Green’s Theorem

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Objectives:

1.: Know the statement of Green’s Theorem.
2.: Understand the required orientation of the curve in the statement of Green’s Theorem.
3.: Be able to use any technique to compute a line integral.

Recap Video

First, take a look at this recap video going over Green’s theorem.

Examples

The following video shows an example using Green’s theorem.

Let $C$ be the line segment from $(-2,-2)$ to $(2,2)$ followed by the counterclockwise portion of $x^2+y^2=8$ from $(2,2)$ to $(-2,-2)$ , evaluate $\int _C (xy^2-y^3)dx+(x^2y-y^2+\sin (y^3))dy.$

Note that the integrand for the double integral before the answer should be $3r^3\sin ^2(\theta )$ , not $3r^2\sin ^2(\theta )$ . The answer might change once this is changed.

This next video illustrates how to close the curve and then use Green’s theorem.

If $C$ is the curve $y = 1-|x|$ from $x = -1$ to $x=1$ , and $\textbf {F} = \left \langle x^2-y,x^2+y+\sin (y^2) \right \rangle$ , evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

Problems

Let $\textbf {F} = \left \langle x-y,x+y \right \rangle$ and let $C$ be the triangle with vertices $(0,0)$ , $(4,0)$ , and $(0,6)$ , oriented counterclockwise. Evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

We will eventually use Green’s Theorem, but it will be good to see a possible thought process for line integrals. Notice $\frac {\partial F_2}{\partial x} - \frac {\partial F_1}{\partial y} = \answer {2} \neq 0,$ so $\textbf {F}$ isis not conservative. Therefore, we cannot use FTLI or its consequences.

Instead, we will see if we can use Green’s theorem. We have three things to check:

The curve $C$ isis not closed.
The components of $\textbf {F}$ have continuous partials on the closed region enclosed by $C$ .
The curve $C$ isis not positively oriented.

Since all three conditions are satisfied, we can use Green’s theorem. If $D$ is the closed region enclosed by $C$ , Green’s theorem says $\int _C \textbf {F} \cdot d\textbf {r} = \iint _D \left (\frac {\partial F_2}{\partial x} - \frac {\partial F_1}{\partial y}\right )dA = \iint _D \answer {2}dA.$ This integral is equal to:

Twice the length of $C$ . Twice the area of $D$ .

Therefore, the answer is: $\answer {24}$ .

Compute the line integral $\displaystyle \int _C \left \langle xy,y \right \rangle \cdot d\textbf {r}$ , where $C$ is the unit circle, oriented clockwise.

Let $\textbf {F} = \left \langle xy,y \right \rangle$ . Notice $\frac {\partial F_2}{\partial x} - \frac {\partial F_1}{\partial y} = \answer {-x} \neq 0,$ so $\textbf {F}$ isis not conservative. Therefore, we cannot use FTLI or its consequences.

Instead, we will see if we can use Green’s theorem. We have three things to check:

The curve $C$ isis not closed.
The components of $\textbf {F}$ have continuous partials on the closed region enclosed by $C$ .
The curve $C$ isis not positively oriented.

We can fix the third condition by adding a negative in the statement of Green’s theorem to correct the orientation. With this, Green’s theorem says $\int _C \textbf {F} \cdot d\textbf {r} = -\iint _D \left (\frac {\partial F_2}{\partial x} - \frac {\partial F_1}{\partial y}\right )dA = \iint _D \answer {x}dA.$ To set this integral up, we can go to polar coordinates. In polar, we get $\int _D xdA = \int _{\theta =0}^{\theta = \answer {2\pi }} \int _{r = \answer {0}}^{r = \answer {1}} \answer {r^2\cos (\theta )}dr\,d\theta .$ This integral evaluates to $\answer {0}$ .

Let $C$ be the boundary of the unit square $0 \leq x \leq 1$ , $0 \leq y \leq 1$ , oriented counterclockwise. Then $\int _C \left \langle y^2,x^2 \right \rangle \cdot d\textbf {r} = \answer {0}.$

First, we will check whether the vector field $\textbf {F} = \left \langle y^2,x^2 \right \rangle$ is conservative. The components of $\textbf {F}$ have continuous partials everywhere, so the first condition is satisfied. The mixed partials $\frac {\partial F_2}{\partial x} = \answer {2x},\quad \frac {\partial F_1}{\partial y} = \answer {2y},$ so $\textbf {F}$ isis not conservative. So we cannot use the FTLI.

Next, we can try Green’s Theorem. There are three things to check:

Closed curve: $C$ isis not closed.
Orientation: $C$ isis not properly oriented.
Vector Field: $\textbf {F}$ have continuous partials in the region enclosed by $C$ .

Therefore, we can use Green’s Theorem, which says:

$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& \iint_D \left(\frac{\partial F_2}{\partial x}- \frac{\partial F_1}{\partial y} \right)dA \\ &=& \iint_D \answer{2x-2y}\,dA. \end{eqnarray*}$

Now the only decision to make is whether to do the double integral in Cartesian or polar. Which would be better in this case?

Cartesian Polar

Therefore,

$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& \iint_D \left(\frac{\partial F_2}{\partial x}- \frac{\partial F_1}{\partial y} \right)dA \\ &=& \iint_D (2x-2y)\,dA \\ &=& \int_{\answer{0}}^{\answer{1}} \int_{\answer{0}}^{\answer{1}} (2x-2y)\,dy\,dx \\ &=& \answer{0}. \end{eqnarray*}$

Let $C$ be the line segment from $(0,0)$ to $(1,1)$ followed by the portion of $y=x^2$ from $(1,1)$ to $(0,0)$ . If $\textbf {F} = \left \langle x,x^2 \right \rangle$ , then $\int _C \textbf {F} \cdot d\textbf {r} = \answer {\frac {1}{6}}.$

First, we will check whether the vector field $\textbf {F} = \left \langle x,x^2 \right \rangle$ is conservative. The components of $\textbf {F}$ have continuous partials everywhere, so the first condition is satisfied. The mixed partials $\frac {\partial F_2}{\partial x} = \answer {2x},\quad \frac {\partial F_1}{\partial y} = \answer {0},$ so $\textbf {F}$ isis not conservative. So we cannot use the FTLI.

Next, we can try Green’s Theorem. There are three things to check:

Closed curve: $C$ isis not closed.
Orientation: $C$ isis not properly oriented.
Vector Field: $\textbf {F}$ have continuous partials in the region enclosed by $C$ .

Therefore, we can use Green’s Theorem after adding a negative sign to fix the orientation problem. We then get

$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& -\iint_D \left(\frac{\partial F_2}{\partial x}- \frac{\partial F_1}{\partial y} \right)dA \\ &=& -\iint_D \answer{2x}\,dA. \end{eqnarray*}$

Now the only decision to make is whether to do the double integral in Cartesian or polar. Which would be better in this case?

Cartesian Polar

Therefore,

$\begin{eqnarray*} \int_C \textbf{F} \cdot d\textbf{r} &=& -\iint_D \left(\frac{\partial F_2}{\partial x}- \frac{\partial F_1}{\partial y} \right)dA \\ &=& -\iint_D 2x\,dA \\ &=& -\int_{\answer{0}}^{\answer{1}} \int_{\answer{x^2}}^{\answer{x}} 2x\,dy\,dx \\ &=& \answer{-\frac{1}{6}}. \end{eqnarray*}$

Let $\textbf {F} = \left \langle -y^2,x+y+\sin (y^2) \right \rangle$ , and $C$ is the portion of $y=x^2$ from $(-1,1)$ to $(1,1)$ , evaluate $\displaystyle \int _C \textbf {F} \cdot d\textbf {r}$ .

Since $\frac {\partial F_2}{\partial x} -\frac {\partial F_1}{\partial y} = \answer {1+2y} \neq 0,$ we know that $\textbf {F}$ isis not conservative. Since $C$ isn’t closed, we cannot use Green’s theorem (yet). However, to do this integral directly would be annnoying because of the $\sin (y^2)$ term in the $y$ -component (try parametrizing $C$ and doing this integral directly).

So we are left with trying to use Green’s theorem. However, since $C$ isn’t closed, we must close it ourselves. We can do this by adding in, for example, the curve $C'$ which is the line segment from $(-1,1)$ to $(1,1)$ . Then the combined curve consisting of $C$ and $C'$ is closed. We still have to check the other two conditions:

The components of $\textbf {F}$ have continuous partials on the closed region enclosed by $C$ and $C'$ .
The curve $C$ positively oriented and the curve $C'$ isis not positively oriented.

So we will have to account for the orientation in the statement of Green’s theorem. The theorem gives $\int _C \textbf {F} \cdot d\textbf {r} - \int _{C'} \textbf {F} \cdot d\textbf {r} = \iint _D \answer {1+2y}dA,$ where $D$ is the region enclosed by $C$ and $C'$ . (Notice the $-$ sign in the second integral to fix the orientation.) We now have two integrals to compute:

To compute $\displaystyle \iint _D \answer {1+2y}dA$ , we can set up the iterated integrals. In the order $dy\,dx$ , it becomes $\iint _D (1+2y)dA = \int _{x=\answer {-1}}^{x=\answer {1}} \int _{y = \answer {x^2}}^{y = \answer {1}} \answer {1+2y}dy\,dx = \answer {44/15}.$
To compute $\displaystyle \int _{C'} \textbf {F} \cdot d\textbf {r}$ , we can parametrize $C'$ as $\textbf {r}(t) = \left \langle t,1 \right \rangle$ for $\answer {-1} \leq t \leq \answer {1}$ . Then $\textbf {r}'(t) = \left \langle \answer {1},\answer {0} \right \rangle ,$ and $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {-1},\answer {t+1+\sin (1)} \right \rangle .$ Therefore, $\int _{C'} \textbf {F} \cdot d\textbf {r} = \int _{-1}^1 \answer {-1}dt = \answer {-2}.$

Putting all this together, we get $\int _C \textbf {F} \cdot d\textbf {r} = \answer {14/15}.$

Let $\textbf {F} = \left \langle \sin (x)+y,3x+y \right \rangle$ , and $C$ is the line segment from $(0,0)$ to $(2,2)$ , followed by the line segment from $(2,2)$ to $(2,4)$ , followed again by the line segment from $(2,4)$ to $(0,6)$ , then $\int _C \textbf {F} \cdot d\textbf {r} = \answer {34}.$

First, we will check whether the vector field $\textbf {F} = \left \langle \sin (x)+y,3x+y \right \rangle$ is conservative. The components of $\textbf {F}$ have continuous partials everywhere, so the first condition is satisfied. The mixed partials $\frac {\partial F_2}{\partial x} = \answer {3},\quad \frac {\partial F_1}{\partial y} = \answer {1},$ so $\textbf {F}$ conservative. So we cannot use the FTLI.

Next, we can try Green’s Theorem. There are three things to check. The first is whether $C$ is closed. Notice $C$ closed, so we cannot use Green’s Theorem yet. The last alternative would be to do the integral directly, but this would mean doing three separate integrals, which we don’t want to do. So we can close the curve ourselves and use Green’s Theorem.

Let $C'$ be the line segment from $(0,0)$ to $(0,6)$ , so that $C$ and $C'$ together make a closed curve. For Green’s Theorem, $C$ is oriented, and $C'$ is oriented. We will fix the latter by adding a negative. Lastly, the components of $\textbf {F}$ have continuous partials on the enclosed region $D$ . Therefore, $\int _C \textbf {F} \cdot d\textbf {r} - \int _{C'} \textbf {F} \cdot d\textbf {r} = \iint _D \answer {2}\,dA.$ We now have two pieces to compute:

The double integral. The shortcut would be to notice that $\iint _D 2\,dA = 2\text {area}(D).$ The other way would be to set up the double integral, noting Cartesian is easier here. Which order would be the easiest to compute the double integral in?
$dy\,dx$ $dx\,dy$
The setup is: $\int _{\answer {0}}^{\answer {2}} \int _{\answer {x}}^{\answer {6-x}}2\,dy\,dx = \answer {16}.$
The line integral over $C'$ : Parametrize $C'$ as $\textbf {r}(t) = \left \langle \answer {0},6t \right \rangle$ for $\answer {0} \leq t \leq \answer {1}$ . Then $\textbf {F}(\textbf {r}(t)) = \left \langle \answer {6t},\answer {6t} \right \rangle ,$ $\textbf {r}'(t) = \left \langle \answer {0},\answer {6} \right \rangle .$ Therefore,
$\begin{eqnarray*} \int_{C'} \textbf{F} \cdot d\textbf{r} &=& \int_0^1 \textbf{F}(\textbf{r}(t)) \cdot \textbf{r}'(t)\,dt \\ &=& \int_0^1 \answer{36t} \,dt \\ &=& \answer{18}. \end{eqnarray*}$

Putting everything together, we get the answer for the line integral over $C$ to be $\answer {34}$ .