The area between the curves `x=1-y^(2)` and `x=y^(2)-1` is

Options

A. `(4)/(3)`
B. `(1)/(3)`
C. `(8)/(3)`
D. `(5)/(3)`

Answer

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

### Step 1: Identify the curves and their intersections
The given curves are:
1. \( x = 1 - y^2 \) (which opens to the left)
2. \( x = y^2 - 1 \) (which opens to the right)

To find the points of intersection, we set the equations equal to each other:
\[ 1 - y^2 = y^2 - 1 \]
Rearranging gives:
\[ 2y^2 = 2 \quad \Rightarrow \quad y^2 = 1 \quad \Rightarrow \quad y = 1 \text{ or } y = -1 \]

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

### Step 2: Set up the integral for the area
The area \( A \) between the curves can be found using the formula:
\[ A = \int_{y_1}^{y_2} (f(y) - g(y)) \, dy \]
where \( f(y) \) is the right curve and \( g(y) \) is the left curve. In our case:
- \( f(y) = y^2 - 1 \)
- \( g(y) = 1 - y^2 \)

The limits of integration are from \( y = -1 \) to \( y = 1 \).

### Step 3: Calculate the area
Now we can set up the integral:
\[ A = \int_{-1}^{1} [(y^2 - 1) - (1 - y^2)] \, dy \]
This simplifies to:
\[ A = \int_{-1}^{1} [y^2 - 1 - 1 + y^2] \, dy = \int_{-1}^{1} [2y^2 - 2] \, dy \]
Factoring out the constant:
\[ A = 2 \int_{-1}^{1} (y^2 - 1) \, dy \]

### Step 4: Evaluate the integral
Now we evaluate the integral:
\[ \int (y^2 - 1) \, dy = \frac{y^3}{3} - y \]
Evaluating from \( -1 \) to \( 1 \):
\[ \left[ \frac{(1)^3}{3} - (1) \right] - \left[ \frac{(-1)^3}{3} - (-1) \right] = \left[ \frac{1}{3} - 1 \right] - \left[ -\frac{1}{3} + 1 \right] = \left[ \frac{1}{3} - 1 \right] - \left[ -\frac{1}{3} + 1 \right] = \left[ \frac{1}{3} - 1 \right] + \left[ \frac{1}{3} - 1 \right] = \left[ \frac{1}{3} - 1 + \frac{1}{3} - 1 \right] = \left[ \frac{2}{3} - 2 \right] = \left[ \frac{2 - 6}{3} \right] = -\frac{4}{3} \]
Thus, the area becomes:
\[ A = 2 \left( -\frac{4}{3} \right) = \frac{8}{3} \]

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