Do I have the right bounds and function for this integral?

Question

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?

Answer 1

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$$\int_0^{2\pi}\int_0^{\pi/2}\int_0^{4\cos\varphi}\rho^5\sin(\varphi)\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$$

Answer 2

The region of integration is the sphere of radius $2$ centered at $(0,0,2)$

So in spherical coordinates, the change of variables is

$x = r \sin \theta \cos \phi$

$y = r \sin \theta \sin \phi$

$z = 2 + r \cos \theta$

The differential volume is $ r^2 \sin \theta dr d\theta d\phi $ and the limits of integration are $ r \in [0, 2], \theta \in [0, \pi] , \phi \in [0, 2 \pi] $

The integrand is $(x^2 + y^2 + z^2)^{\dfrac{3}{2}} = (4 + 4 r \cos \theta + r^2)^{\dfrac{3}{2} } $

Hence, the required integral is

$ I = \displaystyle \int_{0}^{2} \int_0^\pi \int_0^{2 \pi} (4 + 4 r \cos \theta + r^2)^{\dfrac{3}{2} } r^2 \sin \theta d \phi d\theta d r $

Since the integrand does not depend on $\phi$, this becomes

$ I = 2 \pi \displaystyle \int_{0}^{2} \int_0^\pi (4 + 4 r \cos \theta + r^2)^{\dfrac{3}{2} } r^2 \sin \theta d\theta d r $