Answer
$\dfrac{4\pi}{3}$
Work Step by Step
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, $=\dfrac{4\pi}{3}$
