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

Answer

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

### Step 1: Understand the curves
1. The first inequality \( y^2 \leq 2x \) represents a parabola that opens to the right. The equation \( y^2 = 2x \) can be rewritten as \( y = \pm \sqrt{2x} \).
2. The second inequality \( x^2 + y^2 \leq 4x \) represents a circle. Rearranging gives \( (x - 2)^2 + y^2 \leq 4 \), which is a circle centered at \( (2, 0) \) with a radius of \( 2 \).

### Step 2: Identify the relevant region
Since we are interested in the region where \( x \geq 0 \) and \( y \leq 0 \), we will only consider the lower half of the parabola and the lower half of the circle.

### Step 3: Find the intersection points
To find the area of the region, we need to determine the points where the parabola intersects the circle. We set the equations equal to each other:
- From \( y^2 = 2x \), we substitute \( y^2 \) into the circle's equation:
\[ x^2 + 2x - 4x = 0 \implies x^2 - 2x = 0 \implies x(x - 2) = 0 \]
Thus, \( x = 0 \) and \( x = 2 \).

### Step 4: Set up the integral
The area can be calculated using the integral from \( x = 0 \) to \( x = 2 \) of the difference between the upper curve (circle) and the lower curve (parabola):
\[ \text{Area} = \int_{0}^{2} \left( \sqrt{4 - x^2 + 4x - 4} - (-\sqrt{2x}) \right) dx \]

### Step 5: Simplify the expression
The expression simplifies to:
\[ \text{Area} = \int_{0}^{2} \left( \sqrt{4 - (x-2)^2} + \sqrt{2x} \right) dx \]

### Step 6: Calculate the integral
1. **Integral of the circle part**:
\[ \int_{0}^{2} \sqrt{4 - (x-2)^2} \, dx \]
This represents the area of a quarter circle of radius 2.

2. **Integral of the parabola part**:
\[ \int_{0}^{2} \sqrt{2x} \, dx = \int_{0}^{2} (2^{1/2} x^{1/2}) \, dx = \frac{2}{3} x^{3/2} \Big|_0^2 = \frac{2}{3} (2^{3/2}) = \frac{4\sqrt{2}}{3} \]

### Step 7: Combine the areas
The area of the quarter circle is \( \frac{\pi \cdot 2^2}{4} = \pi \). Thus, the total area is:
\[ \text{Total Area} = \pi - \frac{4\sqrt{2}}{3} \]

### Final Answer
The area of the region is:
\[ \text{Area} = \pi - \frac{8}{3} \]