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

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- 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.

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