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

Options

A. 2 sq. unit
B. 4 sq. unit
C. `6` sq. unit
D. `8` sq. unit

Explanation

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 given curve 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 intersection points
To find the points of intersection between the curve and the line \(y = 6\), we substitute \(y = 6\) into the equation of the curve:

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

Thus, the points of intersection are \((3, 6)\).

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

\[ A = \int_{0}^{3} (y_1 - y_2) \, dx \]

Where:
- \(y_1\) is the upper function (the line \(y = 6\))
- \(y_2\) is the lower function (the curve \(y = \sqrt{12x}\))

### Step 4: Express the area in terms of the integral
Thus, we can express the area as:

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

### Step 5: 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:

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

Calculating the second integral:

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

Now, integrating \(x^{1/2}\):

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

Thus,

\[ \int_{0}^{3} x^{1/2} \, dx = \frac{2}{3} x^{3/2} \bigg|_{0}^{3} = \frac{2}{3} (3^{3/2} - 0) = \frac{2}{3} (3\sqrt{3}) = 2\sqrt{3} \]

So,

\[ \int_{0}^{3} \sqrt{12x} \, dx = \sqrt{12} \cdot 2\sqrt{3} = 2\sqrt{36} = 12 \]

### Step 6: Combine the results
Now substituting back into the area formula:

\[ A = 18 - 12 = 6 \]

### Final Answer
The area enclosed between the curves \(y^2 = 12x\) and the lines \(x = 0\) and \(y = 6\) is:

\[ \boxed{6} \text{ square units} \]