convert triple integral $\int_{-6}^{6}\int_{-\sqrt{36-x^2}}^{\sqrt{36-x^2}}\int_{x^2+y^2}^{36}x\,dz\,dy\,dx$ to spherical coordinates I need to convert the following integral from rectangular to spherical coordinates
$$
\int_{-6}^{6}\int_{-\sqrt{36-x^2}}^{\sqrt{36-x^2}}\int_{x^2+y^2}^{36}x\,dz\,dy\,dx
$$
I am able to convert it to cylindrical with ease:
$$
\int_{0}^{2\pi}\int_{0}^{6}\int_{r^2}^{36}r^2 cos(\theta)\,dz\,dr\,d\theta
$$
but for spherical coordinates, i can't get my head around it without a picture, but I am having a hard time picturing the actual shape.
Any suggestions?
 A: Notice that the desired region is bounded by the paraboloid $z = x^2 + y^2$ and the plane $z = 36$. It's easy to see that $\theta$ can range from $0$ to $2\pi$. Now consider what happens when we slice the region at some fixed angle $\theta$ so that we are in the $rz$-half-plane. Then our cross-section is bounded by $z = r^2$, $z = 36$, and $r = 0$. It's easy to see that $\phi$ can range from $0$ to $\pi/2$. Now if we slice this cross-section at some fixed angle $\phi$, notice that there will be two types of upper boundaries for our $\rho$, depending on the value of $\phi$. To figure out the threshold, we determine the intersection points of the two boundaries:
\begin{align*}
r^2 = z = 36 &\iff r = 6 \\
\tan\phi = \frac{\rho\sin \phi}{\rho\cos\phi} = \frac{r}{z} = \frac{6}{36} &\iff \phi = \arctan(1/6)
\end{align*}
Transforming the bounds to spherical coordinates, we obtain:
$$
z = 36 \iff \rho\cos\phi = 36 \iff \rho = 36\sec\phi \\
z = r^2 \iff \rho\cos\phi = \rho^2\sin^2\phi \iff \rho = \csc\phi\cot\phi
$$
Thus, we obtain:
$$
\int_0^{2\pi} \int_0^{\arctan(1/6)} \int_0^{36\sec\phi} (\rho\sin\phi\cos\theta)\rho^2\sin\phi \, d\rho \, d\phi \, d\theta \\
+ \int_0^{2\pi} \int_{\arctan(1/6)}^{\pi/2} \int_0^{\csc\phi\cot\phi} (\rho\sin\phi\cos\theta)\rho^2\sin\phi \, d\rho \, d\phi \, d\theta
$$
