# How do you find the region inside cardioid r=(alpha)(1+sin(theta)), (alpha)>0?

Oct 16, 2015

$A = \frac{3 \pi {\alpha}^{2}}{2}$

#### Explanation:

$A = \int \int r \mathrm{dr} d \theta$

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

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

$A = {\alpha}^{2} / 2 {\int}_{0}^{2 \pi} \left(1 + 2 \sin \theta + {\sin}^{2} \theta\right) d \theta$

$A = {\alpha}^{2} / 2 {\int}_{0}^{2 \pi} \left(1 + 2 \sin \theta + {\sin}^{2} \theta\right) d \theta$

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

$A = {\alpha}^{2} / 2 \left(\theta - 2 \cos \theta + \frac{\theta}{2} - \frac{1}{4} \sin 2 \theta\right) {|}_{0}^{2 \pi} \implies$

$A = {\alpha}^{2} / 2 \left(3 \pi - 2 + 2\right)$

$A = \frac{3 \pi {\alpha}^{2}}{2}$