Answer
$$\frac{27 \pi}{2}$$
Work Step by Step
The integral is given by
\[
\begin{aligned}
\iiint_{R} d z d y d x &=\int_{-3 / 2}^{3 / 2} \int_{-\sqrt{9-4 x^{2}}}^{\sqrt{9-4 x^{2}}} \int_{0}^{y+3} d z d y d x \\
&=\int_{-3 / 2}^{3 / 2} \int_{-\sqrt{9-4 x^{2}}}^{\sqrt{9-4 x^{2}}}(3+y) d y d x \\
&=\int_{-3 / 2}^{3 / 2} 6 \sqrt{-4 x^{2}+9} d x \\
&=\frac{27 \pi}{2}
\end{aligned}
\]
