\(dA = \frac{1}{2}{r^2}d\theta \)

\(dA = \frac{1}{2}{\left( {2aCos\theta } \right)^2}d\theta \)

Let \(r = 0\)

\(\cos \theta = 0\)

\(\theta = \frac{\pi }{2},\frac{{3\pi }}{2}\)

Since it's symmetrical along the x axis. The limits can also be from 0 to \(\frac{\pi }{2}\) and multiply the area twice.

\(A = \frac{1}{2}\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {4{a^2}{{\cos }^2}\theta d\theta } \)

\(A = 2{a^2}\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\cos }^2}\theta d\theta } \)

\({\cos ^2}\theta d\theta = \frac{1}{2}\left( {1 + \cos 2\theta } \right)\)

\(A = 2{a^2}\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\frac{1}{2}\left( {1 + \cos 2\theta } \right)d\theta } \)

\(A = {a^2}\left[ {\theta + \frac{1}{2}\sin 2\theta } \right]_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}}\)

\(A = {a^2}\left[ {\left( {\frac{\pi }{2} + 0} \right) - \left( {\frac{{ - \pi }}{2} + 0} \right)} \right]\)

\(A = \pi {a^2}\) square unitsAnswer