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Messy double integral

  1. May 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate

    [itex]\int \int x dx dy[/itex]

    Over the area defined by [itex]1 \leq x^{2} + 4y^{2} \leq 9[/itex]

    2. Relevant equations



    3. The attempt at a solution

    First we'll do the sub:

    u = x + y
    v = sqrt(3)y

    Which gives us the area [itex]1 \leq u^{2} + v^{2} \leq 9, u,v\geq0 [/itex]

    and the integral

    [itex]\sqrt{3} \int \int u - \frac{1}{\sqrt{3}} v du dv[/itex]

    Now we will switch to polar:

    [itex]u^{2} + v^{2} = ρ^{2}[/itex]
    [itex]u = ρ cos(θ)[/itex]
    [itex]v = ρ sin(θ)[/itex]

    Since the functional determinant is ρ, this gives us

    [itex]\sqrt{3} \int \int ρ(ρ cos(θ) - \frac{1}{\sqrt{3}} ρ sin(θ))dρ dθ[/itex]

    Since [itex] 1 \leq u^{2} + v^{2} \leq 9[/itex] and [itex]x,y \geq 0[/itex]:
    [itex]1 \leq ρ \leq 3, 0 \leq θ \leq \frac{\pi}{2}[/itex]

    [itex]\sqrt{3} /3 \int [ρ^{3} cos(θ) - \frac{1}{\sqrt{3}} ρ^{3} sin(θ)] dθ[/itex]

    [itex]\sqrt{3} /3 \int [27 cos(θ) - \frac{27}{\sqrt{3}} sin(θ) - cos(θ) + \frac{1}{\sqrt{3}} sin(θ)] dθ[/itex]

    [itex]\sqrt{3} /3 \int 26 cos(θ) - \frac{26}{\sqrt{3}} sin(θ) dθ[/itex]

    Evaluating this between 0 and pi/2 nets something like 26 sqrt(3) blabla. Something terribly off. What am I doing wrong? Been wrestling for hours with the same problem now.
     
  2. jcsd
  3. May 14, 2013 #2
    Why not u=x, v=2y?
     
  4. May 14, 2013 #3
    Goddamnit.

    Because the region is bounded by x^2 + 4y^2 + 2xy. Typo...
     
  5. May 14, 2013 #4

    haruspex

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    I don't see any previous mention of [itex]x,y \geq 0[/itex]. Are you sure that's part of the boundary? And the map to a range for θ is not that simple - remember ρ and θ relate to x and y via u and v.
     
  6. May 14, 2013 #5
    Yeah, x and y are both positive, which means u and v are too.

    I am guessing we already identified the issue though. I think I dropped the 2xy somewhere when trying to express the integrand x in u and v... Gonna check.
     
  7. May 14, 2013 #6

    Ray Vickson

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    Could you please re-write the whole question exactly? I can't figure out what you mean here.
     
  8. May 14, 2013 #7

    haruspex

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    You didn't answer my second point, that the region 1 ≤ ρ ≤ 3, 0 ≤ θ ≤ π/2 is the same as u ≥ 0, v ≥ 0, but it's not the same as x ≥ 0, y ≥ 0. E.g. consider x = -1, y = 1.
     
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