Line integrals and maximums

Never knew you could find maximums of line integrals but here's my question.

Find the positively oriented simple closed curve C for which the value of the line integral ∫ (along C) of (y^3 – y)dx – 2x^3 dy is a maximum.

By Green's Theorem, this equals
∫∫ [(∂/∂x)(-2x^3) - (∂/∂y)(y^3 - y)] dA = ∫∫ [1 - 3(2x^2 + y^2)] dA.

By applying 'elliptical coordinates' x = (1/√2) r cos θ and y = r sin θ, this reduces to maximizing
(assuming a nice closed curve; in this case, a circle r = a for some a > 0)

∫(θ = 0 to 2π) ∫(r = 0 to a) (1 - 3r^2) * (1/√2) r dr dθ (remember the jacobian!)
= (2π/√2) ∫(r = 0 to a) (r - 3r^3) dr
= (π√2) (r^2/2 - 3r^4/4) {for r = 0 to a}
= π√2 (a^2/2 - 3a^4/4).

To maximize this:
Let g(a) = a^2/2 - 3a^4/4.

So, g'(a) = a - 3a^3.
Setting this equal to 0 yields a = 1/√3 (assuming a > 0).
It's easy to check that this yields the maximum.

So, the curve with maximal line integral is (by the given change of coordinates)
x = (1/√2) (1/√3) cos θ = (1/√6) cos θ and y = (1/√3) sin θ
==> 6x^2 + 3y^2 = 1, an ellipse.

I hope this helps!