Question:
The area of the loop of the curve $ay²=x²(a-x)$ is
The area of the loop of the curve $ay²=x²(a-x)$ is
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The area of the loop of the curve $ay(a-x)$ is
Updated On: Apr 15, 2026
- $\frac{8a^{2}}{15}sq~unit$
- $\frac{4a^{2}}{15}sq~unit$
- $\frac{2a^{2}}{15}sq~unit$
- None of these
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The loop of $ay^2 = x^2(a-x)$ occurs between $x=0$ and $x=a$. Total area is $2\int_{0}^{a} y \, dx$.
Step 2: Analysis
$y = x\sqrt{\frac{a-x}{a}}$. Area $= 2 \int_{0}^{a} x\sqrt{\frac{a-x}{a}} \, dx$.
Step 3: Evaluation
Using substitution $x = a \sin^2 \theta$ leads to the integral $4a^2 \int_{0}^{\pi/2} \sin^3 \theta \cos^2 \theta \, d\theta$.
Step 4: Conclusion
Solving the trigonometric integral results in $\frac{8a^2}{15}$.
Final Answer: (a)
The loop of $ay^2 = x^2(a-x)$ occurs between $x=0$ and $x=a$. Total area is $2\int_{0}^{a} y \, dx$.
Step 2: Analysis
$y = x\sqrt{\frac{a-x}{a}}$. Area $= 2 \int_{0}^{a} x\sqrt{\frac{a-x}{a}} \, dx$.
Step 3: Evaluation
Using substitution $x = a \sin^2 \theta$ leads to the integral $4a^2 \int_{0}^{\pi/2} \sin^3 \theta \cos^2 \theta \, d\theta$.
Step 4: Conclusion
Solving the trigonometric integral results in $\frac{8a^2}{15}$.
Final Answer: (a)
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