Answer
$$\int_{0}^{2} \int_{3 y^{2}-6 y}^{2 y-y^{2}} 3 y d x d y=16$$
Work Step by Step
Given$$\int_{0}^{2} \int_{3 y^{2}-6 y}^{2 y-y^{2}} 3 y d x d y$$
So, we get
\begin{align}
\int_{0}^{2} \int_{3 y^{2}-6 y}^{2 y-y^{2}} 3 y d x d y&=\int_{0}^{2}[3 x y]_{3 y^{2}-6 y}^{2y-y^2} d y\\
&= 3\int_{0}^{2} y\left(2y-y^2-\left(3 y^{2}-6 y\right)\right) d y\\
&=3 \int_{0}^{2}\left(8 y^{2}-4 y^{3}\right) d y\\
&=\left[3\left(\frac{8}{3} y^{3}-y^{4}\right)\right]_{0}^{2}\\
&= 3\left(\frac{8}{3} 2^{3}-2^{4}\right) \\
&=16
\end{align}
