Quote Originally Posted by galactus View Post
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}

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

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