Question

Find the area of the region that lies inside the circle r = 3sin$\theta$ and outside the cardioid r = 1+sin$\theta$

Answer 1

The cardioid intersects with the circle

$$3\sin\theta=1+\sin \theta$$

$$\sin \theta=\dfrac{1}{2}$$

The cardioid intersects with the circle at $(\dfrac{3}{2},\dfrac{\pi}{6}),(\dfrac{3}{2},\dfrac{5\pi}{6})$ and the pole.

The area of interest has been shaded above.

To find the area of a polar curve, we use

$$A=\dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}((3\sin \theta)^2-(1+\sin \theta)^2)d\theta$$

$$=\dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}(8\sin^2 \theta-2\sin\theta-1)d\theta$$

$$=\dfrac{1}{2}\displaystyle\int_{\pi/6}^{5\pi/6}(3-4\cos2 \theta-2\sin\theta)d\theta$$

$$=\dfrac{1}{2}[3\theta-2\sin2\theta+2\cos \theta]\begin{matrix} 5\pi/6\\ \pi/6 \end{matrix}$$

$$=\dfrac{1}{2}(\dfrac{5\pi}{2}+\sqrt{3}-\sqrt{3}-\dfrac{\pi}{2}+\sqrt{3}-\sqrt{3})$$

$$=\pi$$

$\pi$ square units.