Answer
$\dfrac{\pi h^3 }{6}$
Work Step by Step
The Washer method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq g(x) \geq 0$ on the interval $[m,n]$, then the volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
Now, $V=\pi \int_{-\sqrt {R^2-r^2}}^{\sqrt {R^2-r^2}} [(\sqrt {R^2-y^2})^2-1^2] \ dy \\ = \pi [(R^2-r^2)-\dfrac{y^3}{3}]_{-\sqrt {R^2-r^2}}^{\sqrt {R^2-r^2}} \\=\dfrac{4}{3}\pi (R^2-r^2)^{3/2} $
Since, $\dfrac{h}{2}=\sqrt {R^2-r^2}$
Thus, we have: $V=\dfrac{\pi h^3 }{6}$
