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

$\begin{array}{1 1} 6\pi sq.units \\12\pi sq.units. \\ 18\pi sq.units. \\ 8\pi sq.units.\end{array}$

Toolbox:
• Area of the region bounded by the ellipse is 4 times the area bounde by it in the first quadrant.
• If it is bounded by the minor axis and the lines and x=a,then the area of the region $A=\int_0^a\sqrt{1-\frac{x^2}{a^2}}$
The ellipse is bounded along the minor axis and hence its shape is as shown in the fig.
It is also symmetrical about both the x-axis and y-axis.
Hence the required area is 4 times the area bounded by it in the Ist quadrant and x-axis and x=0 and x=2.
Hence $A=4\int_0^2y\;dx.$
$\;\;\;=4\int_0^2\frac{3}{2}\sqrt {4-x^2}dx$
$\;\;\;=6\int_0^2\sqrt {4-x^2}dx$
on integrating we get,
$\;\;\;=6\begin{bmatrix}\frac{x}{2}\sqrt {4-x^2}+\frac{4}{2}\sin^{-1}(\frac{x}{2})\end{bmatrix}_0^2$
on applying the limits we get,
$\;\;\;=6\begin{bmatrix}\frac{2}{2}\sqrt {4-4}+2\sin^{-1}(1)\end{bmatrix}$
$\;\;\;=6\begin{bmatrix}2\frac{\pi}{2}\end{bmatrix}=6\pi$sq.units.
Hence the required area is $6\pi$ sq.units.
reshown Dec 20, 2013 by yamini.v