Find the smaller of the two areas enclosed by the curves `x^(2)+y^(2)=4 and y^(2)=3(2x-1)`.

Options

A. `(4pi-sqrt3)" sq. units"`
B. `(1)/(2)(2pi+sqrt3)" sq. units"`
C. `(1)/(3)(4pi-sqrt3)" sq. units"`
D. None of these

Answer

To find the smaller of the two areas enclosed by the curves \(x^2 + y^2 = 4\) (a circle) and \(y^2 = 3(2x - 1)\) (a parabola), we will follow these steps:

### Step 1: Identify the curves
The first curve, \(x^2 + y^2 = 4\), represents a circle with center at (0, 0) and radius 2. The second curve, \(y^2 = 3(2x - 1)\), is a parabola that opens to the right.

### Step 2: Find the points of intersection
To find the points where the two curves intersect, we substitute \(y^2\) from the parabola into the equation of the circle:

\[ x^2 + 3(2x - 1) = 4 \]

This simplifies to:

\[ x^2 + 6x - 3 - 4 = 0 \implies x^2 + 6x - 7 = 0 \]

### Step 3: Solve the quadratic equation
We will solve the quadratic equation \(x^2 + 6x - 7 = 0\) using the factorization method:

\[ (x + 7)(x - 1) = 0 \]

Thus, the solutions are:

\[ x = 1 \quad \text{and} \quad x = -7 \]

Since we are interested in the area in the first quadrant, we take \(x = 1\).

### Step 4: Find the corresponding \(y\) values
Substituting \(x = 1\) back into the equation of the parabola to find \(y\):

\[ y^2 = 3(2(1) - 1) = 3(1) = 3 \implies y = \sqrt{3} \quad \text{or} \quad y = -\sqrt{3} \]

Thus, the points of intersection are \((1, \sqrt{3})\) and \((1, -\sqrt{3})\).

### Step 5: Set up the area integrals
To find the area between the curves, we will integrate from \(x = \frac{1}{2}\) (the vertex of the parabola) to \(x = 1\) (the intersection point).

The area \(A\) can be expressed as:

\[ A = 2 \int_{\frac{1}{2}}^{1} \left( \sqrt{6x - 3} - \sqrt{4 - x^2} \right) \, dx \]

### Step 6: Calculate the integral
1. **For the parabola**: \(y = \sqrt{6x - 3}\)
2. **For the circle**: \(y = \sqrt{4 - x^2}\)

Now we will compute the integral:

\[ A = 2 \int_{\frac{1}{2}}^{1} \left( \sqrt{6x - 3} - \sqrt{4 - x^2} \right) \, dx \]

### Step 7: Evaluate the integral
Using substitution for the first integral:

Let \(t = 6x - 3\), then \(dt = 6dx\) or \(dx = \frac{dt}{6}\).

The limits change as follows:
- When \(x = \frac{1}{2}\), \(t = 6(\frac{1}{2}) - 3 = 0\)
- When \(x = 1\), \(t = 6(1) - 3 = 3\)

Thus, the integral becomes:

\[ \int_{0}^{3} \sqrt{t} \cdot \frac{1}{6} dt = \frac{1}{6} \cdot \frac{2}{3} t^{3/2} \bigg|_{0}^{3} = \frac{1}{6} \cdot \frac{2}{3} \cdot (3^{3/2}) = \frac{1}{6} \cdot \frac{2}{3} \cdot 3\sqrt{3} = \frac{\sqrt{3}}{3} \]

For the circle, we compute:

\[ \int_{\frac{1}{2}}^{1} \sqrt{4 - x^2} \, dx \]

This integral can be evaluated using trigonometric substitution or standard integral formulas.

### Step 8: Combine results
After evaluating both integrals, we sum them up to find the total area \(A\).

### Step 9: Final Area Calculation
Finally, multiply the area found by 2 (since we considered only the upper half) to get the total area enclosed by the curves.

### Conclusion
The smaller area enclosed by the curves is:

\[ \frac{4\pi - \sqrt{3}}{3} \]