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Quick double integral question

  1. Nov 27, 2012 #1

    Zondrina

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    Homework Helper

    1. The problem statement, all variables and given/known data

    Find the volume of the region R between the surfaces [itex]z = 4x^2 + 2y^2 \space and \space z = 3 + x^2 - y^2[/itex]

    2. Relevant equations



    3. The attempt at a solution

    Okay so I think I have an idea about how to do this one. First I check when the two surfaces intersect, that is when [itex]4x^2 + 2y^2 = 3 + x^2 - y^2[/itex]. So the two surfaces meet when [itex]x^2 + y^2 = 1[/itex] which is a circle of radius 1 with center (0,0).

    So either x or y run from -1 to 1, it doesn't really matter which one I pick here I think, so I'll hold y fixed : [itex]-1 ≤ y ≤ 1[/itex]

    Now solving [itex]x^2 + y^2 = 1[/itex] for x yields [itex]x = ± \sqrt{1-y^2}[/itex] as my upper and lower limits for x.

    Now all that's left to figure out is... what is f? I believe intuitively that f will be the bigger surface minus the smaller surface. So subtracting my two surfaces I get : [itex]3 + x^2 - y^2 - 4x^2 - 2y^2 = 3(1 - x^2 -y^2)[/itex]

    So that my iterated integral becomes :

    [itex]\int_{-1}^{1} \int_{- \sqrt{1-y^2}}^{\sqrt{1-y^2}} 3(1 - x^2 -y^2) \space dxdy[/itex]

    Evaluating that shouldn't be a problem, I'm just hoping I did everything properly while setting it up. If anyone could clarify for me it would be great :)!
     
  2. jcsd
  3. Nov 27, 2012 #2

    Mark44

    Staff: Mentor

    This looks OK. The surface z = 4x2 + 2y2 has a larger z value than the surface z = 3 + x2 - y2 for most points in the domain; namely all of the points that are outside the circle of intersection. The region you're interested in, though, is on or inside the circle. For this region, the second equation has the larger z values.
    I think it would be much easier to convert to polar form. The circle simplifies to r = 1, and the integrand simplifies to 3(1 - r2).
     
  4. Nov 27, 2012 #3

    mfb

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    f(x,y)=1-x2-y2 gives a [strike]sphere[/strike] paraboloid, you can use this to avoid integrals at all (and add the prefactor later). Otherwise, polar coordinates are useful.

    Edit: Oh, missed the missing square root.
     
    Last edited: Nov 27, 2012
  5. Nov 27, 2012 #4

    Zondrina

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    I don't believe we're allowed to use polar coordinates yet ( wish I could ), but I would see why it would make things easier.

    So my integrand is correct, but incorrect? A little confused with the feedback.
     
  6. Nov 27, 2012 #5

    Mark44

    Staff: Mentor

    No, it doesn't. The graph of this function is a cone. If the equation were z2 = 1-x2-y2, then the graph would be a sphere.
     
  7. Nov 27, 2012 #6

    Mark44

    Staff: Mentor

    I believe your integrand is correct. I don't believe anyone said it was incorrect.
     
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