Determine the area enclosed by the two curves given by `y^(2)=x+1` and `y^(2)=-x+1`

Options

A. `(8)/(3)` sq units
B. `(4)/(3)` sq units
C. `(9)/(4)` sq units
D. `(7)/(3)` sq units

Explanation

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

### Step 1: Identify the curves
The equations given are:
1. \( y^2 = x + 1 \) (which can be rewritten as \( x = y^2 - 1 \))
2. \( y^2 = -x + 1 \) (which can be rewritten as \( x = 1 - y^2 \))

These are parabolas that open to the right and to the left, respectively.

### Step 2: Find the points of intersection
To find the points where the two curves intersect, we set the two equations equal to each other:
\[ y^2 - 1 = 1 - y^2 \]
Rearranging gives:
\[ 2y^2 = 2 \implies y^2 = 1 \implies y = 1 \text{ or } y = -1 \]
Now substituting \( y = 1 \) and \( y = -1 \) back into either equation to find \( x \):
- For \( y = 1 \): \( x = 1^2 - 1 = 0 \)
- For \( y = -1 \): \( x = (-1)^2 - 1 = 0 \)

Thus, the points of intersection are \( (0, 1) \) and \( (0, -1) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( y = -1 \) to \( y = 1 \) can be calculated using the formula:
\[ A = \int_{y_1}^{y_2} (x_{\text{right}} - x_{\text{left}}) \, dy \]
Here, \( x_{\text{right}} = 1 - y^2 \) and \( x_{\text{left}} = y^2 - 1 \).

Thus, the area becomes:
\[ A = \int_{-1}^{1} \left((1 - y^2) - (y^2 - 1)\right) \, dy \]
This simplifies to:
\[ A = \int_{-1}^{1} (1 - y^2 - y^2 + 1) \, dy = \int_{-1}^{1} (2 - 2y^2) \, dy \]

### Step 4: Calculate the integral
Now we can factor out the 2:
\[ A = 2 \int_{-1}^{1} (1 - y^2) \, dy \]
Now we compute the integral:
\[ \int (1 - y^2) \, dy = y - \frac{y^3}{3} \]
Evaluating from -1 to 1:
\[ A = 2 \left[ \left(1 - \frac{1^3}{3}\right) - \left(-1 + \frac{(-1)^3}{3}\right) \right] \]
Calculating this gives:
\[ = 2 \left[ \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{-1}{3}\right) \right] = 2 \left[ \frac{2}{3} + \frac{2}{3} \right] = 2 \cdot \frac{4}{3} = \frac{8}{3} \]

### Final Answer
The area enclosed by the two curves is:
\[ \boxed{\frac{8}{3}} \text{ square units} \]