Calculus III 15.04 Green’s Theorem

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15.04 Green's Theorem[1]

  • Use Green’s Theorem to evaluate a line integral.
  • Green's Theorem variants.

Green's Theorem

Figure 15.4.1

Green's Theorem named after the English mathematician George Green (1793–1841).[2] This theorem states that the value for a double integral over a simply connected plane region \(R\) is determined by the value for a line integral around \(R\)'s boundary.

A curve \(C\) given by \(\textbf{r} = x(t)\textbf{i}+y(t)\textbf{j}\), where \(a \leqslant t \leqslant b\), is simple when it does not cross itself -- that is, \(\textbf{r}(c) \ne \textbf{r}(d)\) for all \(c\) and \(d\) in the open interval \((a,b)\). A connected plane region \(R\) is simply connected when every simple closed curve in \(R\) encloses only points that are in \(R\), as shown in Figure 15.4.1. Informally, a simply connected region is coterminous with no separate parts or holes.

Theorem 15.4.1 Green's Theorem

\(R\) is vertically simple.
Figure 15.4.2

Let \(R\) be a simply connected region with a piecewise smooth boundary \(C\), oriented counterclockwise, \(C\) is traversed once so that the region \(R\) always lies to the left. If \(M\) and \(N\) have continuous first partial derivatives in an open region containing then \(R\), then

$$\int_{C} M \: dx + N \: dy = \int_{R} \int \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) \: dA.$$

Proof This proof applies to a region that is both vertically and horizontally simple, as shown in Figure 15.4.2.

$$\int_{C} M \: dx $$ $$=\int_{C_{1}} M \: dx + \int_{C_{2}} M \: dx $$
$$= \int_{a}^{b} M(x,f_{1}(x)) \: dx + \int_{b}^{a} M(x,f_{2}(x)) \: dx $$
$$= \int_{a}^{b} [M(x,f_{1}(x))- M(x,f_{2}(x))] \: dx$$

The \(\partial M\) produces

$$\int_{R} \int \frac{\partial M}{\partial y} \: dA $$ $$=\int_{a}^{b} \int_{f_{1}(x)}^{f_{2}(x)} \frac{\partial M}{\partial y} \: dy \: dx $$
$$=\left. \int_{a}^{b} M(x,y) \right]_{f_{1}(x)}^{f_{2}(x)} \: dx $$
$$= \int_{a}^{b} [M(x,f_{2}(x))- M(x,f_{1}(x))] \: dx.$$

Consequently,

$$ \int_{C} M \: dx = - \int_{R} \int \frac{\partial M}{\partial y} \: dA .$$

The same method will show that

$$\int_{C} N \: dy = \int_{R} \int \frac{\partial N}{\partial x} \: dA.$$

Adding both equals the theorem's concluding statement.

Square Half.jpg

An integral sign with a circle is sometimes used to indicate a line integral around a simple closed curve, as shown below. To indicate the boundary orientation, an arrow can be used. For instance, in the second integral, the arrow indicates the boundary is oriented clockwise. The third integral indicates the boundary is oriented counterclockwise.

$$ \textbf{1. } \oint_{C} M \:dx + N \: dy \:\:\:\: \textbf{2. } \rlap{\mkern5.5mu\circlearrowright}\int_{C} M \:dx + N \: dy \:\:\:\: \textbf{3. } \rlap{\mkern5.5mu\circlearrowleft}\int_{C} M \:dx + N \: dy $$

Example 15.4.1 Using Green's Theorem

\(C\) is simple and closed. The region \(R\) always lies to \(C\)'s left.
Figure 15.4.3

Use Green’s Theorem to evaluate the line integral

$$\int_{C} y^{3} \: dx + (x^{3}+3xy^{2}) \: dy $$

where \(C\) is the path from \((0,0)\) to \((1,1)\) along the graph for \(y=x^{3}\) and from \((1,1)\) to \((0,0)\) along the graph for \(y=x\), as shown in Figure 15.4.3.
Solution Because \(M=y^{3}\) and \(N=x^{3}+3xy^{2}\), it follows that

$$ \frac{\partial N}{\partial x} = 3x^{2}+3y^{2} \text{ and } \frac{\partial M}{\partial y} =3y^{2} .$$

Applying Green's Theorem produces

$$\int_{C} y^{3} \: dx + (x^{3}+3xy^{2}) \: dy $$ $$= \int_{R} \int \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) \: dA $$
$$=\int_{0}^{1} \int_{x^{3}}^{x} [(3x^{2}+3y^{2})- 3y^{2}] \:dx \: dy $$
$$= \int_{0}^{1} \int_{x^{3}}^{x} 3x^{2} \:dx \: dy $$
$$=\left. \int_{0}^{1} 3x^{2}y \right]_{x^{3}}^{x} \:dx$$
$$= \int_{0}^{1} (3x^{3}-3x^{5}) \:dx $$
$$= \left[\frac{3x^{4}}{4}-\frac{x^{6}}{2} \right]_{0}^{1} = \frac{1}{4}.$$
Square Half.jpg

Green’s Theorem cannot be applied to every line integral. Among other restrictions stated in Theorem 15.4.1, the curve must be simple and closed. When Green’s Theorem does apply, however, it can save time. To see this, try using the techniques described in Section 15.2 to evaluate the line integral in Example 15.4.1. To do this, you would need to write the line integral as

$$\int_{C} y^{3} \: dx + (x^{3}+3xy^{2}) \: dy$$ $$=\int_{C_{1}} y^{3} \: dx + (x^{3}+3xy^{2}) \: dy + \int_{C_{2}} y^{3} \: dx + (x^{3}+3xy^{2}) \: dy$$

where \(C_{1}\) is the cubic path given by

\(\textbf{r}(t)=t\textbf{i}+t^{3}\textbf{j}\)

from \(t=0\) to \(t=1\), and \(C_{2}\) is the line segment given by

\(\textbf{r}(t)=(1-t)\textbf{i}+(1-t)\textbf{j}\)

from \(t=0\) to \(t=1\).

Example 15.4.2 Using Green's Theorem to Calculate Work

Figure 15.4.4

While subject to the force

\(\textbf{F}(x,y)=y^{3}\textbf{i}+(x^{3}+3xy^{2})\textbf{j}\)

a particle travels once around the circle with radius 3, as shown in Figure 15.4.4. Use Green's Theorem to find the work done by \(\textbf{F}\).
Solution Example 15.4.1 proved that by Green's Theorem

$$\int_{C} y^{3} \: dx + (x^{3}+3xy^{2}) \: dy = \int_{R} \int 3x^{2} \:dA.$$

In polar coordinates, using \(x=\cos \theta\) and \(dA=r \:dr \: d\theta\), the work done is

$$W$$ $$= \int_{R} \int 3x^{2} \:dA$$
$$= \int_{0}^{2\pi} \int_{0}^{3} 3(r \cos \theta)^{2} r \: dr \: d\theta$$
$$= 3 \int_{0}^{2\pi} \int_{0}^{3} r^{3} \cos^{2} \theta \: dr \: d\theta $$
$$= \left. 3 \int_{0}^{2\pi} \frac{r^{4}}{4} \cos^{2} \theta \right]_{0}^{3} \: d \theta $$
$$=3 \int_{0}^{2\pi} \frac{81}{4} \cos^{2} \theta \: d \theta $$
$$=\frac{243}{8} \int_{0}^{2\pi} (1+\cos 2\theta) \: d \theta $$
$$=\frac{243}{8} \left[ \theta+\frac{\sin 2 \theta}{2} \right]_{0}^{2\pi}= \frac{243\pi}{4}. $$
Square Half.jpg

When evaluating line integrals over closed curves, remember that for conservative vector fields the value for the line integral is zero. This is easily seen from Green’s Theorem:

$$\int_{C} M \: dx + N \: dy = \int_{R} \int \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) \: dA = 0.$$

Example 15.4.3 Using Green's Theorem and Conservative Vector Fields

\(C\) is closed
Figure 15.4.5

Evaluate the line integral

$$\int_{C} y^{3} \: dx + 3xy^{2} \: dy $$

where \(C\) is the path shown in Figure 15.4.5.
Solution From this line integral, \(M=y^{3}\) and \(N=3xy^{2}\). Therefore,

$$\frac{\partial M}{\partial y}=3y^{2} \text{ and } \frac{\partial N}{\partial x}=3y^{2}.$$

This means the vector field \(\textbf{F}=M\textbf{i}+N\textbf{j}\) is conservative. Because \(C\) is closed the conclusion is

$$ \int_{C}y^{3} \: dx + 3xy^{2} \: dy =0.$$

Example 15.4.4 Using Green's Theorem in Polar Coordinates

\(C\) is piecewise smooth.
Figure 15.4.6

Evaluate

$$\int_{C} (\arctan x + y^{2}) \: dx + (e^{y}-x^{2}) \: dy$$

where \(C\) is the path enclosing the annular region shown in Figure 15.4.6.
Solution In polar coordinates, \(R\) is given by \(1 \leqslant r \leqslant 3\) for \(0 \leqslant \theta \leqslant \pi\). Moreover,

$$ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} $$ \(=-2x-2y \)
\(= -2(r \cos \theta + r \sin \theta). \)

Applying Green's Theorem produces

$$\int_{C} (\arctan x + y^{2}) \: dx + (e^{y}-x^{2}) \: dy$$ $$=\int_{R} \int -2(x+y) \: dA $$
$$= \int_{0}^{\pi} \int_{1}^{3} -2r(\cos \theta + \sin \theta)r \: dr \: d\theta $$
$$= \left. \int_{0}^{\pi} -2r(\cos \theta + \sin \theta) \frac{r^{3}}{3} \right]_{1}^{3} \: d\theta$$
$$= \int_{0}^{\pi} \left( -\frac{52}{3}\right) (\cos \theta + \sin \theta) \: d\theta$$
$$= -\frac{52}{3} \left[ \vphantom{\frac{1}{2}} \sin \theta - \cos \theta\right]_{0}^{\pi} = -\frac{104}{3}.$$

Theorem 15.4.2 Line Integral for Area

If \(R\) is a plane region bounded by a piecewise smooth simple closed curve \(C\), oriented counterclockwise, then \(R\)'s area is given by

$$A=\frac{1}{2} \int_{C} x \: dy - y \: dx.$$

Proof Let

$$M=-\frac{y}{2} \text{ and } N=\frac{x}{2},$$

applying Green's Theorem produces,

$$ \int_{C} M \: dx + N \: dy$$ $$= \int_{R} \int \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \:dA $$
$$= \int_{R} \int 1 \: dA$$
= area for region \(R\).

Example 15.4.5 Finding Area by a Line Integral

Figure 15.4.7

Use a line integral to find the area for the ellipse

$$\left( \frac{x^{2}}{a^{2}}\right) +\left( \frac{y^{2}}{b^{2}}\right) = 1,$$

as shown in Figure 15.4.7.
Solution The counterclockwise orientation can be expressed by letting \(x= a \cos t\) and \(y= b \sin t\), \( 0 \leqslant t \leqslant 2\pi\). This makes the area

$$ A$$ $$= \frac{1}{2} \int_{0}^{2\pi} [(a \cos t )(b \cos t) \:dt - (b \sin t)(-a \sin t) \: dt] $$
$$= \frac{ab}{2} \int_{0}^{2\pi} (\cos^{2}t + \sin^{2} t) \: dt$$
$$= \frac{ab}{2}\left[ \vphantom{\frac{1}{2}} t \right]_{0}^{2\pi}=\pi a b.$$

Example 15.4.6 Green’s Theorem Extended to a Region with a Hole

Figure 15.4.8

Figure 15.4.9

Let \(R\) be the region inside the ellipse

$$\left( \frac{x^{2}}{9}\right) +\left( \frac{y^{2}}{4}\right) = 1$$

and outside the circle \(x^{2}+y^{2}=1\). Evaluate the line integral

$$ \int_{C} 2xy \: dx + (x^{2}+2x)\:dy $$

where \(C=C_{1} +C_{2}\) is the boundary for \(R\), as shown in Figure 15.4.8.
Solution Green’s Theorem can be extended to cover some regions that are not simply connected. In this case introduce the line segments \(C_{3}\) and \(C_{4}\), as shown in Figure 15.4.9. Note that because the curves \(C_{3}\) and \(C_{4}\) have opposite orientations, the line integrals over them cancel. Applying Green's Theorem to the region \(R\) using the boundary \(C_{1}+C_{2}+C_{4}+C_{4}\) to produce

$$\int_{C} 2xy \: dx + (x^{2}+2x)\:dy$$ $$ \int_{R} \int \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \:dA = $$
$$=\int_{R} \int (2x+2-2x) \: dA $$
$$=2 \int_{R} \int \:dA $$
$$=2(R\text{area for }R) $$
$$=2(\pi ab - \pi r^{2}) $$
$$=2[\pi(3)(2)-\pi (1^{2})]= 10 \pi. $$
Square Half.jpg

The Theorem 15.1.1 proof was for one direction. Green's Theorem can be used to prove it for the other direction. Let \(\textbf{F}(x,y)= M \textbf{i} + N\textbf{j}\) be defined on an open disk \(R\). Show that if \(M\) and \(N\) have continuous first partial derivatives and \(\frac{\partial M}{\partial y} =\frac{\partial N}{\partial x}\), then \(\textbf{F}\) is conservative. Let \(C\) be a closed path forming the boundary for a connected region lying in \(R\). Then, using the fact that \(\frac{\partial M}{\partial y} =\frac{\partial N}{\partial x}\), apply Green's Theorem to conclude that

$$\int_{C} \textbf{F} \cdot \: d\textbf{r}=\int_{C} M \: dx + N \: dy = \int_{R} \int \left(\frac{\partial N}{\partial x}- \frac{\partial M}{\partial y} \right)\:dA=0.$$

This is equivalent to showning that \(\textbf{F} \) is conservative in Theorem 15.3.3.

Green's Theorem Variants

Figure 15.4.10

A vector field \( \textbf{F}\) in the plane can be expressed as

\( \textbf{F}(x,y,z)=M\textbf{i}+N\textbf{j}+O\textbf{k}\)

so that the curl for \( \textbf{F}\), as described in Section 15.1, is given by

$$ \text{curl } \textbf{F}= \nabla \times \textbf{F}$$ $$=\begin{vmatrix}\textbf{i} & \textbf{j} &\textbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ M & N & O\end{vmatrix}$$
$$=- \frac{\partial N}{\partial z} \textbf{i} + \frac{\partial M}{\partial z} \textbf{j}+ \left( \frac{\partial N}{\partial x} -\frac{\partial M}{\partial y} \right) \textbf{k}.$$

Consequently,

\((\text{curl } \textbf{F}) \cdot \textbf{k}\) $$=\left[ - \frac{\partial N}{\partial z} \textbf{i} + \frac{\partial M}{\partial z} \textbf{j}+ \left( \frac{\partial N}{\partial x} -\frac{\partial M}{\partial y} \right) \textbf{k}\right] \cdot \textbf{k} $$
$$= \frac{\partial N}{\partial x} -\frac{\partial M}{\partial y}.$$

With appropriate conditions on \(\textbf{F}\), \(C\), and \(R\), the vector form for Green's Theorem is

$$ \int_{C} \textbf{F} \cdot \: d\textbf{r}$$ $$=\int_{R} \int \left( \frac{\partial N}{\partial x} -\frac{\partial M}{\partial y} \right) \: dA $$
$$=\int_{R} \int (\text{curl } \textbf{F}) \cdot \textbf{k} \: dA. $$     Vector field variant

Applying this variant to vectors in three-dimensions produces Stoke's Theorem, which is discussed in Section 15.8.

The outward unit normal vector \(\textbf{N}\) assumes the same conditions for \(\textbf{F}\), \(C\), and \(R\). Using the arc length parameter \(s\) for \(C\),

\(\textbf{r}=x(s)\textbf{i}+y(s)\textbf{j}\).

A unit tangent vector \(\textbf{T}\) to curve \(C\) is given by

\(\textbf{r}^{\prime}(s)=x^{\prime}(s)\textbf{i}+y^{\prime}(s)\textbf{j}\).

Then \(\textbf{N}\) can be expressed as

\(\textbf{N}=y^{\prime}(s)\textbf{i}+x^{\prime}(s)\textbf{j}.\)

Applying Green's Theorem to

\(\textbf{F}(x,y)=M\textbf{i}+N\textbf{j},\)

produces

$$\int_{C} \textbf{F} \cdot \textbf{N} \:ds $$ $$=\int_{a}^{b} (M\textbf{i}+N\textbf{j}) \cdot (y^{\prime}(s)\textbf{i}+x^{\prime}(s)\textbf{j}) \: ds $$
$$= \int_{a}^{b} \left(M \frac{dy}{ds}-N \frac{dx}{ds} \right) \: ds$$
$$= \int_{C} M \: dy - N \: dx $$
$$= \int_{C} -N \: dx + M \: dy$$
$$= \int_{R} \int \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y}\right) \: dA$$     Green's Theorem
$$=\int_{R} \int \text{div }\textbf{F} \: dA. $$     Outward normal unit vector variant

The Divergence Theorem extends this to three-dimensions and is discussed in Section 15.7 and Section 15.8.

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Parent Article: Calculus III 15 Vector Analysis