limits of integration in non spherical coordinates I have been bashing my head against this for a solid 2 hours now and I have no idea where to even start. I have found stuff for spherical coordinates but I don't know those so I am not sure if they are right. The question is:

Suppose the solid W in the figure is a cone centered about the positive z-axis with its vertex at the origin, a 90∘ angle at its vertex, and topped by a sphere radius 6. Find the limits of integration for an iterated integral of the form: ∫(B to A)∫(D to C)∫(F to E) dzdydx.

Thanks in advance
 A: By a sketch 

we see that by cylindrical coordinates the set up should be 
$$\int_0^{6\frac{\sqrt2} 2}\, dz\int_0^{2\pi}\, d\theta\int_0^{z} \, dr+\int_{6\frac{\sqrt2} 2}^{6+6\frac{\sqrt2} 2}\, dz\int_0^{2\pi}\, d\theta\int_0^{\sqrt{6^2-\left(z-6\sqrt 2\right)^2}} \, dr$$
A: I think the problem might mean a sphere of radius $6$ $\textit{centered at the origin}$ since that is the most common set up for intro multivariable calculus problems  but I may be incorrect, only a picture of the figure in question would resolve that mystery.
In this case, we would integrate $z$ first as it goes from the cone to the sphere. Then the figure squishes down to its projection (shadow) in the $xy$ plane created by the intersection of 
$$z^2 = x^2 + y^2$$
$$x^2 + y^2 + z^2 = 36$$
since that is where the figure is the widest. Solving we get 
$$x^2 + y^2 = 18$$
So that leaves us with the bounds 
$$\int_{-\sqrt{18}}^{\sqrt{18}} \int_{-\sqrt{18-x^2}}^{\sqrt{18-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{36-x^2-y^2}} f(x,y,z)dzdydx$$
