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

Options

A. `pi - (4sqrt2)/(3)`
B. `(pi)/2 - (2sqrt2)/(3)`
C. `pi - 4/3`
D. `pi - 8/3`

Answer

To find the area of the region defined by the inequalities \( y^2 \leq 2x \) and \( x^2 + y^2 \leq 4x \), with the constraints \( x \geq 0 \) and \( y \leq 0 \), we will follow these steps:

### Step 1: Identify the curves
The first inequality \( y^2 \leq 2x \) represents a parabola that opens to the right. The equation can be rewritten as:
\[ y = \pm \sqrt{2x} \]
The vertex of this parabola is at the origin (0,0).

The second inequality \( x^2 + y^2 \leq 4x \) can be rearranged to:
\[ x^2 - 4x + y^2 \leq 0 \]
Completing the square for the \( x \) terms gives:
\[ (x - 2)^2 + y^2 \leq 4 \]
This represents a circle centered at (2, 0) with a radius of 2.

### Step 2: Determine the intersection points
To find the points where the parabola intersects the circle, we set \( y^2 = 2x \) into the circle's equation:
\[ x^2 + 2x \leq 4x \]
This simplifies to:
\[ x^2 - 2x \leq 0 \]
Factoring gives:
\[ x(x - 2) \leq 0 \]
The solutions to this inequality are \( 0 \leq x \leq 2 \).

Next, we find the corresponding \( y \) values for \( x = 0 \) and \( x = 2 \):
- For \( x = 0 \):
\[ y^2 = 2(0) \implies y = 0 \]
- For \( x = 2 \):
\[ y^2 = 2(2) \implies y = \pm 2 \]

Thus, the intersection points are \( (0, 0) \), \( (2, 2) \), and \( (2, -2) \).

### Step 3: Sketch the region
We are interested in the area in the first quadrant (where \( x \geq 0 \) and \( y \leq 0 \)). The relevant area is bounded by the parabola and the circle from \( x = 0 \) to \( x = 2 \).

### Step 4: Calculate the area
The area can be calculated by integrating the difference between the upper curve (the circle) and the lower curve (the parabola) from \( x = 0 \) to \( x = 2 \).

1. The equation of the circle in terms of \( y \):
\[ y = -\sqrt{4x - x^2} \]
2. The equation of the parabola:
\[ y = -\sqrt{2x} \]

The area \( A \) is given by:
\[ A = \int_{0}^{2} \left(-\sqrt{4x - x^2} - (-\sqrt{2x})\right) dx \]
This simplifies to:
\[ A = \int_{0}^{2} \left(-\sqrt{4x - x^2} + \sqrt{2x}\right) dx \]

### Step 5: Evaluate the integral
We can split this integral into two parts:
\[ A = \int_{0}^{2} -\sqrt{4x - x^2} \, dx + \int_{0}^{2} \sqrt{2x} \, dx \]

**Calculating the first integral:**
Using the substitution \( u = 4x - x^2 \):
\[ \int -\sqrt{4x - x^2} \, dx = \text{(requires trigonometric substitution or numerical methods)} \]

**Calculating the second integral:**
\[ \int_{0}^{2} \sqrt{2x} \, dx = \int_{0}^{2} \sqrt{2} \sqrt{x} \, dx = \sqrt{2} \cdot \left[\frac{2}{3} x^{3/2}\right]_{0}^{2} = \sqrt{2} \cdot \frac{2}{3} (2^{3/2}) = \frac{4\sqrt{2}}{3} \]

### Final Area Calculation
Combining both results, we find the total area of the region.