The area ( in square units ) enclosed by `| y | - x^2 = 1 ` and `x^(2) + y^(2) =1 ` is

Options

A. 2
B. zero
C. infinite
D. None of these

Answer

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

### Step 1: Rewrite the equations
The equation \( |y| - x^2 = 1 \) can be rewritten as:
\[ |y| = 1 + x^2 \]
This means we have two cases:
1. \( y = 1 + x^2 \) (for \( y \geq 0 \))
2. \( y = - (1 + x^2) \) (for \( y < 0 \))

The second equation \( x^2 + y^2 = 1 \) represents a circle with center at the origin (0,0) and radius 1.

### Step 2: Analyze the curves
- The curve \( y = 1 + x^2 \) is a parabola that opens upwards and intersects the y-axis at (0, 1).
- The curve \( y = - (1 + x^2) \) is a parabola that opens downwards and intersects the y-axis at (0, -1).
- The circle \( x^2 + y^2 = 1 \) intersects the y-axis at (0, 1) and (0, -1).

### Step 3: Find intersection points
To find the intersection points of the curves, we need to set \( y = 1 + x^2 \) and \( y = - (1 + x^2) \) equal to the circle equation.

1. For \( y = 1 + x^2 \):
\[ x^2 + (1 + x^2)^2 = 1 \]
Expanding this gives:
\[ x^2 + (1 + 2x^2 + x^4) = 1 \implies x^4 + 3x^2 + 1 - 1 = 0 \implies x^4 + 3x^2 = 0 \]
Factoring out \( x^2 \):
\[ x^2(x^2 + 3) = 0 \]
This gives \( x^2 = 0 \) (i.e., \( x = 0 \)) as the only real solution. Substituting \( x = 0 \) back gives \( y = 1 \).

2. For \( y = - (1 + x^2) \):
\[ x^2 + (-(1 + x^2))^2 = 1 \]
Expanding this gives:
\[ x^2 + (1 + 2x^2 + x^4) = 1 \implies x^4 + 3x^2 + 1 - 1 = 0 \implies x^4 + 3x^2 = 0 \]
Again, this gives \( x = 0 \) and substituting back gives \( y = -1 \).

### Step 4: Determine the area
The curves intersect at the points (0, 1) and (0, -1). The parabola \( y = 1 + x^2 \) and the circle do not enclose any area since they are tangent at these points. Thus, there is no bounded region between the curves.

### Conclusion
The area enclosed by the curves \( |y| - x^2 = 1 \) and \( x^2 + y^2 = 1 \) is:
\[ \text{Area} = 0 \text{ square units.} \]