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Double integral over a region bounded by an ellipse

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Evaluate. ∫∫D x2 dAxy, bounded by 5x2 + 4xy + y2 = 1

    2. Relevant equations

    ∫∫D H(x,y) dAxy = ∫∫D H(u,v)[itex]\frac{\delta(x,y)}{\delta(u,v)}[/itex]dAuv

    3. The attempt at a solution

    So I understand I'm supposed to find a change of variables to transform the ellipse into a circle in the uv-plane, and then transform the circle into a square in polar coordinates, which has clearly defined boundaries.

    I have no problem taking the integral or making a change of variables, but I'm not sure what to pick for my uv-plane. Clearly, I want to end up with u2 + v2 = 1, but how do I figure out what to define my uv-plane as? The part that's throwing me off is the 4xy term.
     
  2. jcsd
  3. Dec 5, 2011 #2

    HallsofIvy

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    The fact that the equation of the ellipse has an "xy" term means it is rotated. There are several different ways of determining the rotation. Essentially, you will get y= ax as one of the principal axes of the ellipse and taking u= y- ax will make that line the u-axis.
     
  4. Dec 5, 2011 #3
    What would be the best method of finding the principal axes?
     
  5. Dec 5, 2011 #4
    Sorry, I can't seem to find an edit button. By "best" method, I mean that this is a question that would typically appear on an exam and it seems like applying the principal axis theorem, making two change of variables and then taking two integrals is at least 30 minutes of work. On an exam that is 2 and a half hours long it seems to be a bit time-inefficient and stupid way of doing this question.

    The reason I raise this point is because I spent 30 minutes doing some 3-D graphing question on the midterm and ended up not having enough time to finish even though the rest of the exam was somewhat straight-foward and quite easy.

    Surely there's an easy way to find an equation for a principal axis (and then I can just take its orthogonal compliment to find the other one)? I can see by observation that (-1, 2) is an integer solution, but there's nothing telling me that it lies on the u-axis.
     
  6. Dec 6, 2011 #5
    Ain't no way I could do that on a test without the text book, Mathematica, coffee, and an hour or so

    You got:

    [tex]5x^2+4xy+y^2=1[/tex]

    The xy term says it's rotated so use the rotation forumulas:

    [tex]x=u\cos(\alpha)-v\sin(\alpha)[/tex]
    [tex]y=u\sin(\alpha)+v\cos(\alpha)[/tex]

    where

    [tex]\cot(2\alpha)=\frac{5-1}{4}[/tex]

    so [itex]\alpha=\pi/8[/itex]

    Draw both of them, it's an ellipse rotated by pi/8

    make those substitutions and simplify I get:

    [tex](3+2\sqrt{2})u^2+(3-2\sqrt{2})v^2=1[/tex]

    so I guess if you want to make it a circle, let:

    [tex]a=3-2\sqrt{2}[/tex]
    [tex]b=3+2\sqrt{2}[/tex]
    [tex]w=\sqrt{b/a}u[/tex]

    and I get:

    [tex]w^2+v^2=1/b[/tex]

    Is it still time to drop and add?
     
  7. Dec 10, 2011 #6
    Sorry to revive a somewhat dead thread, but I thought about this and we can just complete the square. I spent like half an hour typing out the itex of the full solution and to my dismay, somehow hit the back button on my browser. So I'll just post the hard part. I thought I should post my solution in case anyone else is having trouble with a similar problem.

    Rewrite
    [itex]5x^{2} + 4xy + y^{2} = 1[/itex]
    [itex]= 5x^{2} + y^{2} + 4xy + 4x^{2} - 4x^{2}[/itex]
    [itex]= x^{2} + (y-2x)^{2} = 1[/itex]

    So we can take (u, v) = (x, y-2x) clearly, the jacobian is 1 and to polar coordinates it's also quite easy.

    Hence we are left to integrate on the rectangle, 0 <= r <= 1, 0 <= theta <= 2pi. We have x = u = rcos(theta). Thus our integral becomes:

    02pi01 r3cos(theta)2dr dtheta
     
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