Chapter 6: Applications of Definite Integrals - Section 6.1 - Volumes Using Cross-Sections - Exercises 6.1 - Page 322: 38

$\dfrac{108 \pi}{5}$

Area $=R^2\pi- r^2 \pi=\pi (12+4x-9x^2+x^4)$ We integrate the integral to calculate the volume as follows: $V= \pi \times \int_{-1}^{2} (12+4x-9x^2+x^4) dx$ Now, $V= \pi (24+8-24+(32/5)+12-2-3+\dfrac{1}{5})$ or, $= \pi (15+\dfrac{33}{5})$ or, $=\dfrac{108 \pi}{5}$