Find the area bounded by `x=2y-y^2

Explanation

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

### Step 1: Understand the curve
The equation \( x = 2y - y^2 \) is a quadratic equation in terms of \( y \). To analyze it, we can rewrite it in standard form:
\[ y^2 - 2y + x = 0 \]
This is a parabola that opens to the left.

### Step 2: Find the points of intersection with the y-axis
To find the area, we need to determine the points where the curve intersects the y-axis. This occurs when \( x = 0 \):
\[ 0 = 2y - y^2 \]
Factoring gives:
\[ y(2 - y) = 0 \]
Thus, \( y = 0 \) and \( y = 2 \) are the points of intersection.

### Step 3: Set up the integral for the area
The area \( A \) bounded by the curve can be found by integrating with respect to \( y \) from \( y = 0 \) to \( y = 2 \):
\[ A = \int_{0}^{2} (2y - y^2) \, dy \]

### Step 4: Calculate the integral
Now we will compute the integral:
\[ A = \int_{0}^{2} (2y - y^2) \, dy \]
Breaking it down:
\[ A = \int_{0}^{2} 2y \, dy - \int_{0}^{2} y^2 \, dy \]

Calculating each integral:
1. For \( \int 2y \, dy \):
\[ \int 2y \, dy = y^2 \quad \text{(evaluated from 0 to 2)} \]
\[ = 2^2 - 0^2 = 4 \]

2. For \( \int y^2 \, dy \):
\[ \int y^2 \, dy = \frac{y^3}{3} \quad \text{(evaluated from 0 to 2)} \]
\[ = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} \]

Putting it all together:
\[ A = 4 - \frac{8}{3} \]
To combine these, convert 4 to a fraction:
\[ 4 = \frac{12}{3} \]
Thus,
\[ A = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \]

### Step 5: Conclusion
The area bounded by the curve \( x = 2y - y^2 \) is:
\[ \boxed{\frac{4}{3}} \text{ square units} \]

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