Question

Find the area of the region bounded by the ellipse $\frac{x2}{4}+\frac{y^2}{9}=1$

Answer 1

The given equation of the ellipse can be represented as

$\frac{x2}{4}+\frac{y^2}{9}=1$

$⇒y=3√1-\frac{x2}{4}...(1)$

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

∴Area bounded by ellipse=4×Area OAB

$∴Area of OAB=∫_0^2ydx$

$=∫_0^2 3√1-\frac{x^2}{4}dx [Using(1)]$

$=\frac{3}{2}∫_0^2√4-x^2dx$

$=\frac{3}{2}[\frac{x}{2}√4-x^2+\frac{4}{2}sin-\frac{x}{2}]_0^2$

$=\frac{3}{2}[\frac{2π}{2}]$

$=\frac{3π}{2}$

Therefore,area bounded by the ellipse =4$×\frac{3π}{2}$=6πunits.