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Homework Help: Volume Problem (Can someone check my work?)

  1. Aug 27, 2005 #1
    "Find the volume of the solid that lies within the sphere [tex]x^2+y^2+z^2=4[/tex], above the x-y plane, and below the cone [tex]z=\sqrt{x^2+y^2}[/tex]."

    I set this up in polar coordinates as follows:

    [tex]V=\int_{0}^{2\pi}\int_{0}^{\sqrt{2}}r^2drd\theta+\int_{0}^{2\pi}\int_{\sqrt{2}}^{2}r\sqrt{4-r^2}drd\theta[/tex]

    and then solved it coming up with [tex]V=\frac{8\pi\sqrt{2}}{3}[/tex].

    Is this the correct way to set this up? Also, is there a way I could do this with one double integral rather than adding two of them together?

    Thanks for your help.
     
    Last edited: Aug 27, 2005
  2. jcsd
  3. Aug 28, 2005 #2

    HallsofIvy

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    I think that will work but the way I would analyze it is this: On the sphere, [itex]z= \sqrt{4- x^2-y^2}= \sqrt{4- r^2}[/itex]. On the cone, [itex]z= \sqrt{x^2+ y^2}= r[/itex]. So the "height" at each point is [itex]\sqrt{4- r^2}- r[/itex]. You want to integrate that, times [itex]dA= rdrd\theta[/itex], over the disk inside the intersection of the two figures projected down to the xy-plane.
    The sphere and cone intersect when [itex]\sqrt{4- r^2}= r[/itex] or [itex]4- r^2= r^2[/tex] so r2= 2 or [itex]r= \sqrt{2}[/itex] just as you have.

    The volume is [tex]\int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta[/itex].


    Personally, I would put it in spherical coordinates and do it as a single triple integral:
    [itex]\theta[/itex] goes from 0 to [itex]2\pi[/itex], [itex]\phi[/itex] goes from 0 (vertical) to [itex]\frac{\pi}{4}[/itex] (the cone) and [itex]\rho[/itex] goes from 0 to 2 (the sphere). Integrate [itex]dV= \rho sin(\phi)d\theta d\phi d\rho[/itex] over that.
     
    Last edited by a moderator: Aug 28, 2005
  4. Aug 28, 2005 #3
    Ok that makes sense. Thanks for your help HallsofIvy :smile:
     
  5. Aug 28, 2005 #4

    TD

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    Ivy, I computed your approach as well and I don't find the same as your first solution, and they're both different from apmcavoy's answer.

    I get:

    [tex]\int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta = \frac{-8\,\left( -2 + {\sqrt{2}} \right) \,\pi }{3}[/tex]

    While:

    [tex]\int\limits_0^{2} {\int\limits_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle 4$}}} {\int\limits_0^{2 \pi} {\rho \sin \left( \phi \right)} } } d\theta d\phi d\rho = -2\,\left( -2 + {\sqrt{2}} \right) \,\pi[/tex]

    Any idea what's wrong?
     
  6. Aug 28, 2005 #5
    I may be wrong, but doesn't [tex]\int_{\theta= 0}^{2\pi}\int_{r=0}^{\sqrt{2}}\left(\sqrt{4- r^2}- r\right)rdrd\theta[/tex] represent the volume above the cone and within the sphere?
     
  7. Nov 3, 2005 #6
    try changing from psin to p^2sin. :)
     
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