Question

Using Integration find the area enclosed by ellipse $4x^2+9y^2=36$

Answer 1

$4x^2+9y^2=36$
$\therefore \frac{4x^2}{36}+\frac{9y^2}{36}=1$
$\therefore \frac{x^2}{9}+\frac{y^2}{4}=1$
Here $x=3$ and $b=2$
$\therefore a>b$
$\therefore$ Given equation represents ellipse symmetric about X-axis.
$\therefore$ Required area $=4$(Area of the region OAB in first quadrant)
$=4\left|\stackrel{a}{\underset{0}{\int }}ydx\right|$
Now $\frac{y^2}{4}=1-\frac{x^2}{9}$
$=\frac{9-x^2}{9}$
$\therefore y^2=\frac{4}{9}(9-x^2)$
$\therefore y=\frac{2}{3}\sqrt{3^2-x^2}$
($\because$ In first quadrant $y>0$)
$\therefore \stackrel{a}{\underset{0}{\int }}ydx=\frac{2}{3}\stackrel{3}{\underset{0}{\int }}\sqrt{3^2-x^2}dx$
$=\frac{2}{3}{\{\frac{x\sqrt{3^2-x^2}}{2}+\frac{3^2}{2}{\sin }^{-1}\left(\frac{x}{3}\right)\}}_0^3$
$=\frac{2}{3}\{(0+\frac{9}{2}{\sin }^{-1}(1\left)\right)-(0+0\left)\right\}$
$=\frac{2}{3}\left(\frac{9}{2}\right)\frac{\pi }{2}$
$\therefore \stackrel{a}{\underset{0}{\int }}ydx=\frac{3\pi }{2}$
$\therefore$ Total area $=4\left(\frac{3\pi }{2}\right)$
$=6\pi$ sq. Unit.