Question

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

Answer 1

We know that the given curve $\frac{x^2}{4}+\frac{y^2}{9}=1$ is ellipse and whose vertex $(0,0)$, major axis is ${}^{\prime }{y}^{\prime }$ and minor axis ${}^{\prime }{x}^{\prime }$.
$\frac{y^2}{9}=1-\frac{x^2}{4}$
$\Rightarrow y=\frac{3}{2}\sqrt{4-x^2}$.......(i)
Area of given area is $=4\times$ (Area of shaded)
$=4\times {\int }_0^2|y|dx$ [From equation(i)]
$=4\times \frac{3}{2}{\int }_0^2\sqrt{4-x^2}dx=6{\int }_0^2\sqrt{2^2-x^2}dx$
$\int \sqrt{a^2-x^{2}dx}=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}$
$=6{[\frac{x}{2}\sqrt{2^2-x^2}+\frac{2^2}{2}{\sin }^{-1}\frac{x}{2}]}_0^2$
$=6[\frac{2}{2}\sqrt{2^2-2^2}+2{\sin }^{-1}\frac{2}{2}-\frac{0}{2}\sqrt{2^2-0}-2{\sin }^{-1}\frac{0}{2}]$
$=6[0+2{\sin }^{-1}1-0-0]=6\times 2\times \frac{\pi }{2}=6\pi$ sq.units