Answer
$$\int_{-1}^{5} \int_{0}^{3 y}\left(3+x^{2}+\frac{1}{4} y^{2}\right) d x d y=1629$$
Work Step by Step
Given$$\int_{-1}^{5} \int_{0}^{3 y}\left(3+x^{2}+\frac{1}{4} y^{2}\right) d x d y$$
So, we get
\begin{align}
&\int_{-1}^{5} \int_{0}^{3 y}\left(3+x^{2}+\frac{1}{4} y^{2}\right) d x d y\\
&=\int_{-1}^{5}\left(3 x+\frac{x^{3}}{3}+\frac{1}{4} x y^{2}\right]_{0}^{3 y} d y\\
&=\int_{-1}^{5}\left[9 y+9 y^{3}+\frac{3}{4} y^{3}\right] d y\\
&=\int_{-1}^{5}\left[9 y+\frac{39}{4} y^{3}\right] d y\\
&=\left[\frac{9}{2} y^{2}+\frac{39}{16} y^{4}\right]_{-1}^{5}\\
&=\left(\frac{9}{2}(25)+\frac{39}{16}(625)\right)-\left(\frac{9}{2}+\frac{39}{16}\right)\\
&=1629
\end{align}
