Question

Find by integration, the are of the ellipse $16x^2+25y^2=400$.

Answer 1

$16x^2+25y^2=400$

$\Rightarrow$$y=y=\sqrt{\frac{400-16x^2}{25}}$

$\frac{x^2}{25}+\frac{y^2}{16}=1$

Area of ellipse $=4{\int }_0^{5}ydx=\frac{4}{5}{\int }_0^5\sqrt{400-16x^2}dx$

$\frac{4}{5}{\int }_0^5\sqrt{16(25-x^2)}dx=\frac{16}{5}{\int }_0^5\sqrt{5^2-x^2}dx$

$=\frac{16}{5}{[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}{\sin }^{-1}\left(\frac{x}{5}\right)]}_0^5=\frac{16}{5}[\frac{25}{2}.\frac{\pi }{2}]=20\pi$