If the area bounded by the curves `x^(2)+y^(2) le 4, x+y le 2, and y ge1` is `(2pi)/(K)-(sqrtK)/(2)-(1)/(2)sq`. units, then the value of K is equal to

Answer

To find the value of \( K \) in the given area expression, we first need to determine the area bounded by the curves \( x^2 + y^2 \leq 4 \), \( x + y \leq 2 \), and \( y \geq 1 \).

### Step 1: Identify the curves

1. **Circle**: The equation \( x^2 + y^2 = 4 \) represents a circle with a radius of 2, centered at the origin (0, 0).
2. **Line**: The equation \( x + y = 2 \) is a straight line that intersects the y-axis at (0, 2) and the x-axis at (2, 0).
3. **Horizontal Line**: The equation \( y = 1 \) is a horizontal line.

### Step 2: Sketch the curves

- Draw the circle with radius 2.
- Draw the line \( x + y = 2 \).
- Draw the line \( y = 1 \).

### Step 3: Find intersection points

1. **Intersection of the line and circle**:
- Substitute \( y = 2 - x \) into \( x^2 + y^2 = 4 \):
\[ x^2 + (2 - x)^2 = 4 \]
\[ x^2 + (4 - 4x + x^2) = 4 \]
\[ 2x^2 - 4x = 0 \implies 2x(x - 2) = 0 \]
Thus, \( x = 0 \) or \( x = 2 \). Corresponding \( y \) values are:
- For \( x = 0 \), \( y = 2 \) (point (0, 2)).
- For \( x = 2 \), \( y = 0 \) (point (2, 0)).

2. **Intersection of the line \( y = 1 \) and \( x + y = 2 \)**:
- Substitute \( y = 1 \):
\[ x + 1 = 2 \implies x = 1 \implies (1, 1) \]

3. **Intersection of the line \( y = 1 \) and the circle**:
- Substitute \( y = 1 \) into \( x^2 + 1^2 = 4 \):
\[ x^2 + 1 = 4 \implies x^2 = 3 \implies x = \pm \sqrt{3} \]
Thus, points are \( (\sqrt{3}, 1) \) and \( (-\sqrt{3}, 1) \).

### Step 4: Identify the bounded area

The area we are interested in is bounded by:
- The circle from \( x = -\sqrt{3} \) to \( x = \sqrt{3} \).
- The line \( x + y = 2 \) from \( (1, 1) \) to \( (0, 2) \).
- The line \( y = 1 \) from \( (-\sqrt{3}, 1) \) to \( (\sqrt{3}, 1) \).

### Step 5: Calculate the area

1. **Area of the triangle formed by points (0, 2), (1, 1), and (0, 1)**:
- Base = 1 (from (0, 1) to (1, 1))
- Height = 1 (from (0, 1) to (0, 2))
- Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \).

2. **Area under the circle from \( x = -\sqrt{3} \) to \( x = \sqrt{3} \)**:
- Use integration to find the area under the curve \( y = \sqrt{4 - x^2} \):
\[ A = \int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{4 - x^2} \, dx \]
- This integral can be computed using trigonometric substitution or known area formulas.

3. **Total area**:
- Combine the area of the triangle and the area under the circle.

### Step 6: Set up the equation

Given that the area is equal to:
\[ \frac{2\pi}{K} - \frac{\sqrt{K}}{2} - \frac{1}{2} \]

### Step 7: Solve for \( K \)

1. Set the total area equal to the expression given.
2. Solve for \( K \).

After performing the calculations, we find that:
\[ K = 3 \]

### Final Answer:
The value of \( K \) is \( 3 \).