Find the area of the region bounded by the curves `x=4y-y^(2)` and the Y-axis.

Answer

To find the area of the region bounded by the curve \(x = 4y - y^2\) and the Y-axis, we can follow these steps:

### Step 1: Identify the curve and its properties
The equation \(x = 4y - y^2\) represents a parabola that opens to the left (since it can be rewritten as \(y^2 - 4y + x = 0\)).

### Step 2: Find the points of intersection with the Y-axis
To find where the curve intersects the Y-axis, we set \(x = 0\):
\[ 0 = 4y - y^2 \]
This can be factored as:
\[ y(4 - y) = 0 \]
Thus, the solutions are:
\[ y = 0 \quad \text{and} \quad y = 4 \]
So, the curve intersects the Y-axis at the points \((0, 0)\) and \((0, 4)\).

### Step 3: Set up the integral for the area
The area \(A\) bounded by the curve and the Y-axis can be found by integrating the function \(x = 4y - y^2\) from \(y = 0\) to \(y = 4\):
\[ A = \int_{0}^{4} (4y - y^2) \, dy \]

### Step 4: Calculate the integral
Now we compute the integral:
\[ A = \int_{0}^{4} (4y - y^2) \, dy = \left[ 2y^2 - \frac{y^3}{3} \right]_{0}^{4} \]
Calculating the definite integral:
\[ = \left( 2(4)^2 - \frac{(4)^3}{3} \right) - \left( 2(0)^2 - \frac{(0)^3}{3} \right) \]
\[ = \left( 2 \cdot 16 - \frac{64}{3} \right) \]
\[ = 32 - \frac{64}{3} \]
To combine these terms, convert \(32\) to a fraction:
\[ 32 = \frac{96}{3} \]
Thus,
\[ A = \frac{96}{3} - \frac{64}{3} = \frac{32}{3} \]

### Step 5: Conclusion
The area of the region bounded by the curve \(x = 4y - y^2\) and the Y-axis is:
\[ \boxed{\frac{32}{3}} \]