The area bounded by the curve `y^(2)(2a-x)=x^(3)` and the line x = 2a is

Options

A. `3pia^(2)` sq units
B. `(3pia^(2))/(2)` sq units
C. `(3pia^(2))/(4)` sq unit
D. `(6pia^(2))/(5)` sq units

Explanation

To find the area bounded by the curve \( y^2(2a - x) = x^3 \) and the line \( x = 2a \), we will follow these steps:

### Step 1: Rewrite the equation of the curve
We start with the equation of the curve:
\[ y^2(2a - x) = x^3 \]
From this, we can express \( y^2 \):
\[ y^2 = \frac{x^3}{2a - x} \]
Thus, we have:
\[ y = \sqrt{\frac{x^3}{2a - x}} \]

### Step 2: Determine the limits of integration
The area we want to find is bounded between \( x = 0 \) and \( x = 2a \).

### Step 3: Set up the integral for the area
The area \( A \) can be expressed as:
\[ A = \int_{0}^{2a} y \, dx = \int_{0}^{2a} \sqrt{\frac{x^3}{2a - x}} \, dx \]

### Step 4: Use substitution
To simplify the integral, we will use the substitution \( x = 2a \sin^2 \theta \). Then, we differentiate:
\[ dx = 4a \sin \theta \cos \theta \, d\theta \]
When \( x = 0 \), \( \theta = 0 \) and when \( x = 2a \), \( \theta = \frac{\pi}{2} \).

### Step 5: Change the limits and the integrand
Substituting \( x = 2a \sin^2 \theta \) into the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{(2a \sin^2 \theta)^3}{2a - 2a \sin^2 \theta}} \cdot 4a \sin \theta \cos \theta \, d\theta \]
This simplifies to:
\[ A = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{8a^3 \sin^6 \theta}{2a \cos^2 \theta}} \cdot 4a \sin \theta \cos \theta \, d\theta \]

### Step 6: Simplify the integral
Further simplifying:
\[ A = 4a \int_{0}^{\frac{\pi}{2}} \sqrt{4a^2 \sin^6 \theta \cdot \frac{1}{\cos^2 \theta}} \cdot \sin \theta \cos \theta \, d\theta \]
\[ = 4a \int_{0}^{\frac{\pi}{2}} 4a \sin^4 \theta \, d\theta \]
\[ = 16a^2 \int_{0}^{\frac{\pi}{2}} \sin^4 \theta \, d\theta \]

### Step 7: Evaluate the integral
Using the reduction formula or known integral:
\[ \int_{0}^{\frac{\pi}{2}} \sin^n \theta \, d\theta = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2} \]
For \( n = 4 \):
\[ \int_{0}^{\frac{\pi}{2}} \sin^4 \theta \, d\theta = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16} \]

### Step 8: Final area calculation
Substituting back:
\[ A = 16a^2 \cdot \frac{3\pi}{16} = 3\pi a^2 \]

Thus, the area bounded by the curve and the line \( x = 2a \) is:
\[ \boxed{3\pi a^2} \]