What is the volume of the solid where the base is the triangular region with vertices at (0,0), (0,2) and (3,0). The cross sections perpendicular to the y-axis are semicircles.
My set up is: $$V = \int_0^2 \frac{\pi}{2} (\frac{-3y}{4} + \frac{3}{2})^2 dy$$
Yielding $$ = \frac{3\pi}{4} $$
Is my set up correct?
Let's have a plot of what is given:
The radius of that arbitrary section is $r=\frac{6-3y}{4}=\frac{3}{2}-\frac{3y}{4}$ so the volume of that section is $V_{dy}=\pi r^2/2$ and so $$V=\int_0^2\frac{\pi}2r^2dy$$