Answer
$\dfrac{53}{20}$
Work Step by Step
$$Volume=\int_{0}^{1} \int_{y+1}^{4-2y} x^2y dx \space dy =\int_{0}^{1} [\dfrac{x^3y}{3}]_{y+1}^{4-2y} dy =\int_0^1 (\dfrac{y}{3}) \times [(4-2y)^3-(y+1)^3] dy =\int_0^1 -3y^4+15y^3-33y^2+21 y dy =[\dfrac{-3}{5} y^5+\dfrac{15}{4} y^4 -11y^3+\dfrac{21}{2} y^2|_0^1 =\dfrac{53}{20}$$
