Finding the Area Enclosed by a Cardioid

Use double integration to find the area enclosed by the cardioid \(r=a(1-\cos \theta)\).

Solution The cardioid \(r=a(1-\cos \theta)\) is shown in Figure 31. Using symmetry, the area enclosed by the cardioid is twice the area of the shaded region \(R\) in Figure 31. So, \[ \begin{equation*} A=2\displaystyle\iint\limits_{\kern-3ptR}r\,dr\,d\theta \end{equation*} \]

where the region \(R\) is given by \(0\leq r\leq a( 1-\cos \theta )\) and \(0\leq \theta \leq \pi.\) Then \begin{eqnarray*} A &=&2\int_{0}^{\pi }\int_{0}^{a(1-\cos \theta )}r\,dr\,d\theta =2\int_{0}^{\pi }\left[ \dfrac{r^{2}}{2}\right] _{0}^{a( 1-\cos \theta ) }d\theta\\[4pt] &=&\int_{0}^{\pi }a^{2}( 1-\cos \theta ) ^{2}\,d\theta =a^{2}\int_{0}^{\pi }( 1-2\cos \theta +\cos ^{2}\theta )\, d\theta\\[4pt] &=&a^{2}\!\left[ \theta -2\sin \theta +\left( \dfrac{\theta }{2}+\dfrac{\sin (2\theta) }{4}\right) \right] _{0}^{\pi }=\dfrac{3}{2}\pi a^{2} \hbox{ square units} \end{eqnarray*}