Example 10. Calculate the volume of the body formed by rotating the astroid $x=a \cos ^{3} t, y=a \sin ^{3} t$ around the line $y=-a$, where $a>0$.
Solution. We will use the symmetry of the astroid relative to the coordinate axes (Fig. 2.10). The desired volume $V$ is the difference between the volumes $V_{1}-V_{2}$ of two bodies. The first is obtained by rotating the arc $A B C$ of the astroid, and the second by rotating the arc $A D C$ around the line $y=-a$. In this case,

$$
\begin{aligned}
& V_{1}=2 \pi \int_{0}^{a}\left(y_{1}+a\right)^{2} d x \\
& V_{2}=2 \pi \int_{0}^{a}\left(y_{2}+a\right)^{2} d x
\end{aligned}
$$

where $y_{1}=y=a \sin ^{3} t, t \in\left[0 ; \frac{\pi}{2}\right], y_{2}=y=$ $=a \sin ^{3} t, t \in\left[\frac{3 \pi}{2} ; 2 \pi\right]$.

Thus, the problem reduces to the calculation of two integrals. The first can be represented as

$$
\begin{gathered}
I_{1}=\int_{0}^{a}\left(y_{1}+a\right)^{2} d x=\int_{\pi / 2}^{0}\left(a \sin ^{3} t+a\right)^{2} d\left(a \cos ^{3} t\right)= \\
=3 a^{2} \int_{0}^{\pi / 2}\left(\sin ^{3} t+1\right)^{2} \cos ^{2} t \sin t d t
\end{gathered}
$$

The second integral, due to the evenness of the cosine and the oddness of the sine, can be brought to the form

$$
\begin{aligned}
& I_{2}=\int_{0}^{a}\left(y_{2}+a\right)^{2} d x=\int_{3 \pi / 2}^{2 \pi}\left(a \sin ^{3} t+a\right)^{2} d\left(a \cos ^{3} t\right)= \\
&=3 a^{3} \int_{0}^{\pi / 2}\left(1-\sin ^{3} t\right)^{2} \cos ^{2} t \sin t d t
\end{aligned}
$$

It was noted above that the desired volume is expressed through the difference of these integrals: $V=2 \pi\left(I_{1}-I_{2}\right)$. After squaring in the integrands, it is not difficult to obtain the following expression for the difference $I_{1}-I_{2}$: $I_{1}-I_{2}=3 a^{3} \int_{0}^{\pi / 2} 4 \sin ^{3} t \cos ^{2} t \sin t d t$. Thus, $I_{1}-I_{2}=3 a^{3} \int_{0}^{\pi / 2} \sin ^{2} 2 t \sin ^{2} t d t=\frac{3}{4} a^{3} \int_{0}^{\pi / 2}(1-\cos 4 t)(1-\cos 2 t) d t=$
$=\frac{3}{4} a^{3} \int_{0}^{\pi / 2}[1-\cos 4 t-\cos 2 t+1 / 2(\cos 6 t+\cos 2 t)] d t=\frac{3}{8} \pi a^{3}$, since the integrals of the cosines give zero. Thus, $V=2 \pi\left(I_{1}-I_{2}\right)=\frac{3}{4} \pi^{2} a^{3}$.

Answer. $\frac{3}{4} \pi^{2} a^{3}$.