Solution:

The area of the smaller part of the circle $x^{2}+y^{2}=a^{2}$ cutt off by the line $x=\frac{a}{\sqrt{2}}$ is the area of the ABCDA

It can be observed that the area $A B C D$ is symmetrical about $x$-axis.
$\therefore$ Area $\mathrm{ABCD}=2 \times$ Area $\mathrm{ABC}$

$\begin{array}{l}
=\int_{a}^{a} y d x \\
=\int_{a}^{a} \sqrt{a^{2}-x^{2}} d x
\end{array}$

$\begin{array}{l}
=\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right]_{\frac{a}{\sqrt{2}}}^{a} \\
=\left[\frac{a^{2}}{2}\left(\frac{\pi}{2}\right)-\frac{a}{2 \sqrt{2}} \sqrt{a^{2}-\frac{a^{2}}{2}}-\frac{a^{2}}{2} \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right]
\end{array}$

$=\frac{a^{2} \pi}{4}-\frac{a}{2 \sqrt{2}} \cdot \frac{a}{\sqrt{2}}-\frac{a^{2}}{2}\left(\frac{\pi}{4}\right)$
$=\frac{a^{2} \pi}{4}-\frac{a^{2}}{4}-\frac{a^{2} \pi}{8}$
$=\frac{a^{2}}{4}\left[\pi-1-\frac{\pi}{2}\right]$
$=\frac{a^{2}}{4}\left[\frac{\pi}{2}-1\right]$
$\Rightarrow \text { Area } A B C D=2\left[\frac{a^{2}}{4}\left(\frac{\pi}{2}-1\right)\right]=\frac{a^{2}}{2}\left(\frac{\pi}{2}-1\right)$

As a result, the area of smaller part of the circle, $x^{2}+y^{2}=a^{2}$, cut off by the line, $x=\frac{a}{\sqrt{2}}$. is $\frac{a^{2}}{2}\left(\frac{\pi}{2}-1\right)$ units.