Evaluate the volume of $$A_r=\{(x,y,z): x^2+y^2+z^2\le r^2;\, x^2+y^2+z^2\le 2zr \}.$$
I use spherical coordinates so i found $\theta \in [0,\pi], \phi \in [0,2\pi]$ but for $\rho $ i have two choise: $ \rho \le r$ or $\rho \le 2r \cos \theta$. Have I to divide the range of $\theta$ ?
$$\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{r} {\rho}^2 \sin \theta\dots+\int_{0}^{2 \pi} \int_{\pi/3}^{\pi} \int_{0}^{2r\cos \theta}....?$$
Hint. The inequality $x^2+y^2+z^2\le 2zr$ can be written as $$x^2+y^2+(z-r)^2\leq r^2.$$ Therefore $A_r$ is the intersection of two balls of radius $r$: one is centered at $(0,0,0)$ and the other at $(0,0,r)$.
In spherical coordinates you need to split the evaluation into two parts: $$V=\int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{0}^{r} {\rho}^2 \sin \theta d\rho d\theta d\phi +\int_{0}^{2 \pi} \int_{\pi/3}^{\pi/2} \int_{0}^{2r\cos \theta}{\rho}^2 \sin \theta d\rho d\theta d\phi.$$ Note that in your formula, in the second integral, the range of $\theta$ is from $\pi/3$ to $\pi/2$ (not $\pi$).
We may also use cylindrical coordinates: $$V=\int_{\rho=0}^{\sqrt{3}r/2}\left( \int_{z=r-\sqrt{r^2-\rho^2}}^{z=\sqrt{r^2-\rho^2}}\left(\int_{\theta=0}^{2\pi}d\theta\right) dz\right)\rho d\rho.$$