Question

Find the area of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, (a > b) by the method of integration and hence find the area of the ellipse $\frac{x^2}{16}+\frac{9}{b^2}=1$.

Answer 1

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\frac{y^2}{b^2}=\frac{a^2-x^2}{a^2}\Rightarrow y=\pm \sqrt{a^2-x^2}$

Area$={\int }_{-a}^{a}ydx$

$={\int }_{-a}^a\frac{b}{a}\sqrt{a^2-x^2}dx=\frac{b}{a}\times [\frac{x}{2}\sqrt{a^2-x^2}+\frac{a}{2}{\sin }^{-1}\frac{x}{a}{]}_{-a}^a=\frac{b}{a}\times [0+\frac{a}{2}{\sin }^{-1}\frac{a}{a}-0-\frac{a}{2}{\sin }^{-1}\frac{a}{-a}]$

$=\frac{b}{a}\times [\frac{a}{2}\times \pi +\frac{a}{2}\times \pi ]$

$=\frac{b}{a}\times \pi a^2=\pi ab$

$\therefore$ area of ellipse $=\pi ab$

in the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1a=4,b=9\therefore area=\pi ab=\pi \times 4\times 9=36\pi sq.unit$