Finding the Area Enclosed by a Line, Circle, and the X-axis

The problem asks for the area of a specific region in the first quadrant. This region is bounded by three curves:

• The line \({\rm{x}} = \sqrt 3 {\rm{\;y}}\)
• The circle x\(^2\) + y\(^2\) = 4
• The x-axis

Analyzing the Given Equations

Let's understand each boundary:

• The line \({\rm{x}} = \sqrt 3 {\rm{\;y}}\): This can be rewritten as \({\rm{y}} = \frac{1}{\sqrt 3}{\rm{\;x}}\). This is a linear equation passing through the origin (0,0). The slope of the line is \(m = \frac{1}{\sqrt 3}\). The angle \(\theta\) that this line makes with the positive x-axis is given by \(\tan \theta = m\). \[ \tan \theta = \frac{1}{\sqrt 3} \] For angles in the first quadrant, this implies \(\theta = 30^\circ\), which is equal to \(\frac{\pi}{6}\) radians.
• The circle x\(^2\) + y\(^2\) = 4: This is the standard equation of a circle centered at the origin (0,0) with radius \(r = \sqrt{4} = 2\).
• The x-axis: This is the line y=0.

Identifying the Region of Interest

We are looking for the area in the first quadrant. This means \(x \ge 0\) and \(y \ge 0\). The region is enclosed by:

• The x-axis (y=0) as the lower boundary.
• The line \({\rm{y}} = \frac{1}{\sqrt 3}{\rm{\;x}}\) as one side boundary.
• The circle x\(^2\) + y\(^2\) = 4 as the outer boundary.

Since the line \({\rm{y}} = \frac{1}{\sqrt 3}{\rm{\;x}}\) passes through the origin and makes an angle of \(\frac{\pi}{6}\) with the positive x-axis, the region described is a sector of the circle centered at the origin with radius 2. The sector is bounded by the x-axis (angle 0 radians) and the line \({\rm{y}} = \frac{1}{\sqrt 3}{\rm{\;x}}\) (angle \(\frac{\pi}{6}\) radians).

The angle of this sector is the difference between the angles made by the line and the x-axis, which is \(\frac{\pi}{6} - 0 = \frac{\pi}{6}\) radians.

Calculating the Area of the Circular Sector

The area of a circular sector with radius \(r\) and central angle \(\theta\) (in radians) is given by the formula:

\[ \text{Area} = \frac{1}{2} r^2 \theta \] In this case, the radius of the circle is \(r = 2\), and the central angle of the sector is \(\theta = \frac{\pi}{6}\) radians.

Substituting these values into the formula:

\[ \text{Area} = \frac{1}{2} (2)^2 \left(\frac{\pi}{6}\right) \]

\[ \text{Area} = \frac{1}{2} (4) \left(\frac{\pi}{6}\right) \]

\[ \text{Area} = 2 \times \frac{\pi}{6} \]

\[ \text{Area} = \frac{2\pi}{6} \]

\[ \text{Area} = \frac{\pi}{3} \]

Thus, the area of the region in the first quadrant enclosed by the x-axis, the line \({\rm{x}} = \sqrt 3 {\rm{\;y}}\), and the circle x\(^2\) + y\(^2\) = 4 is \(\frac{\pi}{3}\).

Comparing with Options

Let's check the calculated area against the given options:

The calculated area matches Option 1.

Revision Table: Key Concepts for Area Calculation

Additional Information: Alternative Approach (Integration)

While the sector method is the most straightforward here, one could also approach this using integration. However, this would involve splitting the area into parts or using polar coordinates.

• In Cartesian coordinates, you would need to integrate from x=0 to the x-coordinate of the intersection of the line and the circle, integrating the line function from the x-axis. Then, from that x-coordinate to x=2 (radius), you would integrate the circle function from the x-axis. This approach is complex because the upper boundary changes.
• Using polar coordinates might be more suitable for integration in this case. The circle is \(r=2\). The x-axis is \(\theta=0\). The line \(x = \sqrt{3}y\) is \(r \cos \theta = \sqrt{3} r \sin \theta\). Since r=0 (origin) is excluded, we can divide by r: \(\cos \theta = \sqrt{3} \sin \theta\), or \(\tan \theta = \frac{1}{\sqrt{3}}\), which is \(\theta = \frac{\pi}{6}\). The area in polar coordinates is given by \( \frac{1}{2} \int_{\theta_1}^{\theta_2} r(\theta)^2 d\theta \). Here, \(r(\theta) = 2\) and the angle ranges from \(\theta_1 = 0\) to \(\theta_2 = \frac{\pi}{6}\). \[ \text{Area} = \frac{1}{2} \int_{0}^{\pi/6} (2)^2 d\theta \] \[ \text{Area} = \frac{1}{2} \int_{0}^{\pi/6} 4 d\theta \] \[ \text{Area} = 2 \int_{0}^{\pi/6} d\theta \] \[ \text{Area} = 2 [\theta]_{0}^{\pi/6} \] \[ \text{Area} = 2 \left(\frac{\pi}{6} - 0\right) \] \[ \text{Area} = 2 \times \frac{\pi}{6} = \frac{\pi}{3} \]

Both the sector area formula and integration in polar coordinates confirm the result \(\frac{\pi}{3}\). This illustrates that understanding the geometry of the region can often simplify area calculations significantly.