1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Messy double integral

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


    [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

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted