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Finding the mass of a solid, using Spherical Coordinates.

  1. Jul 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the mass of the solid bounded from above z = √(25 - x2-y2) and below from z = 4, if its density is δ = k(x^2 + y^2 + z^2)^(-1/2).

    2. Relevant equations
    m = ∫∫∫δdV

    3. The attempt at a solution
    ajjZKoy.png
    The plane z = 4 is transformed into ρcosφ = 4, that is, ρ = 4secφ. And x^2 + y^2 + z^2 = 25 is ρ = 5. θ goes from 0 to 2π. But I'm struggling finding the limits for φ (the azimuth). I think φ must go from 0 to π/2, but I can't get the correct answer (which is kπ). The book suggests that I must express the upper limit for φ as an inverse cosine, but why is that? From the figure I can see φ goes from 0 to 90°.

    Thanks in advance
     
  2. jcsd
  3. Jul 30, 2015 #2

    HallsofIvy

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    Yes, [itex]\rho[/itex], the radial variable, goes from 0 to 5 and [itex]\theta[/itex], the "longitude" goes from 0 to [itex]2\pi[/itex]. [itex]\phi[/itex], the "co-latitude" goes from 0 to arcsin(1/5) since line making angle [itex]\phi[/itex] with the z-axis, to a point on the hemi-sphere with z= 1, is the hypotenuse of a right triangle with "opposite side" 1.
     
  4. Jul 30, 2015 #3
    Wait, ρ goes from 0 to 5? I tought ρ went from 4secφ to 5. And why do you use z = 1?
     
  5. Jul 30, 2015 #4

    LCKurtz

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    You are correct for ##\rho## and ##\phi## from 0 to ##\arcsin(\frac 3 5)## or ##\arccos(\frac 4 5)##.
     
  6. Jul 30, 2015 #5
    Thanks both of you for your help. The result it's correct :)
     
  7. Jul 31, 2015 #6

    HallsofIvy

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    I used z= 1 because I misread your post, thinking the base was 4 places below the top, at z= 1, rather than at z= 4. Of course, you should use z= 4, not z= 1.
     
  8. Jul 31, 2015 #7
    Thank you, and thanks for the idea of the triangle.
     
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