The area bounded by `|x| =1-y ^(2) and |x| +|y| =1` is:

Options

A. `1/3`
B. `1/2`
C. `2/3`
D. `1`

Answer

To find the area bounded by the curves \( |x| = 1 - y^2 \) and \( |x| + |y| = 1 \), we will follow these steps:

### Step 1: Understand the curves
The first curve \( |x| = 1 - y^2 \) represents two parabolas:
- \( x = 1 - y^2 \) for \( x \geq 0 \)
- \( x = - (1 - y^2) \) for \( x < 0 \)

The second curve \( |x| + |y| = 1 \) represents a square with vertices at \( (1, 0) \), \( (0, 1) \), \( (-1, 0) \), and \( (0, -1) \).

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

1. For \( x = 1 - y^2 \):
- The intersection with \( x + y = 1 \):
\[ 1 - y^2 + y = 1 \implies -y^2 + y = 0 \implies y(y - 1) = 0 \]
Thus, \( y = 0 \) or \( y = 1 \). The corresponding \( x \) values are \( x = 1 \) (for \( y = 0 \)) and \( x = 0 \) (for \( y = 1 \)).

2. For \( x = - (1 - y^2) \):
- The intersection with \( -x + y = 1 \):
\[ -(-1 + y^2) + y = 1 \implies 1 - y^2 + y = 1 \implies -y^2 + y = 0 \]
Thus, \( y = 0 \) or \( y = 1 \). The corresponding \( x \) values are \( x = -1 \) (for \( y = 0 \)) and \( x = 0 \) (for \( y = 1 \)).

### Step 3: Set up the area integral
The area bounded by these curves can be calculated by integrating the difference between the upper curve and the lower curve. The area in the first quadrant can be calculated and then multiplied by 4 (due to symmetry).

In the first quadrant, the curves are:
- Upper curve: \( x = 1 - y^2 \)
- Lower curve: \( x = 1 - y \)

The area \( A \) can be expressed as:
\[ A = 4 \int_0^1 \left( (1 - y^2) - (1 - y) \right) dy \]

### Step 4: Simplify the integral
\[ A = 4 \int_0^1 (y - y^2) dy \]

### Step 5: Evaluate the integral
Calculating the integral:
\[ \int (y - y^2) dy = \frac{y^2}{2} - \frac{y^3}{3} \]
Evaluating from 0 to 1:
\[ = \left[ \frac{1^2}{2} - \frac{1^3}{3} \right] - \left[ \frac{0^2}{2} - \frac{0^3}{3} \right] = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \]

Thus, the area \( A \) becomes:
\[ A = 4 \cdot \frac{1}{6} = \frac{2}{3} \]

### Final Result
The area bounded by the curves is \( \frac{2}{3} \).

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