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 curves defined by 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 first curve is \( y = \sqrt{2 - x^2} \), which represents the upper half of a circle with radius \(\sqrt{2}\) centered at the origin.
2. The second curve, \( y^2 = x \), can be rewritten as \( y = \sqrt{x} \) and \( y = -\sqrt{x} \), representing a sideways parabola.
3. The third curve, \( \sqrt{y} = -x \), can be rewritten as \( y = x^2 \) for \( x \leq 0 \), representing another parabola.

### Step 2: Sketch the curves
Draw the curves on a coordinate plane to visualize the area of interest. The curve \( y = \sqrt{2 - x^2} \) will intersect the x-axis at \( x = -\sqrt{2} \) and \( x = \sqrt{2} \). The parabola \( y = x^2 \) opens upwards, and the parabola \( y = \sqrt{x} \) opens to the right.

### Step 3: Find intersection points
To find the area bounded by these curves, we need to determine the points where they intersect:
1. Set \( \sqrt{2 - x^2} = \sqrt{x} \):
\[ 2 - x^2 = x \implies x^2 + x - 2 = 0 \]
Factoring gives:
\[ (x - 1)(x + 2) = 0 \implies x = 1 \text{ or } x = -2 \]
Thus, the points of intersection are \( (1, 1) \) and \( (-2, 2) \).

2. Set \( \sqrt{2 - x^2} = x^2 \):
\[ 2 - x^2 = x^4 \implies x^4 + x^2 - 2 = 0 \]
Let \( u = x^2 \):
\[ u^2 + u - 2 = 0 \]
Factoring gives:
\[ (u - 1)(u + 2) = 0 \implies u = 1 \text{ or } u = -2 \]
Thus, \( x^2 = 1 \implies x = 1 \text{ or } x = -1 \).

### Step 4: Set up the integrals
The area \( A \) can be computed as:
\[ A = \int_{-2}^{-1} (\sqrt{2 - x^2} - x^2) \, dx + \int_{-1}^{1} (\sqrt{2 - x^2} - \sqrt{x}) \, dx \]

### Step 5: Evaluate the integrals
1. **First Integral**:
\[ A_1 = \int_{-2}^{-1} (\sqrt{2 - x^2} - x^2) \, dx \]
Use the formula for integrating \( \sqrt{a^2 - x^2} \).

2. **Second Integral**:
\[ A_2 = \int_{-1}^{1} (\sqrt{2 - x^2} - \sqrt{x}) \, dx \]
Evaluate this integral using appropriate techniques.

### Step 6: Combine results
Combine the results from both integrals to find the total area.

### Final Result
After evaluating the integrals, we find the total area bounded by the curves.

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