I know that the limits of integration in spherical coordinates are these, but I can't find the reason why that 2 appears,but I can't find a way to go further, and evaluate r in a $a \sin \theta$,
$$V = \int_0^{2\pi}\int_0^{a\sin{\theta}}\int_{-\sqrt{a^{2} - r^{2}}}^{\sqrt{a^2-r^{2}}}r\mathrm{d}z\mathrm{d}r\mathrm{d}\theta = 2\int_0^{2\pi}\int_0^{a\sin{\theta}}r\sqrt{a^2-r^2}\mathrm{d}r\mathrm{d}\theta$$
On
The 2 comes from the limits of integration: $\sqrt{a^2-r^2} - (-\sqrt{a^2-r^2}) = 2\sqrt{a^2-r^2}$
As for proceeding, here's a hint: $$ 2\int_0^{2\pi}\int_0^{a\sin{\theta}}r\sqrt{a^2-r^2}\mathrm{d}r\mathrm{d}\theta = \int_0^{2\pi}\int_0^{a\sin{\theta}}\sqrt{a^2-r^2}(2r\mathrm{d}r)\mathrm{d}\theta $$ Does a possible $u$-substitution come to mind?
The number two appears because you have to consider the integral corresponding to the upper and lower parts of the sphere, which are equal due to the symmetry of the problem.
As far as I have understood, you are not applying spherical change of coordinates. Instead, you are applying the cylindrical change of coordinates.