The sphere and cone meet in a cylinder,
[tex]x^2 + y^2 + \left(\sqrt{3x^2+3y^2}\right)^2 = 4x^2+4y^2 = 4 \implies x^2+y^2=1[/tex]
Then the volume of the region is given in Cartesian coordinates by
[tex]\displaystyle \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{\sqrt{3x^2+3y^2}}^{\sqrt{4-x^2-y^2}} dz \, dy \, dx[/tex]
which reduces to the double integral
[tex]\displaystyle \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \left(\sqrt{4-x^2-y^2} - \sqrt{3x^2+3y^2}\right) \, dy \, dx[/tex]
Now convert to polar coordinates. We're integrating over the unit circle, so the integral in polar is
[tex]\displaystyle \int_0^{2\pi} \int_0^1 \left(\sqrt{4-r^2} - \sqrt{3r^2}\right) r \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_0^1 \left(r \sqrt{4-r^2} - \sqrt3\,r^2\right) \, dr \, d\theta}[/tex]
