## Calculus (3rd Edition)

$\dfrac{1748 \pi }{35}$
The volume of a region can be calculated as: $V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$ where, $R_{outside}=x^3+2-(-2)=x^3+4$ and $R_{inside}=4-x^2-(-2)=6-x^2$ Now, $V=\pi \int_1^2 [(x^3+4)^2-(6-x^2)^2] \ dx =\pi \int_1^2 [x^6+8x^3+16-36+12x^2-x^4] \ dx \\=\pi \int_1^2 [x^6+8x^3+12x^2-x^4-20] \ dx \\=\pi [\dfrac{x^7}{7}+2x^4 +4x^3-\dfrac{x^5}{5}-20 x]_1^2 \\=\pi [\dfrac{1256}{35}+\dfrac{492}{35}] \\=\dfrac{1748 \pi }{35}$