Understanding the Problem: Finding the Area

The question asks for the area of a specific region located in the first quadrant of the coordinate plane. This region is defined by three boundaries:

• The x-axis (where \(y=0\)).
• The vertical line \(x = \sqrt{3}\).
• The circle with the equation \(x^2 + y^2 = 4\).

The circle \(x^2 + y^2 = 4\) is centered at the origin (0,0) and has a radius of \(\sqrt{4} = 2\).

Identifying the Boundaries and Intersection Points

Let's find where these boundaries intersect in the first quadrant (\(x \ge 0, y \ge 0\)).

• Circle and x-axis: Set \(y=0\) in \(x^2 + y^2 = 4\). \(x^2 + 0^2 = 4 \Rightarrow x^2 = 4 \Rightarrow x = 2\) (since in the first quadrant). Intersection point: \((2, 0)\).
• Line \(x = \sqrt{3}\) and x-axis: This is simply the point on the x-axis where \(x = \sqrt{3}\). Intersection point: \((\sqrt{3}, 0)\).
• Line \(x = \sqrt{3}\) and Circle: Set \(x = \sqrt{3}\) in \(x^2 + y^2 = 4\). \((\sqrt{3})^2 + y^2 = 4 \Rightarrow 3 + y^2 = 4 \Rightarrow y^2 = 1 \Rightarrow y = 1\) (since in the first quadrant). Intersection point: \((\sqrt{3}, 1)\).

These intersection points help define the region's corners or limits.

Defining the Region for Area Calculation

The region in the first quadrant enclosed by the x-axis, the line \(x=\sqrt{3}\), and the circle is bounded by:

• The x-axis from \(x = \sqrt{3}\) to \(x = 2\).
• The vertical line segment \(x = \sqrt{3}\) from \(y = 0\) to \(y = 1\).
• The arc of the circle \(x^2 + y^2 = 4\) (specifically \(y = \sqrt{4-x^2}\) in the first quadrant) from the point \((\sqrt{3}, 1)\) to \((2, 0)\).

This region can be visualized as the area under the curve \(y = \sqrt{4-x^2}\) from \(x = \sqrt{3}\) to \(x = 2\), bounded below by the x-axis.

Calculating the Area Using Integration

The area of this region can be calculated by integrating the function \(y = \sqrt{4-x^2}\) with respect to \(x\) from \(x = \sqrt{3}\) to \(x = 2\).

The integral we need to evaluate is: \(\int_{\sqrt{3}}^{2} \sqrt{4-x^2} dx\).

We use the standard integral formula for \(\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\arcsin\left(\frac{x}{a}\right) + C\). Here, \(a^2 = 4\), so \(a=2\).

So, \(\int \sqrt{4-x^2} dx = \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\arcsin\left(\frac{x}{2}\right) + C = \frac{x}{2}\sqrt{4-x^2} + 2\arcsin\left(\frac{x}{2}\right) + C\).

Now, we evaluate the definite integral:

\(\int_{\sqrt{3}}^{2} \sqrt{4-x^2} dx = \left[\frac{x}{2}\sqrt{4-x^2} + 2\arcsin\left(\frac{x}{2}\right)\right]_{\sqrt{3}}^{2}\)

Evaluate at the upper limit \(x = 2\):

\(\left(\frac{2}{2}\sqrt{4-2^2} + 2\arcsin\left(\frac{2}{2}\right)\right) = \left(1\sqrt{4-4} + 2\arcsin(1)\right) = (1\sqrt{0} + 2 \times \frac{\pi}{2}) = (0 + \pi) = \pi\)

Evaluate at the lower limit \(x = \sqrt{3}\):

\(\left(\frac{\sqrt{3}}{2}\sqrt{4-(\sqrt{3})^2} + 2\arcsin\left(\frac{\sqrt{3}}{2}\right)\right) = \left(\frac{\sqrt{3}}{2}\sqrt{4-3} + 2 \times \frac{\pi}{3}\right) = \left(\frac{\sqrt{3}}{2}\sqrt{1} + \frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} + \frac{2\pi}{3}\)

Subtract the lower limit value from the upper limit value:

Area \( = \pi - \left(\frac{\sqrt{3}}{2} + \frac{2\pi}{3}\right) = \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3}\)

Combine the terms with \(\pi\):

Area \( = \left(\pi - \frac{2\pi}{3}\right) - \frac{\sqrt{3}}{2} = \left(\frac{3\pi}{3} - \frac{2\pi}{3}\right) - \frac{\sqrt{3}}{2} = \frac{\pi}{3} - \frac{\sqrt{3}}{2}\)

Calculating the Area Using Geometry

Alternatively, we can use geometric shapes. Consider the sector of the circle formed by the origin (O), the point \(A(\sqrt{3}, 1)\), and the point \(B(2, 0)\). The point \(A(\sqrt{3}, 1)\) on the circle of radius 2 corresponds to an angle \(\theta\) such that \(\cos \theta = \frac{\sqrt{3}}{2}\) and \(\sin \theta = \frac{1}{2}\). This angle is \(\theta = \frac{\pi}{6}\) radians or 30 degrees with the positive x-axis. The point \(B(2, 0)\) corresponds to an angle of 0 radians with the positive x-axis.

The area of the sector OAB is given by \(\frac{1}{2}r^2\theta\), where \(r=2\) and the angle between the two points A and B relative to the origin is \(\frac{\pi}{6} - 0 = \frac{\pi}{6}\).

Area of sector OAB \( = \frac{1}{2}(2)^2\left(\frac{\pi}{6}\right) = \frac{1}{2}(4)\left(\frac{\pi}{6}\right) = 2 \times \frac{\pi}{6} = \frac{\pi}{3}\).

The region whose area we want is bounded by the arc AB, the line segment \(AC\) on \(x=\sqrt{3}\) (where \(C\) is \((\sqrt{3}, 0)\)), and the line segment \(CB\) on the x-axis. This region is the area of the sector OAB minus the area of the triangle OAC, where O=(0,0), A\((\sqrt{3}, 1)\), and C\((\sqrt{3}, 0)\).

Triangle OAC is a right-angled triangle with vertices O(0,0), C\((\sqrt{3}, 0)\) along the x-axis, and A\((\sqrt{3}, 1)\). The base of the triangle can be taken as the segment OC along the x-axis, and the height is the segment AC along the line \(x=\sqrt{3}\).

• Base \(OC = \sqrt{3} - 0 = \sqrt{3}\).
• Height \(AC = 1 - 0 = 1\).

Area of triangle OAC \( = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{3} \times 1 = \frac{\sqrt{3}}{2}\).

The required area is Area(Sector OAB) - Area(Triangle OAC).

Required Area \( = \frac{\pi}{3} - \frac{\sqrt{3}}{2}\).

Both the integration method and the geometric method yield the same result.

Comparing with Options

The calculated area is \(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\).

Let's look at the given options:

Our calculated area matches Option 1.

Revision Table: Key Concepts for Area Calculation

Additional Information: Understanding Regions and Boundaries

When a question asks for the area of a region "enclosed by" several curves, it typically refers to the finite area bounded by segments of those curves. Visualizing the graph of the functions and lines is essential to correctly identify which parts of the curves form the boundary and what the limits of integration should be.

In this problem, the phrase "enclosed by the x-axis the line \(x = \sqrt 3 \) and the circle" defines a specific bounded region. The boundaries are segments of \(y=0\), \(x=\sqrt{3}\), and \(x^2+y^2=4\). The region lies in the first quadrant. By finding the intersection points, we identify the vertices of this enclosed region. These vertices are \((\sqrt{3}, 0)\), \((\sqrt{3}, 1)\), and \((2, 0)\). The boundary consists of the line segment from \((\sqrt{3}, 0)\) to \((\sqrt{3}, 1)\), the circular arc from \((\sqrt{3}, 1)\) to \((2, 0)\), and the line segment from \((2, 0)\) back to \((\sqrt{3}, 0)\).

The initial mention of the line \(x = \sqrt{3}y\) (\(y = \frac{1}{\sqrt{3}}x\)) is relevant because its intersection with the circle \(x^2+y^2=4\) in the first quadrant is \((\sqrt{3}, 1)\), which is one of the key points defining the boundary of the requested region. This line passes through the origin and the point \((\sqrt{3}, 1)\), making an angle of \(\pi/6\) with the x-axis. This connection is useful for the geometric method involving sectors.