Area bounded by the region consisting of points `(x,y)` satisfying `y le sqrt(2-x^(2)), y^(2)ge x, sqrt(y)ge -x` is

Options

A. `pi/2`
B. `pi`
C. `2pi`
D. `pi//4`

Answer

To find the area bounded by the region consisting of points \((x,y)\) satisfying the inequalities \(y \leq \sqrt{2 - x^2}\), \(y^2 \geq x\), and \(\sqrt{y} \geq -x\), we will follow these steps:

### Step 1: Identify the curves
1. The curve \(y = \sqrt{2 - x^2}\) represents the upper half of a circle with radius \(\sqrt{2}\) centered at the origin.
2. The curve \(y^2 = x\) represents a parabola that opens to the right.
3. The curve \(\sqrt{y} = -x\) can be rewritten as \(y = x^2\), which is a parabola that opens upwards.

### Step 2: Find the points of intersection
To find the area, we first need to determine the points of intersection of these curves.

1. Set \(y = \sqrt{2 - x^2}\) equal to \(y = x^2\):
\[ \sqrt{2 - x^2} = x^2 \]
Squaring both sides:
\[ 2 - x^2 = x^4 \]
Rearranging gives:
\[ x^4 + x^2 - 2 = 0 \]
Let \(u = x^2\):
\[ u^2 + u - 2 = 0 \]
Factoring:
\[ (u - 1)(u + 2) = 0 \]
Thus, \(u = 1\) (since \(u = x^2\) must be non-negative). Therefore, \(x^2 = 1\) gives \(x = \pm 1\).

2. Now, substitute \(x = 1\) into \(y = x^2\):
\[ y = 1^2 = 1 \quad \text{and} \quad y = \sqrt{2 - 1^2} = \sqrt{1} = 1 \]
So, one point of intersection is \((1, 1)\).

3. For \(x = -1\):
\[ y = (-1)^2 = 1 \quad \text{and} \quad y = \sqrt{2 - (-1)^2} = \sqrt{1} = 1 \]
So, another point of intersection is \((-1, 1)\).

### Step 3: Determine the area
Now we need to find the area between the curves from \(x = -1\) to \(x = 1\).

1. The area \(A\) can be calculated using the integral:
\[ A = \int_{-1}^{1} \left( \sqrt{2 - x^2} - x^2 \right) \, dx \]

### Step 4: Compute the integral
1. First, compute the integral of \(\sqrt{2 - x^2}\):
\[ \int \sqrt{2 - x^2} \, dx = \frac{x}{2} \sqrt{2 - x^2} + 1 \cdot \sin^{-1}\left(\frac{x}{\sqrt{2}}\right) + C \]
Evaluating from \(-1\) to \(1\):
\[ \left[ \frac{1}{2} \sqrt{2 - 1^2} + \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \right] - \left[ \frac{-1}{2} \sqrt{2 - (-1)^2} + \sin^{-1}\left(\frac{-1}{\sqrt{2}}\right) \right] \]
This simplifies to:
\[ \left[ \frac{1}{2} \cdot 1 + \frac{\pi}{4} \right] - \left[ -\frac{1}{2} \cdot 1 - \frac{\pi}{4} \right] = 1 + \frac{\pi}{4} + \frac{1}{2} + \frac{\pi}{4} = 1 + 1 + \frac{\pi}{2} = 2 + \frac{\pi}{2} \]

2. Now compute the integral of \(x^2\):
\[ \int x^2 \, dx = \frac{x^3}{3} \bigg|_{-1}^{1} = \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3} \]

3. Therefore, the area \(A\) is:
\[ A = \left( 2 + \frac{\pi}{2} \right) - \frac{2}{3} \]
Simplifying gives:
\[ A = 2 + \frac{\pi}{2} - \frac{2}{3} = \frac{6}{3} + \frac{2}{3} - \frac{2}{3} + \frac{\pi}{2} = \frac{6}{3} + \frac{\pi}{2} = 2 + \frac{\pi}{2} \]

### Final Area
Thus, the area bounded by the curves is:
\[ \boxed{2 + \frac{\pi}{2}} \]