Question

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

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

Hint:

For a parabola of the form \(y^2 = 4ax\), expressing \(x\) in terms of \(y\) simplifies area calculations when bounded by vertical lines.

Answer 1

The Correct Option is A

Concept: To find the area bounded between a parabola and a vertical line, it is convenient to express \(x\) in terms of \(y\) and integrate with respect to \(y\). Given curve: \[ y^2 = 4x \quad \Rightarrow \quad x = \frac{y^2}{4} \] Area between curves: \[ A = \int (x_{right} - x_{left})\,dy \]

Step 1: Find the points of intersection. The line \(x = 3\) intersects the parabola when: \[ y^2 = 4(3) = 12 \] \[ y = \pm 2\sqrt{3} \] Thus the limits are: \[ -2\sqrt{3} \le y \le 2\sqrt{3} \]

Step 2: Set up the integral. Right boundary: \[ x = 3 \] Left boundary: \[ x = \frac{y^2}{4} \] \[ A = \int_{-2\sqrt{3}}^{2\sqrt{3}} \left(3 - \frac{y^2}{4}\right) dy \]

Step 3: Evaluate the integral. \[ A = \int_{-2\sqrt{3}}^{2\sqrt{3}} 3\,dy - \int_{-2\sqrt{3}}^{2\sqrt{3}} \frac{y^2}{4} dy \] First integral: \[ = 3(4\sqrt{3}) = 12\sqrt{3} \] Second integral: \[ \int_{-2\sqrt{3}}^{2\sqrt{3}} \frac{y^2}{4} dy = \frac{1}{4}\left[\frac{y^3}{3}\right]_{-2\sqrt{3}}^{2\sqrt{3}} \] \[ (2\sqrt{3})^3 = 24\sqrt{3} \] \[ = \frac{1}{12}(48\sqrt{3}) = 4\sqrt{3} \] Therefore, \[ A = 12\sqrt{3} - 4\sqrt{3} = 8\sqrt{3} \] \[ \boxed{A = 8\sqrt{3}\ \text{sq. units}} \]