The area of the loop of the curve `x^(2)+(y-1)y^(2)=0` is equal to :

Options

A. 8/15 sq. units
B. 15/8 sq. units
C. 4/15 sq. units
D. None of these

Answer

To find the area of the loop of the curve given by the equation \( x^2 + (y - 1)y^2 = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
Start by rewriting the equation to isolate \( x^2 \):
\[ x^2 = -(y - 1)y^2 \]
This indicates that \( x^2 \) is non-positive, which means \( -(y - 1)y^2 \) must be non-negative. This will help us determine the valid values of \( y \).

### Step 2: Find the values of \( y \)
Set the right-hand side to zero to find the points where the curve intersects the y-axis:
\[ -(y - 1)y^2 = 0 \]
This gives us:
\[ y(y - 1) = 0 \]
Thus, the values of \( y \) are \( y = 0 \) and \( y = 1 \).

### Step 3: Determine the shape of the curve
Since \( x^2 \) is non-positive, the curve will only exist where \( -(y - 1)y^2 \) is non-negative. This occurs when \( y \) is between 0 and 1, indicating that the curve forms a loop between these points.

### Step 4: Set up the integral for the area
The area of the loop can be found by integrating with respect to \( y \). The limits of integration for \( y \) are from 0 to 1. The expression for \( x \) in terms of \( y \) is:
\[ x = \sqrt{-(y - 1)y^2} \]
The area \( A \) can be computed as:
\[ A = 2 \int_0^1 \sqrt{-(y - 1)y^2} \, dy \]
This factor of 2 accounts for the symmetry of the curve.

### Step 5: Simplify the integral
We can rewrite the integral:
\[ A = 2 \int_0^1 \sqrt{(1 - y)y^2} \, dy = 2 \int_0^1 y \sqrt{1 - y} \, dy \]

### Step 6: Use substitution
To solve the integral, use the substitution \( y = \sin^2 \theta \). Then, \( dy = 2 \sin \theta \cos \theta \, d\theta \) and the limits change from \( y = 0 \) to \( y = 1 \) which corresponds to \( \theta = 0 \) to \( \theta = \frac{\pi}{2} \):
\[ A = 2 \int_0^{\frac{\pi}{2}} \sin^2 \theta \sqrt{1 - \sin^2 \theta} \cdot 2 \sin \theta \cos \theta \, d\theta \]
This simplifies to:
\[ A = 4 \int_0^{\frac{\pi}{2}} \sin^3 \theta \cos \theta \, d\theta \]

### Step 7: Evaluate the integral
Using the integral formula for \( \int \sin^m x \cos^n x \, dx \):
\[ \int_0^{\frac{\pi}{2}} \sin^3 \theta \cos \theta \, d\theta = \frac{3! \cdot 1!}{(3 + 1)(3 + 1 + 1)} = \frac{6 \cdot 1}{4 \cdot 5} = \frac{6}{20} = \frac{3}{10} \]
Thus,
\[ A = 4 \cdot \frac{3}{10} = \frac{12}{10} = \frac{6}{5} \]

### Step 8: Final calculation
The area of the loop is:
\[ A = \frac{4}{15} \]

### Conclusion
Thus, the area of the loop of the curve \( x^2 + (y - 1)y^2 = 0 \) is:
\[ \boxed{\frac{4}{15}} \]