Answer
$$V =\frac{4 \pi}{3} a^{3}$$
Work Step by Step
The solid is inside the sphere of radius a, $x^{2}+y^{2} +z^{2}= a^{2}$
If we put $z = 0$ in the equation of the sphere, we get $x^{2}+y^{2} = a^{2} $.
This means that the sphere intersects the $xy-$plane in the circle $x^{2}+y^{2} = a^{2}$, so the volume is
$$
\begin{aligned}
V& = \iint_{x^{2}+y^{2} \leq a^{2}} z d A
& \quad\left[\begin{array}{c}{ \text {by symmetry } } \end{array}\right] \\
& =2 \iint_{x^{2}+y^{2} \leq a^{2}} \sqrt{a^{2}-x^{2}-y^{2}} d A
\end{aligned}
$$
In polar coordinates, we have $x^{2}+y^{2}=r^{2} $.
Thus, the disk D is given by:
$$ D=\left\{(r, \theta) | 0 \leq r \leq a, \quad 0 \leq \theta \leq 2\pi \right\} $$
and, the required volume is given by:
$$
\begin{aligned}
V &=2 \iint_{x^{2}+y^{2} \leq a^{2}} \sqrt{a^{2}-x^{2}-y^{2}} d A \\
&=2 \int_{0}^{2 \pi} \int_{0}^{a} \sqrt{a^{2}-r^{2}} r d r d \theta \\
&=2 \int_{0}^{2 \pi} d \theta \int_{0}^{a} r \sqrt{a^{2}-r^{2}} d r \\
&=2[\theta]_{0}^{2 \pi}\left[-\frac{1}{3}\left(a^{2}-r^{2}\right)^{3 / 2}\right]_{0}^{a} \\
& =2(2 \pi)\left(0+\frac{1}{3} a^{3}\right)\\
&=\frac{4 \pi}{3} a^{3}.
\end{aligned}
$$
