### Using Green's theorem to find area

Typically we use Green's theorem as an alternative way to calculate a line integral $\dlint$. If, for example, we are in two dimension, $\dlc$ is a simple closed curve, and $\dlvf(x,y)$ is defined everywhere inside $\dlc$, we can use Green's theorem to convert the line integral into to double integral. Instead of calculating line integral $\dlint$ directly, we calculate the double integral \begin{align*} \iint_\dlr \left(\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}\right) dA \end{align*}

Can we use Green's theorem to go the other direction? If we are given a double integral, can we use Green's theorem to convert the double integral into a line integral and calculate the line integral? If we are given the double integral \begin{align*} \iint_\dlr f(x,y) dA, \end{align*} we can use Green's theorem only if there happens to be a vector field $\dlvf(x,y)$ so that \begin{align*} f(x,y) = \pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}. \end{align*} However, we haven't learned any method to find such a vector field $\dlvf$. So, we aren't likely to use Green's theorem in this direction very often.

There is one important exception to this rule, however, and that is when we are using a double integral to calculate the area of a region $\dlr$. The area of a region $\dlr$ is equal to the double integral of $f(x,y)=1$ over $\dlr$: \begin{align*} \text{Area of $\dlr$} = \iint_\dlr dA = \iint_\dlr 1\, dA. \end{align*} If $f(x,y) =1$, it is easy to find a vector field $\dlvf$ so that \begin{align*} \pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y} = f(x,y) = 1. \end{align*} There are many such vector fields $\dlvf$, but we'll pick the vector field $\dlvf(x,y) = (-y/2, x/2)$. You can confirm that indeed $\displaystyle \pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y} =1$.

In summary, if $\dlc$ is a counterclockwise oriented simple closed curve that bounds a region where you can apply Green's theorem, the area of the region $\dlr$ bounded by $\dlc = \partial \dlr$ is \begin{align*} \text{Area of $\dlr$} = \dlint = \frac{1}{2} \int_\dlc x\,dy - y\,dx, \end{align*} where $\dlvf(x,y) = (-y/2,x/2)$.

#### Example 1

Use Green's Theorem to calculate the area of the disk $\dlr$ of radius $r$ defined by $x^2+y^2 \le r^2$.

**Solution**: Since we know the area of the disk of radius $r$ is
$\pi r^2$, we better get $\pi r^2$ for our answer.

The boundary of $\dlr$ is the circle of radius $r$. We can parametrized it in a counterclockwise orientation using \begin{align*} \dllp(t) = (r\cos t, r\sin t), \quad 0 \le t \le 2\pi. \end{align*} Then \begin{align*} \dllp'(t) = (-r\sin t, r\cos t), \end{align*} and, by Green's theorem, \begin{align*} \text{area of } \dlr &= \iint dA\\ &= \frac{1}{2} \int_\dlc x\,dy - y\,dx\\ &= \frac{1}{2} \int_0^{2\pi} [(r\cos t)(r\cos t) - (r\sin t)(-r\sin t)]dt \\ &= \frac{1}{2} \int_0^{2\pi} r^2 (\sin^2t+\cos^2t) dt = \frac{r^2}{2}\int_0^{2\pi} dt = \pi r^2. \end{align*} Thankfully, our answer agrees with what we know it should be.

#### Example 2

Calculate the area of the region $D$ bounded by the curve $\dlc$ parametrized by $\dllp(t)=\sin 2t\,\vc{i} +\sin t\,\vc{j}$ for $0 \le t \le \pi$. The region and curve are illustrated by the below applet.

*Area inside sinusoidal curve.* The curve $\dlc$ parameterized by $\dllp(t)=(\sin 2t,\sin t)$ for $0 \le t \le \pi$ is the counterclockwise oriented boundary of a region $D$, shown shaded in blue. As you specify $t$ by dragging the green point on the slider, the red point traces out the curve $\dllp(t)$. Alternatively, you can drag the red point around the curve, and the green point on the slider indicates the corresponding value of $t$. One can calculate the area of $D$ using Green's theorem and the vector field $\dlvf(x,y)=(-y,x)/2$.

**Solution**:
We'll use Green's theorem to calculate the area bounded by the curve.
Since $\dlc$ is a counterclockwise oriented boundary of $D$, the area is just the line integral of the vector field $\dlvf(x,y) = \frac{1}{2}(-y,x)$ around the curve $\dlc$ parametrized by $\dllp(t)$.

To integrate around $\dlc$, we need to calculate the derivative of the parametrization $\dllp'(t)=2\cos 2t \,\vc{i}+\cos t\,\vc{j}$. Then, by Green's theorem, we turn the double integral defining the area into line integral and calculate the area. \begin{align*} \text{area of } \dlr &= \iint dA\\ &= \dlint = \plint{0}{\pi}{\dlvf}{\dllp}\\ &= \frac{1}{2} \int_0^{\pi} ( -\sin t, \sin 2t) \cdot ( 2\cos 2t,\cos t)dt\\ &= \frac{1}{2} \int_0^{\pi} ( -\sin t, 2 \sin t \cos t ) \cdot (2\cos^2 t - 2\sin^2 t,\cos t)dt\\ &= \int_0^{\pi} \sin^3t dt =\int_0^{\pi} (\sin t-\sin t\,\cos^2t)\, dt\\ &= \Bigl[-\cos t -\frac{1}{3}\cos^3 t\Bigr]_0^{\pi} = 1+\frac{1}{3}=\frac{4}{3} \end{align*}

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