Find the area of the region bounded by the curve `y^(2)=9x" and " y=3x`.

Explanation

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

### Step 1: Identify the curves
The first curve is \( y^2 = 9x \), which is a rightward-opening parabola. The second curve is \( y = 3x \), which is a straight line.

### Step 2: Find the points of intersection
To find the points where these two curves intersect, we set \( y = 3x \) into the equation of the parabola:

\[ (3x)^2 = 9x \]

This simplifies to:

\[ 9x^2 = 9x \]

Rearranging gives:

\[ 9x^2 - 9x = 0 \]

Factoring out \( 9x \):

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

Thus, \( x = 0 \) or \( x = 1 \).

### Step 3: Find corresponding \( y \) values
Now we find the \( y \) values for these \( x \) values using \( y = 3x \):

- For \( x = 0 \):
\[ y = 3(0) = 0 \quad \Rightarrow \quad (0, 0) \]

- For \( x = 1 \):
\[ y = 3(1) = 3 \quad \Rightarrow \quad (1, 3) \]

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) can be expressed as:

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

Here, the upper curve is \( y = 3\sqrt{x} \) (from \( y^2 = 9x \)) and the lower curve is \( y = 3x \). Thus, we have:

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

### Step 5: Simplify the integral
We can factor out the constant \( 3 \):

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

### Step 6: Evaluate the integral
Now we calculate the integral:

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

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

Thus, we have:

\[ A = 3 \left[ \left( \frac{2}{3} x^{3/2} - \frac{x^2}{2} \right) \right]_{0}^{1} \]

Evaluating from 0 to 1:

\[ = 3 \left[ \left( \frac{2}{3}(1)^{3/2} - \frac{(1)^2}{2} \right) - \left( \frac{2}{3}(0)^{3/2} - \frac{(0)^2}{2} \right) \right] \]

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

### Step 7: Simplify the expression
To combine \( \frac{2}{3} \) and \( \frac{1}{2} \), we find a common denominator (which is 6):

\[ \frac{2}{3} = \frac{4}{6}, \quad \frac{1}{2} = \frac{3}{6} \]

Thus:

\[ \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \]

Now substituting back:

\[ A = 3 \cdot \frac{1}{6} = \frac{1}{2} \]

### Final Answer
The area of the region bounded by the curves is:

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