The area (in sq. units) of the smaller part of the circle \(x^2+y^2=\mathrm{a}^2\) cut off by the line \(x=\frac{\mathrm{a}}{\sqrt{2}}\) is

Substitute \(x=\frac{\mathrm{a}}{\sqrt{2}}\) in \(x^2+y^2=\mathrm{a}^2\), we get \(\frac{\mathrm{a}^2}{2}+y^2=\mathrm{a}^2 \Rightarrow y= \pm \frac{\mathrm{a}}{\sqrt{2}}\)
\(\therefore \quad\) Required area
\(
\begin{aligned}
& =2 \int_{\frac{a}{\sqrt{2}}}^a \sqrt{a^2-x^2} d x \\
& =2\left[\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]_{\frac{a}{\sqrt{2}}}^2 \\
& =2\left\{\left[0+\frac{a^2}{2} \times \frac{\pi}{2}\right]-\left[\frac{a}{2 \sqrt{2}} \sqrt{a^2-\frac{a^2}{2}}+\frac{a^2}{2} \times \frac{\pi}{4}\right]\right\} \\
& =2\left[\frac{a^2 \pi}{4}-\frac{a^2}{4}-\frac{a^2 \pi}{8}\right] \\
& =\frac{a^2}{2}\left|\pi-1-\frac{\pi}{2}\right| \\
& =\frac{a^2}{2}\left|\frac{\pi}{2}-1\right|
\end{aligned}
\)