Answer
$\int_{\pi/2}^{2\pi}\int_{0}^{\pi/2} \int_1^{2}f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi d\rho d\phi d\theta$
Work Step by Step
In the spherical coordinates system, we have $x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
and
$\rho=\sqrt {x^2+y^2+z^2}$;
and $ \phi =\cos^{-1}(\dfrac{z}{\rho})$; $ \theta=\cos^{-1} (\dfrac{x}{\rho \sin \phi})$
In the cylindrical coordinate system, we have
$r^2=x^2+y^2 ; \theta=\arctan(\dfrac{y}{x})$
and $x=r \cos \theta; y=r \sin \theta, z=z$
Consider the given integral $\iiint f(x,y,z) dz r dr d\theta=\iiint f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi d\phi d\rho d\theta$
Now, use the boundaries in order to get the triple integration.
$\iiint f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi d\phi d\rho d\theta= \int_{\pi/2}^{2\pi}\int_{0}^{\pi/2} \int_1^{2}f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi d\rho d\phi d\theta$
