Answer to Question #204677 in Calculus for hayrunisa

Question

Calculate the area of the region inside the cardioid r = a(1 + sin Ɵ) outside the r = a sinƟ circle and above the polar ray. Show by drawing.

Answer 1

$r=a(1+\sin \theta)$

$r=a\sin \theta$

$0\leq \theta \leq \pi$

The area of the circle

$A_1=\dfrac{1}{2}\displaystyle\int_{0}^{\pi}(a\sin \theta)^2d\theta=\dfrac{a^2}{4}\displaystyle\int_{0}^{\pi}(1-\cos(2 \theta))d\theta$

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

The area of the part of the cardioid

$A_2=\dfrac{1}{2}\displaystyle\int_{0}^{\pi}(a(1+\sin \theta))^2d\theta$

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

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

$=a^2(\dfrac{3\pi}{4}+2)(units^2)$

$Area=A_2-A_1=\dfrac{3\pi a^2}{4}+2a^2-\dfrac{\pi a^2}{4}$

$=(\dfrac{\pi }{2}+2)a^2\ (units^2)$