# Green’s Theorem (by Example) • By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

## Green’s Theorem for Simply Connected Regions

Green’s Theorem is named after the mathematician George Green.

Theorem. (Green’s Theorem) Let $R$ be a simply connected region with a piecewise smooth boundary curve $C$ oriented counterclockwise and let ${F}= M {i}+ N {j}+0 {k}$ be a continuously differentiable vector field on $R,$ then \begin{align*} \oint_C M dx+Ndy= \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)d A. \end{align*}

Example. Use Green’s theorem to evaluate the line integral $$\oint_C4 y \, dx-3x \, dy$$ around the curve $C$ defined by the ellipse $2x^2+y^2=4$ oriented counterclockwise.

Solution. Let ${F}(x,y)=4y {i}+(-3x){j}$, then $M(x,y)=4y$ which is continuously differentiable over the ellipse as well as $N(x,y)=-3x.$ Therefore ${F}$ and ${R}$ satisfy the hypothesis of Green’s theorem, and \begin{align*} \oint_C 4 y \, dx-3x \, dy & =\oint_C M \, dx+N \, dy \\ & = \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)d A \\ & = \iint_R (-3-4)d A \\ & = \iint_R -7d A \\ & =-7 \iint_R d A \\ & =-7(2)\left(\sqrt{2}\right)\pi. \end{align*}

Example. Use Green’s theorem to evaluate the line integral $$\oint_C4x y \, dx$$ around the curve $C$ defined by the unit circle oriented clockwise.

Solution. Let ${F}(x,y)=4x y {i}+0 {j}$, then $M(x,y)=4x y$ which is continuously differentiable over the unit circle as well as $N(x,y)=0.$ Therefore ${F}$ and $R$ satisfy the hypothesis of Green’s theorem and \begin{align*} \oint_C4x y \, dx & =\oint_C M \, dx+N \, dy \\ & =-\iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right) \,d A \\ & =-\iint_R (-4x) \, d A \\ & \\ & =4\iint_R x \, d A \\ & =4\int _0^{2\pi }\int _0^1 r^2 \cos \theta drd\theta =0 \end{align*}

Example. Use Green’s theorem to evaluate the line integral $$\oint_Cy^2dx+xdy$$ around the curve $C$ defined by the square with vertices $(0,0),$ $(2,0),$ $(2,2),$ $(0,2)$ oriented counterclockwise.

Solution. Let ${F}(x,y)=y^2 {i}+x {j}$, then $M(x,y)=y^2$ which is continuously differentiable over the square as well as $N(x,y)=x.$ Therefore ${F}$ and $R$ satisfy the hypothesis of Green’s theorem and \begin{align*} \oint_C y^2 \, dx+x \, dy & =\oint_C M \, dx+N \, dy \\ & = \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)d A \\ & =\int _0^2\int _0^2(1-2y) dy dx \\ & =\int_0^2 (-2) \, dx =-4. \end{align*}

Example. Find the work done an object moves in the force field $${F}(x,y) = y^2 {i}+x^2 {j}$$ once counterclockwise around the circular path $x^2+y^2=2.$

Solution. Let ${F}(x,y)=y^2 {i}+x^2 {j}$, then $M(x,y)=y^2$ which is continuously differentiable over the circle as well as $N(x,y)=x^2.$ Let $R$ be the region bounded by the curve $x^2+y^2=2.$ Then ${F}$ and $R$ satisfy the hypothesis of Green’s theorem and \begin{align*} W & =\oint_C {F}\cdot d{R} \\ & =\oint_C \left(y^2 {i}+x^2 {j}\right)\cdot ( dx {i}+dy {j}) \\ & =\oint_C \left(y^2dx +x^2dy\right) \\ & =\oint_C M \, dx+N\,dy \\ & = \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)\, d A \\ & =2 \iint_R (x-y) \, dA \\ & =2\int _0^{2\pi }\int _0^{\sqrt{2}}r^2(\cos \theta -\sin \theta )\, dr \, d\theta \\ & =\frac{4\sqrt{2}}{3}\int_0^{2\pi } (\cos \theta -\sin \theta ) \, \, d\theta =0. \end{align*}

Example. Find the work done an object moves in the force field $${F}(x,y)=\left(x+2y^2\right) {j}$$ once counterclockwise around the circular path $(x-2)^2+y^2=1.$

Solution. Let ${F}(x,y)=0 {i}+\left(x+2y^2\right) {j}.$ Then $M(x,y)=0$ which is continuously differentiable over the circle as well as $N(x,y)=x+2y^2.$ Let $R$ be the region bounded by the curve $(x-2)^2+y^2=1.$ Then ${F}$ and $R$ satisfy the hypothesis of Green’s theorem and \begin{align*} W& =\oint_C {F}\cdot d{R} \\ & =\oint_C \left(x+2y^2\right){j}\cdot ( dx {i}+dy {j}) \\ & =\oint_C \left(x+2y^2\right) dy \\ & =\oint_C M dx +N dy \\ & = \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)d A \\ & = \iint_R dA =\pi. \end{align*}

Example. Evaluate the closed line integral \begin{equation} \oint_C \frac{-y dx+(x-1)dy}{(x-1)^2+y^2} \end{equation} where $C$ is any Jordan curve whose interior does not contain the point $(1,0)$ transversed counterclockwise.

Solution. Let $${F}(x,y)=\frac{-y }{(x-1)^2+y^2}{i}+ \frac{x-1 }{(x-1)^2+y^2} {j},$$ then $$M(x,y)=\frac{-y }{(x-1)^2+y^2}$$ which is continuously differentiable over the circle as well as $$N(x,y)=\frac{x-1 }{(x-1)^2+y^2}.$$ Let $R$ be the region bounded by the given curve $C.$ Then ${F}$ and $R$ satisfy the hypothesis of Green’s theorem and \begin{align*} & \oint_C \frac{-y dx+(x-1)dy}{(x-1)^2+y^2} \\ & \qquad =\oint_C \frac{-y }{(x-1)^2+y^2}dx+\frac{x-1}{(x-1)^2+y^2}dy \\ & \qquad = \iint_R \left(\partial _x \left(\frac{x-1 }{(x-1)^2+y^2}\right)-\partial _y \left(\frac{-y }{(x-1)^2+y^2}\right)\right)dA \\ & \qquad = \iint_R 0 \, dA =0. \end{align*}

## Green’s Theorem for Doubly-Connected Regions

Definition. A Jordan curve is a closed curve $C$ that does not intersect itself and a simply connected region $R$ has the property that it is connected and the interior of every Jordan curve $C$ in $R$ also lies in $R.$

Theorem. (Green’s Theorem for Doubly-Connected Regions) Let $R$ be a doubly-connected region with a piecewise smooth outer boundary curve $C_1$ oriented counterclockwise and a piecewise smooth inner boundary curve $C_2$ oriented clockwise and let ${F}= M {i}+ N {j}+0 {k}$ be a continuously differentiable vector field on $R,$ then \begin{equation} \oint_{C_1} M \, dx+N \, dy+\oint_{C_2} M \, dx+N \, dy= \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right) \, d A.\end{equation}

Example. Evaluate the closed line integral \begin{equation} \oint_C \frac{x dx+ydy}{x^2+y^2} \end{equation} where $C$ is any Jordan curve whose interior contains the point $(0,0)$ transversed counterclockwise

Solution. Let $C_1$ be a circle centered at $(0,0)$ with radius $r$ so small that all of $C_1$ is contained within $C.$ Let $C_1$ be oriented clockwise and let $R$ be the region between $C_1$ and $C.$ Then by Green’s theorem for doubly-connected regions, \begin{align*} & \oint_C \frac{x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy+\oint_{C_1} \frac{x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy \\ & \qquad = \iint_R \left(\frac{\partial }{\partial x}\left(\frac{y}{x^2+y^2}\right)-\frac{\partial }{\partial y}\left(\frac{x}{x^2+y^2}\right)\right)d A \\ & \qquad = \iint_R \left(\frac{-2 x y}{x^2+y^2}-\frac{-2 x y}{x^2+y^2}\right)d A =0.\end{align*} Thus, \begin{equation} \oint_C \frac{x dx+ydy}{x^2+y^2}=-\oint_{C_1} \frac{x dx+ydy}{x^2+y^2} \end{equation} which is something that can be easily evaluated, using say, the parametrization $C_1: x=r \sin \theta , y=r \cos \theta; 0\leq \theta \leq 2\pi .$ Therefore \begin{align*} \oint_C \frac{x dx+ydy}{x^2+y^2} & =-\oint_{C_1} \frac{x dx+ydy}{x^2+y^2} \\ & = \int_0^{2\pi } \frac{(r \sin \theta ) (r \cos \theta )+(r \cos \theta )(-r \sin \theta )}{\sin ^2\theta +\cos ^2\theta } \, d\theta \\ & =0. \end{align*}

Example. Evaluate the closed line integral \begin{equation} \oint_C\left(x^2y dx-y^2xdy\right) \end{equation} where $C$ is the boundary of the region between the $x$-axis and the semicircle $y=\sqrt{a^2-x^2},$ traversed counterclockwise (including the $x$-axis).

Solution. Using Green’s Theorem, we find \begin{align*} \oint_C\left(x^2y dx-y^2xdy\right) & =\int \int \left(-y^2\right)-\left(x^2\right)dA \\ & =-\int \int \left(x^2+y^2\right) \, dA \\ & =-\int _0^{\pi }\int _0^ar^3drd\theta \\ & =-\int_0^{\pi } \frac{a^4}{4} \, d\theta \\ & =-\frac{a^4 \pi }{4} \end{align*}

Example. Evaluate the closed line integral \begin{equation} \oint_C\frac{-(y+2)dx+(x-1)dy}{(x-1)^2+(y+2)^2}\end{equation} where $C$ is any Jordan curve whose interior does not contain the point $(1,-2).$

Solution. Green’s theorem applies because we are using any Jordan curve which does not contain the point $(1,-2).$ \begin{align*} & \oint_C\frac{-(y+2)dx+(x-1)dy}{(x-1)^2+(y+2)^2} \\ & \quad =\oint_C-\frac{(y+2)}{(x-1)^2+(y+2)^2} dx+\frac{(x-1)}{(x-1)^2+(y+2)^2}dy \\ & \quad =\iint_D \left(\partial _x \left(\frac{(x-1)}{(x-1)^2+(y+2)^2}\right)\right)-\left(\partial _y \left(-\frac{(y+2)}{(x-1)^2+(y+2)^2}\right)\right)dA \\ & \quad =\iint_D\left(\frac{-x^2+2 x+y^2+4 y+3}{\left(x^2-2 x+y^2+4 y+5\right)^2}\right)-\left(\frac{-x^2+2 x+y^2+4 y+3}{\left(x^2-2 x+y^2+4 y+5\right)^2}\right) \, dA \\ & =0 \end{align*}

Example. Evaluate the closed line integral \begin{equation} \oint_C\left[(x-3y)dx+\left(2x-y^2\right)dy\right] \end{equation} where $A$ is the region $D$ enclosed by a Jordan curve $C.$

Solution. Green’s theorem applies and \begin{align*} & \oint_C(x-3y)dx+\left(2x-y^2\right)dy \\ & \qquad =\iint_D \left(\partial _x \left(2x-y^2\right)\right)-\left(\partial _y (x-3y)\right)dA \\ & \qquad =\iint_D (2+3)dA \\ & \qquad =5A. \end{align*}

Example. Use a line integral to find the area enclosed by the region $R$ defined by the circle $x^2+y^2=4$

Solution. We can parametrize the circle by $x=2 \cos t$ and $y=2 \sin t$ for $0\leq t\leq 2\pi .$ Then the area is \begin{align*} & \frac{1}{2} \oint_C x dy-ydx \\ & = \frac{1}{2}\int _0^{2\pi } [4\cos (t)\cos (t)+4 \sin (t)\sin (t)] \, dt \\ & =4\pi. \end{align*} We can parametrize the line segments by \begin{align*} & C_1: x=t, y=t; 0\leq t\leq 1 \\ & C_2: x=1-t, y=1+t; 0\leq t\leq 1 \\ & C_3: x=0, y=2(1-t); 0\leq t\leq 1 \end{align*} Then the area is \begin{align*} \oint_C x dy & =\int {C_1} x dy+\int {C_2} x dy+\int {C_3} x dy \\ & =\int_0^1 t \, dt+\int_0^1 (1-t) \, dt+\int _0^10(-2)dt \\ & =\frac{1}{2}+\frac{1}{2}+0 \\ & =1. \end{align*}

Example. Use a line integral to find the area enclosed by the region $R$ defined by the curve $C: x=\cos ^3t,$ $y=\sin^3 t$ for $0\leq t\leq 2\pi$

Solution. The required area is \begin{align*} A& =\oint_C x dy =\int _0^{2\pi }\cos ^3t \left(3 \sin ^2 t \cos t\right)dt \\ & =\int _0^{2\pi }\cos ^4 t \left(1-\cos ^2 t \right) dt =\int _0^{2\pi }\left(\cos ^4 t -\cos ^6 t \right) dt \\ & =3\int _0^{2\pi }\left(\cos ^4 t -\cos ^6 t \right) dt \\ & =3\int _0^{2\pi }\left(\cos ^4(t)-\frac{5}{6}\cos ^4(t)\right) dt \\ & =\frac{1}{2}\int _0^{2\pi }\cos ^4(t) dt \\ &=\frac{1}{2}\left(\frac{3}{4}\right)\int _0^{2\pi }\cos ^2(t) dt \\ & =\frac{1}{2}\left(\frac{3}{4} \right) \left(\frac{1}{2}\right)\int _0^{2\pi } dt \\ & =\frac{3 \pi }{8}.\end{align*} by using the reduction formula, \begin{align*} \int \cos ^nx \, dx=\frac{\cos ^{n-1}x \sin x}{n}+\frac{n-1}{n}\int \cos ^{n-2}xdx. \end{align*}

## Exercises on Green’s Theorem

Exercise. Use Green’s theorem to evaluate the closed line integral \begin{equation} \oint_C y^2 dx+x^2 dy\end{equation} where $C$ is the boundary of the square with vertices $(0,0),$ $(1,0),$ $(1,1),$ and $(0,1)$ traversed counterclockwise.

Exercise. Use Green’s theorem to evaluate the closed line integral
\begin{equation}\oint_C4x y dx \end{equation} where $C$ is the boundary of the circle $x^2+y^2=1$ traversed clockwise.

Exercise. Use Green’s theorem to evaluate the closed line integral \begin{equation}\oint_C x \sin x dx-e^{y^2} dy\end{equation} where $C$ is the boundary of the triangle with vertices $(-1,-1),$ $(1,-1),$ and $(2,5)$ traversed counterclockwise.

Exercise. Use Green’s theorem to evaluate the closed line integral \begin{equation}\oint_C \sin x \cos y \, dx+\cos x \sin y \, dy \end{equation}
where $C$ is the boundary of the square with vertices $(0,0),$ $(2,0),$ $(2,2),$ and $(0,2)$ traversed clockwise.

Exercise. Use Green’s theorem to evaluate a closed line integral that represents the area enclosed by the region defined by the curve $x^2+y^2=4.$

Exercise. Use Green’s theorem to evaluate a closed line integral that represents the area enclosed by the region defined by the trapezoid with vertices
$(0,0),$ $(4,0),$ $(1,3),$ and $(0,3).$

Exercise. Evaluate the closed line integral \begin{equation} \oint_C\frac{x dx+ydy}{x^2+y^2}\end{equation} where $C$ is any piecewise smooth Jordan curve enclosing the origin, traversed counterclockwise.

Exercise. Evaluate the closed line integral \begin{equation}\oint_C x^2y \, dx-y^2x \, dy, \end{equation} where $C$ is the boundary of the region between the $x$-axis and the semicircle $y=\sqrt{a^2-x^2},$ traversed counterclockwise (including the $x$-axis).

Exercise. Evaluate \begin{equation}\oint_C\frac{-(y+2)dx+(x-1)dy}{(x-1)^2+(y+2)^2}\end{equation} where $C$ is any Jordan curve whose interior does not contain the point $(1,-2).$

Exercise. If $C$ is a Jordan curve, show that \begin{equation} \oint_C (x-3y)dx+\left(2x-y^2\right)dy =5A\end{equation} where $A$ is the region $D$ enclosed by $C.$

Exercise. Suppose ${F}=M(x,y)\, {i}+N(x,y)\, {j}$ is continuously differentiable in a doubly-connected region $R$ and that $$\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}$$ throughout $R.$ How many distinct values of $I$ are there for the integral\begin{equation} I=\oint_C M(x,y)\, dx+N(x,y)\, dy
\end{equation} where $C$ is a piecewise smooth Jordan curve in $R?$

Exercise. Evaluate the line integral \begin{equation}\oint_C x^2y \, dx-y^2x \, dy,\end{equation} where $C$ is the boundary of the region between the $x$-axis and the semicircle $y=\sqrt{a^2-x^2},$ transversed counterclockwise (including the $x$-axis).

Exercise. Find the work done if an object moves in the force field ${F}(x,y)=y^2 {i}+x^2 {j}$ once counterclockwise around the circular path $x^2+y^2=2.$