Answer
The area of the region enclosed by the asteroid:
$$
x=a \cos^{3}\theta, y=a \sin^{3}\theta,
$$
is given by
$$
\begin{split}
A&=4 \int_{0}^{a} y d x\\
&=4 \int_{x / 2}^{0} a \sin ^{3} \theta\left(-3 a \cos ^{2} \theta \sin \theta\right) d \theta \\
&=\frac{3}{8} \pi a^{2} .
\end{split}
$$
Work Step by Step
The area of the region enclosed by the asteroid:
$$
x=a \cos^{3}\theta, y=a \sin^{3}\theta,
$$
is given by
$$
\begin{split}
A&=4 \int_{0}^{a} y d x\\
&=4 \int_{x / 2}^{0} a \sin ^{3} \theta\left(-3 a \cos ^{2} \theta \sin \theta\right) d \theta \\
&=12 a^{2} \int_{0}^{\pi / 2} \sin ^{4} \theta \cos ^{2} \theta d \theta . \\
&=12 a^{2} \int_{0}^{\pi / 2} \int \sin ^{4} \theta \cos ^{2} \theta d \theta \\ &=12 a^{2} \int_{0}^{\pi / 2} \sin ^{2} \theta\left(\frac{1}{4} \sin ^{2} 2 \theta\right) d \theta\\
&=12 a^{2} \frac{1}{8} \int_{0}^{\pi / 2} (1-\cos 2 \theta) \sin ^{2} 2 \theta d \theta \\
&=12 a^{2} \frac{1}{8} \int_{0}^{\pi / 2} \left[\frac{1}{2}(1-\cos 4 \theta)-\sin ^{2} 2 \theta \cos 2 \theta\right] d \theta \\
&=12 a^{2} \left[ \frac{1}{16} \theta-\frac{1}{4} \sin 4 \theta-\frac{1}{48} \sin ^{3} 2 \theta\right]_{0}^{\pi / 2} \\
& =12 a^{2} \left[ \frac{\pi}{32} \right]\\
&=\frac{3}{8} \pi a^{2} .
\end{split}
$$
