The area included between the parabolas `y^(2)=4ax and x^(2)=4by` is

Options

A. `8/3 ab`
B. `16/3 ab`
C. `4/3 ab`
D. None

Answer

To find the area included between the parabolas \( y^2 = 4ax \) and \( x^2 = 4by \), we will follow these steps:

### Step 1: Find the intersection points of the parabolas
We have the equations:
1. \( y^2 = 4ax \)
2. \( x^2 = 4by \)

To find the intersection points, we can substitute \( y \) from the second equation into the first equation. From the second equation, we can express \( y \) as:
\[ y = \frac{x^2}{4b} \]

Now, substituting this into the first equation:
\[ \left(\frac{x^2}{4b}\right)^2 = 4ax \]
This simplifies to:
\[ \frac{x^4}{16b^2} = 4ax \]
Multiplying both sides by \( 16b^2 \) gives:
\[ x^4 = 64ab^2x \]
Rearranging this, we get:
\[ x^4 - 64ab^2x = 0 \]
Factoring out \( x \):
\[ x(x^3 - 64ab^2) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad x^3 = 64ab^2 \implies x = 4a^{1/3}b^{2/3} \]

### Step 2: Find the corresponding \( y \) values
For \( x = 0 \):
\[ y = 0 \]

For \( x = 4a^{1/3}b^{2/3} \):
Substituting this value into the equation \( y = \frac{x^2}{4b} \):
\[ y = \frac{(4a^{1/3}b^{2/3})^2}{4b} = \frac{16a^{2/3}b^{4/3}}{4b} = 4a^{2/3}b^{1/3} \]

Thus, the points of intersection are:
1. \( (0, 0) \)
2. \( \left(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3}\right) \)

### Step 3: 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 \( x = 4a^{1/3}b^{2/3} \).

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

Thus, the area \( A \) is given by:
\[ A = \int_{0}^{4a^{1/3}b^{2/3}} \left(\frac{x^2}{4b} - \sqrt{4ax}\right) \, dx \]

### Step 4: Compute the integral
Calculating the integral:
\[ A = \int_{0}^{4a^{1/3}b^{2/3}} \left(\frac{x^2}{4b} - 2\sqrt{a}\sqrt{x}\right) \, dx \]

Calculating the first term:
\[ \int \frac{x^2}{4b} \, dx = \frac{1}{4b} \cdot \frac{x^3}{3} = \frac{x^3}{12b} \]
Calculating the second term:
\[ \int 2\sqrt{a}\sqrt{x} \, dx = 2\sqrt{a} \cdot \frac{2}{3} x^{3/2} = \frac{4\sqrt{a}}{3} x^{3/2} \]

Now evaluating from \( 0 \) to \( 4a^{1/3}b^{2/3} \):
\[ A = \left[\frac{(4a^{1/3}b^{2/3})^3}{12b} - \frac{4\sqrt{a}}{3}(4a^{1/3}b^{2/3})^{3/2}\right] \]

Calculating each term:
1. The first term:
\[ \frac{(4^3)(a^{1})b^{2}}{12b} = \frac{64ab^2}{12b} = \frac{16ab}{3} \]
2. The second term:
\[ \frac{4\sqrt{a}}{3}(4^{3/2}(a^{1/3})^{3/2}(b^{2/3})^{3/2}) = \frac{4\sqrt{a}}{3}(8a^{1/2}b) = \frac{32a^{1/2}b}{3} \]

Thus, the area is:
\[ A = \frac{16ab}{3} - \frac{32a^{1/2}b}{3} \]

### Final Result
The area included between the parabolas is:
\[ A = \frac{16}{3} ab \, \text{square units} \]