Answer
$\dfrac{\pi}{9} a^3 $
Work Step by Step
We can define the region $E$ using the point of intersection as follows:
$E=\left\{ (\rho, \theta, \phi) | 0 \leq \rho \leq a, 0 \leq \theta \leq \pi/6, 0 \leq \phi \leq \pi \right\}$
Therefore,
$V=\iint_{E} dV=\int_0^{\pi/6} \int_0^{\pi} \int_0^a \rho^2 \sin \phi d \rho d \theta d \phi$
or, $=[-\cos\phi]_0^{\pi} [ \theta]{0}^{\pi/6} [\dfrac{\rho^3}{3}]_0^a $
or, $= \dfrac{\pi}{6}\times (2) \times (\dfrac{a^3}{3}-0)$
Thus, we have
$E=\dfrac{\pi}{9} a^3 $
