Question

The area enclosed between the curves \(y^2 = 2x\) and \(x^2 = 2y\) is

• A. 3/4 sq unit
• B. 4/3 sq unit
• C. 1/2 sq unit
• D. 4/3 sq unit

Hint:

Choose integration variable wisely—here integrating w.r.t \(y\) simplifies calculations.

Answer 1

The Correct Option is B

Concept: Find intersection points and integrate between curves.

Step 1: Find points of intersection.
\[ y^2 = 2x,\quad x^2 = 2y \] Substitute \(x = \frac{y^2}{2}\) into second: \[ \left(\frac{y^2}{2}\right)^2 = 2y \Rightarrow \frac{y^4}{4} = 2y \Rightarrow y(y^3 - 8)=0 \Rightarrow y=0,2 \] \[ x=0,2 \]

Step 2: Express curves in terms of \(y\).
\[ x = \frac{y^2}{2}, \quad x = \sqrt{2y} \]

Step 3: Area between curves.
\[ A = \int_0^2 \left(\sqrt{2y} - \frac{y^2}{2}\right) dy \]

Step 4: Evaluate.
\[ = \int_0^2 (2^{1/2} y^{1/2} - \frac{y^2}{2}) dy \] \[ = \left[ \frac{2^{1/2}\cdot 2}{3} y^{3/2} - \frac{y^3}{6} \right]_0^2 \] \[ = \frac{2\sqrt{2}}{3} (2^{3/2}) - \frac{8}{6} = \frac{2\sqrt{2}\cdot 2\sqrt{2}}{3} - \frac{4}{3} = \frac{8}{3} - \frac{4}{3} = \frac{4}{3} \]