Answer
$$\iint \limits_{R} 4 x y^{3} d A=0$$
Work Step by Step
Given $$\iint \limits_{R} 4 x y^{3} d A, \quad R=[-1,1] \times[-2,2]$$
So, we have
\begin{aligned}
I&=\iint \limits_{R} 4 x y^{3} d A\\
&=4 \int_{-2}^{2}\int_{-1}^{1} x y^{3} d x d y=4 \int_{-2}^{2} y^{3}\left[\frac{x^{2}}{2}\right]_{-1}^{1} d y\\
&=4 \int_{-2}^{2} y^{3}\left(\frac{1}{2}-\frac{1}{2}\right) d y\\
&=4 \int_{-2}^{2} y^{3} [0 ]d y\\
&=0
\end{aligned}
