Answer
$\dfrac{4 \pi (R^2-r^2)^{3/2}}{3}$
Work Step by Step
The shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(y) \ dy$
$V=2\pi \int_{r}^{R} (x) [2\sqrt {R^2-x^2}] \ dx$
Let us apply the substitution method:
$a=R^2-x^2 \implies da=-2x dx$
Now, $V= 2 \pi \int_0^{R^2-r^2} (a^{1/2}) \ da \\= 2\pi [\dfrac{2a^{3/2}}{3}]_0^{R^2-r^2} \\=\dfrac{4 \pi (R^2-r^2)^{3/2}}{3}$
