Area bounded by the loop of the curve
`ay^(2)=x^2(a- x)` is equal to

Options

A. `4/15a^(2)`
B. `8/15a^(2)`
C. `16/15a^(2)`
D. None

Answer

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

### Step 1: Rearranging the Equation
The given equation can be rearranged to express \( y \) in terms of \( x \):
\[ y^2 = \frac{x^2(a - x)}{a} \]
Taking the square root gives:
\[ y = \pm \sqrt{\frac{x^2(a - x)}{a}} = \pm \frac{x\sqrt{a - x}}{\sqrt{a}} \]

### Step 2: Finding Points of Intersection
To find the points where the curve intersects the x-axis (where \( y = 0 \)):
\[ \frac{x^2(a - x)}{a} = 0 \]
This gives us:
\[ x^2 = 0 \quad \text{or} \quad a - x = 0 \]
Thus, the points of intersection are \( x = 0 \) and \( x = a \).

### Step 3: Setting Up the Integral
The area \( A \) bounded by the loop can be found by integrating the positive part of \( y \) from \( 0 \) to \( a \) and then doubling it:
\[ A = 2 \int_0^a y \, dx = 2 \int_0^a \frac{x\sqrt{a - x}}{\sqrt{a}} \, dx \]

### Step 4: Simplifying the Integral
We can factor out the constant \( \frac{2}{\sqrt{a}} \):
\[ A = \frac{2}{\sqrt{a}} \int_0^a x\sqrt{a - x} \, dx \]

### Step 5: Substituting for Integration
To solve the integral \( \int_0^a x\sqrt{a - x} \, dx \), we can use the substitution \( t = a - x \) which implies \( x = a - t \) and \( dx = -dt \). The limits change from \( x = 0 \) to \( x = a \) which translates to \( t = a \) to \( t = 0 \):
\[ \int_0^a x\sqrt{a - x} \, dx = \int_a^0 (a - t)\sqrt{t} (-dt) = \int_0^a (a - t)\sqrt{t} \, dt \]
This can be split into two integrals:
\[ \int_0^a a\sqrt{t} \, dt - \int_0^a t\sqrt{t} \, dt \]

### Step 6: Evaluating the Integrals
1. For \( \int_0^a a\sqrt{t} \, dt \):
\[ = a \cdot \left[ \frac{2}{3} t^{3/2} \right]_0^a = a \cdot \frac{2}{3} a^{3/2} = \frac{2}{3} a^{5/2} \]

2. For \( \int_0^a t\sqrt{t} \, dt \):
\[ = \left[ \frac{2}{5} t^{5/2} \right]_0^a = \frac{2}{5} a^{5/2} \]

Combining these results:
\[ \int_0^a x\sqrt{a - x} \, dx = \frac{2}{3} a^{5/2} - \frac{2}{5} a^{5/2} = \left( \frac{10}{15} - \frac{6}{15} \right) a^{5/2} = \frac{4}{15} a^{5/2} \]

### Step 7: Final Area Calculation
Substituting back into the area formula:
\[ A = \frac{2}{\sqrt{a}} \cdot \frac{4}{15} a^{5/2} = \frac{8}{15} a^2 \]

Thus, the area bounded by the loop of the curve is:
\[ \boxed{\frac{8a^2}{15}} \]