Thank you for reporting, we will resolve it shortly
Q.
The ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is divided into two parts by the line $2x = a$. The area of the smaller part is
Application of Integrals
Solution:
We have ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, with centre $\left(0,0\right)$
Required area = area of shaded region
$A=2\cdot\int_{\frac{a}{2}}^{a}y\, dx$
$=\frac{2b}{a} \int_{\frac{a}{2}}^{a}\sqrt{a^{2}-x^{2}} \,dx$,
Put $x = a \,sin\, \theta$
$ \Rightarrow dx=a\,cos\,\theta\,d \,\theta$
$\Rightarrow A=2ab\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} cos^{2}\,\theta\,d \,\theta$
$=ab \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\left(1+cos\,2\theta\right)\, d\theta$
$=ab\left[\theta+\frac{sin\,2\theta}{2}\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$
$= ab\left[\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right]$ sq. units
