Find the area of the region bounded by the curve `y^(2)=4x" and " x^(2)=4y`.

Explanation

To find the area of the region bounded by the curves \( y^2 = 4x \) and \( x^2 = 4y \), we will follow these steps:

### Step 1: Identify the curves
The first curve is \( y^2 = 4x \), which is a rightward-opening parabola. The second curve is \( x^2 = 4y \), which is an upward-opening parabola.

### Step 2: Find the points of intersection
To find the points where the curves intersect, we can solve the equations simultaneously.

1. From \( y^2 = 4x \), we can express \( x \) in terms of \( y \):
\[ x = \frac{y^2}{4} \]

2. Substitute \( x \) into the second equation \( x^2 = 4y \):
\[ \left(\frac{y^2}{4}\right)^2 = 4y \]
\[ \frac{y^4}{16} = 4y \]
\[ y^4 - 64y = 0 \]
\[ y(y^3 - 64) = 0 \]
This gives us \( y = 0 \) or \( y^3 = 64 \) (thus \( y = 4 \)).

3. Now, substituting back to find \( x \):
- For \( y = 0 \):
\[ x = \frac{0^2}{4} = 0 \]
- For \( y = 4 \):
\[ x = \frac{4^2}{4} = 4 \]

Thus, the points of intersection are \( (0, 0) \) and \( (4, 4) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 4 \) can be calculated using:
\[ A = \int_{0}^{4} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]
Here, \( y_{\text{upper}} \) is from the curve \( y^2 = 4x \) and \( y_{\text{lower}} \) is from \( x^2 = 4y \).

1. From \( y^2 = 4x \), we have:
\[ y = 2\sqrt{x} \]

2. From \( x^2 = 4y \), we have:
\[ y = \frac{x^2}{4} \]

Thus, the area can be expressed as:
\[ A = \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} \right) \, dx \]

### Step 4: Evaluate the integral
Now we will compute the integral:
\[ A = \int_{0}^{4} 2\sqrt{x} \, dx - \int_{0}^{4} \frac{x^2}{4} \, dx \]

1. Calculate \( \int 2\sqrt{x} \, dx \):
\[ \int 2\sqrt{x} \, dx = \frac{2 \cdot x^{3/2}}{3/2} = \frac{4}{3} x^{3/2} \]
Evaluating from 0 to 4:
\[ \left[ \frac{4}{3} x^{3/2} \right]_{0}^{4} = \frac{4}{3} (4^{3/2}) - 0 = \frac{4}{3} \cdot 8 = \frac{32}{3} \]

2. Calculate \( \int \frac{x^2}{4} \, dx \):
\[ \int \frac{x^2}{4} \, dx = \frac{1}{4} \cdot \frac{x^3}{3} = \frac{x^3}{12} \]
Evaluating from 0 to 4:
\[ \left[ \frac{x^3}{12} \right]_{0}^{4} = \frac{4^3}{12} - 0 = \frac{64}{12} = \frac{16}{3} \]

### Step 5: Combine the results
Now, we combine the results:
\[ A = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \]

### Final Answer
The area of the region bounded by the curves \( y^2 = 4x \) and \( x^2 = 4y \) is:
\[ \boxed{\frac{16}{3}} \]