Do I have the right bounds and function for this integral? Find the integral of function $f(x,y,z)=(x^2+y^2+z^2)^{3/2}$ inside the sphere $(z-2)^2+x^2+y^2=4$
My approach: by changing it to spherical coordinates, we have $0\le\rho\le2\cos\varphi$ $(0\le\theta\le2\pi,\ 0\le\varphi\le\frac\pi2)$, and the function becomes $f(\rho,\theta,\varphi)=ρ^3$, which, when multplied by the Jacobian, $ρ^2\sin\varphi$, becomes $ρ^5\sin\varphi$. In other words, we are integrating this function in the above $\rho,\theta,\varphi$ bounds. Is this correct？Why would Wolfram alpha's calculator give a much larger value when integrated in Cartesian coordinates?

No, the bounds are not correct. Note that\begin{align}x^2+y^2+(z-2)^2\leqslant4&\iff x^2+y^2+z^2\leqslant4z\\&\iff\rho^2\leqslant4\rho\cos\varphi\\&\iff\rho\leqslant4\cos\varphi.\end{align}So, it should be$$\int_0^{2\pi}\int_0^{\pi/2}\int_0^{4\cos\varphi}\rho^5\sin(\varphi)\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$$
