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
The conversion of rectangular coordinates to spherical coordinates is given as:
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
Here, $\rho=\sqrt {x^2+y^2+z^2}$; $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \theta=\cos^{-1}[\dfrac{x}{\rho \sin \phi}]$
Now, $\iiint f(x,y,z) dz r dr d\theta=\iiint f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi d\phi d\rho d\theta$
Plug the boundaries , then we get the triple integration as follows;
$\iiint f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi d\phi d\rho d\theta= \int_{\pi/2}^{2\pi}\int_{0}^{\pi/2} \int_1^{2}f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi d\rho d\phi d\theta$
