Revision history [back]

I feel like it is a great start. Now if you were to set the two equations, $8-r^2$ and $r^2$, equal to each other you'll get $2r^2=8$ or $r^2=4$, which is $r=2$. So your radial boundary will go from 0 to 2 and seeing as it is a non-restricted 3d domain, $\theta$ will travel from $0$ to $2\pi$.

I feel like it is a great start. Now if you were to set the two equations, $8-r^2$ and $r^2$, equal to each other you'll get $2r^2=8$ or $r^2=4$, which is $r=2$. So your radial boundary will go from 0 to 2 and seeing as it is a non-restricted 3d domain, $\theta$ will travel from $0$ to $2\pi$.$2\pi$.

I feel like it is a great start. Now if you were to set the two equations, $8-r^2$ and $r^2$, equal to each other you'll get $2r^2=8$ or $r^2=4$, which is $r=2$. So your radial boundary will go from 0 to 2 and seeing as it is a non-restricted 3d domain, $\theta$ will travel from $0$ to $2\pi$.

Ending up with: $$\int_0^{2\pi}\int_0^2\int_{r^2}^{8-r^2}r^2dzdrd\theta$$$$\int_0^{2\pi}\int_0^2\int_{r^2}^{8-r^2}r^2rdzdrd\theta$$

I feel like it is a great start. Now if you were to set the two equations, $8-r^2$ and $r^2$, equal to each other you'll get $2r^2=8$ or $r^2=4$, which is $r=2$. So your radial boundary will go from 0 to 2 and seeing as it is a non-restricted 3d domain, $\theta$ will travel from $0$ to $2\pi$.

Ending up with: $$\int_0^{2\pi}\int_0^2\int_{r^2}^{8-r^2}r^2rdzdrd\theta$$

Edit: Sorry, mistake fixed. Thanks SPS.SMS.