## Calculus (3rd Edition)

$\dfrac{\pi}{3} [1-\dfrac{1}{(1+H)^3}]$
The area $(A)$ of each horizontal cross-section is given by: $A(y)=\pi r^2 =\pi [(1+y)^{-2}]^2=\pi (1+y)^{-4}$ Thus, the volume of the solid is: $V=\pi \int_{0}^{H} (1+y)^{-4} \ dy \\=-\dfrac{\pi}{3} [(1+y)^{-3}]_0^{H}\\=\dfrac{\pi}{3} [1-(1+H)^{-3}] \\=\dfrac{\pi}{3} [1-\dfrac{1}{(1+H)^3}]$