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Finding the volume of a sphere sliced by a plane

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the volume of the region above the plane z = 2 and inside the sphere x2 +y2 +z2 = 9.


    2. Relevant equations


    3. The attempt at a solution
    I am trying to use a spherical triple integral. I am mainly having trouble finding ∅.
    This is what I have been trying ∫arccos(2/3)02∏32 ρ2sin∅dpdθp∅
    The answer is 8∏/3. I am pretty sure there are errors in my setup
     
  2. jcsd
  3. Apr 20, 2012 #2

    sharks

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    You should plot the graph to help you visualize the volume and its boundaries.
    Using spherical coordinates:
    [tex]\iiint {\rho}^2 \sin \phi \,.d\rho d\phi d\theta[/tex]

    Then describe the region (the crucial step in determining the limits):
    For [itex]\phi[/itex] and [itex]\theta[/itex] fixed, [itex]\rho[/itex] varies from [itex]\frac{2}{\cos \phi}[/itex] to 3
    For [itex]\theta[/itex] fixed, [itex]\phi[/itex] varies from 0 to [itex]\cos^{-1} \frac{2}{3}[/itex]
    [itex]\theta[/itex] varies from 0 to 2∏

    In order to get the exact answer, you need to work with the inverse angles accordingly, without using their decimal counterparts, which will obviously give you a decimal answer.
     
    Last edited: Apr 20, 2012
  4. Apr 20, 2012 #3
    Would you mind explaining where you got 2/cos∅ from? Also, can you recommend a good utility to graph these functions?
    Thanks!
     
  5. Apr 20, 2012 #4

    sharks

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    What is the definition of z in terms of spherical coordinates?
    Example, x in terms of spherical coordinates is given by: [itex]\rho \sin \phi \cos \theta[/itex]

    Personally, i just draw them on paper, and i recommend you do the same, unless you're allowed to bring special graphing equipments with you in the exams.
    Maybe you could try this: http://www.wolframalpha.com/
    Although, i think someone else might be better able to advise you on finding the best software for graphing in 3D.
     
  6. Apr 21, 2012 #5

    ehild

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    The integration would be much easier in cylindrical coordinates. The volume element can be the thin slice of thickness dz at height z with radius ρ=√(9-z2).


    ehild
     

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  7. Apr 21, 2012 #6

    sharks

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    Using ehild's suggestion, you would get:

    [tex]\iiint rdrd\theta dz[/tex]

    Description of region:
    For r and θ fixed, z varies from z=[itex]\sqrt{(9-r^2)}[/itex] to z=3.
    For θ fixed, r varies from r=0 to r=√5
    θ varies from 0 to 2∏

    You should get the same answer.
     
  8. Apr 21, 2012 #7

    ehild

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    I meant the integral
    [tex]V=\int_2^3{\int_0^{\sqrt{9-z^2}}{r \int_0^{2\pi}{d\theta}dr}dz}[/tex]
    which is equivalent to integrating the volume element dV=ρ2πdz with respect to z from z=2 to z=3 where ρ=√(9-z^2)

    [tex]V=\pi\int_2^3{(9-z^2)dz}[/tex]

    ehild
     
    Last edited: Apr 21, 2012
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