You want to find the volume inside the hemisphere [tex]x^2+y^2+z^2=4[/tex] (i.e. inside the sphere but above the plane [tex]z=0[/tex]) and outside the cylinder [tex]x^2+y^2=1[/tex]. Call this region [tex]R[/tex].
In cylindrical coordinates, we have
[tex]\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_1^2\int_0^{\sqrt{4-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]\displaystyle=2\pi\int_1^2 r\sqrt{4-r^2}\,\mathrm dr[/tex]
[tex]\displaystyle=-\pi\int_3^0\sqrt u\,\mathrm du[/tex]
(where [tex]u=4-r^2[/tex])
[tex]\displaystyle=\pi\int_0^3\sqrt u\,\mathrm du[/tex]
[tex]=\dfrac{2\pi}3u^{3/2}\bigg|_0^3=2\sqrt3\,\pi[/tex]
