Answer
$2 \pi $
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= \pi\int_0^{\pi/4} \sec^2 x \ dx +\pi \int_{\pi/4}^{\pi/2} \csc^2 x \ dx \\=\pi [\tan x]_0^{\pi/4} +\pi [-\cot x]_{\pi/4}^{\pi/2} \\= \pi+\pi [0+\dfrac{1}{\tan (\dfrac{\pi}{4})}] \\=2 \pi $
