Area lying in the first quadrant and bounded by the circle `x^(2)+y^(2)=4` the line `x=sqrt(3)y` and x-axis , is

Options

A. `pi`
B. `pi//2`
C. `pi//3`
D. `pi//4`

Answer

To find the area lying in the first quadrant and bounded by the circle \(x^2 + y^2 = 4\), the line \(x = \sqrt{3}y\), and the x-axis, we will follow these steps:

### Step 1: Identify the Circle and Line
The equation of the circle is given by:
\[ x^2 + y^2 = 4 \]
This represents a circle centered at the origin (0,0) with a radius of 2.

The equation of the line is:
\[ x = \sqrt{3}y \]

### Step 2: Find Points of Intersection
To find the points of intersection between the circle and the line, we substitute \(x = \sqrt{3}y\) into the circle's equation:
\[ (\sqrt{3}y)^2 + y^2 = 4 \]
This simplifies to:
\[ 3y^2 + y^2 = 4 \implies 4y^2 = 4 \implies y^2 = 1 \implies y = 1 \quad (\text{since we are in the first quadrant}) \]
Now substituting \(y = 1\) back into the line equation to find \(x\):
\[ x = \sqrt{3}(1) = \sqrt{3} \]
Thus, the point of intersection is \((\sqrt{3}, 1)\).

### Step 3: Set Up the Area Integral
The area bounded by the circle, the line, and the x-axis can be computed by integrating the difference between the upper curve (circle) and the lower curve (line) from \(x = 0\) to \(x = \sqrt{3}\), and then from \(x = \sqrt{3}\) to \(x = 2\) (the rightmost point of the circle).

1. From \(x = 0\) to \(x = \sqrt{3}\), the upper curve is the line \(y = \frac{x}{\sqrt{3}}\).
2. From \(x = \sqrt{3}\) to \(x = 2\), the upper curve is the circle \(y = \sqrt{4 - x^2}\).

The area \(A\) can be expressed as:
\[ A = \int_0^{\sqrt{3}} \frac{x}{\sqrt{3}} \, dx + \int_{\sqrt{3}}^{2} \sqrt{4 - x^2} \, dx \]

### Step 4: Calculate the First Integral
Calculating the first integral:
\[ \int_0^{\sqrt{3}} \frac{x}{\sqrt{3}} \, dx = \frac{1}{\sqrt{3}} \int_0^{\sqrt{3}} x \, dx = \frac{1}{\sqrt{3}} \left[\frac{x^2}{2}\right]_0^{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3})^2}{2} = \frac{1}{\sqrt{3}} \cdot \frac{3}{2} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]

### Step 5: Calculate the Second Integral
Calculating the second integral:
\[ \int_{\sqrt{3}}^{2} \sqrt{4 - x^2} \, dx \]
Using the formula for the area of a quarter circle, we can find this area directly. The area of a quarter circle of radius 2 is:
\[ \frac{1}{4} \pi (2^2) = \pi \]
However, we need to subtract the area of the triangle formed by the line:
\[ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{3} \times 1 = \frac{\sqrt{3}}{2} \]
Thus, the area from \(\sqrt{3}\) to \(2\) is:
\[ \int_{\sqrt{3}}^{2} \sqrt{4 - x^2} \, dx = \frac{\pi}{3} - \frac{\sqrt{3}}{2} \]

### Step 6: Combine Areas
Combining both areas:
\[ A = \frac{\sqrt{3}}{2} + \left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \]

### Final Answer
Thus, the area lying in the first quadrant and bounded by the circle, the line, and the x-axis is:
\[ \boxed{\frac{\pi}{3}} \]