Hi,

My goal is to set up a triple integral that representshe the volume under the plane $\displaystyle z = 6$, and inside the paraboloid $\displaystyle z = x^2 + y^2$. I attempted to do this using cylindrical coodinates, and this is what I got:

$\displaystyle

\int_0^{2\pi} \int_0^{\sqrt{6}}\int_{r^2}^{6} 1dzrdrd\theta

$

does it this look about right?

Thanks for any help

james