Question

Find the area of the region bounded by the curve \(y^2=4x\) and the line \(x=3\).

• A. \(4\sqrt{3}\) sq. units
• B. \(6\sqrt{3}\) sq. units
• C. \(8\sqrt{3}\) sq. units
• D. \(12\sqrt{3}\) sq. units

Hint:

For curves of the form \(y^2 = ax\), remember the graph is symmetric about the \(x\)-axis. So the total area between the curve and a vertical boundary can be computed using \[ \text{Area} = 2\int y\,dx. \] This avoids calculating upper and lower parts separately.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to calculate the area enclosed between a rightward-opening parabola \(y^2=4x\) and a vertical line \(x=3\).
Because the parabola is symmetric about the x-axis, we can find the area in the first quadrant and double it.
Step 2: Key Formula or Approach:
The total area bounded by a curve symmetric about the x-axis and a vertical line from \(x_1\) to \(x_2\) is given by:
\[ \text{Area} = 2 \int_{x_1}^{x_2} y\,dx \] Step 3: Detailed Solution:
First, express \(y\) in terms of \(x\) for the upper half of the curve:
\[ y^2 = 4x \] \[ y = \sqrt{4x} = 2\sqrt{x} \quad \text{(considering the positive root for the first quadrant)} \] The region is bounded between \(x = 0\) (the vertex of the parabola) and the line \(x = 3\).
Set up the definite integral for the total area:
\[ \text{Area} = 2 \int_{0}^{3} 2\sqrt{x}\,dx \] \[ \text{Area} = 4 \int_{0}^{3} x^{1/2}\,dx \] Now, evaluate the integral:
\[ \text{Area} = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{3} \] \[ \text{Area} = 4 \left( \frac{2}{3} \right) \left[ x^{3/2} \right]_{0}^{3} \] \[ \text{Area} = \frac{8}{3} \left( 3^{3/2} - 0 \right) \] Simplify the numeric expression:
\[ 3^{3/2} = 3\sqrt{3} \] \[ \text{Area} = \frac{8}{3} \times 3\sqrt{3} \] \[ \text{Area} = 8\sqrt{3} \] Step 4: Final Answer:
The total bounded area is \(8\sqrt{3}\) sq. units.