\mathbf{r}. $$
It is possible to do this by "transferring" flux to a simpler boundary.

Example 3

Figure 10

If

$$ \rb[F](x,y) = \frac{-y\,\ihat + x\,\jhat}{x^2 + y^2} $$
show that

$$ \displaystyle\int_C \rb[F]\cdot\rb = 2\pi $$
for any choice of boundary $C$.

Solution

This path is irregular and the integral is likely difficult to compute. Let us introduce a counter-clockwise oriented circle $C'$ of radius $a$ centered at the origin. We will choose $a$ small enough so that the circle and its boundary $C'$ lies in the interior of $D$.

Figure 11

In our previous example we determined that the positively oriented boundary for region $D$ with the disk removed is $C\cup(-C')$. Thus the general version of Green's Theorem yields

$$ \begin{align*} \displaystyle\int_C P\,dx + Q\,dy\ +\ \displaystyle\int_{-C'} P\,dx + Q\,dy &= \displaystyle\iint_D \left(\,\pypx{Q} - \pypx[y]{P}\,\right)\,dA \\ \\ &= \displaystyle\iint_D \left[\,\frac{y^2-x^2}{\left(x^2+y^2\right)^2} - \frac{y^2-x^2}{\left(x^2+y^2\right)^2}\,\right]\,dA = 0 \end{align*} $$
This gives us

$$ \displaystyle\int_C P\,dx + Q\,dy = -\displaystyle\int_{-C'} P\,dx + Q\,dy = \displaystyle\int_{C'} P\,dx + Q\,dy $$
Using the parameterization for the circle of radius $a$, $\rb(t) = a\cos(t)\ihat + a\sin(t)\jhat$ we obtain

$$ \begin{align*} \displaystyle\int_C \rb[F]\cdot d\rb &= \displaystyle\int_{C'} \rb[F]\cdot d\rb = \displaystyle\int_0^{2\pi} \rb[F](\rb(t))\cdot\rb'(t)\,dt \\ \\ &= \displaystyle\int_0^{2\pi} \frac{(-a\sin(t))(-a\sin(t)) + (a\cos(t))(a\cos(t))}{a^2\cos^2(t) + a^2\sin^2(t)}\,dt \\ \\ &= \displaystyle\int_0^{2\pi}\,dt = 2\pi \end{align*} $$
This derivation shows us that for every piecewise simple closed curve containing the origin in the enclosed domain,

$$ \displaystyle\int_C \rb[F]\cdot\rb = 2\pi $$ when

$$ \rb[F](x,y) = \frac{-y\,\ihat + x\,\jhat}{x^2 + y^2}. $$

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