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 can 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:
\[ y^2 - 3y = x \]

### Step 2: Complete the square
To make it easier to analyze, we complete the square for the left-hand side:
\[ y^2 - 3y + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 = x \]
This simplifies to:
\[ \left(y - \frac{3}{2}\right)^2 = x + \frac{9}{4} \]
This shows that the parabola opens to the right with vertex at \( \left(-\frac{9}{4}, \frac{3}{2}\right) \).

### Step 3: Identify the intersection points with the Y-axis
To find the points where the parabola intersects the Y-axis, we set \( x = 0 \):
\[ 0 = y^2 - 3y \]
Factoring gives:
\[ y(y - 3) = 0 \]
Thus, \( y = 0 \) and \( y = 3 \). Therefore, the intersection points are \( (0, 0) \) and \( (0, 3) \).

### Step 4: Set up the integral for the area
The area \( A \) enclosed by the parabola and the Y-axis can be found using the integral:
\[ A = \int_{0}^{3} (x \text{ in terms of } y) \, dy \]
From our earlier manipulation, we have \( x = y^2 - 3y \). Thus, we can write:
\[ A = \int_{0}^{3} (y^2 - 3y) \, dy \]

### Step 5: Evaluate 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. \(\int_{0}^{3} y^2 \, dy = \left[\frac{y^3}{3}\right]_{0}^{3} = \frac{27}{3} = 9\)
2. \(\int_{0}^{3} 3y \, dy = 3\left[\frac{y^2}{2}\right]_{0}^{3} = 3 \cdot \frac{9}{2} = \frac{27}{2}\)

Putting it all together:
\[ A = 9 - \frac{27}{2} = \frac{18}{2} - \frac{27}{2} = -\frac{9}{2} \]
Since area cannot be negative, we take the absolute value:
\[ A = \frac{9}{2} \]

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