Question

(b) Find the area of the region inside the cardioid r=2 + 2

Answer 1

$$r=2+2\sin \theta$$

$$A=\displaystyle\int_{0}^{2\pi}\dfrac{1}{2}r^2d\theta=\displaystyle\int_{0}^{2\pi}\dfrac{1}{2}(2+2\sin \theta)^2d\theta$$

$$=\displaystyle\int_{0}^{2\pi}(2+4\sin\theta+2\sin^2\theta)d\theta$$

$$=\displaystyle\int_{0}^{2\pi}(2+4\sin\theta+1-\cos(2\theta))d\theta$$

$$=[3\theta-4\cos\theta-\dfrac{1}{2}\sin(2\theta)]\begin{matrix} 2\pi\\ 0 \end{matrix}$$

$$=6\pi-4-0-(0-4-0)=6\pi$$

$Area=6\pi$ square units.