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.