Area (in square units) of the region bounded by the curve `y^(2)=4x,`y-axis and the line `y=3` , is

Options

A. 2
B. `9//4`
C. `6sqrt(3)`
D. none of these

Answer

To find the area of the region bounded by the curve \( y^2 = 4x \), the y-axis, and the line \( y = 3 \), we can follow these steps:

### Step 1: Understand the equations
The equation \( y^2 = 4x \) represents a parabola that opens to the right. The line \( y = 3 \) is a horizontal line.

### Step 2: Find the points of intersection
To find the points where the line intersects the parabola, substitute \( y = 3 \) into the parabola's equation:

\[ (3)^2 = 4x \implies 9 = 4x \implies x = \frac{9}{4} \]

Thus, the points of intersection are \( \left(\frac{9}{4}, 3\right) \) and the point on the y-axis is \( (0, 3) \).

### Step 3: Set up the integral for the area
The area \( A \) can be calculated using the integral of the function from the y-axis (where \( y = 0 \)) to the line \( y = 3 \). The area can be expressed as:

\[ A = \int_{0}^{3} x \, dy \]

From the equation of the parabola, we can express \( x \) in terms of \( y \):

\[ x = \frac{y^2}{4} \]

### Step 4: Substitute and compute the integral
Now substitute \( x \) into the integral:

\[ A = \int_{0}^{3} \frac{y^2}{4} \, dy \]

### Step 5: Factor out constants and integrate
Factor out \( \frac{1}{4} \):

\[ A = \frac{1}{4} \int_{0}^{3} y^2 \, dy \]

Now compute the integral:

\[ \int y^2 \, dy = \frac{y^3}{3} \]

Thus,

\[ A = \frac{1}{4} \left[ \frac{y^3}{3} \right]_{0}^{3} \]

### Step 6: Evaluate the definite integral
Now evaluate from 0 to 3:

\[ A = \frac{1}{4} \left( \frac{3^3}{3} - \frac{0^3}{3} \right) = \frac{1}{4} \left( \frac{27}{3} \right) = \frac{1}{4} \cdot 9 = \frac{9}{4} \]

### Conclusion
The area of the region bounded by the curve \( y^2 = 4x \), the y-axis, and the line \( y = 3 \) is

\[ \boxed{\frac{9}{4}} \text{ square units.} \]