#### Answer

$\dfrac{2 \pi}{3}$

#### Work Step by Step

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