Find the area of the smaller part of the circle `x^2+y^2=a^2`cut off by the line `x=a/(sqrt(2))`

Options

A. `a^4[pi/5 - 1/3]`
B. `a^2[pi/4 - 1/2]`
C. `a^2[pi/3 - 1/4]`
D. `a^3[pi/3 - 1/2]`

Explanation

To find the area of the smaller part of the circle \( x^2 + y^2 = a^2 \) cut off by the line \( x = \frac{a}{\sqrt{2}} \), we can follow these steps:

### Step 1: Understand the Circle and the Line
The equation of the circle is \( x^2 + y^2 = a^2 \) with center at the origin (0,0) and radius \( a \). The line \( x = \frac{a}{\sqrt{2}} \) is a vertical line that intersects the circle.

### Step 2: Find the Points of Intersection
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:

\[ y^2 = a^2 - \frac{a^2}{2} = \frac{a^2}{2} \]

Thus, \( y = \pm \frac{a}{\sqrt{2}} \). 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 3: Set Up the Integral for Area Calculation
The area of the smaller part of the circle can be calculated by finding the area above the line \( x = \frac{a}{\sqrt{2}} \) and below the circle, then doubling that area (due to symmetry).

The area \( A \) can be expressed as:

\[ A = 2 \int_{\frac{a}{\sqrt{2}}}^{a} y \, dx \]

Where \( y = \sqrt{a^2 - x^2} \) from the circle's equation.

### Step 4: Calculate the Integral
Substituting for \( y \):

\[ A = 2 \int_{\frac{a}{\sqrt{2}}}^{a} \sqrt{a^2 - x^2} \, dx \]

### Step 5: Solve the Integral
To solve this integral, we can use the formula:

\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]

Evaluating from \( \frac{a}{\sqrt{2}} \) to \( a \):

\[ \left[ \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) \right]_{\frac{a}{\sqrt{2}}}^{a} \]

Calculating at the upper limit \( x = a \):

\[ = \frac{a}{2} \cdot 0 + \frac{a^2}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4} \]

Calculating at the lower limit \( x = \frac{a}{\sqrt{2}} \):

\[ = \frac{\frac{a}{\sqrt{2}}}{2} \sqrt{a^2 - \left(\frac{a}{\sqrt{2}}\right)^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \]

\[ = \frac{a}{2\sqrt{2}} \cdot \sqrt{a^2 - \frac{a^2}{2}} + \frac{a^2}{2} \cdot \frac{\pi}{4} \]

\[ = \frac{a}{2\sqrt{2}} \cdot \frac{a}{\sqrt{2}} + \frac{a^2 \pi}{8} = \frac{a^2}{4} + \frac{a^2 \pi}{8} \]

### Step 6: Combine the Results
Now, substituting back into the area formula:

\[ A = 2 \left( \frac{\pi a^2}{4} - \left(\frac{a^2}{4} + \frac{a^2 \pi}{8}\right) \right) \]

Simplifying gives:

\[ A = 2 \left( \frac{\pi a^2}{4} - \frac{a^2}{4} - \frac{a^2 \pi}{8} \right) \]

\[ = 2 \left( \frac{2\pi a^2}{8} - \frac{2a^2}{8} - \frac{a^2 \pi}{8} \right) = 2 \left( \frac{(2\pi - 2 - \pi)a^2}{8} \right) \]

\[ = \frac{(2 - \pi)a^2}{4} \]

Thus, the area of the smaller part of the circle cut off by the line is:

\[ \text{Area} = \frac{a^2}{2} \left( \frac{\pi}{2} - 1 \right) \]