Answer
$64 \pi$
Work Step by Step
We can write the equation of the region as:
$\iiint_{R} 4+5x^2yz^2 \ dV=\iiint_{R} 4 \ dV+\iiint_{R} 5x^2yz^2 \ dV(1)$
But, $\iiint_{R} 5x^2yz^2 \ dV=0$ because the region inside the sphere is symmetrical, so the integral of an odd function will be zero.
Thus, equation (1) becomes:
$\iiint_{R} 4+5x^2yz^2 \ dV=\iiint_{R} 4 \ dV+0=\iiint_{R} 4 \ dV$
Here, $dV$ refers to the volume $V$ of the cylinder and is given by: $V=\pi r^2h$
So, $\iiint_{R} 4 \ dV= 4 (\pi r^2 h) =(4) \times [\pi (2)^2 (4)] =64 \pi$
