What is the area enclosed between the curve `y^(2)=12x` and the lines x=0 and y=6?

Options

A. 2 sq. units
B. 4 sq. units
C. 6 sq. units
D. 8 sq. units

Answer

To find the area enclosed between the curve \( y^2 = 12x \) and the lines \( x = 0 \) and \( y = 6 \), we will follow these steps:

### Step 1: Identify the curves and lines
The curve given is \( y^2 = 12x \), which is a rightward-opening parabola. The lines are \( x = 0 \) (the y-axis) and \( y = 6 \) (a horizontal line).

### Step 2: Find the points of intersection
To find the area, we need to determine the points where the curve intersects the line \( y = 6 \). We substitute \( y = 6 \) into the equation of the curve:

\[ 6^2 = 12x \]
\[ 36 = 12x \]
\[ x = \frac{36}{12} = 3 \]

So, the point of intersection is \( (3, 6) \).

### Step 3: Set up the integral for the area
The area \( A \) enclosed between the curve and the line can be expressed as:

\[ A = \int_{0}^{3} (y_{\text{top}} - y_{\text{bottom}}) \, dx \]

Here, \( y_{\text{top}} = 6 \) (the line) and \( y_{\text{bottom}} = \sqrt{12x} \) (the curve expressed in terms of \( y \)). Thus, the area can be written as:

\[ A = \int_{0}^{3} \left( 6 - \sqrt{12x} \right) \, dx \]

### Step 4: Calculate the integral
Now we will compute the integral:

\[ A = \int_{0}^{3} 6 \, dx - \int_{0}^{3} \sqrt{12x} \, dx \]

Calculating the first integral:

\[ \int_{0}^{3} 6 \, dx = 6x \Big|_{0}^{3} = 6(3) - 6(0) = 18 \]

Calculating the second integral:

\[ \int_{0}^{3} \sqrt{12x} \, dx = \int_{0}^{3} 2\sqrt{3} \sqrt{x} \, dx = 2\sqrt{3} \int_{0}^{3} x^{1/2} \, dx \]

Now, we calculate \( \int_{0}^{3} x^{1/2} \, dx \):

\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]

Evaluating this from 0 to 3:

\[ \frac{2}{3} \left( 3^{3/2} - 0 \right) = \frac{2}{3} \cdot 3\sqrt{3} = 2\sqrt{3} \]

Thus, we have:

\[ \int_{0}^{3} \sqrt{12x} \, dx = 2\sqrt{3} \cdot 2\sqrt{3} = 12 \]

### Step 5: Combine the results
Now we can find the area:

\[ A = 18 - 12 = 6 \]

### Final Answer
The area enclosed between the curve \( y^2 = 12x \) and the lines \( x = 0 \) and \( y = 6 \) is \( 6 \) square units.

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