Find the area of the region bounded by the ellipse \(\frac{x2}{4}+\frac{y^2}{9}=1\)
Solution and Explanation
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.
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