find the volume of the region bounded above by the the sphere x^2+y^2+z^2=100 an

Question

find the volume of the region bounded above by the the sphere x^2+y^2+z^2=100 and below by the cone z=(x^2+y^2)^1/2. Complete, detailed and show all path of geting to the answer, thanks.

Explanation / Answer

Using cylindrical coordinates, x^2 + y^2 + z^2 = 9 ==> r^2 + z^2 = 9 ==> z = v(9 - r^2), since z > 0 in this region 4z = v5 (x^2 + y^2) ==> z = (v5/4) r^2. These intersect when v(9 - r^2) = (v5/4) r^2 ==> 9 - r^2 = (5/16) r^4 ==> 5r^4 + 16r^2 - 144 = 0 ==> (5r^2 + 36) (r^2 - 4) = 0 ==> r = 2 (since we take r to be positive and real). Hence, the volume ??? 1 dV equals ?(? = 0 to 2p) ?(r = 0 to 2) ?(z = (v5/4) r^2 to v(9 - r^2)) 1 * (r dz dr d?) = 2p ?(r = 0 to 2) r [v(9 - r^2) - (v5/4) r^2] dr = 2p ?(r = 0 to 2) [rv(9 - r^2) - (v5/4) r^3] dr = 2p [(-1/3)(9 - r^2)^(3/2) - (v5/16) r^4] {for r = 0 to 2} = 2p [((-1/3) 5^(3/2) - v5) + 9] = 2p [9 - (8/3)v5].

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