Originally Posted by galactus
Here's the polar set up:

$\int_{0}^{\frac{\pi}{2}}\int_{0}^{2}\int_{0}^{rsin {\theta}}r^{3}sin^{2}{\theta} \;\ dzdrd{\theta}$

Next, try spherical.
I don't follow your limits for the polar set up:

z is bounded by 0 and y (rsinθ)
r is bounded by 0 and 2
... shouldnt θ be bounded by 0 and π? I thought it was the top-half of the circle (not the part in the 1st quadrant)

and then...
$yz = rsin{\theta}z r dz dr d{\theta}$

Originally Posted by galactus
Then use the negative portion, you'll get the same result.
This?

$-\sqrt{4-x^{2}} <= y <= 0$