I am trying to solve a question from Schaum's Calculus but I have been unable to.
Find the volume common to $\rho^2 + z^2 = a^2$ and $\rho = asin(\theta)$.
I have tried several things and the one I feel is more close to being correct was the following:
$\rho^2 + z^2 = a^2$ is a sphere of radius $a$.
$\rho = asin(\theta)$ is a cylinder.
$Volume=4I$, where $I= \int_0^{\pi/2} \int_0^{asin(\theta)} (a^2 - \rho^2)^{1/2} d\rho d\theta $.
I got the result $I=a^2 \pi^2/8$, or $V=a^2 \pi^2/2$. The answer is supposed to be $2(3\pi-4)a^2/9$.
I am probably making a mistake in the set up of the integral but I cannot spot it. Any tips?
Thank you.