(879,#21) Using spherical coordinates, find the volume of the solid that lies above the cone phi = Pi/3 and below the sphere rho = 4 cos(phi).

Solution: The sphere is centered at (0,0, 2) and is of radius 2. The solid is generated by letting phi vary from 0 to Pi/3, and then rotating about the z-axis. The solid looks like this:

The volume integral is

It value is 10 Pi:

lprint( int(rho^2 *sin(phi), rho) );
lprint( int(rho^2 *sin(phi), rho = 0..4*cos(phi)) );
1/3*rho^3*sin(phi)
64/3*cos(phi)^3*sin(phi)

lprint( int(int(rho^2 *sin(phi), rho = 0..4*cos(phi)), phi) );
lprint( int(int(rho^2 *sin(phi), rho = 0..4*cos(phi)), phi = 0..Pi/3) );

-16/3*cos(phi)^4
5

lprint( int(int(int(rho^2 *sin(phi), rho = 0..4*cos(phi)), phi = 0..Pi/3), t ));
lprint( int(int(int(rho^2 *sin(phi), rho = 0..4*cos(phi)), phi = 0..Pi/3), t = 0..2*Pi));
5*t
10*Pi

by:sh