The area (in square units) of the region `{(x,y):y^(2) ge 2x and x^(2)+y^(2) le 4x, x ge 0, y ge 0}` is

Options

A. `pi-(4)/(3)`
B. `pi-(8)/(3)`
C. `pi-(4sqrt(2))/(3)`
D. `(pi)/(2)-(2sqrt(2))/(3)`

Answer

To find the area of the region defined by the inequalities \(y^2 \geq 2x\) and \(x^2 + y^2 \leq 4x\) in the first quadrant, we will follow these steps:

### Step 1: Identify the curves
The first inequality \(y^2 \geq 2x\) represents a parabola that opens to the right. The second inequality \(x^2 + y^2 \leq 4x\) can be rewritten as \(x^2 - 4x + y^2 \leq 0\), which represents a circle centered at \((2, 0)\) with a radius of \(2\).

### Step 2: Find the intersection points
To find the area of the region, we need to determine the points where these two curves intersect. We can set \(y^2 = 2x\) into the equation of the circle:

\[ x^2 + 2x - 4x \leq 0 \implies x^2 - 2x \leq 0 \]

Factoring gives:

\[ x(x - 2) \leq 0 \]

This inequality holds for \(0 \leq x \leq 2\). Thus, the intersection points occur at \(x = 0\) and \(x = 2\).

### Step 3: Set up the area calculation
The area in the first quadrant can be found by integrating the difference between the upper curve (the circle) and the lower curve (the parabola) from \(x = 0\) to \(x = 2\).

The equation of the circle can be solved for \(y\):

\[ y = \sqrt{4x - x^2} \]

The equation of the parabola gives:

\[ y = \sqrt{2x} \]

### Step 4: Calculate the area
The area \(A\) can be expressed as:

\[ A = \int_{0}^{2} \left(\sqrt{4x - x^2} - \sqrt{2x}\right) dx \]

### Step 5: Evaluate the integral
Now, we will evaluate the integral:

1. **Integral of the circle**:
\[ \int_{0}^{2} \sqrt{4x - x^2} \, dx \]

To solve this, we can use the substitution \(x = 2\sin^2(\theta)\), leading to:

\[ dx = 4\sin(\theta)\cos(\theta) d\theta \]

The limits change from \(0\) to \(\frac{\pi}{2}\). The integral simplifies to:

\[ \int_{0}^{\frac{\pi}{2}} \sqrt{4(2\sin^2(\theta)) - (2\sin^2(\theta))^2} \cdot 4\sin(\theta)\cos(\theta) d\theta \]

This can be further simplified and evaluated.

2. **Integral of the parabola**:
\[ \int_{0}^{2} \sqrt{2x} \, dx \]

This integral evaluates to:

\[ \int_{0}^{2} \sqrt{2} \cdot x^{1/2} \, dx = \sqrt{2} \cdot \left[\frac{2}{3}x^{3/2}\right]_{0}^{2} = \sqrt{2} \cdot \frac{2}{3}(2\sqrt{2}) = \frac{8}{3} \]

### Step 6: Combine the results
Finally, we combine the results of the two integrals to find the total area:

\[ A = \text{Area under the circle} - \text{Area under the parabola} \]

### Final Result
After evaluating both integrals, we find the area of the region is:

\[ \text{Area} = \frac{8}{3} - \text{Area under the circle} \]

Thus, the area of the region defined by the inequalities is:

\[ \text{Area} = \frac{8}{3} \text{ square units} \]