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Where is the error?

  1. Mar 26, 2015 #1
    1. The problem statement, all variables and given/known data
    Problem statement.PNG

    2. Relevant equations
    Spherical Coordinates.PNG
    And the volume element dV expressed in spherical coordinates.

    3. The attempt at a solution
    1.PNG

    Can you spot where I've made an error?
     
    Last edited: Mar 26, 2015
  2. jcsd
  3. Mar 26, 2015 #2

    SammyS

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    What's you question?

    Oh! I see.

    Usually the question is posted in the text of the thread.
     
  4. Mar 26, 2015 #3
    Where is the error? * (Where have I made an error)

    -Ok! Edited*
     
  5. Mar 26, 2015 #4

    SammyS

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    I don't see the error after going through what you did twice.

    What is the correct answer?
     
  6. Mar 26, 2015 #5

    Ray Vickson

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    Why do you think you have made an error?
     
  7. Mar 26, 2015 #6

    haruspex

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    There is a slightly easier way. By symmetry you can change the x to z in the integrand.
     
  8. Mar 27, 2015 #7
    My study friends have gotten a different result than I. (Pi/8). And my intuition tells me that by integrating over radius=1, I should get something that looks like 1/8 the volume of a sphere of radius 1 (Pi/6). Would someone like to clear this up? I'd like to know if I'm doing something wrong.
     
    Last edited: Mar 27, 2015
  9. Mar 27, 2015 #8

    haruspex

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    Sure but the integrand, x, is on average less than 1/2.
    You can check your answer by looking up the mass centre of a hemisphere.
     
  10. Mar 27, 2015 #9
    I turned in the assigment. Giving it another shot tonight.
     
  11. Mar 27, 2015 #10

    Ray Vickson

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    Do it the easy way: ##V = \int_0^1 x A(x) \, dx##, where ##A(x)## is the area of the slice at ##x## parallel to the ##yz## plane. The slice is a quarter circle of radius ##\sqrt{1-x^2}##, so ##A(x) = \frac{1}{4} \pi (1-x^2)##. This gives ##V = \pi/16##, as you obtained.
     
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