Area using Green's theorem

Green's theorem: ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy for a simple closed curve C traversed counter-clockwise.

Area formula

Choosing P = −y/2, Q = x/2 gives ∂Q/∂x − ∂P/∂y = 1, so

Area = ∬_R dx dy = ½ ∮_C (x dy − y dx)

The curve (x/a)^(2/3) + (y/b)^(2/3) = 1 (a stretched astroid)

Parametrize: x = a cos³t, y = b sin³t, 0 ≤ t ≤ 2π. This satisfies cos²t + sin²t = 1.

dx = −3a cos²t sin t dt
dy =  3b sin²t cos t dt

x dy − y dx = a cos³t · 3b sin²t cos t dt + b sin³t · 3a cos²t sin t dt
            = 3ab sin²t cos²t (cos²t + sin²t) dt
            = 3ab sin²t cos²t dt = (3ab/4) sin²2t dt

Area = ½ ∫₀^{2π} (3ab/4) sin²2t dt
     = (3ab/8) ∫₀^{2π} (1 − cos 4t)/2 dt
     = (3ab/8) · π

Answer: Area = 3πab/8 square units. (For a = b this reduces to the astroid area 3πa²/8.)