## Greenâ€™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{equation} \oint_C M dx+Ndy= \iint_R \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} \right)d A. \end{equation}

**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 & =\iint_D \left(\partial _x \left(2x-y^2\right)\right)-\left(\partial _y (x-3y)\right)dA \\ & =\iint_D (2+3)dA =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{equation} \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{equation} \item 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} \end{align*} by using the reduction formula, \begin{equation} \int \cos ^nx \, dx=\frac{\cos ^{n-1}x \sin x}{n}+\frac{n-1}{n}\int \cos ^{n-2}xdx. \end{equation}

## 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 $x^2+y^2=2.$