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