OBJECTIVE

When you finish this section, you should be able to:

1. Use Green’s Theorem to find a line integral (p. 1009)
2. Use Green’s Theorem to find area (p. 1010)
3. Use Green’s Theorem with Multiply-Connected Regions (p. 1011)

ORIGINS

George Green (1793–1841) was a miller and self-taught mathematician who also worked with applications in electricity, magnetism, and fluid flow.

RECALL

A region $R$ is closed if it contains all its boundary points.

THEOREM Green’s Theorem

Let $C$ be a piecewise-smooth, simple closed curve with a counter clockwise orientation and let $R$ be the simply connected, closed region consisting of $C$ and its interior. Let the functions $P$ and $Q$ have continuous first-order partial derivatives on some open region that contains $R$. Then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \oint_{C}(P\,dx+Q\,dy)=\iint\limits_{R}\left( \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right) dx\,dy}}$$

where the line integral is taken around $C$ in the counterclockwise direction.

NEED TO REVIEW?

$x$-simple and $y$-simple regions are discussed in Section 14.2, pp. 913 and 915.

Proof

It is sufficient to show that $$\oint_{C}P(x,y)\,dx=-\iint\limits_{R}\dfrac{\partial P}{\partial y}\,dx\,dy \qquad \hbox{and}\qquad \oint_{C}Q(x,y)\,dy=\iint\limits_{R}\dfrac{\partial Q}{\partial x}\,dx\,dy$$

since the result follows from adding these two equations.

Suppose the region $R$ is both $x$-simple and $y$-simple. Then since it is $x$-simple as shown in Figure 32, the boundary of $R$ consists of two smooth curves, $C_{1}$: $y=g_{1}(x)$ and $C_{2}$: $y=g_{2}(x)$, where $g_{1}(x)\leq g_{2}(x)$, $a\leq x\leq b$.

Figure 32

To show $\oint_{C}P(x,y)\,dx=-\iint\limits_{R}\dfrac{\partial P}{\partial y} \,dx\,dy$, we begin with the double integral. Since $R$ is $x$-simple, by Fubini’s Theorem, we have $$\begin{eqnarray*} -\iint\limits_{R}\dfrac{\partial P}{\partial y}\,dx\,dy& =&-\int_{a}^{b}\left[\int_{g_{1}(x)}^{g_{2}(x)}\dfrac{\partial P}{\partial y}\,dy\right] dx=-\int_{a}^{b}\big[ P(x,y)\big] _{g_{1}(x)}^{g_{2}(x)}\,dx \\ & =&-\int_{a}^{b}[ P(x,g_{2}(x))-P(x,g_{1}(x))]\,dx=\int_{a}^{b}P(x,g_{1}(x))\,dx+\int_{b}^{a}P(x,g_{2}(x))\,dx \\ &=&\int_{C_{1}}P(x,y)\,dx+\int_{C_{2}}P(x,y)\,dx =\oint_{C}P(x,y)\,dx \end{eqnarray*}$$

To show $\oint_{C}Q(x,y)\,dy=\iint\limits_{R}\dfrac{\partial Q}{\partial x} \,dx\,dy,$ use the fact that the region $R$ is also $y$-simple. The derivation is left as an exercise (Problem 49).

NEED TO REVIEW?

Fubini’s Theorems are discussed in Section 14.2, pp. 913 and 915.

Use Green’s Theorem to find the line integral $$\begin{equation*} \oint_{C}[ ( -2xy+y^{2}) \,dx+x^{2}\,dy] \end{equation*}$$

where $C$ is the boundary of the region $R$ enclosed by $y=4x$ and $y=2x^{2}$.

Figure 33

Solution Figure 33 illustrates the curve $C$ and the region $R$. $C$ is a piecewise-smooth closed curve and $R$ is both simply connected and closed. We let $$\begin{equation*} P(x,y)=-2xy+y^{2}\qquad \hbox{and}\qquad Q(x,y)=x^{2} \end{equation*}$$

Since $P$ and $Q$ have continuous first-order partial derivatives in $R,$ we use Green’s Theorem. Then $$\begin{eqnarray*} \oint_{C}(P\,dx+Q\,dy)& =&\iint\limits_{R}\left( \dfrac{\partial Q}{\partial x }-\dfrac{\partial P}{\partial y}\right) \,dx\,dy \\ \oint_{C}[(-2xy+y^{2})\,dx+x^{2}\,dy]& =&\iint\limits_{R}\left[ \dfrac{ \partial }{\partial x}(x^{2})-\dfrac{\partial }{\partial y}(-2xy+y^{2})\right] \,dx\,dy\\ &=&\iint\limits_{R}(4x-2y)\,dx\,dy \underset{\underset{\color{#0066A7}{R\hbox{ is } x\hbox{-simple}}}{\color{#0066A7}{\uparrow}}} {=} \int_{0}^{2} \int_{2x^{2}}^{4x}(4x-2y)\,dy\,dx \\ & =&\int_{0}^{2}\big[ 4xy-y^{2}\big] _{2x^{2}}^{4x}\,dx = \int_{0}^{2}[(16x^{2}-16x^{2})-(8x^{3}-4x^{4})]\,dx \\ & =&\left[ -2x^{4}+\dfrac{4}{5}x^{5}\right] _{0}^{2}=-32+\dfrac{128}{5}=\dfrac{ -32}{5} \end{eqnarray*}$$

NOW WORK

Problem 9.

Use Green’s Theorem to find the line integral $$\begin{equation*} \oint_{C}[ ( e^{-x^{2}}+y^{2}) \,dx+( \ln y-x^{2}) \,dy] \end{equation*}$$

Figure 34

where $C$ is the square illustrated in Figure 34.

Solution  Recall that $\int e^{-x^{2}}dx$ cannot be expressed in terms of elementary functions, so finding the integral directly is impossible. Since both the curve $C$ and the region $R$ enclosed by $C$ satisfy the conditions of Green’s Theorem, we use it to find the line integral. If $P(x,y)=e^{-x^{2}}+y^{2}$ and $Q(x,y)=\ln y-x^{2}$ then $\dfrac{\partial P}{ \partial y}=2y$ and $\dfrac{\partial Q}{\partial x}=-2x$. Since $P$ and $Q$ have continuous first-order derivatives in $R$, by Green’s Theorem, We have $$\begin{eqnarray*} &&\oint_{C}[(e^{-x^{2}}+y^{2})\,dx+(\ln y-x^{2})]\,dy =\iint\limits_{R}(-2x-2y)\,dx\,dy \\ &&\hspace{10.4pc}=-2\int_{1}^{2}\int_{0}^{1}(x+y) \,dx\,dy=-2\int_{1}^{2}\left[ \dfrac{x^{2}}{2}+xy\right] _{0}^{1}\,dy \\ &&\hspace{9.1pc}\underset{\color{#0066A7}{R\hbox{ is } y\hbox{-simple}}}{\color{#0066A7}{\uparrow}}\\ &&\hspace{10.4pc}=-2\int_{1}^{2}\left( \dfrac{1}{2}+y\right) dy=-2\left[ \dfrac{1}{2}y+ \dfrac{y^{2}}{2}\right] _{1}^{2}=-4 \end{eqnarray*}$$

NOW WORK

Problem 11.

THEOREM Green’s Theorem for Area

If $C$ is a piecewise-smooth, simple closed curve with a counterclockwise orientation, then the area $A$ of the simply connected closed region $R$ enclosed by $C$ is given by $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ A=\dfrac{1}{2}\oint_{C}(-y\,dx+x\,dy) }} \tag{1}$$

Proof

In Green’s Theorem, let $P=P(x,y)=-y$ and $Q=Q(x,y)=x$. Then $$\oint_{C}(-y\,dx+x\,dy)=\iint\limits_{R}\left[ \dfrac{\partial }{\partial x} (x)-\dfrac{\partial }{\partial y}(-y)\right] \,dx\,dy=\iint\limits_{R}2\,dx \,dy=2A$$

RECALL

$\iint\limits_{R}\,dx\,dy$ equals the area of the region $R$.

Use Green’s Theorem to find the area of the region enclosed by the ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\qquad a>0\quad b>0$$

Solution We express the ellipse using the parametric equations $x=a\;\cos t$, $y=b\;\sin t$, $0\leq t\leq 2\pi$. Then $dx=-a\;\sin t\,dt$ and $dy=b\;\cos t\,dt$. Now we use (1) to find the area. $$\begin{eqnarray*} A& =&\dfrac{1}{2}\oint_{C}(-y\,dx+x\,dy)=\dfrac{1}{2}\int_{0}^{2\pi }[-b\sin t(-a\sin t\,dt)+a\cos t(b\cos t\,dt)] \\ & =&\dfrac{1}{2}\int_{0}^{2\pi }ab(\sin ^{2}t+\cos ^{2}t)\,dt=\dfrac{1}{2}(ab)\int_{0}^{2\pi }dt=\dfrac{1}{2}ab(2\pi )=\pi ab \end{eqnarray*}$$

NOW WORK

Problem 25.

THEOREM Green’s Theorem for Multiply-Connected Regions

Let $C_{1}, C_{2}, \ldots , C_{n}$ be $n$ piecewise-smooth, simple closed curves for which:

• No two curves $C_{i}$ and $C_{j},$ $i\neq j$, $i,$ $j=1,$ $2, \ldots , n$, intersect.
• Each of the curves $C_{i}$, $i=1, 2, \ldots , n$ lies in the interior of a piecewise-smooth, simple closed curve $C$.
• Any curve $C_{i}$ is exterior to every curve $C_{j}$, $i\neq j$, $i$, $j=1,$ $2, \ldots , n$.

Let $R$ denote the region consisting of the interior of $C$ and its boundary, less the interior of each of the curves $C_{1},$ $C_{2}, \ldots , C_{n}$. Let functions $P$ and $Q$ have continuous first-order partial derivatives throughout $R$. Then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \iint\limits_{R}\left( \dfrac{\partial Q}{\partial x}- \dfrac{\partial P}{\partial y}\right) dx\,dy=\oint_{C}(P\,dx+Q\,dy) - \sum\limits_{i=1}^{n}\oint_{C_{i}}(P \,dx+Q\,dy) }}$$

Proof

We construct line segments $\overline{AB}$ and $\overline{ED}$ that join the curves $C$ and $C_{1}$. This decomposes $R$ into two regions $R_{1}$ and $R_{2}$, with boundaries $\Gamma _{1}$ and $\Gamma _{2}$, respectively.

Since $R_{1}$ and $R_{2}$ each satisfy the conditions of Green’s Theorem, we have $$\begin{eqnarray*} \iint\limits_{R}\left( \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{ \partial y}\right) dx\,dy &=&\iint\limits_{R_{1}}\left( \dfrac{\partial Q}{ \partial x}-\dfrac{\partial P}{\partial y}\right) dx\,dy+\iint\limits_{R_{2}}\left( \dfrac{\partial Q}{\partial x}-\dfrac{ \partial P}{\partial y}\right) dx\,dy \\ &=&\oint_{\Gamma _{1}}(P\,dx+Q\,dy)+\oint_{\Gamma _{2}}(P\,dx+Q\,dy) \\ &=&\left[ \int_{DA}(P\,dx+Q\,dy)+\int_{\overline{AB}}(P\,dx+Q\,dy)+ \int_{BE}(P\,dx+Q\,dy)+\int_{\overline{ED}}(P\,dx+Q\,dy)\right] \\ &&+\left[ \int_{\overline{DE}}(P\,dx+Q\,dy)+\int_{EB}(P\,dx+Q\,dy)+\int_{ \overline{BA}}(P\,dx+Q\,dy)+\int_{AD}(P\,dx+Q\,dy)\right] \end{eqnarray*}$$

Along the line segments $\overline{AB}$ and $\overline{ED}$, $$\int_{\overline{AB}}(P\,dx+Q\,dy)=-\int_{\overline{BA}}(P\,dx+Q\,dy)\quad\hbox{and}\quad \int_{\overline{ED}}(P\,dx+Q\,dy)=-\int_{\overline{DE} }(P\,dx+Q\,dy)$$

So, $$\begin{eqnarray*} \iint\limits_{R}\left( \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{ \partial y}\right) dx\,dy &=&\oint_{C}(P\,dx+Q\,dy)+\oint_{-C_{1}}(P\,dx+Q\,dy) \\ &=&\oint_{C}(P\,dx+Q\,dy)-\oint_{C_{1}}(P\,dx+Q\,dy) \end{eqnarray*}$$

THEOREM

Let $P$ and $Q$ be functions that have continuous first-order partial derivatives throughout an open, connected set $R$. Let $C_{1}$ and $C_{2}$ denote two piecewise-smooth, simple closed curves, each lying entirely in $R$ . Suppose the curve $C_{1}$ can be obtained by continuously deforming the curve $C_{2}$ so that all the intermediate deformed curves lie entirely in $R$. If $\dfrac{\partial P }{\partial y}=\dfrac{\partial Q}{\partial x}$ throughout $R$, then $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{C_{1}}(P\,dx+Q\,dy)=\int_{C_{2}}(P\,dx+Q\,dy) }}$$

Find $\oint_{\!\!\!\!C}\left( \dfrac{-y}{x^{2}+y^{2}}\,dx+\dfrac{x}{x^{2}+y^{2}} \,dy\right)$ if:

Solution For $P=\dfrac{-y}{x^{2}+y^{2}}$ and $Q=\dfrac{x}{ x^{2}+y^{2}}$, we have $$\dfrac{\partial P}{\partial y}=\dfrac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}\qquad \hbox{and} \qquad \dfrac{\partial Q}{\partial x}=\dfrac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}$$

Notice that, except at the point $(0,0),$ the partial derivatives $\dfrac{ \partial P}{\partial y}$ and $\dfrac{\partial Q}{\partial x}$ are continuous.

(a) It is difficult to find the line integral along $C$. Moreover, because $x^{2/3}+y^{2/3}=1$ contains the point $(0,0)$ in its interior, we cannot use Green’s Theorem as we did in Example 2 to change the line integral to a double integral. However, since $\dfrac{ \partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ throughout $R,$ we can use a different curve that leads to an easier line integral. For example, suppose we replace $x^{2/3}+y^{2/3}=1$ by the unit circle $C_{1}$: $x^{2}+y^{2}=1,$ noting that the conditions required for $\int_{C}( P\,dx+Q\,dy)$ to equal $\int_{C_{1}}( P\,dx+Q\,dy)$ are met. See Figure 38.

Figure 38

On $C_{1},$ $x=\cos \theta$, $y=\sin \theta$, $0\leq \theta \leq 2\pi$. Then $dx=-\sin \theta d\theta$, $dy=\cos \theta d\theta$, and $$\begin{eqnarray*} \oint_{C}\left( \dfrac{-y}{x^{2}+y^{2}}\,dx+\dfrac{x}{x^{2}+y^{2}}\,dy\right) &=&\oint_{C_{1}}\left( \dfrac{-y}{x^{2}+y^{2}}\,dx+\dfrac{x}{x^{2}+y^{2}} \,dy\right)\\ &=&\int_{0}^{2\pi }(\sin ^{2}\theta +\cos ^{2}\theta )\,d\theta =2\pi \end{eqnarray*}$$

(b) In this case, the interior of the curve $C$ does not contain the origin. Since $$\begin{equation*} \dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x} \end{equation*}$$

in the simply connected region $R$ enclosed by $C$, it follows that $\mathbf{ F}=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}$ is a conservative vector field. Since $C$ is closed, $$\oint_{C}\left( \dfrac{-y}{x^{2}+y^{2}}\,dx+\dfrac{x}{x^{2}+y^{2}}\,dy\right)= \oint_{C}\mathbf{F}\,{\bf\cdot}\, d\mathbf{r} =0$$

NOW WORK

Problem 23.

THEOREM Converse of the Conservative Vector Field Theorem

Let $\mathbf{F}=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}$ be a vector field whose components $P$ and $Q$ have continuous first-order partial derivatives throughout an open, simply connected region $R$. If $\dfrac{\partial P}{ \partial y}=\dfrac{\partial Q}{\partial x}$ throughout $R$, then $\mathbf{F}$ is a conservative vector field.

Proof

Let $C_{1}$ and $C_{2}$ be two piecewise-smooth curves in $R$ joining $A$ to $(x,y)$. Assume that these curves do not intersect or coincide except at $A$ and $(x,y)$.* The curves $C_{1}$ and $C_{2}$ form the boundary of a region $D$ that is a subset of $R$, since $R$ is simply connected. See Figure 39.

Since the conditions for Green’s Theorem are satisfied on the region $D,$ we have $$\begin{eqnarray*} \iint\limits_{D}\left( \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{ \partial y}\right) \,dx\,dy& =&\int_{C_{2}+( -C_{1})}(P\,dx+Q\,dy)\\ &=&\int_{C_{2}}(P\,dx+Q\,dy)+\int_{-C_{1}}(P\,dx+Q\,dy) \\ & =&\int_{C_{2}}(P\,dx+Q\,dy)-\int_{C_{1}}(P\,dx+Q\,dy) \end{eqnarray*}$$

Since $\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$ throughout $R$, then $\iint\limits_{D}\left( \dfrac{\partial Q}{\partial x }-\dfrac{\partial P}{\partial y}\right) dx\,dy=0$.

Therefore, $$\begin{equation*} \int_{C_{1}}(P\,dx+Q\,dy)=\int_{C_{2}}( P\,dx+Q\,dy) \end{equation*}$$ showing that the line integral is independent of the path, and $\mathbf{F}$ is conservative.

NEED TO REVIEW?

The properties of double integrals are discussed in Section 14.2, pp. 917-918.