Question

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

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

Hint:

If a curve contains \(y^2\) (like \(y^2 = 4x\)), it is symmetric about the \(x\)-axis. When finding the enclosed area, integrate for the positive branch and multiply the result by \(2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The curve $y^2 = 4x$ is a parabola opening to the right. The area required is the region between the vertex $(0,0)$ and the vertical line $x=3$.
Step 2: Key Formula or Approach:
The total area $A$ is symmetric about the x-axis, so we calculate the area in the first quadrant and multiply by 2. \[ \text{Area} = 2 \int_{0}^{3} y \, dx \]
Step 3: Detailed Explanation:
From $y^2 = 4x$, we get $y = 2\sqrt{x}$ for the upper branch. \[ \text{Area} = 2 \int_{0}^{3} 2\sqrt{x} \, dx \] \[ \text{Area} = 4 \int_{0}^{3} x^{1/2} \, dx \] \[ \text{Area} = 4 \left[ \frac{x^{3/2}}{3/2} \right]_{0}^{3} = 4 \cdot \frac{2}{3} [x\sqrt{x}]_{0}^{3} \] \[ \text{Area} = \frac{8}{3} (3\sqrt{3} - 0) = 8\sqrt{3} \] Note: Comparing $8\sqrt{3}$ with options, it equals $\frac{24\sqrt{3}}{3}$. If the question meant only the area in the first quadrant, it would be $4\sqrt{3} = \frac{12\sqrt{3}}{3}$ (Option D). Standard interpretation of "bounded by the curve" usually implies the total symmetric area. However, if restricted to the first quadrant, (D) is the result. Given the choices, (C) $\frac{8\sqrt{3}}{3}$ is likely a typo for $8\sqrt{3}$ or represents the integral $\int_0^3 \sqrt{4x}dx = 4\sqrt{3}$. Let's provide the calculation for the full area.
Step 4: Final Answer:
The total bounded area is $8\sqrt{3}$.