To find the area of the smaller part of the circle defined by the equation \(x^2 + y^2 = a^2\) that is cut off by the line \(x = \frac{a}{\sqrt{2}}\), we can follow these steps:
### Step 1: Find the Points of Intersection
We need to find the points where the line intersects the circle. Substitute \(x = \frac{a}{\sqrt{2}}\) into the circle's equation:
\[
\left(\frac{a}{\sqrt{2}}\right)^2 + y^2 = a^2
\]
This simplifies to:
\[
\frac{a^2}{2} + y^2 = a^2
\]
Rearranging gives:
\[
y^2 = a^2 - \frac{a^2}{2} = \frac{a^2}{2}
\]
Taking the square root, we find:
\[
y = \pm \frac{a}{\sqrt{2}}
\]
Thus, the points of intersection are \(\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)\) and \(\left(\frac{a}{\sqrt{2}}, -\frac{a}{\sqrt{2}}\right)\).
### Step 2: Calculate the Area of the Sector
The area of the sector of the circle formed by the angle subtended at the center can be calculated using the formula for the area of a sector:
\[
\text{Area of Sector} = \frac{1}{2} r^2 \theta
\]
where \(r = a\) and \(\theta\) is the angle in radians. To find \(\theta\), we note that the line \(x = \frac{a}{\sqrt{2}}\) corresponds to an angle of \(\frac{\pi}{4}\) radians in the first quadrant. Therefore, the total angle for the sector is:
\[
\theta = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}
\]
Thus, the area of the sector is:
\[
\text{Area of Sector} = \frac{1}{2} a^2 \cdot \frac{\pi}{2} = \frac{a^2 \pi}{4}
\]
### Step 3: Calculate the Area of the Triangle
Next, we calculate the area of the triangle formed by the points \(\left(0, 0\right)\), \(\left(\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right)\), and \(\left(\frac{a}{\sqrt{2}}, -\frac{a}{\sqrt{2}}\right)\). The base of the triangle is the vertical distance between the two points on the line, which is:
\[
\text{Base} = \frac{a}{\sqrt{2}} - \left(-\frac{a}{\sqrt{2}}\right) = \frac{2a}{\sqrt{2}} = a\sqrt{2}
\]
The height of the triangle is the horizontal distance from the origin to the line, which is:
\[
\text{Height} = \frac{a}{\sqrt{2}}
\]
Thus, the area of the triangle is:
\[
\text{Area of Triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times a\sqrt{2} \times \frac{a}{\sqrt{2}} = \frac{a^2}{2}
\]
### Step 4: Calculate the Area of the Smaller Segment
The area of the smaller segment of the circle is the area of the sector minus the area of the triangle:
\[
\text{Area of Segment} = \text{Area of Sector} - \text{Area of Triangle}
\]
Substituting the values we found:
\[
\text{Area of Segment} = \frac{a^2 \pi}{4} - \frac{a^2}{2}
\]
To combine these terms, we can express \(\frac{a^2}{2}\) with a common denominator:
\[
\frac{a^2}{2} = \frac{2a^2}{4}
\]
Thus, we have:
\[
\text{Area of Segment} = \frac{a^2 \pi}{4} - \frac{2a^2}{4} = \frac{a^2 (\pi - 2)}{4}
\]
### Final Answer
The area of the smaller part of the circle cut off by the line \(x = \frac{a}{\sqrt{2}}\) is:
\[
\text{Area} = \frac{a^2 (\pi - 2)}{4}
\]
