The area of the region bounded by the curves `y=x^(2) " and " x=y^(2)` is

Options

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

Answer

To find the area of the region bounded by the curves \( y = x^2 \) and \( x = y^2 \), we will follow these steps:

### Step 1: Find Points of Intersection
First, we need to determine the points where the two curves intersect. We have:
1. \( y = x^2 \)
2. \( x = y^2 \)

Substituting \( y = x^2 \) into \( x = y^2 \):
\[ x = (x^2)^2 = x^4 \]
Rearranging gives:
\[ x^4 - x = 0 \]
Factoring out \( x \):
\[ x(x^3 - 1) = 0 \]
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 2: Set Up the Integral
Next, we need to express the area between the curves. We will integrate with respect to \( x \) from \( 0 \) to \( 1 \). The upper curve is \( y = \sqrt{x} \) (from \( x = y^2 \)) and the lower curve is \( y = x^2 \).

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

### Step 3: Evaluate the Integral
Now, we will evaluate the integral:
\[ A = \int_{0}^{1} (\sqrt{x} - x^2) \, dx \]
This can be split into two separate integrals:
\[ A = \int_{0}^{1} \sqrt{x} \, dx - \int_{0}^{1} x^2 \, dx \]

Calculating the first integral:
\[ \int \sqrt{x} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Evaluating from \( 0 \) to \( 1 \):
\[ \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3}(1) - \frac{2}{3}(0) = \frac{2}{3} \]

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

### Step 4: Combine the Results
Now, substituting back into the area formula:
\[ A = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \]

### Final Answer
The area of the region bounded by the curves \( y = x^2 \) and \( x = y^2 \) is:
\[ \boxed{\frac{1}{3}} \]