Find the volume between the cone yx2z2 and the sphere x2y2z2

Solution

The sphere is given by x^2 + y^2 + z^2 = 25, and the cone is given by (z - 1)^2 = x^2 + y^2. In cylindrical coordinates, these are given as r^2 + z^2 = 25 and z = r + 1, respectively. These intersect when (z - 1)^2 + z^2 = 25 ==> z = 4 (since z > 0 via r). This yields the xy-region x^2 + y^2 = 3^2. Therefore, the volume equals ??? 1 dV = ?(? = 0 to 2p) ?(r = 0 to 3) ?(z = r+1 to v(25 - r^2)) 1 * r dz dr d? = 2p ?(r = 0 to 3) r [v(25 - r^2) - (r+1)] dr = p ?(r = 0 to 3) [2r v(25 - r^2) - 2r^2 - 2r] dr = p [(-2/3)(25 - r^2)^(3/2) - 2r^3/3 - r^2] {for r = 0 to 3} = 41p/3.