Area Between Two Polar Curves - Calculus 2

Find the area of the region inside $r = 3$ but outside $r = 3 - 3\cos\theta$ .
Answer: $\frac{9\pi}{2} - \frac{27}{4}$ The cardioid $r = 3 - 3\cos\theta = 3(1 - \cos\theta)$ intersects the circle $r = 3$ when $3 = 3 - 3\cos\theta$ , giving $\cos\theta = 0$ , so $\theta = \pi/2, 3\pi/2$ . The area inside the circle but outside the cardioid is $\frac{1}{2}\int_{\pi/2}^{3\pi/2}[3^2 - (3-3\cos\theta)^2]d\theta = \frac{1}{2}\int_{\pi/2}^{3\pi/2}[9 - 9(1-\cos\theta)^2]d\theta = \frac{9}{2}\int_{\pi/2}^{3\pi/2}[1 - (1-2\cos\theta+\cos^2\theta)]d\theta = \frac{9}{2}\int_{\pi/2}^{3\pi/2}(2\cos\theta - \cos^2\theta)d\theta$ . Evaluating this integral gives $\frac{9\pi}{2} - \frac{27}{4}$ .
The area enclosed by one loop of $r^2 = 4\cos(2\theta)$ is:
Answer: $2$ The lemniscate $r^2 = 4\cos(2\theta)$ exists when $\cos(2\theta) \geq 0$ , which occurs for $-\pi/4 \leq \theta \leq \pi/4$ and $3\pi/4 \leq \theta \leq 5\pi/4$ . One loop corresponds to $-\pi/4 \leq \theta \leq \pi/4$ where $r = 2\sqrt{\cos(2\theta)}$ . The area of one loop is $\frac{1}{2}\int_{-\pi/4}^{\pi/4}r^2 d\theta = \frac{1}{2}\int_{-\pi/4}^{\pi/4}4\cos(2\theta)d\theta = 2\int_{-\pi/4}^{\pi/4}\cos(2\theta)d\theta = 2[\frac{1}{2}\sin(2\theta)]_{-\pi/4}^{\pi/4} = [\sin(2\theta)]_{-\pi/4}^{\pi/4} = \sin(\pi/2) - \sin(-\pi/2) = 1 - (-1) = 2$ .