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Homework Help: Calc3 Homework problem

  1. May 4, 2005 #1
    I'm having a problem seeing where I went wrong with this question...

    S is the solid bounded above by the sphere [tex]x^2 + y^2 + z^2 = 8[/tex]
    and below by the paraboloid [tex]x^2 + y^2 = 2z[/tex]. Solve this using spherical coordinates

    The correct answer is [tex]\displaystyle{\frac{4\Pi}{3}}[/tex][tex](8\sqrt{2}-7)[/tex]

    Here is the solution that I achieved....

    [tex]V=\int\int\int dV[/tex]

    After drawing the 3d graph, I decided to just take the volume of one octant and multiply that by 4 (since the graph is only symmetrical about the z-axis and lies above the xy plane.

    [tex]4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\sqrt{2}\thet }\;\rho^2\;\sin\phi\;d\rho\;d\theta\;d\phi[/tex]

    [tex]4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\thet }\;{\frac{\rho^3}{3}}\;\sin\phi\;\mid_{0}^{\2\sqrt{2}}\;d\theta\;d\phi[/tex]

    [tex]\frac{64\sqrt{2}}{3}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\thet }\;\sin\phi\;d\theta\;d\phi[/tex]





    Where is the -7?
    I don't see any place I went wrong at.
  2. jcsd
  3. May 4, 2005 #2
    Maybe I'm misreading but on step two, when you cube [itex] \sqrt{2}[/itex] and multiply by 4, how do you get [itex] \frac{64\sqrt{2}}{3} [/tex]. I'm getting an eight of that.

    Also, your bounds need to reflect that the solid isnt a sphere, your integral is the volume of a sphere of radius [itex] 2\sqrt{2} [/itex]. You should account for the lower bound of the solid in your p limit.
  4. May 4, 2005 #3


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    Where do you take the lower boundary (the paraboloid) into account? As whozum said, you have calculated the volume of the sphere only. Are you really required to use spherical coordinates? Looks to me like cylindrical coordinates would be much easier.

    If you are required to use spherical coordinates, then you want to realize that the sphere and paraboloid intersect where x2+y2+(x2+y2)2/4= ρ2cos2φ+ ρ4sin4φ= 8. You need to determine the value of φ corresponding to that. For φ past that value, ρ will go from 0 to whatever is correct for the paraboloid.
    Last edited by a moderator: May 4, 2005
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