Area using Green's theorem

Green's theorem: ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy.

Choose P = −y/2, Q = x/2. Then ∂Q/∂x − ∂P/∂y = 1/2 + 1/2 = 1, so

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

For the ellipse x²/a² + y²/b² = 1 use x = a cos t, y = b sin t, 0 ≤ t ≤ 2π (anticlockwise):

dx = −a sin t dt,   dy = b cos t dt
x dy − y dx = (a cos t)(b cos t) dt − (b sin t)(−a sin t) dt
            = ab (cos²t + sin²t) dt = ab dt

A = ½ ∫₀^2π ab dt = ½ · ab · 2π = πab

Answer: Area of the ellipse = πab.