Find the area enclosed by the cardioid $r=1-\sin \theta$.

Solution Look again at the cardioid in Figure 51(a). The region enclosed by the cardioid is swept out beginning with the ray $\theta =0$ and ending with the ray $\theta =2\pi$. So, the limits of integration are $0$ and $2\pi ,$ and the area $A$ is $$A=\int_{0}^{2\pi }\frac{1}{2}(1-\sin \theta )^{2}~d\theta =\frac{1}{2} \left[ \frac{3}{2}\theta +2 \cos \theta -\frac{1}{4}~\sin ( 2\theta ) \right] _{0}^{2\pi }=\frac{3\pi }{2}$$