Question

​The area of the bounded by the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1\mathrm{is}\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

To find: area enclosed by the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$



The whole area enclosed by the given ellipse = 4(Area of the region bounded by AOBA)

Thus,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{ellipse}=4{\int }_0^{5}ydx \\ =4{\int }_0^5\left(4\sqrt{1-\frac{x^2}{25}}\right)dx \\ =\frac{16}{5}{\int }_0^5\left(\sqrt{25-x^2}\right)dx \\ =\frac{16}{5}{\left(\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}{\sin }^{-1}\frac{x}{5}\right)}_0^5 \\ =\frac{16}{5}\left[\left(\frac{5}{2}\sqrt{25-25}+\frac{25}{2}{\sin }^{-1}\frac{5}{5}\right)-\left(\frac{0}{2}\sqrt{25-0^2}+\frac{25}{2}{\sin }^{-1}\frac{0}{5}\right)\right] \\ =\frac{16}{5}\left[\left(0+\frac{25}{2}{\sin }^{-1}1\right)-\left(0+\frac{25}{2}{\sin }^{-1}0\right)\right] \\ =\frac{16}{5}\left[\left(\frac{25}{2}\frac{\mathrm{\pi }}{2}\right)-\left(0\right)\right] \\ =\frac{16}{5}\left(\frac{25\mathrm{\pi }}{4}\right) \\ =20\mathrm{\pi } \\ \mathrm{Thus},\mathrm{Area}=20\mathrm{\pi }\mathrm{sq}.\mathrm{units} \end{gathered}$$


Hence, the area of the bounded by the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1\mathrm{is}\underline{20\mathrm{\pi }}.$