The area of the loop of the curve `a y^2=x^2(a-x)` is `4a^2s qdotu n i t s` (b) `(8a^2)/(15)s qdotu n i t s` `(16 a^2)/9s qdotu n i t s` (d) None of these

Options

A. `4a^(2)` sq. units
B. `(8a^(2))/(15)` sq. units
C. `(16a^(2))/(9)` sq. units
D. None of these

Explanation

To find the area of the loop of the curve given by the equation \( ay^2 = x^2(a - x) \), we will follow these steps:

### Step 1: Understand the Curve
The equation \( ay^2 = x^2(a - x) \) describes a curve. We need to find the area of the loop formed by this curve.

### Step 2: Identify the Points of Intersection
To find the points where the curve intersects the x-axis, we set \( y = 0 \):
\[ ay^2 = x^2(a - x) \implies 0 = x^2(a - x) \]
This gives us \( x = 0 \) and \( x = a \). Thus, the loop is formed between \( x = 0 \) and \( x = a \).

### Step 3: Set Up the Area Integral
The area \( A \) of the loop can be calculated by integrating the function for \( y \) from \( x = 0 \) to \( x = a \). We first express \( y \) in terms of \( x \):
\[ y = \sqrt{\frac{x^2(a - x)}{a}} \]
The area above the x-axis is given by:
\[ A = 2 \int_0^a y \, dx = 2 \int_0^a \sqrt{\frac{x^2(a - x)}{a}} \, dx \]

### Step 4: Simplify the Integral
We can simplify the integral:
\[ A = 2 \int_0^a \frac{x \sqrt{a - x}}{\sqrt{a}} \, dx = \frac{2}{\sqrt{a}} \int_0^a x \sqrt{a - x} \, dx \]

### Step 5: Use Substitution
To evaluate the integral, we can use the substitution \( u = a - x \), which gives \( du = -dx \) and changes the limits:
- When \( x = 0 \), \( u = a \)
- When \( x = a \), \( u = 0 \)

Thus, the integral becomes:
\[ \int_0^a x \sqrt{a - x} \, dx = \int_a^0 (a - u) \sqrt{u} (-du) = \int_0^a (a - u) \sqrt{u} \, du \]

### Step 6: Expand and Integrate
Now we expand the integrand:
\[ \int_0^a (a\sqrt{u} - u^{3/2}) \, du = a \int_0^a \sqrt{u} \, du - \int_0^a u^{3/2} \, du \]

Calculating these integrals:
1. \( \int_0^a \sqrt{u} \, du = \frac{2}{3} u^{3/2} \bigg|_0^a = \frac{2}{3} a^{3/2} \)
2. \( \int_0^a u^{3/2} \, du = \frac{2}{5} u^{5/2} \bigg|_0^a = \frac{2}{5} a^{5/2} \)

Putting it all together:
\[ \int_0^a (a\sqrt{u} - u^{3/2}) \, du = a \cdot \frac{2}{3} a^{3/2} - \frac{2}{5} a^{5/2} = \frac{2}{3} a^{5/2} - \frac{2}{5} a^{5/2} \]

### Step 7: Combine and Finalize the Area
Now, we combine the terms:
\[ \frac{2}{3} a^{5/2} - \frac{2}{5} a^{5/2} = \left(\frac{10}{15} - \frac{6}{15}\right) 2 a^{5/2} = \frac{4}{15} a^{5/2} \]
Thus:
\[ A = \frac{2}{\sqrt{a}} \cdot \frac{4}{15} a^{5/2} = \frac{8}{15} a^2 \]

### Conclusion
The area of the loop of the curve is:
\[ \boxed{\frac{8a^2}{15}} \text{ square units} \]