Answer
$$
12
$$
Work Step by Step
The integration area is limited to $ 1 = x / 4 + y / 6 $ in the first quadrant. Using the rules of integrals, we obtain:
\[
\begin{aligned}
&\int_{0}^{4} \int_{0}^{6-\frac{2}{3} x} z d y d x=\iint_{R} z d A \\
&=\int_{0}^{4} \int_{0}^{6-\frac{2}{3} x}\left(3-\frac{3}{4} x-\frac{1}{2} y\right) d y d x \\
&=\int_{0}^{4}\left(\frac{9 x^{2}}{16}-\frac{9 x}{2}+9\right) d x \\
&=\left[9x+\frac{3 x^{3}}{16}-\frac{9 x^{2}}{4}\right]_{0}^{4} \\
&=12
\end{aligned}
\]
