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Triple integral volume

  1. Dec 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the object, defined by these inequalities(?): x^2+y^2+z^2≤4, (x-1)^2+y^2≥1, (x+1)^2+y^2≥1


    2. Relevant equations



    3. The attempt at a solution

    First we draw the object, and realize that it's a sphere with 2 circles in it with radius 1 at (-1,0) and (1,0). Our object is defined outside these circles, so I imagine we could consider the square with 2 cylinders drilled out. If we can find the volume of one of these cylinders from 0 to z we are done. (Vsphere-4*Vcylinder)
    Is this the way to go or do we have other solutions? This one doesnt give me right answer...
     
  2. jcsd
  3. Dec 13, 2012 #2

    HallsofIvy

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    I suspect that you have calculated the volume of the cylinders incorrectly. The top and bottom are pieces of the sphere not planes. Yes, this is a sphere of radius 2 and so its volume, ignoring the cylinders for the moment, is [itex](16/3)\pi[/itex].

    By symmetry, it is sufficient to find volume of one cylinder, then subtract twice that from the volume of the sphere. Also, I think I would be inclined to shift the coordinate system so that the center of one cylinder is at the origin: if we let x'= x- 1, y'= y, z'= z, then one cylinder has equation [itex]x'^2+ y'^2= 1[/itex] and the sphere has equation [itex](x'+1)^2+ y'^2+ z'^2= x'^2+ 2x'+ 1+ y'^2+ z'^2= 4[/itex], [itex](x'^2+ y'^2)+ 2x'+ z'^2= 3[/itex].

    Now, convert to cylindrical coordinates: [itex]x'= rcos(\theta)[/itex], [itex]y'= r sin(\theta)[/itex], [itex]z'= z[/itex] so the equation of the sphere becomes [itex]r^2+ 2r cos(\theta)+ z^2= 3[/itex] so that [itex]z= \pm (3- r^2- 2r cos(\theta))^{1/2}[/itex]. Then the volume of the cylinder is
    [tex]\int_{\theta= 0}^{2\pi}\int_{r= 0}^1\int_{z= -(3- r^2- 2rcos(\theta))^{1/2}}^{(3- r^2- 2rcos(\theta))^{1/2}} r cos(\theta)dzdrd\theta[/tex]
     
  4. Dec 14, 2012 #3
    Hi, thx for answering. This was kinda the vision I had, tho i was unfamiliar with cylindrical coordinates. I dont understand where cos(θ) comes from in the integrand? The jacobian from the substitution would just be r, no? Or have I missed something fundamental here? The double integral gets kinda messy after we've integrated with respect to z, in my mind.

    And I dont seem to get the correct answer, which should be 128/9.
     
    Last edited: Dec 14, 2012
  5. Dec 14, 2012 #4

    SammyS

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    You are right regarding the Jacobian in your above post.

    Yes.

    You are to find the volume enclosed by the sphere, but external to the two cylinders.

    So, find the portion of one cylinder enclosed by the sphere and above z = 0. Then use your expression, (Vsphere-4*Vcylinder) .

    What is the equation of the circle, (x-1)2+y2 = 1, in polar coordinates?

    (I haven't worked the problem, but I suspect the answer is more like 128π/9 .)
     
  6. Dec 14, 2012 #5
    In polar coordinates it would be r^2-2rcosθ=0 ⇔ r=2cosθ. So it will go from 2cosθ and follow the sphere upwards... nah Im having a hard time visualizing it :(

    Yes, the answer is probably a small typo then.
     
  7. Dec 14, 2012 #6

    SammyS

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    I would integrate over z first: from 0 to √(4-r2)

    Then r goes from 0 to 2cos(θ) .

    Then θ : from -π/2 to π/2 .
     
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