the area included between the curve `xy^(2)=a^(2)(a-x)` and y -axis is -

Options

A. `(pia^(2))/(2)`
B. `2pia^(2)`
C. `pi a^(2)`
D. None of these

Explanation

To find the area included between the curve \( xy^2 = a^2(a - x) \) and the y-axis, we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation of the curve:
\[ xy^2 = a^2(a - x) \]
We can express \( y^2 \) in terms of \( x \):
\[ y^2 = \frac{a^2(a - x)}{x} \]
Taking the square root gives us:
\[ y = \sqrt{\frac{a^2(a - x)}{x}} = a \sqrt{\frac{a - x}{x}} \]

### Step 2: Determining the Limits of Integration
Next, we need to find the limits of integration. The curve intersects the y-axis when \( x = 0 \). To find the other limit, we set \( y = 0 \):
\[ 0 = a \sqrt{\frac{a - x}{x}} \]
This occurs when \( a - x = 0 \) or \( x = a \). Thus, our limits of integration are from \( x = 0 \) to \( x = a \).

### Step 3: Setting Up the Integral for Area
The area \( A \) between the curve and the y-axis can be expressed as:
\[ A = \int_0^a y \, dx \]
Substituting for \( y \):
\[ A = \int_0^a a \sqrt{\frac{a - x}{x}} \, dx \]

### Step 4: Substituting for Simplification
To simplify the integral, we can use the substitution \( x = a \sin^2 \theta \). Then, \( dx = 2a \sin \theta \cos \theta \, d\theta \). The limits change as follows:
- When \( x = 0 \), \( \theta = 0 \)
- When \( x = a \), \( \theta = \frac{\pi}{2} \)

### Step 5: Changing the Integral
Substituting into the integral:
\[ A = \int_0^{\frac{\pi}{2}} a \sqrt{\frac{a - a \sin^2 \theta}{a \sin^2 \theta}} \cdot 2a \sin \theta \cos \theta \, d\theta \]
This simplifies to:
\[ A = 2a^2 \int_0^{\frac{\pi}{2}} \sqrt{\frac{1 - \sin^2 \theta}{\sin^2 \theta}} \sin \theta \cos \theta \, d\theta \]
\[ = 2a^2 \int_0^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta} \sin \theta \cos \theta \, d\theta \]
\[ = 2a^2 \int_0^{\frac{\pi}{2}} \cos^2 \theta \, d\theta \]

### Step 6: Evaluating the Integral
Using the identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \):
\[ A = 2a^2 \int_0^{\frac{\pi}{2}} \frac{1 + \cos(2\theta)}{2} \, d\theta \]
\[ = a^2 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{2}} = a^2 \left[ \frac{\pi}{2} + 0 - 0 \right] = \frac{\pi a^2}{2} \]

### Step 7: Total Area
Since we only calculated the area in one quadrant, we need to double this result:
\[ \text{Total Area} = 2 \cdot \frac{\pi a^2}{2} = \pi a^2 \]

### Final Answer
Thus, the area included between the curve and the y-axis is:
\[ \boxed{\pi a^2} \]