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Double integral and polar cordinates other problem.

  1. Nov 16, 2012 #1
    If we have to find the volume, written in polar cordinates, inside this sphere X2+y2+z2=16 and outside this cylinder x2+y2=4

    How should I approach this?
    Could I take the sphere function and reqrite in polar cordinates z=√(16-X2-y2) which is the same as z=√(16-r2)

    But then I have to make r depend on a function of the cylinder right?

    so x2+y2=4 ---> r2=4 so r=2 this must be the boundaries of the cylinder..

    Now I get a bit confused. Do we subtract the sphere function from the cylinder function? Or do we make the cylinder function a function that r depends on?
     
  2. jcsd
  3. Nov 16, 2012 #2

    SammyS

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    What does r have to be to be outside the cylinder?
     
  4. Nov 17, 2012 #3
    Hmm then this would make sense 2<r<4 because the cylinder has a radius of 2, so i guess the theta would go from 0 to 2pi ?
     
  5. Nov 17, 2012 #4

    SammyS

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    Yes.

    What is the function you integrate, i.e., what is your integrand?
     
  6. Nov 17, 2012 #5
    The function we want to integrate must be the sphere, right?
    X^2+y^2+z^2=16 so if we isolate z we must get z=√(16-X^2-y^2) and written in polar form
    z=√(16-r^2) So this is the function we integrate?
     
  7. Nov 17, 2012 #6

    SammyS

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    That's the equation for the top half of the sphere, the half for which z is positive.
     
  8. Nov 17, 2012 #7
    I see, so if 2<r<4 and 0< theta< 2pi
    Hmm so we could add two double integrals? one of the lower, and one of the upper sphere?
     
  9. Nov 17, 2012 #8

    SammyS

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    ... or multiply the first integral by 2, because of the symmetry involved.

    Do you know what dxdy , the element of area, is in polar coordinates?
     
  10. Nov 17, 2012 #9
    yes that's a smart solution :) The polar form of dxdy must be r*dr*dθ Right?

    I assume it means the rectangles delta r times delta θ multiplied by the length r (a bit hard for me to visualize)
     
  11. Nov 17, 2012 #10
    So 2*∫(∫((√(16-r^2) )*r,r,2,4),θ,0,2∏) This must be the solution? :)
     
  12. Nov 17, 2012 #11

    SammyS

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    2*∫∫(√(16-r^2) )*r*dr*dθ; where r goes from 2 to 4, θ goes from 0 to 2π .

    In my opinion, you should always include the differentials when writing an integral.
     
  13. Nov 18, 2012 #12
    Ahh yes a good idea. Thank you very much for your help :)
     
  14. Nov 18, 2012 #13
    Hmm but is there an easier way? Because (√16-r2)r is not that easy to do by hand, and id love to be able to do this by hand :)
     
  15. Nov 18, 2012 #14

    SammyS

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    [STRIKE]Multiply through by r .[/STRIKE]

    Added in Edit:

    Putting the parentheses in the correct place gives
    ( √(16-r2) )r​

    Do integration by substitution.

    What is the derivative of 16-r2 ?
     
    Last edited: Nov 18, 2012
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