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Volume (Double Integral)

  1. Nov 10, 2012 #1
    Find the exact volume of the solid between the paraboloids [itex]z=2x ^{2}+y ^{2}[/itex] and [itex]z=8-x ^{2}-2y ^{2}[/itex] and inside the cylinder [itex]x ^{2}+y ^{2}=1[/itex].

    I really don't know how to set this up. Would it be something like ∫∫(2x^2+y^2)-(8-x^2-2y^2)dA + ∫∫(x^2+y^2-1)dA ?

    If so, how would I find the bounds of integration?
  2. jcsd
  3. Nov 10, 2012 #2
    Just like integrating areas between two curves in the plane, you need to determine additional information. First, which surface is on top and which surface is below? Rigorously, which function's values are greater in the region we are interested in? This is important for which volume is subtracted from which, as we want the volume of a solid to be positive. If you are not familiar with the equations of paraboloids, graph some cross sections to get a picture of the surface. If you have access to a 3d-graphing system, use that to verify your answer.
    Next, where do the surfaces intersect within the region of interest, if anywhere? If they intersect within the region you are integrating over (the circle x^2 + y^2 = 1), you may need to break up the volume integral, as the volume changes sign if the surface that used to be below crosses the second surface and is now on top (just like when curves cross each other when doing area integrals).
    Finally, use your information to determine the region in the xy-plane over which you will integrate the volume under the surfaces. If they do not intersect, you are simply integrating over the circular region x^2+y^2 = 1, which covers a cylindrical region of volume. Once you set up your integral correctly, you may want to change to a more convenient system of coordinates for integrating over a circular region, such as cylindrical coordinates.
    Last edited: Nov 10, 2012
  4. Nov 10, 2012 #3

    Simon Bridge

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    What he said :)

    If you already know how to find the volume under a surface z=f(x,y) inside a cylinder, the this problem is likely a matter of doing exactly that for each surface and finding the difference. The likely gotcha is if the surfaces themselves intersect inside the cylinder.

    It is often easier to change coordinates to suit the symmetry.
  5. Nov 10, 2012 #4
    Thanks guys! z=8-x^2-2y^2 is on top

    So would it be... ∫∫(8-x^2-2y^2)-(2x^2+y^2)dA + ∫∫(x^2+y^2-1)dA ?

    I get how to solve the function for the volume, but typically I would find the bounds by setting the surfaces equal to each other to find intersection points and such, but setting them equal...
    I get 9=4y^2 + 4x^2
  6. Nov 10, 2012 #5

    Simon Bridge

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    So the intersection is a circle at a constant z?! (you should always relate your equations to geometry.)
    1. Check to see if any part of that intersection lies inside x^2+y^2=1 (the cylinder).

    if none of it does - then the limits are entirely determined by the bounds of the cylinder instead of the intersection curve. (You realize you can do a volume integral between any limits you like - you are not restricted to the region between two surfaces?)
  7. Nov 11, 2012 #6
    Alright! So, I found that...
    [itex]\int\limits_{-1}^{1}\int\limits_{-\sqrt{1-x ^{2}}}^{\sqrt{1-x ^{2}}}8-3x ^{2}-3y ^{2}dydx[/itex]
    I found the square root part because [itex]y=\sqrt{1-x ^{2}}[/itex]

    Is this correct, so all I have to do now is solve?

    Thank you
  8. Nov 11, 2012 #7

    Simon Bridge

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    That looks like the right shape - you understand that I'm not going to actually do it unless someone pays me :)

    Did you check to see if any of the intersection of the surfaces was inside the cylinder in question?
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