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Volume of a rectangle through a sphere

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data

    1. The problem statement, all variables and given/known data
    Suppose that a square hole with sides of length 2 is cut symmetrically through
    the center of a sphere of radius 2. Show that the volume removed is given by
    Screen_Shot_2011_11_13_at_11_12_35_PM.png
    where
    Screen_Shot_2011_11_13_at_11_12_49_PM.png


    I'm not sure how to approach this, but I figure you can express the sphere in terms of f(x,y,z) = [itex]x^2+y^2+z^2=4[/itex]
    then I express z in terms of x and y
    [itex]\sqrt{4-y^2-x^2}[/itex]

    But now I'm guessing I need to set up a double integral. Looking at bounds, -2<x<2 and -1<y<1,

    [itex]\int_{-2}^{2}\int_{-1}^{1}y\sqrt{4-x^2-y^2}dydx[/itex]

    Is this the right way to go???
    How then would I show it is given by the above integral?
     
  2. jcsd
  3. Nov 13, 2011 #2

    Dick

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    You are thinking about the hole going parallel to the z axis, right? Did you mean the integrand to be 2*sqrt(4-x^2-y^2)? And why should the x limits be different from the y limits if you want to cut out a square hole? The original question is a little confusing. You could just put F(x)=V. But I suspect they just want you do the dy integration and then use symmetry.
     
  4. Nov 14, 2011 #3

    HallsofIvy

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  5. Nov 14, 2011 #4


    Show, yeah, that's right. Now you livin' in the big house:

    [tex]\text{myblackarea(x)}=\int_{-1}^{1} 2\sqrt{4-x^2-y^2}dy[/tex]

    [tex]\text{myvolume}=\int_{0}^{1} \text{2 myblackarea(x)}dx[/tex]

    And if you take the time to learn how to do this, you'll never have a problem with these integrals ever again. My point is, the act of creating the plot, cultivates an intutitive understanding of the underlying mathematics. So tell your teacher I said ask everyone in class to draw this picture even if it takes them 6 hours to figure out how.
     

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    Last edited: Nov 14, 2011
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