Answer
$$27 \pi$$
Work Step by Step
Volume under the function $ z = f (x, y) $ and over the region $ R $ is given by
\[
\iint_{R} f(x, y) d A=V
\]
\[
\iint_{R} -x+3 d A=V
\]
Using polar coordinates, we can write
\[
\begin{array}{l}
=\int_{0}^{2 \pi} \int_{0}^{3}(3-r \cos \theta) r d r d \theta \\
=\int_{0}^{2 \pi} \int_{0}^{3} 3 r-r^{2} \cos \theta d r d \theta \\
=\int_{0}^{2 \pi}\left[\frac{3 r^{2}}{2}-\frac{r^{3}}{3} \cos \theta\right]_{0}^{3} d \theta \\
\quad=\int_{0}^{2 \pi} \frac{27}{2}-9 \cos \theta d \theta \\
=\left[\frac{27 \theta}{2}-9 \sin \theta\right]_{0}^{2 \pi}=27 \pi
\end{array}
\]
