Question:
The area included between the curves $y^{2} = 2x$ and $x^{2} = 2y$ is
The area included between the curves $y^{2} = 2x$ and $x^{2} = 2y$ is
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Area between two curves $= \displaystyle\int_a^b(y_{\text{upper}} - y_{\text{lower}})\,dx$. Always identify the upper and lower curves before integrating.
Updated On: May 2, 2026
- $\dfrac{4}{3}$ sq units
- $\dfrac{3}{4}$ sq units
- $\dfrac{1}{3}$ sq units
- $\dfrac{2}{3}$ sq units
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Find intersection points, then integrate the difference of the upper and lower curves.
Step 2: Detailed Explanation:
Curves intersect at $(0,0)$ and $(2,2)$. For $0 \le x \le 2$: upper curve $y = \sqrt{2x}$, lower $y = \dfrac{x^2}{2}$.
Area $= \displaystyle\int_0^2\!\left(\sqrt{2x} - \frac{x^2}{2}\right)dx = \left[\frac{\sqrt{2}\cdot 2x^{3/2}}{3} - \frac{x^3}{6}\right]_0^2 = \frac{2\sqrt{2}}{3}(2\sqrt{2}) - \frac{8}{6} = \frac{8}{3} - \frac{4}{3} = \frac{4}{3}$.
Step 3: Final Answer:
Area $= \dfrac{4}{3}$ sq units.
Find intersection points, then integrate the difference of the upper and lower curves.
Step 2: Detailed Explanation:
Curves intersect at $(0,0)$ and $(2,2)$. For $0 \le x \le 2$: upper curve $y = \sqrt{2x}$, lower $y = \dfrac{x^2}{2}$.
Area $= \displaystyle\int_0^2\!\left(\sqrt{2x} - \frac{x^2}{2}\right)dx = \left[\frac{\sqrt{2}\cdot 2x^{3/2}}{3} - \frac{x^3}{6}\right]_0^2 = \frac{2\sqrt{2}}{3}(2\sqrt{2}) - \frac{8}{6} = \frac{8}{3} - \frac{4}{3} = \frac{4}{3}$.
Step 3: Final Answer:
Area $= \dfrac{4}{3}$ sq units.
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