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

Explanation

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
We start with the equations of the parabolas:
1. \( y^2 = 2px \) (Equation 1)
2. \( x^2 = 2py \) (Equation 2)

To find the points of intersection, we can express \( x \) in terms of \( y \) from Equation 1:
\[ x = \frac{y^2}{2p} \]
Now, substitute this expression for \( x \) into Equation 2:
\[ \left(\frac{y^2}{2p}\right)^2 = 2py \]
This simplifies to:
\[ \frac{y^4}{4p^2} = 2py \]
Multiplying both sides by \( 4p^2 \) gives:
\[ y^4 = 8p^3y \]
Rearranging this, we have:
\[ y^4 - 8p^3y = 0 \]
Factoring out \( y \):
\[ y(y^3 - 8p^3) = 0 \]
This gives us two solutions:
1. \( y = 0 \)
2. \( y^3 = 8p^3 \) which implies \( y = 2p \)

Now, we have the \( y \)-coordinates of the points of intersection: \( y = 0 \) and \( y = 2p \).

### Step 2: Find the Corresponding \( x \)-Coordinates
To find the corresponding \( x \)-coordinates, we substitute \( y = 2p \) back into Equation 1:
\[ y^2 = 2px \implies (2p)^2 = 2px \implies 4p^2 = 2px \implies x = \frac{4p^2}{2p} = 2p \]
Thus, the points of intersection are \( (0, 0) \) and \( (2p, 2p) \).

### Step 3: Set Up the Integral for Area
The area \( A \) between the curves can be calculated using the integral:
\[ A = \int_{0}^{2p} \left( \text{Upper curve} - \text{Lower curve} \right) \, dy \]
From the equations, the upper curve is given by \( y^2 = 2px \) or \( y = \sqrt{2px} \), and the lower curve is given by \( x^2 = 2py \) or \( y = \frac{x^2}{2p} \).

### Step 4: Express \( x \) in Terms of \( y \)
From \( y^2 = 2px \):
\[ x = \frac{y^2}{2p} \]
From \( x^2 = 2py \):
\[ y = \frac{x^2}{2p} \]

### Step 5: Calculate the Area
Now we can express the area as:
\[ A = \int_{0}^{2p} \left( \sqrt{2p \cdot \frac{y^2}{2p}} - \frac{(\frac{y^2}{2p})^2}{2p} \right) \, dy \]
This simplifies to:
\[ A = \int_{0}^{2p} \left( y - \frac{y^4}{8p^3} \right) \, dy \]

### Step 6: Evaluate the Integral
Calculating the integral:
\[ A = \int_{0}^{2p} y \, dy - \int_{0}^{2p} \frac{y^4}{8p^3} \, dy \]
Calculating each part:
1. \( \int_{0}^{2p} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{2p} = \frac{(2p)^2}{2} = 2p^2 \)
2. \( \int_{0}^{2p} \frac{y^4}{8p^3} \, dy = \frac{1}{8p^3} \left[ \frac{y^5}{5} \right]_{0}^{2p} = \frac{1}{8p^3} \cdot \frac{(2p)^5}{5} = \frac{32p^5}{40p^3} = \frac{4p^2}{5} \)

Thus, the area becomes:
\[ A = 2p^2 - \frac{4p^2}{5} = \frac{10p^2}{5} - \frac{4p^2}{5} = \frac{6p^2}{5} \]

### Final Result
The area of the region bounded by the two parabolas is:
\[ \boxed{\frac{6p^2}{5}} \]