If the area enclosed by the curve `y=sqrt(x)` and `x=-sqrt(y)` , the circle `x^2+y^2=2` above the x-axis is `A ,` then the value of `(16)/piA` is___

Explanation

To solve the problem, we need to find the area \( A \) enclosed by the curves \( y = \sqrt{x} \), \( x = -\sqrt{y} \), and the circle \( x^2 + y^2 = 2 \) above the x-axis. Then, we will compute the value of \( \frac{16}{\pi A} \).

### Step 1: Understand the curves and their intersections
1. The circle \( x^2 + y^2 = 2 \) has a radius of \( \sqrt{2} \).
2. The curve \( y = \sqrt{x} \) is defined for \( x \geq 0 \).
3. The curve \( x = -\sqrt{y} \) is defined for \( y \geq 0 \) and represents the left half of the parabola.

### Step 2: Find the intersection points
To find the intersection points of the curves and the circle, we can substitute \( y = \sqrt{x} \) into the equation of the circle:
\[ x^2 + (\sqrt{x})^2 = 2 \implies x^2 + x = 2 \implies x^2 + x - 2 = 0 \]
Factoring the quadratic:
\[ (x - 1)(x + 2) = 0 \]
Thus, \( x = 1 \) or \( x = -2 \). Since \( y = \sqrt{x} \) is only defined for \( x \geq 0 \), we take \( x = 1 \):
\[ y = \sqrt{1} = 1 \]
So one intersection point is \( (1, 1) \).

Now, substituting \( y = -\sqrt{y} \) into the circle's equation:
\[ (-\sqrt{y})^2 + y^2 = 2 \implies y + y^2 = 2 \implies y^2 + y - 2 = 0 \]
Factoring gives us:
\[ (y - 1)(y + 2) = 0 \]
Thus, \( y = 1 \) or \( y = -2 \). Again, we take \( y = 1 \), leading to \( x = -1 \):
\[ (-1, 1) \]
So the second intersection point is \( (-1, 1) \).

### Step 3: Calculate the area \( A \)
The area enclosed by the curves and the circle above the x-axis can be calculated by finding the area of the circle in the first quadrant and subtracting the areas under the curves.

1. **Area of the circle in the first quadrant:**
The area of the full circle is:
\[ \text{Area} = \pi r^2 = \pi ( \sqrt{2} )^2 = 2\pi \]
The area in the first quadrant is:
\[ \frac{1}{4} \times 2\pi = \frac{\pi}{2} \]

2. **Area under the curves:**
The area under \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 1 \):
\[ \int_0^1 \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^1 = \frac{2}{3} \]
The area under \( x = -\sqrt{y} \) from \( y = 0 \) to \( y = 1 \):
\[ \int_0^1 -\sqrt{y} \, dy = -\left[ \frac{2}{3} y^{3/2} \right]_0^1 = -\frac{2}{3} \]
The total area under the curves is:
\[ A_{\text{curves}} = \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \]

3. **Total area \( A \):**
The area \( A \) is:
\[ A = \text{Area of circle in first quadrant} - A_{\text{curves}} = \frac{\pi}{2} - \frac{4}{3} \]

### Step 4: Calculate \( \frac{16}{\pi A} \)
Now we need to compute:
\[ \frac{16}{\pi A} = \frac{16}{\pi \left( \frac{\pi}{2} - \frac{4}{3} \right)} \]
This simplifies to:
\[ \frac{16}{\frac{\pi^2}{2} - \frac{4\pi}{3}} = \frac{16 \cdot 6}{3\pi^2 - 8\pi} = \frac{96}{3\pi^2 - 8\pi} \]

### Final Answer
After simplification, we find that:
\[ \frac{16}{\pi A} = 8 \]