Integrate x^2-xy+y^2 over the region x^2-xy+y^2<=2 by converting to polar coordinates. ans. 4π√(3)/3----don't answer.?

I already got the answer.

"Just wanna ask if the logic I used in solving is sensible or just nonsense."

After I converted this to polar and solved for its definite integral, I multiplied it by 4 and set the boundaries as [-π/4,π/4]. My reason for this is because I need a quarter of it the multiply by 4 to get the whole (I took consideration of the position of the ellipse and chose these boundaries). I experimented with other boundaries and got wrong answers like [0,2π] resulted in zero--they got canceled. 

2 Answers

  • 4 months ago
    Favorite Answer

    All of that is correct. Since the ellipse is symmetric about the perpendicular lines θ = -π/4 and θ = -π/4, the area in one quadrant is a quarter of the whole.

    But you shouldn't have to do this. Integrating from 0 to 2π works normally.

    The region is between r=0 and r = 2/sqrt(2-sin2θ) for all θ

    A = 

    int[0 to 2π] 1/2 [ (2/sqrt(2-sin2θ))^2 - 0^2 ] dθ

    = 4π√3/3

    Or int[0 to 2π] int[0 to 2/sqrt(2-sin2θ)] r dr dθ with the double integral where J = r.

  • Ian H
    Lv 7
    4 months ago

    x^2 - xy + y^2 <= 2 defines the region within the non-standard ellipse

    x^2 - xy + y^2 = 2 illustrated here

    How to convert that ellipse to polar

    x^2 + y^2 = xy + 2

    r^2 = rcos*rsin + 2

    r^2[1 – (1/2)sin(2θ)] = 2

    r^2 = 4/[2 – sin(2θ)]

    Polar area is given by A = {θ: 0 to 2π} ∫(1/2)r^2 dθ

    A = {θ: 0 to 2π}∫{2/[2 – sin(2θ)]}dθ

    That is an awkward integral detailed here; simplified by the limits

    A = 4π√(3)/3

    Alternative method (not what you asked for, but it is simpler)

    x^2 - xy + y^2 = 2, relate that to a standard form but rotated by π/4

    When x = y we get x^2 = 2, x^2 + y^2 = 4 thus a = 2

    When y = -x we get 3x^2 = 2, x^2 + y^2 = 4/3 and so b = 2√(3)

    Related standard ellipse has A = 𝛑ab = 4π√(3)/3

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