If the area bounded by `y^(2)=4ax and x^(2)=4ay` is `(64)/(3)` square units, then the positive value of a is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the positive value of \( a \) given that the area bounded by the curves \( y^2 = 4ax \) and \( x^2 = 4ay \) is \( \frac{64}{3} \) square units, we can follow these steps:

### Step 1: Identify the curves
The equations given are:
1. \( y^2 = 4ax \) (a rightward-opening parabola)
2. \( x^2 = 4ay \) (an upward-opening parabola)

### Step 2: Find the points of intersection
To find the points where these curves intersect, we can substitute \( y \) from the first equation into the second equation.

From \( y^2 = 4ax \), we have:
\[ y = \sqrt{4ax} \]

Substituting into \( x^2 = 4ay \):
\[ x^2 = 4a(\sqrt{4ax}) \]
Squaring both sides:
\[ x^4 = 16a^2x \]
Rearranging gives:
\[ x^4 - 16a^2x = 0 \]
Factoring out \( x \):
\[ x(x^3 - 16a^2) = 0 \]
Thus, \( x = 0 \) or \( x^3 = 16a^2 \).

From \( x^3 = 16a^2 \):
\[ x = 2\sqrt[3]{2a} \]

### Step 3: Find the corresponding \( y \) values
Using \( x = 2\sqrt[3]{2a} \) in \( y^2 = 4ax \):
\[ y^2 = 4a(2\sqrt[3]{2a}) = 8a\sqrt[3]{2a} \]
Thus,
\[ y = \sqrt{8a\sqrt[3]{2a}} = 2\sqrt{2a}\sqrt[3]{\sqrt{2a}} = 2\sqrt{2a} \cdot (2a)^{1/6} = 2 \cdot 2^{1/2} \cdot a^{1/2} \cdot a^{1/6} = 2^{3/2} a^{2/3} \]

### Step 4: Set up the area integral
The area \( A \) between the curves can be found using the formula:
\[ A = \int_{0}^{2\sqrt[3]{2a}} \left( \sqrt{4ax} - \frac{x^2}{4a} \right) dx \]

### Step 5: Calculate the area
1. The integral becomes:
\[ A = \int_{0}^{2\sqrt[3]{2a}} \left( \sqrt{4ax} - \frac{x^2}{4a} \right) dx \]

2. Calculate the first integral:
\[ \int \sqrt{4ax} \, dx = \frac{2}{3}(4a)^{1/2} x^{3/2} = \frac{2\sqrt{4a}}{3} x^{3/2} \]

3. Calculate the second integral:
\[ \int \frac{x^2}{4a} \, dx = \frac{1}{4a} \cdot \frac{x^3}{3} = \frac{x^3}{12a} \]

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

5. Simplifying gives the area:
\[ A = \frac{16a^{2}}{3} \]

### Step 6: Set the area equal to \( \frac{64}{3} \)
Setting the area equal to \( \frac{64}{3} \):
\[ \frac{16a^2}{3} = \frac{64}{3} \]

### Step 7: Solve for \( a \)
Multiply both sides by \( 3 \):
\[ 16a^2 = 64 \]
Divide by \( 16 \):
\[ a^2 = 4 \]
Taking the positive square root:
\[ a = 2 \]

### Final Answer
The positive value of \( a \) is \( \boxed{2} \).