Question

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

Answer 1

Equaion of given ellipse-

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

For $y=0\Rightarrow x=\pm 3$

Area of ellipse $=$ Area of $ABCD=2\times$ Area of $ABC$

Therefore,

Area of ellipse $=2{\int }_{-3}^{3}ydx$

From given equation of ellipse-

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

$\frac{y^2}{4}=1-\frac{x^2}{9}$

$\Rightarrow y^2=\frac{4}{9}(9-x^2)$

$y=\pm \frac{2}{3}\sqrt{9-x^2}$

Since $ABC$ is positive.

Therefore,

$y=\frac{2}{3}\sqrt{9-x^2}$

$\therefore$ Area of ellipse $=2{\int }_{-3}^3\left(\frac{2}{3}\sqrt{9-x^2}\right)dx=\frac{4}{3}{\int }_{-3}^3\sqrt{9-x^2}=\frac{4}{3}{[\frac{1}{2}x\sqrt{9-x^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{x}{3}\right)]}_{-3}^3=\frac{4}{3}[(\frac{1}{2}\left(3\sqrt{9-9}\right)+\frac{9}{2}{\sin }^{-1}\left(\frac{3}{3}\right))-(\frac{1}{2}(-3\sqrt{9-9})+\frac{9}{2}{\sin }^{-1}\left(\frac{-3}{3}\right))]=\frac{4}{3}\times 9{\sin }^{-1}\left(1\right)=6\pi$

Hence the area of ellipse is $6\pi$.