Find the area of the region bounded by the parabola `y^(2)=2 px` and `x^(2)=2 py`.

Answer

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

### Step 1: Find the points of intersection
To find the points of intersection of the two parabolas, we can express \( y \) from the first equation and substitute it into the second equation.

From \( y^2 = 2px \), we have:
\[ y = \sqrt{2px} \]

Substituting \( y \) into the second equation \( x^2 = 2py \):
\[ x^2 = 2p(\sqrt{2px}) \]
Squaring both sides to eliminate the square root gives:
\[ x^4 = 4p^2x \]
Rearranging this equation:
\[ x^4 - 4p^2x = 0 \]
Factoring out \( x \):
\[ x(x^3 - 4p^2) = 0 \]
This gives us \( x = 0 \) or \( x^3 = 4p^2 \). Thus, we find:
\[ x = 0 \quad \text{or} \quad x = \sqrt[3]{4p^2} = 2p \]

### Step 2: Find the corresponding \( y \) values
Now we will find the corresponding \( y \) values for these \( x \) values.

1. For \( x = 0 \):
\[ y^2 = 2p(0) \implies y = 0 \]
So one intersection point is \( (0, 0) \).

2. For \( x = 2p \):
\[ y^2 = 2p(2p) = 4p^2 \implies y = 2p \]
Thus, the second intersection point is \( (2p, 2p) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 2p \) can be calculated using the formula:
\[ A = \int_{0}^{2p} (\text{Upper curve} - \text{Lower curve}) \, dx \]

From the equations:
- The upper curve is given by \( y = \sqrt{2px} \).
- The lower curve is given by \( y = \frac{x^2}{2p} \).

Thus, we have:
\[ A = \int_{0}^{2p} \left( \sqrt{2px} - \frac{x^2}{2p} \right) \, dx \]

### Step 4: Evaluate the integral
We can split the integral into two parts:
\[ A = \int_{0}^{2p} \sqrt{2p} \cdot x^{1/2} \, dx - \int_{0}^{2p} \frac{x^2}{2p} \, dx \]

Calculating the first integral:
\[ \int_{0}^{2p} \sqrt{2p} \cdot x^{1/2} \, dx = \sqrt{2p} \cdot \left[ \frac{2}{3} x^{3/2} \right]_{0}^{2p} = \sqrt{2p} \cdot \frac{2}{3} (2p)^{3/2} \]
Calculating \( (2p)^{3/2} \):
\[ (2p)^{3/2} = 2^{3/2} p^{3/2} = 2\sqrt{2} p^{3/2} \]
Thus:
\[ \int_{0}^{2p} \sqrt{2p} \cdot x^{1/2} \, dx = \sqrt{2p} \cdot \frac{2}{3} \cdot 2\sqrt{2} p^{3/2} = \frac{8p^2}{3} \]

Now for the second integral:
\[ \int_{0}^{2p} \frac{x^2}{2p} \, dx = \frac{1}{2p} \cdot \left[ \frac{x^3}{3} \right]_{0}^{2p} = \frac{1}{2p} \cdot \frac{(2p)^3}{3} = \frac{8p^2}{6} = \frac{4p^2}{3} \]

### Step 5: Combine the results
Now we combine both parts:
\[ A = \frac{8p^2}{3} - \frac{4p^2}{3} = \frac{4p^2}{3} \]

### Final Answer
Thus, the area of the region bounded by the two parabolas is:
\[ \boxed{\frac{4p^2}{3}} \text{ square units} \]