# How do you find region bounded by circle r = 3cosΘ and the cardioid r = 1 + cosΘ?

Oct 25, 2015

$A = \frac{5 \pi}{4}$

#### Explanation:

Lets find the intersection of the curves in the first quadrant:

$3 \cos \theta = 1 + \cos \theta \implies 2 \cos \theta = 1 \implies \cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$

The region is symmetric so we can find the area of the half of it:

$A = 2 \left({\int}_{0}^{\frac{\pi}{3}} d \theta {\int}_{0}^{1 + \cos \theta} r \mathrm{dr} + {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} d \theta {\int}_{0}^{3 \cos \theta} r \mathrm{dr}\right)$

${A}_{1} = \frac{1}{2} {\int}_{0}^{\frac{\pi}{3}} d \theta {r}^{2} {|}_{0}^{1 + \cos \theta} = \frac{1}{2} {\int}_{0}^{\frac{\pi}{3}} d \theta \left(1 + 2 \cos \theta + {\cos}^{2} \theta\right)$

${A}_{1} = \frac{1}{2} {\int}_{0}^{\frac{\pi}{3}} d \theta \left(1 + 2 \cos \theta + \frac{1 + \cos 2 \theta}{2}\right)$

${A}_{1} = \frac{1}{2} \left(\frac{3 \theta}{2} + 2 \sin \theta + \frac{1}{4} \sin 2 \theta\right) {|}_{0}^{\frac{\pi}{3}} = \frac{\pi}{4} + \frac{9 \sqrt{3}}{16}$

${A}_{2} = \frac{1}{2} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} d \theta {r}^{2} {|}_{0}^{3 \cos \theta} =$

$\frac{1}{2} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} d \theta \left(9 {\cos}^{2} \theta\right) = \frac{9}{4} {\int}_{\frac{\pi}{3}}^{\frac{\pi}{2}} d \theta \left(1 + \cos 2 \theta\right) =$

$= \frac{9}{4} \left(\theta + \frac{1}{2} \sin 2 \theta\right) {|}_{\frac{\pi}{3}}^{\frac{\pi}{2}} = \frac{3 \pi}{8} - \frac{9 \sqrt{3}}{16}$

$A = 2 \left(\frac{\pi}{4} + \frac{9 \sqrt{3}}{16} + \frac{3 \pi}{8} - \frac{9 \sqrt{3}}{16}\right) = 2 \frac{5 \pi}{8} = \frac{5 \pi}{4}$