Answer
$A = ab\pi $
Work Step by Step
From the formula (6): \[ A = \oint_C x \, dy = -\oint_C y \, dx \] For the ellipse: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] we can use the parametrization: \[ x = a\cos(t), \quad y = b\sin(t), \quad 0 \leq t \leq 2\pi \] where: \[ \frac{dx}{dt} = -a\sin(t), \quad \frac{dy}{dt} = b\cos(t) \] Using the first integral: \[ A = \int_0^{2\pi} x(t) \, dy(x) = \int_0^{2\pi} a\cos(t)(b\cos(t)) \, dt = ab\int_0^{2\pi} \cos^2(t) \, dt = ab\left[\frac{2t}{2} + \frac{1}{4}\sin(2t)\right]_0^{2\pi} = ab\pi \] Using the second integral: \[ A = -\int_0^{2\pi} y(t) \, dx(t) = -\int_0^{2\pi} b\sin(t)(-a\sin(t)) \, dt = ab\int_0^{2\pi} \sin^2(t) \, dt = ab\left[\frac{2t}{2} - \frac{1}{4}\sin(2t)\right]_0^{2\pi} = ab\pi \] Finally, in both cases: \[ A = ab\pi \]
