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 will follow these steps:

### Step 1: Analyze the Curve
The equation can be rewritten as:
\[ x^2 = -(y - 1)y^2 \]
This indicates that the curve intersects the y-axis at points where \( y = 0 \) and \( y = 1 \).

### Step 2: Identify the Limits of Integration
From the analysis, we see that the critical points are \( y = 0 \) and \( y = 1 \). Therefore, these will be our limits of integration.

### Step 3: Express \( x \) in Terms of \( y \)
From the equation \( x^2 = -(y - 1)y^2 \), we can express \( x \) as:
\[ x = -\sqrt{(y - 1)y^2} \]
This simplifies to:
\[ x = -y\sqrt{y - 1} \]

### Step 4: Set Up the Integral for Area
The area \( A \) of the loop can be calculated as:
\[ A = 2 \int_{0}^{1} -y\sqrt{y - 1} \, dy \]
We take the negative because \( x \) is negative in this region, and we multiply by 2 to account for the symmetry of the loop.

### Step 5: Simplify the Integral
We can rewrite the integral:
\[ A = -2 \int_{0}^{1} y\sqrt{y - 1} \, dy \]
This integral can be simplified further by rewriting \( \sqrt{y - 1} \) as \( \sqrt{-(1 - y)} \).

### Step 6: Change of Variable
Let \( u = 1 - y \), then \( dy = -du \) and the limits change from \( y = 0 \) to \( y = 1 \) into \( u = 1 \) to \( u = 0 \):
\[ A = 2 \int_{1}^{0} (1 - u)\sqrt{u} (-du) = 2 \int_{0}^{1} (1 - u)\sqrt{u} \, du \]

### Step 7: Expand the Integral
Now we can expand the integrand:
\[ A = 2 \int_{0}^{1} (\sqrt{u} - u^{3/2}) \, du \]

### Step 8: Integrate
Now we can integrate term by term:
\[ \int \sqrt{u} \, du = \frac{2}{3}u^{3/2} \quad \text{and} \quad \int u^{3/2} \, du = \frac{2}{5}u^{5/2} \]
Thus,
\[ A = 2 \left[ \frac{2}{3}u^{3/2} - \frac{2}{5}u^{5/2} \right]_{0}^{1} \]

### Step 9: Evaluate the Integral
Evaluating at the limits:
\[ A = 2 \left[ \frac{2}{3}(1) - \frac{2}{5}(1) \right] = 2 \left[ \frac{2}{3} - \frac{2}{5} \right] \]
Finding a common denominator (15):
\[ A = 2 \left[ \frac{10}{15} - \frac{6}{15} \right] = 2 \left[ \frac{4}{15} \right] = \frac{8}{15} \]

### Final Answer
Thus, the area of the loop of the curve is:
\[ \boxed{\frac{8}{15}} \]