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Cylindrial Coordinates to evaluate the integral

  1. Sep 28, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-9-27_22-24-29.png

    2. Relevant equations
    x^2+y^2 = r^2

    3. The attempt at a solution
    I think it is asking me to find the volume of the sphere, which is in the first octant (1/8 of the sphere)

    So I set
    0<= z<= √1-r2
    0 <= r <= 1
    0<=θ<=π/2
     
  2. jcsd
  3. Sep 28, 2016 #2

    Ssnow

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    Hi, remember that if you want the volume of ##D## you must calculate ##\int\int\int_{D}1dxdydz## where ##f(x,y,z)=1## ...
     
  4. Sep 28, 2016 #3
    I think the question wants me to use the du dv
     
  5. Sep 28, 2016 #4

    Ssnow

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    no because ##f(x,y,z)\not=1## ...
     
  6. Sep 28, 2016 #5
    ?
     
  7. Sep 28, 2016 #6

    Ssnow

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    What I want to say is that your coordinates are ok ## x=r\cos{\theta}, y=r\sin{\theta}, z=z ## but evaluating the integral you don't have the volume of the sphere (as you said ...) but simply a result that is the evaluation of the integral ...
     
  8. Sep 28, 2016 #7
    So simply put, do I just plug in the limits:
    0<= z<= √1-r2
    0 <= r <= 1
    0<=θ<=π/2

    and into the zex2+y2+z2

    then the integral will be something like ∫∫∫∫ z dz r dr dθ
     
  9. Sep 28, 2016 #8

    LCKurtz

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    Your integrand will be ##ze^{x^2+y^2+z^2}## in cylindrical coordinates with ##dV = r~dz~dr~d\theta##.
     
  10. Sep 28, 2016 #9
    Okay but where does z goes (it was before ##e^{x^2+y^2+z^2}##)
     
  11. Sep 28, 2016 #10

    LCKurtz

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    ???. Right where I have it.
     
  12. Sep 28, 2016 #11
    I rewrite my thought
    Originally I have limits:
    0<= z<= √1-r2
    0 <= r <= 1
    0<=θ<=π/2

    Then I use them on ∫ ∫ ∫

    replace also ex^2+y^2+z^2 with er2+z2

    Then I have ∫ ∫ ∫ z er2+z2 r dz dr dθ
     
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