1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:


    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Homework Helper

    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]


    [tex]\int\limits_0^{2} {\int\limits_0^{{\raise0.5ex\hbox{$\scriptstyle \pi $}
    \lower0.25ex\hbox{$\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. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook