Find the area of the loop of the curve(x-y)(x2+y2) = 2axy

Answer 1

(x-y)(x^2+y^2) = 2axy
In polar coordinates, that is

r(cosθ-sinθ)(r^2) = 2ar^2 sinθ cosθ
r(cosθ-sinθ) = asin2θ
r = asin2θ/(cosθ-sinθ)

The loop is the area enclosed as θ goes from π/2 to π, so the area is

a = ∫[π/2,π] 1/2 r^2 θ dθ

Now,
= sin^2 2θ / (cosθ-sinθ)^2
= sin^2 2θ/(1-2sinθcosθ)
= sin^2 2θ / (1-sin2θ)
= sin^2 2θ (1+sin2θ) / cos^2 2θ
= tan^2 2θ + (1-cos^2 2θ)(sin2θ)/cos^2 2θ

Now, if we let
u = cos2θ
du = -2sinθ dθ
u(π/2) = -1
u(π) = 1

Now you can see that we have

1/2 ∫[π/2,π] 1 + sec^2 2θ dθ
+ 1/2 ∫[-1,1] (1-u^2)/u^2 (-1/2)du

which are all easy to evaluate.

Answer 2

Sorry - I forgot to include the factor of a on the right side. It will just multiply the final answer, since it's a constant.