Question

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

Answer 1

The given curve is an ellipse with centre at (0, 0) and symmetrical about X-axis and Y-axis.

Area bounded by the ellipse

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

$=4\times {\int }_{x=a}^{x=b}|y|dx=4{\int }_0^2|y|dx=4{\int }_0^2\frac{3}{2}\sqrt{4-x^2}dx(\because \frac{x^2}{4}+\frac{y^2}{9}=1,\therefore |y|=\frac{3}{2}\sqrt{4-x^2})=6{\int }_0^2\sqrt{2^2-x^2}dx=6[\frac{x}{2}\sqrt{4-x^2}+\frac{2^2}{2}si{n}^{-1}(\frac{x}{2}\left){]}_0^2\right[\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\left(\frac{x}{a}\right)]=6(0+2si{n}^{-1}\left(1{)}^{-0}\right)6\times 2\times (\frac{\pi }{2}=6\pi squnit$