Area Enclosed by a Closed Curve

The area enclosed by a closed curve is
$A= \int^{x_2}_{x_1} \int^{\gamma_2}_{\gamma_1} dy dx$

An ellipse centred at the origin has equation
$\frac{x^2}{a^2} +\frac{y^2}{b^2} =1$
.
For this curve we can take
$x_1 = -a, x_2 =a,$
.
Rearranging the equation of the ellipse for
$y$
gives
$y=\pm b \sqrt{1- \frac{x^2}{a^2}}$
so we can take
$\gamma21$
as
$b \sqrt{1- \frac{x^2}{a^2}}$
and
$\gamma_1$
as
$-b \sqrt{1- \frac{x^2}{a^2}}$

The area enclosed by the ellipse is
\begin{aligned} A &= \in^{a}_{-a} \int^{b \sqrt{1- x^2/a^2}}_{-b \sqrt{1- x^2/a^2}} dydx \\ &= \int^{a}_{-a} [y]^{ b \sqrt{1- x^2/a^2}}_{-b \sqrt{1- x^2/a^2}} dx \\ &= 2b \int^{a}_{-a} \sqrt{1- x^2/a^2} dx \\ &= b[x/2 \sqrt{1-x^2/a^2} + a/2 sin^{-1} (x/a)]^a_{-a} \\ &= 2b((0+a/2 sin^{-1} 1 ) - (0-a/2 sin^{-1} -1 )) \\ &= 2b ( \frac{\pi}{2} -(- \frac{\pi}{2})) \\ &= \pi ab \end{aligned}