The area of the region bounded by the parabola `y ^(2) =9x` and the line `y = 3x` is :

Options

A. `1/2` sq. units
B. `1/3` sq. units
C. `1/4` sq. units
D. `2/3` sq. units

Answer

To find the area of the region bounded by the parabola \( y^2 = 9x \) and the line \( y = 3x \), we will follow these steps:

### Step 1: Identify the curves
The first curve is the parabola given by the equation \( y^2 = 9x \). This can be rewritten in standard form as \( y = \pm 3\sqrt{x} \). The second curve is the line \( y = 3x \).

### Step 2: Find the points of intersection
To find the area between the curves, we need to determine where they intersect. We can set \( y = 3x \) into the parabola's equation:

\[ (3x)^2 = 9x \]
\[ 9x^2 - 9x = 0 \]
Factoring out \( 9x \):

\[ 9x(x - 1) = 0 \]

This gives us the solutions \( x = 0 \) and \( x = 1 \). Thus, the points of intersection are \( (0, 0) \) and \( (1, 3) \).

### Step 3: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be found using the integral:

\[ A = \int_{0}^{1} \left( \text{(upper curve)} - \text{(lower curve)} \right) \, dx \]

Here, the upper curve is the parabola \( y = 3\sqrt{x} \) and the lower curve is the line \( y = 3x \). Therefore, we have:

\[ A = \int_{0}^{1} \left( 3\sqrt{x} - 3x \right) \, dx \]

### Step 4: Simplify and integrate
We can factor out the constant \( 3 \):

\[ A = 3 \int_{0}^{1} \left( \sqrt{x} - x \right) \, dx \]

Now, we can integrate each term separately:

\[ A = 3 \left( \int_{0}^{1} x^{1/2} \, dx - \int_{0}^{1} x \, dx \right) \]

Calculating the integrals:

1. For \( \int x^{1/2} \, dx \):
\[ \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2} \]
Evaluating from 0 to 1:
\[ \left[ \frac{2}{3} x^{3/2} \right]_{0}^{1} = \frac{2}{3} - 0 = \frac{2}{3} \]

2. For \( \int x \, dx \):
\[ \int x \, dx = \frac{x^2}{2} \]
Evaluating from 0 to 1:
\[ \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2} \]

### Step 5: Combine the results
Now substituting back into the area expression:

\[ A = 3 \left( \frac{2}{3} - \frac{1}{2} \right) \]

Finding a common denominator (which is 6):

\[ A = 3 \left( \frac{4}{6} - \frac{3}{6} \right) = 3 \left( \frac{1}{6} \right) = \frac{3}{6} = \frac{1}{2} \]

### Final Answer
Thus, the area of the region bounded by the parabola and the line is:

\[ \boxed{\frac{1}{2}} \text{ square units} \]