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Math Help - Spherical Coordinate Triple Integral Problem

  1. #1
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    Spherical Coordinate Triple Integral Problem

    I've been banging my head against the wall for a bit now and can't make any sense of this. I have to use spherical coordinates to find the volume inside the sphere x^2 + y^2 + z^2 = 9, and inside the cone z = (1/sqrt3)(sqrt (x^2 + y^2)).

    I know I have to do a triple integral. Theta will be between 0 and 2pi. Since the radius of the sphere is 3, I think ro (or that p symbol that is basically radius in spherical coordinates) should be between 0 and 3.

    What I am getting stuck on is thi, or the angle between the z and y axis. I tried to figure it out by sticking 1 for x and y in the cone equation and then solving for z, then using the distance formula to figure out ro (the distance from the origin to the point). With that I could then calculate thi since z = ro cos thi....but it ended up being totally off.

    If anyone was able to follow along with this and could lend a helping hand, I would appreciate it SO much. Thanks.
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  2. #2
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    Quote Originally Posted by wolfkrug View Post
    I've been banging my head against the wall for a bit now and can't make any sense of this. I have to use spherical coordinates to find the volume inside the sphere x^2 + y^2 + z^2 = 9, and inside the cone z = (1/sqrt3)(sqrt (x^2 + y^2)).

    I know I have to do a triple integral. Theta will be between 0 and 2pi. Since the radius of the sphere is 3, I think ro (or that p symbol that is basically radius in spherical coordinates) should be between 0 and 3.

    What I am getting stuck on is thi, or the angle between the z and y axis. I tried to figure it out by sticking 1 for x and y in the cone equation and then solving for z, then using the distance formula to figure out ro (the distance from the origin to the point). With that I could then calculate thi since z = ro cos thi....but it ended up being totally off.

    If anyone was able to follow along with this and could lend a helping hand, I would appreciate it SO much. Thanks.
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  3. #3
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    EZ... well the figure would look kinda like an icecream cone with one scoop

    cone: x2 = (1/sqrt3) sqrt(x2+y2); 1/sqrt3 stands for tanphi so phi = pi/6 rad

    its int(0,2pi)int(0,pi/6)int(0,3) rho^2 sin phi drho dphi dtheta
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