Question

Find the area of the region bounded by the curve \(y^2 = 8x\) and the line \(x = 2\).

• A. \( \frac{28}{3} \) sq. units
• B. \( \frac{32}{3} \) sq. units
• C. \( \frac{16}{3} \) sq. units
• D. \( \frac{20}{3} \) sq. units

Hint:

For curves of the form \(y^2 = 4ax\), the upper and lower branches are \(y=\pm\sqrt{4ax}\). The area between the curve and a vertical line \(x=b\) can be found using \[ \int_{0}^{b} (y_{\text{top}}-y_{\text{bottom}})\,dx. \]

Answer 1

The Correct Option is B

Concept: To find the area between a curve and a vertical line, we integrate the difference of the upper and lower functions with respect to \(x\). For the curve: \[ y^2 = 8x \] \[ y = \pm \sqrt{8x} \] The region lies between the upper curve \(y=\sqrt{8x}\) and the lower curve \(y=-\sqrt{8x}\). Hence, \[ \text{Area} = \int (y_{\text{top}} - y_{\text{bottom}})\,dx \]

Step 1: Find the limits of integration. The line is \[ x=2 \] The parabola passes through the origin, so the limits are \[ x=0 \quad \text{to} \quad x=2 \]

Step 2: Set up the integral for the area. \[ A=\int_{0}^{2}\left(\sqrt{8x}-(-\sqrt{8x})\right)dx \] \[ A=\int_{0}^{2}2\sqrt{8x}\,dx \] \[ A=\int_{0}^{2}4\sqrt{2}\sqrt{x}\,dx \]

Step 3: Evaluate the integral. \[ A=4\sqrt{2}\int_{0}^{2}x^{1/2}dx \] \[ A=4\sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_{0}^{2} \] \[ A=\frac{8\sqrt{2}}{3}\left(2^{3/2}\right) \] \[ 2^{3/2}=2\sqrt{2} \] \[ A=\frac{8\sqrt{2}}{3}\times 2\sqrt{2} \] \[ A=\frac{8\times2\times2}{3} \] \[ A=\frac{32}{3} \] Thus, the required area is \[ \boxed{\frac{32}{3}\ \text{sq. units}} \]