Answer
$\dfrac{125}{4}$
Work Step by Step
Consider $Volume=\int_{-3}^{2} \int_{x}^{6-x^2} x^2 \ dy \ dx$
or, $=\int_{-3}^{2} [(x^2y]_{x}^{6-x^2} \ dx$
or, $=\int_{-3}^{2} (6x^2-x^4-x^3) \ dx$
or, $=[ \dfrac{6x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^4}{4}]_{-3}^2$
or, $=(\dfrac{6(2)^3}{3}-\dfrac{6(-3)^3}{3}]+[\dfrac{2^5}{5}-\dfrac{(-3)^5}{5}]-[\dfrac{2^4}{4}-\dfrac{(-3)^4}{4}]$
or, $=\dfrac{125}{4}$
