Area using Green's theorem

Green's theorem

If C is a simple closed curve traversed counter-clockwise, enclosing a region R, and P, Q have continuous first partial derivatives in R, then

∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy

Area formula

Take P = −y/2 and Q = x/2. Then ∂Q/∂x − ∂P/∂y = ½ + ½ = 1, so

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

Area of the astroid x^(2/3) + y^(2/3) = a^(2/3)

Parametric form: x = a cos³t, y = a sin³t, with t from 0 to 2π (this satisfies cos²t + sin²t = 1 after raising to 2/3).

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

Then

x dy − y dx = a cos³t · 3a sin²t cos t dt + a sin³t · 3a cos²t sin t dt
            = 3a² sin²t cos²t (cos²t + sin²t) dt
            = 3a² sin²t cos²t dt
            = (3a²/4) sin²2t dt

Area = ½ ∫₀^{2π} (3a²/4) sin²2t dt
     = (3a²/8) ∫₀^{2π} (1 − cos 4t)/2 dt
     = (3a²/8) · [ t/2 − sin 4t/8 ]₀^{2π}
     = (3a²/8) · π

Answer: Area of the astroid = 3πa²/8 square units.

(Check: the astroid lies inside the circle of radius a, whose area is πa², and 3π/8 ≈ 1.18 < π ✓.)