The area common to the curve `y^(2)=x and x^(2)=y` is …........

Answer

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

### Step 1: Identify the curves
The first curve \( y^2 = x \) is a parabola that opens to the right, while the second curve \( x^2 = y \) is a parabola that opens upwards.

### Step 2: Find the points of intersection
To find the points of intersection, we need to solve the equations simultaneously. We can substitute \( y \) from the second equation into the first equation.

From \( x^2 = y \), we substitute \( y \) in \( y^2 = x \):
\[ (x^2)^2 = x \implies x^4 - x = 0 \implies x(x^3 - 1) = 0 \]
This gives us the solutions:
\[ x = 0 \quad \text{or} \quad x^3 - 1 = 0 \implies x = 1 \]
Thus, the points of intersection are \( (0, 0) \) and \( (1, 1) \).

### Step 3: Set up the integral for the area
The area between the curves from \( x = 0 \) to \( x = 1 \) can be found by integrating the difference of the upper curve and the lower curve.

The upper curve is \( y = \sqrt{x} \) (from \( y^2 = x \)), and the lower curve is \( y = x^2 \) (from \( x^2 = y \)).

### Step 4: Write the integral
The area \( A \) can be expressed as:
\[ A = \int_{0}^{1} (\sqrt{x} - x^2) \, dx \]

### Step 5: Evaluate the integral
Now we will compute the integral:
\[ A = \int_{0}^{1} x^{1/2} \, dx - \int_{0}^{1} x^2 \, dx \]
Calculating each integral separately:
1. For \( \int x^{1/2} \, dx \):
\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Evaluating from 0 to 1:
\[ \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3} (1) - \frac{2}{3} (0) = \frac{2}{3} \]

2. For \( \int x^2 \, dx \):
\[ \int x^2 \, dx = \frac{x^3}{3} \]
Evaluating from 0 to 1:
\[ \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} - 0 = \frac{1}{3} \]

### Step 6: Combine the results
Now, substituting back into the area formula:
\[ A = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

### Final Answer
The area common to the curves \( y^2 = x \) and \( x^2 = y \) is:
\[ \boxed{\frac{1}{3}} \text{ square units} \]