How do you find the area inside of the circle r = 3sin(theta) and outside the cardioid r = 1 + sin(theta)?

Apr 24, 2016

$\pi$

Explanation:

Draw both curves on the same graph paper. Notice that the cardioid intersects with the circle at $\left(\frac{3}{2} , \frac{\pi}{6}\right)$, $\left(\frac{3}{2} , \frac{5 \pi}{6}\right)$ and the pole.

The area of interest has been shaded above.

To find the area of a polar curve, we use

$A = \frac{1}{2} \int {r}^{2} \text{d} \theta$

We find the area of the cardioid and the circle separately on the interval $\frac{\pi}{6} < \theta < \frac{5 \pi}{6}$ and take the difference.

For the circle,

$A = \frac{1}{2} {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} {r}^{2} \text{d} \theta$

$= \frac{1}{2} {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} {\left(3 \sin \left(\theta\right)\right)}^{2} \text{d} \theta$

$= \frac{9}{4} {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} 2 {\sin}^{2} \left(\theta\right) \text{d} \theta$

$= \frac{9}{4} {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} \left(1 - \cos \left(2 \theta\right)\right) \text{d} \theta$

$= \frac{9}{4} {\left[\theta - \sin \frac{2 \theta}{2}\right]}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}}$

$= \frac{9}{4} \left(\left[\frac{5 \pi}{6} + \frac{\sqrt{3}}{4}\right] - \left[\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right]\right)$

$= \frac{3 \pi}{2} + \frac{9 \sqrt{3}}{8}$

For the cardioid,

$A = \frac{1}{2} {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} {r}^{2} \text{d} \theta$

$= \frac{1}{2} {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} {\left(1 + \sin \left(\theta\right)\right)}^{2} \text{d} \theta$

$= \frac{1}{2} {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} \left(1 + 2 \sin \left(\theta\right) + {\sin}^{2} \left(\theta\right)\right) \text{d} \theta$

$= \frac{1}{4} {\int}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}} \left(2 + 4 \sin \left(\theta\right) + 1 - \cos \left(2 \theta\right)\right) \text{d} \theta$

$= \frac{1}{4} {\left[3 \theta - 4 \cos \left(\theta\right) - \sin \frac{2 \theta}{2}\right]}_{\frac{\pi}{6}}^{\frac{5 \pi}{6}}$

$= \frac{1}{4} \left(\left[\frac{5 \pi}{2} + 2 \sqrt{3} + \frac{\sqrt{3}}{4}\right] - \left[\frac{\pi}{2} - 2 \sqrt{3} - \frac{\sqrt{3}}{4}\right]\right)$

$= \frac{\pi}{2} + \frac{9 \sqrt{3}}{8}$

The difference in area is

$\left(\frac{3 \pi}{2} + \frac{9 \sqrt{3}}{8}\right) - \left(\frac{\pi}{2} + \frac{9 \sqrt{3}}{8}\right) = \pi$