Find the volume of the solid bounded inside the sphere x^2+y^2+z^2=9 and outside

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Explanation / Answer

This is best done with cylindrical coordinates. x^2 + y^2 + z^2 = 81 ==> r^2 + x^2 = 81 ==> z = v(81 - r^2), because z is not negative z = 8 v(x^2 + y^2) ==> z = 8r. These intersect when v(81 - r^2) = 8r ==> r = 9/v65. Since this traces a complete circle of radius 9/v65, the volume equals ??? 1 dV = ?(? = 0 to 2p) ?(r = 0 to 9/v65) ?(z = 8r to v(81 - r^2)) 1 * r dz dr d?) = 2p ?(r = 0 to 9/v65) r [v(81 - r^2) - 8r] dr = 2p ?(r = 0 to 9/v65) [r (81 - r^2)^(1/2) - 8r^2] dr = 2p [(-1/3) (81 - r^2)^(3/2) - (8/3) r^3] {for r = 0 to 9/v65} = p(486 - 3888/v65).

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