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 a parabola that opens to the right, given by:
\[ y^2 = 4x \]
The second curve is a parabola that opens upwards, given by:
\[ x^2 = 4y \]

### Step 2: Find the points of intersection
To find the points of intersection, we can substitute \( y \) from one equation into the other. From the first equation, we can express \( x \) in terms of \( y \):
\[ x = \frac{y^2}{4} \]
Now substitute this into the second equation:
\[ \left(\frac{y^2}{4}\right)^2 = 4y \]
This simplifies to:
\[ \frac{y^4}{16} = 4y \]
Multiplying both sides by 16 gives:
\[ y^4 = 64y \]
Rearranging this, we have:
\[ y^4 - 64y = 0 \]
Factoring out \( y \):
\[ y(y^3 - 64) = 0 \]
Thus, \( y = 0 \) or \( y^3 = 64 \) which gives \( y = 4 \).

### Step 3: Find corresponding \( x \) values
Now, we find the corresponding \( x \) values for \( y = 0 \) and \( y = 4 \):
1. For \( y = 0 \):
\[ x = \frac{0^2}{4} = 0 \]
So, one point of intersection is \( (0, 0) \).

2. For \( y = 4 \):
\[ x = \frac{4^2}{4} = 4 \]
So, the other point of intersection is \( (4, 4) \).

### Step 4: Set up the integral for the area
The area between the curves can be found by integrating the difference of the upper curve and the lower curve from \( x = 0 \) to \( x = 4 \).

The upper curve is given by \( y = \sqrt{4x} = 2\sqrt{x} \) (from \( y^2 = 4x \)), and the lower curve is given by \( y = \frac{x^2}{4} \) (from \( x^2 = 4y \)).

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

### Step 5: Evaluate the integral
Now we evaluate the integral:
\[ A = \int_{0}^{4} 2\sqrt{x} \, dx - \int_{0}^{4} \frac{x^2}{4} \, dx \]

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

Calculating the second integral:
\[ \int \frac{x^2}{4} \, dx = \frac{1}{4} \cdot \frac{x^3}{3} = \frac{x^3}{12} \]
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 6: 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:
\[ \boxed{\frac{16}{3}} \text{ square units} \]