Calculate the area enclosed by the parabola `y^(2)=x+3y` and the Y-axis.

Answer

To calculate the area enclosed by the parabola \( y^2 = x + 3y \) and the Y-axis, we will follow these steps:

### Step 1: Rewrite the equation of the parabola
We start with the equation of the parabola:
\[ y^2 = x + 3y \]
Rearranging this gives:
\[ x = y^2 - 3y \]

### Step 2: Find the points of intersection with the Y-axis
To find where the parabola intersects the Y-axis, we set \( x = 0 \):
\[ 0 = y^2 - 3y \]
Factoring this equation:
\[ y(y - 3) = 0 \]
This gives us the points:
\[ y = 0 \quad \text{and} \quad y = 3 \]

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

### Step 4: Calculate the integral
Now we compute the integral:
\[ A = \int_{0}^{3} (y^2 - 3y) \, dy = \int_{0}^{3} y^2 \, dy - \int_{0}^{3} 3y \, dy \]
Calculating each part:
1. For \( \int y^2 \, dy \):
\[ \int y^2 \, dy = \frac{y^3}{3} \Big|_{0}^{3} = \frac{3^3}{3} - 0 = \frac{27}{3} = 9 \]
2. For \( \int 3y \, dy \):
\[ \int 3y \, dy = \frac{3y^2}{2} \Big|_{0}^{3} = \frac{3 \cdot 3^2}{2} - 0 = \frac{27}{2} \]

### Step 5: Combine the results
Now substituting back into the area expression:
\[ A = 9 - \frac{27}{2} \]
To combine these, we convert \( 9 \) into a fraction:
\[ 9 = \frac{18}{2} \]
Thus,
\[ A = \frac{18}{2} - \frac{27}{2} = \frac{18 - 27}{2} = \frac{-9}{2} \]

### Step 6: Take the absolute value
Since area cannot be negative, we take the absolute value:
\[ A = \left| \frac{-9}{2} \right| = \frac{9}{2} \]

### Final Answer
The area enclosed by the parabola \( y^2 = x + 3y \) and the Y-axis is:
\[ \boxed{\frac{9}{2}} \]