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

Concept: The curve \(y^2 = 4x\) represents a parabola opening to the right. Since the equation contains \(y^2\), the curve is symmetric about the \(x\)-axis. To find the area bounded between the parabola and a vertical line, we integrate with respect to \(x\). Because of symmetry, we compute the area above the \(x\)-axis and multiply by \(2\).

Step 1: Express \(y\) in terms of \(x\). From \[ y^2 = 4x \] \[ y = \pm \sqrt{4x} \] Thus, \[ y = \pm 2\sqrt{x} \]

Step 2: Find the limits of integration. The region is bounded by the vertical line \[ x = 3 \] and the parabola starts at the vertex \[ x = 0 \] Thus the limits are: \[ 0 \le x \le 3 \]

Step 3: Set up the area integral. Area of the region: \[ A = 2\int_{0}^{3} y\,dx \] Substitute \(y = 2\sqrt{x}\): \[ A = 2\int_{0}^{3} 2\sqrt{x}\,dx \] \[ A = 4\int_{0}^{3} x^{1/2}\,dx \]

Step 4: Evaluate the integral. \[ \int x^{1/2}dx = \frac{x^{3/2}}{3/2} \] Thus, \[ A = 4\left[\frac{x^{3/2}}{3/2}\right]_{0}^{3} \] \[ A = 4\left[\frac{2}{3}x^{3/2}\right]_{0}^{3} \] \[ A = \frac{8}{3}(3^{3/2}) \] \[ 3^{3/2} = 3\sqrt{3} \] \[ A = \frac{8}{3}(3\sqrt{3}) \] \[ A = 8\sqrt{3} \]