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Triple Integral of a cone bounded by a plane.

  1. Nov 17, 2014 #1
    1. The problem statement, all variables and given/known data

    find the volume using spherical coordinates of the region bounded above by z=9 and below by z=sqrt(x^2+y^2) in the first octant.
    2. Relevant equations


    3. The attempt at a solution
    I found this volume using cartesian and cylindrical coordinates, so I know the answer I am looking for, but I can't for the life of me get it in spherical coordinates.

    The region that I am solving for is simply a cone with a height of 9, correct? At z =9 the height is 9. at z = 9 the radius is also 9. So by the equation for volume of a cone, it would be pi r^2 h/3, which = 9^3/3*pi, then divided by 4 since we just want the first octant, which is a quarter of the cone, right? I am sure I am over complicating this somehow.

    I found rho to be 9/cos(phi) which is 9/cos(pi/4) and (phi = pi/4) and theta is pi/2. So the integral is
    ∫∫∫ ρ2sinφ dρdφdθ. 0≤ρ≤9/cos(π/4), 0≤φ≤π/4, 0≤θ≤π/2

    Am I doing something wrong?
     
  2. jcsd
  3. Nov 17, 2014 #2

    Dick

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    Yes, you are doing something wrong. Shouldn't the upper limit of ##\rho## depend on ##\phi##?
     
  4. Nov 17, 2014 #3
    Yes. I figured it out. thank you, though
     
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