# How do you find the area of the region bounded by the polar curve r=2-sin(theta) ?

Sep 17, 2014

The polar curve $r = 2 - \sin \theta$, $0 \le \theta < 2 \pi$ looks like this. we can find the area $A$ of the enclosed region can be found by

$A = {\int}_{0}^{2 \pi} \setminus {\int}_{0}^{2 - \sin \theta} r \mathrm{dr} d \theta = \frac{9 \pi}{2}$

Let us evaluate the double integral above.

$A = {\int}_{0}^{2 \pi} \setminus {\int}_{0}^{2 - \sin \theta} r \mathrm{dr} d \theta$

$= {\int}_{0}^{2 \pi} {\left[{r}^{2} / 2\right]}_{0}^{2 - \sin \theta} d \theta$

$= \frac{1}{2} {\int}_{0}^{2 \pi} {\left(2 - \sin \theta\right)}^{2} d \theta$

$= \frac{1}{2} {\int}_{0}^{2 \pi} \left(4 - 4 \sin \theta + {\sin}^{2} \theta\right) d \theta$

by ${\sin}^{2} \theta = \frac{1}{2} \left(1 - \cos 2 \theta\right)$,

$= \frac{1}{2} {\int}_{0}^{2 \pi} \left(\frac{9}{2} - 4 \sin \theta - \frac{1}{2} \cos 2 \theta\right) d \theta$

$= \frac{1}{2} {\left[\frac{9}{2} \theta + 4 \cos \theta - \frac{1}{4} \sin 2 \theta\right]}_{0}^{2 \pi}$

$= \frac{1}{2} \left[9 \pi + 4 - 0 - \left(0 + 4 - 0\right)\right] = \frac{9 \pi}{2}$