Please check my substitution for this integral, $$\iiint _V\left(z\right)\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z,$$ where $x^2+y^2+z^2\le 3$, $z\ge 1$.
I am not sure how to do the correct substitution, this is how I've done it since $x^2+y^2+z^2\le 3$ its a sphere with the center in origin and radius $\sqrt{3}$:
$T:x=rsin\phi cos\theta ,\:y=rsin\phi sin\theta \:,\:z=rcos\theta \:$
where $\theta \in \left[0,2\pi \right]$, and $0\le r\le \sqrt{3}$ , $\left|Jt\right|=r^2sin\phi \:$
I am not sure if I found the correct boundary for r and what is the boundary for $\phi$. I plan to integrate in the following order $\mathrm{d}r$ $\mathrm{d}\phi$ $\mathrm{d}\theta$ . No need to evaluate, I just want to find out the boundary for $\phi$ and check if the one for r is correct.
Based on how you have defined x and y in spherical coordinates, you have a typo in z, $z = \rho \cos\phi$ and not $\rho \cos\theta$.
Also the lower limit of $\rho$ is not zero. It is defined by $z = 1$.
$z = 1 \implies \rho \cos\phi = 1$ so $\rho = \sec \phi$.
At intersection of plane $z = 1$ and surface of the sphere, $\rho = \sqrt3, z = 1$.
So $z = 1 = \sqrt3 \cos\phi \implies \phi = \arccos {\left(\dfrac{1}{\sqrt3}\right)}$
So limits are, $ \ \sec \phi \leq \rho \leq \sqrt3, 0 \leq \phi \leq \arccos {\left(\dfrac{1}{\sqrt3}\right)}, 0 \leq \theta \leq 2\pi$.
The order of integral is $d\rho$ first and then $d\phi, d\theta$.
In cylindrical coordinates, if we are going in the order $dr \ dz \ d\theta$,
$0 \leq r \leq \sqrt{3-z^2}, 1 \leq z \leq \sqrt3, 0 \leq \theta \leq 2\pi$