Answer
$\dfrac{\pi r^2 h}{3} $
Work Step by Step
The volume of a region can be calculated as:
Now, $V=\pi \int_{0}^{h} [\dfrac{r}{h}(h-y)]^2 \ dy \\ =\dfrac{r^2\pi}{h^2} \int_{0}^{h} (h-y)^2 \ dy $
Let us apply the substitution method such that: $a=h-y \implies dy=-da$
$V= - \dfrac{r^2\pi}{h^2} \int_{0}^{h} a^2 \ da\\=\dfrac{r^2\pi}{h^2} [\dfrac{(h-y)^3}{3}]_{0}^{h} \\=\dfrac{\pi r^2 h}{3} $
