Question

Find the area of the region bounded by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$

Answer 1

The graph of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is shown below

Consider $\frac{x^2}{16}+\frac{y^2}{9}=1$

$\Rightarrow y^2=9(1-\frac{x^2}{16})\Rightarrow y=3\sqrt{1-\frac{x^2}{16}}$

Required area $=4{\int }_0^{4}ydx$

$=4{\int }_0^{4}3\sqrt{1-\frac{x^2}{16}}dx$

$=12{\int }_0^4\sqrt{1-\frac{x^2}{16}}dx$

$=3{\int }_0^4\sqrt{16-x^2}dx$

$=3{[\frac{x}{2}\sqrt{16-x^2}+8si{n}^{-1}\frac{x}{4}]}_0^4$

$=3[8\times \frac{\pi }{2}-0]$

$=3\left[4\pi \right]$

$=12\pi sq.units$

Hence the required area is $12\pi sq.units$