Aea of the region nclosed between the curves `x=y^2-1` and `x=|y|sqrt(1-y^2)` is

Options

A. 1 sq. units
B. `4//3` sq. units
C. `2//3` sq. units
D. 2 sq. units

Explanation

To find the area of the region enclosed between the curves \( x = y^2 - 1 \) and \( x = |y|\sqrt{1 - y^2} \), we can follow these steps:

### Step 1: Identify the curves
The first curve is a parabola given by:
\[ x = y^2 - 1 \]
The second curve is given by:
\[ x = |y| \sqrt{1 - y^2} \]
This second curve represents a semicircle in the first and fourth quadrants.

### Step 2: Find the points of intersection
To find the area between the curves, we first need to determine the points where they intersect. We set the equations equal to each other:
\[ y^2 - 1 = |y| \sqrt{1 - y^2} \]
Since \( |y| \) can be either \( y \) or \( -y \), we will consider both cases.

#### Case 1: \( y \geq 0 \)
In this case, \( |y| = y \), so we have:
\[ y^2 - 1 = y \sqrt{1 - y^2} \]
Squaring both sides:
\[ (y^2 - 1)^2 = y^2(1 - y^2) \]
Expanding both sides:
\[ y^4 - 2y^2 + 1 = y^2 - y^4 \]
Combining like terms:
\[ 2y^4 - 3y^2 + 1 = 0 \]
Let \( z = y^2 \):
\[ 2z^2 - 3z + 1 = 0 \]
Using the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]
This gives us:
\[ z = 1 \quad \text{or} \quad z = \frac{1}{2} \]
Thus, \( y^2 = 1 \) or \( y^2 = \frac{1}{2} \), leading to:
\[ y = 1, \quad y = \frac{1}{\sqrt{2}} \quad \text{(and } y = -1, \quad y = -\frac{1}{\sqrt{2}} \text{ for } y < 0\text{)} \]

### Step 3: Set up the integral for area
The area \( A \) between the curves from \( y = -1 \) to \( y = 1 \) can be expressed as:
\[ A = 2 \int_0^1 \left( |y| \sqrt{1 - y^2} - (y^2 - 1) \right) dy \]
Since \( |y| = y \) in the interval \( [0, 1] \), we have:
\[ A = 2 \int_0^1 \left( y \sqrt{1 - y^2} - (y^2 - 1) \right) dy \]

### Step 4: Simplify the integral
This simplifies to:
\[ A = 2 \int_0^1 \left( y \sqrt{1 - y^2} + 1 - y^2 \right) dy \]

### Step 5: Evaluate the integral
We can evaluate the integral term by term:
1. For \( \int_0^1 y \sqrt{1 - y^2} \, dy \), we can use the substitution \( u = 1 - y^2 \), leading to \( du = -2y \, dy \).
2. The integral \( \int_0^1 1 \, dy = 1 \).
3. The integral \( \int_0^1 y^2 \, dy = \frac{1}{3} \).

Combining these results will yield the total area.

### Final Step: Calculate the area
After evaluating the integrals and combining the results, we find the area \( A \).