Compute the area enclosed by the loop of the curve `y^(2)=x (x-1)^(2)`

Answer

To compute the area enclosed by the loop of the curve given by the equation \( y^2 = x(x - 1)^2 \), we will follow these steps:

### Step 1: Understand the curve and symmetry
The equation \( y^2 = x(x - 1)^2 \) indicates that the curve is symmetric about the x-axis because \( y^2 \) is always non-negative. This means that for every positive value of \( y \), there is a corresponding negative value of \( y \) at the same \( x \).

### Step 2: Find the points of intersection with the x-axis
To find the area enclosed by the loop, we first need to determine the points where the curve intersects the x-axis. This occurs when \( y = 0 \):
\[ y^2 = 0 \implies x(x - 1)^2 = 0 \]
This gives us:
- \( x = 0 \)
- \( (x - 1)^2 = 0 \implies x = 1 \)

Thus, the points of intersection are \( x = 0 \) and \( x = 1 \).

### Step 3: Set up the integral for the area
The area enclosed by the loop can be calculated using the formula for the area under a curve:
\[ \text{Area} = 2 \int_{0}^{1} y \, dx \]
Since the curve is symmetric, we will calculate the area above the x-axis and then double it.

### Step 4: Express \( y \) in terms of \( x \)
From the equation \( y^2 = x(x - 1)^2 \), we can express \( y \) as:
\[ y = \sqrt{x(x - 1)^2} = (x - 1) \sqrt{x} \quad \text{(for } x \in [0, 1] \text{)} \]

### Step 5: Substitute \( y \) into the integral
Now we can substitute \( y \) into the area integral:
\[ \text{Area} = 2 \int_{0}^{1} (x - 1) \sqrt{x} \, dx \]

### Step 6: Simplify the integral
We can simplify the integral:
\[ \text{Area} = 2 \int_{0}^{1} (x^{3/2} - x^{1/2}) \, dx \]

### Step 7: Evaluate the integral
Now we evaluate the integral:
\[ \int x^{3/2} \, dx = \frac{x^{5/2}}{5/2} = \frac{2}{5} x^{5/2} \]
\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Thus,
\[ \text{Area} = 2 \left[ \left( \frac{2}{5} x^{5/2} - \frac{2}{3} x^{3/2} \right) \bigg|_{0}^{1} \right] \]

### Step 8: Substitute the limits
Evaluating at the limits:
\[ = 2 \left( \left( \frac{2}{5} \cdot 1^{5/2} - \frac{2}{3} \cdot 1^{3/2} \right) - \left( \frac{2}{5} \cdot 0^{5/2} - \frac{2}{3} \cdot 0^{3/2} \right) \right) \]
\[ = 2 \left( \frac{2}{5} - \frac{2}{3} \right) \]

### Step 9: Find a common denominator and simplify
The common denominator for \( \frac{2}{5} \) and \( \frac{2}{3} \) is 15:
\[ \frac{2}{5} = \frac{6}{15}, \quad \frac{2}{3} = \frac{10}{15} \]
Thus,
\[ \frac{2}{5} - \frac{2}{3} = \frac{6}{15} - \frac{10}{15} = -\frac{4}{15} \]
So,
\[ \text{Area} = 2 \left( -\frac{4}{15} \right) = -\frac{8}{15} \]
Since area cannot be negative, we take the absolute value:
\[ \text{Area} = \frac{8}{15} \]

### Final Answer
The area enclosed by the loop of the curve \( y^2 = x(x - 1)^2 \) is:
\[ \text{Area} = \frac{8}{15} \]