Class 12

Math

Calculus

Application of Integrals

Find the area of the region bounded by the ellipse $4x_{2} +9y_{2} =1$.

$4x_{2} +9y_{2} =1$

the standard form of ellipse is given by

$a_{2}x_{2} +b_{2}y_{2} =1$

by comparing we get,

$a=2andb=3$

$⇒y=31−4x_{2} $ .............. (1)

It can be observed that the ellipse is symmetrical about $x$-axis and $y$-axis.

$∴$ Area bounded by ellipse $=4×$ Area $OAB$

$∴$ Area of $OAB$ $=∫_{0}ydx$ $=∫_{0}ydx$

$=∫_{0}31−4x_{2} dx$ [Using (1)]

$=23 ∫_{0}4−x_{2} dx$

$=23 [2x 4−x_{2} +24 sin_{−1}2x ]_{0}$

$=23 [22π ]=23π $

Therefore, area bounded by the ellipse $=4×23π =6π$ units