Find the area enclosed with in the curve
`y^(2)=3x, x=3`

Answer

To find the area enclosed within the curve \( y^2 = 3x \) and the line \( x = 3 \), we can follow these steps:

### Step 1: Understand the curves
The equation \( y^2 = 3x \) represents a parabola that opens to the right, touching the y-axis at the origin (0,0). The line \( x = 3 \) is a vertical line that intersects the parabola.

### Step 2: Find the points of intersection
To find the points where the parabola intersects the line \( x = 3 \), substitute \( x = 3 \) into the parabola's equation:
\[ y^2 = 3(3) = 9 \]
Taking the square root of both sides gives:
\[ y = \pm 3 \]
Thus, the points of intersection are \( (3, 3) \) and \( (3, -3) \).

### Step 3: Set up the integral for the area
The area enclosed between the curve and the line can be calculated by integrating the function representing the upper part of the curve minus the lower part. Since the region is symmetrical about the x-axis, we can calculate the area above the x-axis and then double it.

The upper part of the curve is given by:
\[ y = \sqrt{3x} \]
and the lower part is:
\[ y = -\sqrt{3x} \]

### Step 4: Calculate the area
We will integrate from \( x = 0 \) to \( x = 3 \):
\[ \text{Area} = 2 \int_{0}^{3} \sqrt{3x} \, dx \]

### Step 5: Solve the integral
First, factor out the constant:
\[ \text{Area} = 2 \sqrt{3} \int_{0}^{3} \sqrt{x} \, dx \]
Now, we know that \( \sqrt{x} = x^{1/2} \), so we can integrate:
\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Now, evaluate the integral from 0 to 3:
\[ \int_{0}^{3} \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_{0}^{3} = \frac{2}{3} (3^{3/2} - 0) = \frac{2}{3} (3\sqrt{3}) = 2\sqrt{3} \]

### Step 6: Calculate the total area
Now substitute back into the area formula:
\[ \text{Area} = 2 \sqrt{3} \cdot 2\sqrt{3} = 4 \cdot 3 = 12 \]

### Final Answer
The area enclosed within the curve \( y^2 = 3x \) and the line \( x = 3 \) is \( 12 \) square units.

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