# How do you find the area of the region bounded by the polar curves r=3+2cos(theta) and r=3+2sin(theta) ?

Nov 8, 2014

Let us look at the region bounded by the polar curves, which looks like: Red: $y = 3 + 2 \cos \theta$
Blue: $y = 3 + 2 \sin \theta$
Green: $y = x$

Using the symmetry, we will try to find the area of the region bounded by the red curve and the green line then double it.

$A = 2 {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} \setminus {\int}_{0}^{3 + 2 \cos \theta} r \mathrm{dr} d \theta$

$= 2 {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} {\left[{r}^{2} / 2\right]}_{0}^{3 + 2 \cos \theta} d \theta$

$= {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} \left(9 + 12 \cos \theta + 4 {\cos}^{2} \theta\right) d \theta$

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

$= {\int}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} \left(11 + 12 \cos \theta + 2 \cos 2 \theta\right) d \theta$

$= {\left[11 \theta + 12 \sin \theta + \sin 2 \theta\right]}_{\frac{\pi}{4}}^{\frac{5 \pi}{4}}$

$= \frac{55 \pi}{4} - 6 \sqrt{2} + 1 - \left(\frac{11 \pi}{4} + 6 \sqrt{2} + 1\right)$

$= 11 \pi - 12 \sqrt{2}$

Hence, the area of the region is $11 \pi - 12 \sqrt{2}$.

I hope that this was helpful.