Re: Multiple Integration Help~
I believe the r limits should be 0 to sqrt(2). Seems right.
\(\displaystyle z=r^{2}, \;\ z=\sqrt{6-r^{2}}\)
Because \(\displaystyle \sqrt{6-r^{2}}=r^{2}\), solve for \(\displaystyle r=\pm\sqrt{2}\)
\(\displaystyle \int_{0}^{2\pi}\int_{0}^{\sqrt{2}}\int_{r^{2}}^{\sqrt{6-r^{2}}}rdzdrd{\theta}\)
These things are tedious. Once you have the correct set up, run it thorugh a calculator.
The biggest part of them is setting up the proper integral. Then, most just solve them with tech.
I know I do.
\(\displaystyle \int_{r^{2}}^{\sqrt{6-r^{2}}}{r}dz=r(\sqrt{6-r^{2}}-r^{2})\)
\(\displaystyle \int[r(\sqrt{6-r^{2}}-r^{2})]dr=\frac{-(6-r^{2})^{\frac{3}{2}}}{3}-\frac{r^{4}}{4}\)
Now, use the limits for r, then theta is easy.
