Answer
The volume is $\frac{4\pi a^3}{3}$.
Work Step by Step
A sphere of radius $a$ is obtained by combining upper hemisphere $z_u=\sqrt{a^2-x^2-y^2}$ and lower hemisphere $z_l=-\sqrt{a^2-x^2-y^2}$.
The region where these hemisphere is defined is $R=\{(x,y)|x^2+y^2\leq a^2\}$ or $R=\{(r,\theta)|0\leq r\leq a,0\leq\theta\leq 2\pi\}$.
The volume of the sphere is given by $V=\iint_R z_u-z_ldA$:
Find the volume:
$V=\int_0^a\int_0^{2\pi}(\sqrt{a^2-r^2}-(-\sqrt{a^2-r^2}))rd\theta dr$
$=\int_0^a\int_0^{2\pi}2r\sqrt{a^2-r^2}d\theta dr$
$=\int_0^a2r\sqrt{a^2-r^2}\theta]_0^{2\pi}dr$
$=\int_0^a4\pi r\sqrt{a^2-r^2}dr$
$=-\frac{4\pi (a^2-r^2)^{3/2}}{3}]_0^a$
$=0-(-\frac{4\pi a^3}{3})$
$=\frac{4\pi a^3}{3}$
