**Find the volume of the region above the sphere [tex]x^2+y^2+z^2=4[/tex] and below the**

paraboloid [tex]z = 6-x^2- y^2[/tex].

paraboloid [tex]z = 6-x^2- y^2[/tex].

Ok so the first thing i did was to find out if the two surfaces ever intersect by substituting [tex]x^2+y^2=6-z[/tex] into the first equation and solving for z. I got only complex solutions, hence they never intersect.

If i am picturing the region correctly, i am dealing with the volume of a region who's base is bounded by the equator of the sphere, all the way up to the paraboloid.

My initial thought was to use a cylindrical coordinate system. If i have set up my bounds correctly, the integral i would need to evaluate would be:

[tex]V=\int^{2\pi}_{0}\int^{2}_{0}\int^{6-r^2}_{\sqrt{4-r^2}}dz\:r\:dr\: d\theta=\frac{32\pi}{3}[/tex]

However, according to the solutions, the answer is [tex]\frac{2\pi}{3} (28-3\pi)[/tex]. Im sure that i evaluated my integral correctly because i checked it with Mathematica, but i just cant seem to get the same answer as the solution guide.

Any help would be greatly appreciated,

Thanks,

Dan.