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 \( y^2 = 4x \) is a rightward-opening parabola, and the second curve \( x^2 = 4y \) is an upward-opening parabola.

### Step 2: Find the points of intersection
To find the points of intersection, we need to solve the equations simultaneously.

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

2. Substitute \( x = \frac{y^2}{4} \) 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 \( y^3 = 64 \), so \( y = 4 \).

3. Now, substituting \( y = 0 \) and \( y = 4 \) back into \( x = \frac{y^2}{4} \):
- 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 found using the integral:
\[ A = \int_{0}^{4} (y_{\text{upper}} - y_{\text{lower}}) \, dx \]
Here, \( y_{\text{upper}} \) corresponds to the curve \( y = \sqrt{4x} \) (from \( y^2 = 4x \)), and \( y_{\text{lower}} \) corresponds to the curve \( y = \frac{x^2}{4} \) (from \( x^2 = 4y \)).

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

### Step 5: Simplify and integrate
Now, we simplify the integral:
\[ A = \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} \right) \, dx \]

### Step 6: Calculate the integral
1. Integrate \( 2\sqrt{x} \):
\[ \int 2\sqrt{x} \, dx = \frac{2}{\frac{3}{2}} x^{\frac{3}{2}} = \frac{4}{3} x^{\frac{3}{2}} \]

2. Integrate \( -\frac{x^2}{4} \):
\[ \int -\frac{x^2}{4} \, dx = -\frac{1}{4} \cdot \frac{x^3}{3} = -\frac{x^3}{12} \]

Putting it all together:
\[ A = \left[ \frac{4}{3} x^{\frac{3}{2}} - \frac{x^3}{12} \right]_{0}^{4} \]

### Step 7: Evaluate the definite integral
Now we evaluate at the limits:
1. At \( x = 4 \):
\[ A = \frac{4}{3} \cdot 4^{\frac{3}{2}} - \frac{4^3}{12} \]
\[ = \frac{4}{3} \cdot 8 - \frac{64}{12} \]
\[ = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \]

2. At \( x = 0 \):
\[ A = 0 \]

Thus, the area \( A \) is:
\[ A = \frac{16}{3} \]

### Final Answer
The area of the region bounded by the curves \( y^2 = 4x \) and \( x^2 = 4y \) is \( \frac{16}{3} \) square units.

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