Thomas' Calculus 13th Edition

$\dfrac{4\pi}{3}$
Area $=R^2\pi- r^2 \pi=\pi (y^2+2y)$ We integrate the integral to calculate the volume as follows: $V= \pi \times \int_{0}^{1} (y^2+2y) dy$ Now, $V= \pi [\dfrac{y^3}{3} +y^2]_{0}^{1}$ or, $= \pi (\dfrac{1}{3} +1)_0^1$ or, $=\dfrac{4\pi}{3}$