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Did I set up my integral correctly?

  1. Nov 12, 2015 #1
    1. The problem statement, all variables and given/known data
    use spherical coordinates to find volume inside a sphere of radius 4 and outside sphere of radius 2, and inside/above the cone z=3√(x^2+y^2).

    2. Relevant equations

    x=rcosθ =ρsinφcosθ , y=rsinθ =ρsinφsinθ
    z=ρcosφ
    r= ρsinφ



    3. The attempt at a solution
    replacing z^2 in the sphere of radius 4 with the cone z^2=9(x^2+y^2) I get the radius of the intersection of √( 8/5), using this radius and known ρ ( 4), φ=inverse sin of (r/ρ) i found φ to be 0.322 or around 18.4 degrees.

    V= ∫ 0 to 2 pi ∫0 to 0.322 ∫2 to 4 (ρ^2sinφ) dρdφdθ or shown in the whiteboard :

    Snapshot.jpg
     
  2. jcsd
  3. Nov 12, 2015 #2

    blue_leaf77

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    I agree with your calculation.
     
  4. Nov 12, 2015 #3

    Ray Vickson

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    You should not take the integral from 0 to 0.322, but rather, take it from 0 to ## p = \arcsin \left( \frac{\sqrt{8/5}}{4} \right)##, because that will give you an exact answer in terms of a nifty formula, rather than a decimal approximation.

    I future, could you PLEASE avoid posting all your really messy handwritten work (as you have done in several recent posts)? Just type it out, which you could probably do almost as quickly as you can make images of the handwritten work and upload the files, etc. This Forum greatly prefers typed solutions, although it has so far not forbidden pictures of handwriting. It would be useful to you now, and in the future, to learn how to use LaTeX; but even the type of ASCII plain text you wrote above is acceptable also.
     
    Last edited: Nov 12, 2015
  5. Nov 12, 2015 #4

    LCKurtz

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    I'm guessing the OP is creating those images with a mouse on the "whiteboard". One of the relatively recent "enhancements" of the PF site that don't make it better.
     
  6. Nov 12, 2015 #5
    Ok, thank you, will do. but was my handwriting that bad :frown:
     
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