Conside the line `x=sqrt(3)y` and the circle `x^(2)+y^(2)=4`
What is the area of the region in the first quadrant enclosed by the X-axis, the line `=sqrt(3)y` and the circle?

Options

A. `(pi)/3`
B. `(pi)/6`
C. `(pi)/3-(sqrt(3))/2`
D. None of the above

Answer

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

### Step 1: Identify the equations
We have two equations:
1. The line: \(x = \sqrt{3}y\)
2. The circle: \(x^2 + y^2 = 4\)

### Step 2: Find the intersection points
To find the points where the line intersects the circle, we can substitute the line equation into the circle equation.

Substituting \(x = \sqrt{3}y\) into the circle equation:
\[ (\sqrt{3}y)^2 + y^2 = 4 \]
This simplifies to:
\[ 3y^2 + y^2 = 4 \implies 4y^2 = 4 \implies y^2 = 1 \implies y = 1 \text{ (since we are in the first quadrant)} \]
Now substituting \(y = 1\) back into the line equation to find \(x\):
\[ x = \sqrt{3}(1) = \sqrt{3} \]
Thus, the intersection point in the first quadrant is \((\sqrt{3}, 1)\).

### Step 3: Set up the area integral
The area we want to find is bounded by the line, the circle, and the X-axis. We can split the area into two parts:
1. The area under the line from \(x = 0\) to \(x = \sqrt{3}\).
2. The area under the circle from \(x = \sqrt{3}\) to \(x = 2\).

The area under the line can be expressed as:
\[ \text{Area}_1 = \int_0^{\sqrt{3}} \frac{x}{\sqrt{3}} \, dx \]
The area under the circle can be expressed as:
\[ \text{Area}_2 = \int_{\sqrt{3}}^{2} \sqrt{4 - x^2} \, dx \]

### Step 4: Calculate the area under the line
Calculating \(\text{Area}_1\):
\[ \text{Area}_1 = \int_0^{\sqrt{3}} \frac{x}{\sqrt{3}} \, dx = \frac{1}{\sqrt{3}} \int_0^{\sqrt{3}} x \, dx \]
\[ = \frac{1}{\sqrt{3}} \left[\frac{x^2}{2}\right]_0^{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{(\sqrt{3})^2}{2} = \frac{1}{\sqrt{3}} \cdot \frac{3}{2} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \]

### Step 5: Calculate the area under the circle
Calculating \(\text{Area}_2\):
\[ \text{Area}_2 = \int_{\sqrt{3}}^{2} \sqrt{4 - x^2} \, dx \]
Using the substitution \(x = 2\sin(\theta)\), we have \(dx = 2\cos(\theta) d\theta\) and the limits change accordingly:
- When \(x = \sqrt{3}\), \(\sin(\theta) = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{3}\)
- When \(x = 2\), \(\sin(\theta) = 1 \implies \theta = \frac{\pi}{2}\)

Thus,
\[ \text{Area}_2 = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sqrt{4 - (2\sin(\theta))^2} \cdot 2\cos(\theta) d\theta \]
\[ = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sqrt{4 - 4\sin^2(\theta)} \cdot 2\cos(\theta) d\theta = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 2\sqrt{4(1 - \sin^2(\theta))} \cdot 2\cos(\theta) d\theta \]
\[ = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 4\cos^2(\theta) d\theta \]
Using the identity \(\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}\):
\[ = 2 \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} (1 + \cos(2\theta)) d\theta = 2\left[\theta + \frac{1}{2}\sin(2\theta)\right]_{\frac{\pi}{3}}^{\frac{\pi}{2}} \]
Calculating this gives:
\[ = 2\left[\frac{\pi}{2} + 0 - \left(\frac{\pi}{3} + \frac{1}{2}\sin\left(\frac{2\pi}{3}\right)\right)\right] \]
\[ = 2\left[\frac{\pi}{2} - \frac{\pi}{3} - \frac{\sqrt{3}}{4}\right] = 2\left[\frac{3\pi - 2\pi - \sqrt{3}}{6}\right] = \frac{2(\pi - \sqrt{3})}{6} = \frac{\pi - \sqrt{3}}{3} \]

### Step 6: Total area
The total area \(A\) is:
\[ A = \text{Area}_1 + \text{Area}_2 = \frac{\sqrt{3}}{2} + \frac{\pi - \sqrt{3}}{3} \]

### Step 7: Final simplification
To combine these, we can find a common denominator:
\[ = \frac{3\sqrt{3}}{6} + \frac{2(\pi - \sqrt{3})}{6} = \frac{3\sqrt{3} + 2\pi - 2\sqrt{3}}{6} = \frac{\sqrt{3} + 2\pi}{6} \]

Thus, the area of the region in the first quadrant enclosed by the X-axis, the line \(x = \sqrt{3}y\), and the circle \(x^2 + y^2 = 4\) is:
\[ \boxed{\frac{\sqrt{3} + 2\pi}{6}} \]