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Ellipsoid intersected by cylinder

  1. Jul 26, 2014 #1
    1. The problem statement, all variables and given/known data

    Find volume of the ellipsoid ##x^2 +2(y^2+z^2) \le 10## intersected by the cylinder ##y^2 + z^2 \le 1 ##

    3. The attempt at a solution


    seems like changing to cylindrical coordinates would be best

    so I have

    [tex]\left\{ \begin{array}{cc} r^2cos\theta + 2r^2sin^2\theta +2z^2 \le 10 \\r^2sin^2\theta +z^2 \le 1 \end{array}\right.[/tex]

    [tex]0 \le \theta \le 2\pi[/tex]

    and solving for z [tex]-\sqrt{1-r^2sin^2\theta} \le z \le \sqrt{1-r^2sin^2\theta}[/tex]

    and [tex]-\sqrt{8}sec\theta \le r \le \sqrt{8}sec\theta[/tex]

    [tex]\int_{0}^{2\pi}\Bigg(\int_{-\sqrt{1-r^2sin^2\theta}}^{\sqrt{1-r^2sin^2\theta}}\Bigg(\int_{-\sqrt{8}sec\theta}^{\sqrt{8}sec\theta}rdr\Bigg)dz\Bigg)d\theta[/tex]

    I tried it on Wolfram|Alpha to check but it didn't seem to work.
     
    Last edited: Jul 26, 2014
  2. jcsd
  3. Jul 26, 2014 #2
    Two things, when you set up the boundaries of your integral you have to make sure you eliminate a variable at each step. Therefore you need to do the z integral before the r integral. Otherwise, you will integrate r and then introduce another r which you will not be able to get rid of.

    Second, r is defined from the origin so you integrate from 0 to the maximum value.
     
  4. Jul 26, 2014 #3

    LCKurtz

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    You can use
    Code (Text):
     ##...##
    brackets to enclose tex statements. Presumably that second equation is a typo and means ##y^2 \color{red} + z^2 \le 1##

    Cylindrical coordinates is a good idea, but not in the usual variables. Your cylinder and ellipsoid are both circular in the yz variables. So you want to use cylindrical like variables where the ##r## and ##\theta## variables are used in the ##yz## plane and you treat the ##x## axis as though it were the ##z## axis and integrate that direction first.
     
  5. Jul 26, 2014 #4
    oh yeah, I didn't think of that. thanks
     
  6. Jul 26, 2014 #5
    before I go use up scratch paper. Im going to treat x as z, so should I set ##z = r cos \theta##? so my ##0 \le \theta \le 2\pi## will revolve about the z-axis as opposed to the x axis


    EDIT: better yet after looking at it in grapher I am thinking ##y=rcos\theta ; z = rsin\theta##
     
    Last edited: Jul 26, 2014
  7. Jul 26, 2014 #6
    ##x^2 +2r^2(cos^2\theta + sin^2\theta)=10##

    ##r^2(cos^2\theta + sin^2\theta)=1##

    ##\int_{0}^{2\pi}\Bigg(\int_{-1}^{1}\Bigg(\int_{-\sqrt{10-2r^2}}^{\sqrt{10-2r^2}}dx\Bigg)rdr\Bigg)d\theta##

    EDIT: I think the bounds are still amiss though because I got 0 which cannot be a correct answer for the volume
     
    Last edited: Jul 26, 2014
  8. Jul 26, 2014 #7
    Think about what each of ##\theta, r, x## represent and then have another look at your bounds.
     
  9. Jul 26, 2014 #8
    the ##\theta## bounds makes sense because theres a full revolution the the ##r## bounds are the cylinder so its on the interval [0,1]

    still seems to not work

    EDIT: nevermind I got it.
     
    Last edited: Jul 26, 2014
  10. Jul 26, 2014 #9

    LCKurtz

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    You don't want negative values of ##r##. You get the whole circle for positive ##r## and ##\theta## going all the way around. Much simpler this way, don't you think?
     
  11. Jul 26, 2014 #10
    yeah I wasn't thinking in terms of polar. thanks!


    ##\int_{0}^{2\pi}\Bigg(\int_{0}^{1}\Bigg(\int_{-\sqrt{10-2r^2}}^{\sqrt{10-2r^2}}dx\Bigg)rdr\Bigg)d\theta##
     
  12. Jul 26, 2014 #11

    HallsofIvy

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    You could do this by looking up formulas rather than integrating. The volume of the ellipsoid cut by the cylinder is the volume of the cylinder inside the ellipsoid plus the volume of the two caps at each end of the cylinder. The ellipsoid is bounded by [itex]x^2+ 2(y^2+ z^2)= 10[/itex] and the cylinder by [itex]y^2+ z^2= 1[/itex]. Obviously the cylinder cuts the ellipsoid when [itex]x^2+ 2(y^2+ z^2)= x^2+ 2(1)= 10[/itex], [itex]x^2= 8[/itex], [itex]x= \pm\sqrt{8}= \pm 2\sqrt{2}[/itex].

    So the volume we want is a cylinder with radius 1 and height [itex]2\sqrt{2}- (-2\sqrt{2})= 4\sqrt{2}[/itex] plus the two spherical caps with bases disks of radius 1 and height [itex]\sqrt{10}- 1[/itex].
     
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