The area of smaller part between the circle `x^(2)+y^(2)=4` and the line x=1 is

Options

A. `(4pi)/3-sqrt(3)`
B. `(8pi)/3-sqrt(3)`
C. `(4pi)/3+sqrt(3)`
D. `(5pi)/3+sqrt(3)`

Answer

To find the area of the smaller part between the circle \( x^2 + y^2 = 4 \) and the line \( x = 1 \), we can follow these steps:

### Step 1: Understand the Circle and Line
The equation of the circle \( x^2 + y^2 = 4 \) represents a circle centered at the origin (0, 0) with a radius of 2. The line \( x = 1 \) is a vertical line that intersects the circle.

### Step 2: Find Points of Intersection
To find the points where the line intersects the circle, substitute \( x = 1 \) into the circle's equation:
\[ 1^2 + y^2 = 4 \implies 1 + y^2 = 4 \implies y^2 = 3 \implies y = \pm \sqrt{3} \]
Thus, the points of intersection are \( (1, \sqrt{3}) \) and \( (1, -\sqrt{3}) \).

### Step 3: Set Up the Integral for Area
We need to find the area between the line \( x = 1 \) and the circle from \( x = 1 \) to \( x = 2 \) (the rightmost point of the circle). The area can be calculated as:
\[ \text{Area} = 2 \int_{1}^{2} (y_{\text{circle}} - y_{\text{line}}) \, dx \]
Here, \( y_{\text{circle}} = \sqrt{4 - x^2} \) (the upper half of the circle), and \( y_{\text{line}} = 0 \) (the x-axis).

### Step 4: Write the Integral
The area becomes:
\[ \text{Area} = 2 \int_{1}^{2} \sqrt{4 - x^2} \, dx \]

### Step 5: Evaluate the Integral
To solve the integral \( \int \sqrt{4 - x^2} \, dx \), we can use the formula:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
where \( a = 2 \).

Thus,
\[ \int \sqrt{4 - x^2} \, dx = \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) + C \]

### Step 6: Apply the Limits
Now we apply the limits from 1 to 2:
\[ \text{Area} = 2 \left[ \left( \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1}\left(\frac{x}{2}\right) \right) \bigg|_1^2 \right] \]

Calculating at the upper limit \( x = 2 \):
\[ = \frac{2}{2} \cdot 0 + 2 \sin^{-1}(1) = 0 + 2 \cdot \frac{\pi}{2} = \pi \]

Calculating at the lower limit \( x = 1 \):
\[ = \frac{1}{2} \sqrt{3} + 2 \sin^{-1}\left(\frac{1}{2}\right) = \frac{1}{2} \sqrt{3} + 2 \cdot \frac{\pi}{6} = \frac{1}{2} \sqrt{3} + \frac{\pi}{3} \]

### Step 7: Calculate the Final Area
Now, substituting back into the area expression:
\[ \text{Area} = 2 \left( \pi - \left( \frac{1}{2} \sqrt{3} + \frac{\pi}{3} \right) \right) \]
\[ = 2 \left( \pi - \frac{1}{2} \sqrt{3} - \frac{\pi}{3} \right) \]
\[ = 2 \left( \frac{3\pi}{3} - \frac{\pi}{3} - \frac{1}{2} \sqrt{3} \right) \]
\[ = 2 \left( \frac{2\pi}{3} - \frac{1}{2} \sqrt{3} \right) \]
\[ = \frac{4\pi}{3} - \sqrt{3} \]

### Final Answer
Thus, the area of the smaller part between the circle and the line \( x = 1 \) is:
\[ \text{Area} = \frac{4\pi}{3} - \sqrt{3} \]