Answer
$4\pi$
Work Step by Step
We can write the equation of a region as:
$\iiint_{B} z^3+\sin y +3 \ dV=\iiint_{B} z^3+\sin y \ dV+\iiint_{B} 3 \ dV(1)$
But, $\iiint_{B} z^3+\sin y \ dV=0$ because the region inside the sphere is symmetrical, so the integral of an odd function will be zero.
Thus, equation (1) becomes:
$\iiint_{B} z^3+\sin y +3 \ dV=\iiint_{B} 3 \ dV = 3V$
The volume $V$ of the sphere is:
$V=\dfrac{4}{3}\pi r^3$
So, $\iiint_{B} z^3+\sin y +3 \ dV=3 \times \dfrac{4}{3}\pi r^3=4 \pi r^3=4 \pi (1)^3=4\pi$ $[\because Radius =1]$