The area above the x-axis enclosed by the curves `x^(2)-y^(2) = 0` and `x^(2)+y-2=0` is

Options

A. `(5)/(3)`
B. `(7)/(3)`
C. `(8)/(3)`
D. `(10)/(3)`

Answer

To find the area above the x-axis enclosed by the curves \(x^2 - y^2 = 0\) and \(x^2 + y - 2 = 0\), we will follow these steps:

### Step 1: Identify the curves
The first curve \(x^2 - y^2 = 0\) can be factored as:
\[ (x - y)(x + y) = 0 \]
This gives us the lines:
\[ y = x \quad \text{and} \quad y = -x \]

The second curve \(x^2 + y - 2 = 0\) can be rearranged to:
\[ y = 2 - x^2 \]
This represents a downward-opening parabola.

### Step 2: Find points of intersection
To find the points where these curves intersect, we set \(y = x\) in the equation of the parabola:
\[ x = 2 - x^2 \]
Rearranging gives:
\[ x^2 + x - 2 = 0 \]
Factoring this quadratic equation:
\[ (x + 2)(x - 1) = 0 \]
Thus, the solutions are:
\[ x = -2 \quad \text{and} \quad x = 1 \]
Now, substituting these values back to find \(y\):
- For \(x = 1\), \(y = 1\) (point (1, 1)).
- For \(x = -2\), \(y = -2\) (point (-2, -2)).

### Step 3: Determine the area above the x-axis
The relevant area is between the points where \(y = x\) and \(y = 2 - x^2\) from \(x = 0\) to \(x = 1\).

### Step 4: Set up the integral
The area \(A\) can be calculated using the integral:
\[ A = \int_0^1 \left((2 - x^2) - x\right) \, dx \]

### Step 5: Evaluate the integral
Calculating the integral:
\[ A = \int_0^1 (2 - x^2 - x) \, dx = \int_0^1 (2 - x - x^2) \, dx \]
Now, we can integrate term by term:
\[ = \left[2x - \frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 \]
Evaluating at the limits:
\[ = \left(2(1) - \frac{1^2}{2} - \frac{1^3}{3}\right) - \left(2(0) - \frac{0^2}{2} - \frac{0^3}{3}\right) \]
\[ = 2 - \frac{1}{2} - \frac{1}{3} \]
Finding a common denominator (6):
\[ = \frac{12}{6} - \frac{3}{6} - \frac{2}{6} = \frac{12 - 3 - 2}{6} = \frac{7}{6} \]

### Step 6: Calculate total area
Since the area above the x-axis is symmetric with respect to the y-axis, the total area \(A_{total}\) is:
\[ A_{total} = 2 \times A = 2 \times \frac{7}{6} = \frac{14}{6} = \frac{7}{3} \]

### Final Answer
The area above the x-axis enclosed by the curves is:
\[ \boxed{\frac{7}{3}} \text{ square units} \]