Question

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

Answer 1

The given curve is an ellipse with centre at (0, 0) and symmetrical about X - axis and Y-axis ($\because$ the power of x and y both are even).

Area bounded by the ellipse $=4\times$ (Area of shaded region in the first quadrant only) ($\because$ By symmetry)

$=4\times {\int }_{x=a}^{x=b}ydx=4{\int }_0^{4}ydx=4{\int }_0^4\frac{3}{4}\sqrt{16-x^2}dx$
$(\because \frac{x^2}{16}+\frac{y^2}{9}=1,\therefore y=\frac{3}{4}\sqrt{16-x^2})$
$=3{\int }_0^4\sqrt{4^2-x^2}dx=3[\frac{x}{2}\sqrt{4^2+x^2}+\frac{4^2}{2}si{n}^{-1}(\frac{x}{4}){]}_0^4$
$[\because \int \sqrt{a^2-x^2}dx=\frac{x}{a}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}(\frac{x}{a}\left)\right]$
$=3[2\sqrt{16+16}+8si{n}^{-1}(1)-0-8si{n}^{-1}(0\left)\right]$
$=3[0+8si{n}^{-1}(1)-0]=3\times 8\times \frac{\pi }{2}=12\pi squnit$