Answer
$12 \pi$
Work Step by Step
Consider $Volume=\iint_{x^2+y^2 \leq 4} \int_{0}^{3-y} dz dA$
or, $=\iint_{x^2+y^2 \leq 4} (3-y) dA $
or, $=3 \iint_{x^2+y^2 \leq 4} dA - \iint_{x^2+y^2 \leq 4} y dA$
The inner integral is the integral of the odd continuous function in a symmetric interval so, $\iint_{x^2+y^2 \leq 4} y dA=0$
Thus, $V=(3) (\pi) (2)^2 -0=12 \pi$
