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Triple integral

  1. Aug 8, 2011 #1
    Hi, I have to compute de volume for the region:

    [tex]\{(x,y,z):x^2+y^2+z^2\leq{16};z\geq{1} \}[/tex]

    I've tried to do by two different parametrizations, in spherics and cylindrical coordinates

    For cylindrical coordinates I've made:

    [tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}dz d \theta dr\approx{162.26}[/tex]

    For spherics I've considered [tex]4\sin \phi=\sqrt[ ]{15}\Rightarrow{\phi=Arcsen\left( \displaystyle\frac{\sqrt[ ]{15}}{4} \right)}=u[/tex]

    [tex]3 \displaystyle\int_{0}^{u}\int_{0}^{2\pi}\int_{\cos \phi}^{4}d\rho d \theta d\phi\approx{81.13 }[/tex]

    I've used wolfram to make the calculus, and it gives that for cylindrical coord. I'm having twice the volume that I have for spherical coord. I don't know where I'm committing the mistake. Bye, thanks for helping :)

    Note: I'm multiplying by 3 because I was going to use for Gauss theorem and is the divergence for the vector field given by the problem
     
  2. jcsd
  3. Aug 8, 2011 #2
    wheres your Jacobian for spherical and cylindrical coordinates.
     
  4. Aug 8, 2011 #3
    Thanks :P I forgot about that, I thought I wasn't making any change of variables, but now I see. It's because I've been doing some surface integrals and the jacobian doesn't appear on those kind of integrals, I'm not sure why is that, but I think that role is played by the normal vector, which is pretty much like the jacobian.
     
    Last edited: Aug 8, 2011
  5. Aug 9, 2011 #4
    Well, I added the jacobians, but still giving different, I don't know why. The difference now is more sutil: In cylindrical coord. gives (clic here)

    In spherical coord gives


    I tried to proceed with cylindrical, but the exercise gets more complicated, so I want to be sure that what I'm doing with spherical is ok, so I can proceed that way, but still giving different answers.
     
    Last edited: Aug 9, 2011
  6. Aug 9, 2011 #5
    I think you bound for phi in spherical should be [itex] \phi=arctan(\sqrt{15}) [/itex]
    and not your arcsin that you have.
     
  7. Aug 9, 2011 #6
    why is that? I've reasoned the bound this way:

    attachment.php?attachmentid=37870&stc=1&d=1312912197.png

    Then [tex]4 \sin \phi=\sqrt[]{15}[/tex]
     

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  8. Aug 9, 2011 #7

    vela

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    That gives the same limits because you have, equivalently,
    \begin{align*}
    \cos \phi &= \frac{1}{4} \\
    \sin \phi &= \sqrt{1-\left(\frac{1}{4}\right)^2} = \frac{\sqrt{15}}{4} \\
    \tan \phi &= \frac{\sin\phi}{\cos\phi} = \sqrt{15}
    \end{align*}
    I think the lower bound of [itex]\rho[/itex] is wrong.
     
  9. Aug 9, 2011 #8
    Oh, yes, it says cos phi, but I used secant in wolfram. I'll correct the typo, thanks vela.

    I can´t edit now the first post, but this is the integration I've computed with wolfram:


    [tex]3 \displaystyle\int_{0}^{u}\int_{0}^{2\pi}\int_{\sec \phi}^{4}\rho^2 \sin \phi d\rho d \theta d\phi=81\pi[/tex]

    And for cylindrical:


    [tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}r dz d \theta dr\approx{80\pi}[/tex]
     
  10. Aug 9, 2011 #9
    their equivalent i just did it with tangent. Yours is fine . It looks like your bounds are right but i did both integral and got 80 and 100 . so i got different answers but maybe i made a mistake.
     
  11. Aug 9, 2011 #10

    vela

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    I got [itex]81\pi[/itex] for both using Mathematica.
     
  12. Aug 9, 2011 #11
    Thanks vela, perhaps it's only a bug in wolfram alpha :)
     
  13. Aug 9, 2011 #12
    [tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}r dz d \theta dr=80\pi[/tex]

    This is actually right, I just corroborated it doing it by hand:

    [tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}\int_{1}^{\sqrt[ ]{16-r^2}}r dz d \theta dr=3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}(r\sqrt[ ]{16-r^2}-r)d \theta dr[/tex]
    Using:
    [tex]u=16-r^2\rightarrow du=-2r[/tex]
    [tex]3 \displaystyle\int_{0}^{\sqrt[ ]{15}}\int_{0}^{2\pi}(r\sqrt[ ]{16-r^2}-r)d \theta dr=-3\pi\displaystyle\int_{16}^{0}\sqrt[ ]{u}du-6\pi \left( \frac{r^2}{2} \right ) _0^4=2\pi 16^{3/2}-48\pi=80\pi[/tex]
     
  14. Aug 9, 2011 #13

    vela

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    You got the limits for u wrong and plugged in the wrong limits for the other term.
     
  15. Aug 9, 2011 #14
    Uups you're right. That was the mistake, thank you vela :) I made a big mess with this involving spherical and cylindrical coordinates :P but the thing is that for the volume it's easier to compute it using cylindrical coordinates (unless I think so). But for the surface integrals spheric coordinates seems to be better (gives simpler expressions).

    I wanted to corroborate the Gauss theorem for that region, the problem gives a vector field. I have some doubts about that too, I should make another topic for that, right?
     
  16. Aug 9, 2011 #15

    vela

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    Probably a good idea to start a new thread for that.
     
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