If I have the integral $$ \iiint_S f dV $$ with $S$ as $$x^2+y^2+z^2\leq 18, z\leq \sqrt{x^2+y^2}, z\geq 0$$ and then I make a change to $$ x=r\cos\theta, y=r\sin\theta, z=z$$ is not difficult to see that $$ z=0, z\leq r,\ and\ z\leq \sqrt{18-r^2},\ x=0, \ x=2\pi, \ y=0, \ y=\sqrt{18} $$
but which of the posibilities for $z$ should I take?
$$\int_0^{2\pi}\int_0^{\sqrt{18}}\int_0^{r} f\ r dz\ dr\ d\theta \ \ or \ \ \int_0^{2\pi}\int_0^{\sqrt{18}}\int_0^{\sqrt{18-r^2}} f\ r\ dz\ dr\ d\theta $$
I even start to think that could be $$ \int_0^{2\pi}\int_0^{3\sqrt{2}}\int_z^{\sqrt{18-z^2}} f\ r dr\ dz\ d\theta $$
Could any one tell me which should be or how to identify the the right limits or order?
It's none of them. Since $|z|\leqslant r$ and $|z|\leqslant\sqrt{18-r^2}$, you have$$0\leqslant z\leqslant\min\left\{r,\sqrt{18-r^2}\right\}=\begin{cases}r&\text{ if }0\leqslant r\leqslant3\\\sqrt{18-r^2}&\text{ if }3\leqslant r\leqslant\sqrt{18}.\end{cases}$$So, compute\begin{multline}\int_0^{2\pi}\int_0^3\int_0^rf(r\cos\theta,r\sin\theta,z)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta+\\+\int_0^{2\pi}\int_3^{\sqrt{18}}\int_0^{\sqrt{18-r^2}}f(r\cos\theta,r\sin\theta,z)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.\end{multline}