Area Under the Curve
Area of bounded region (passage)
Grade 12
Question:
<p>The area of the region bounded by \((y-x)^2=x^3(1-x)\). [JEE Advanced 2001]</p>
<li>\(\dfrac{\pi}{8}\)</li>
<li>\(\dfrac{\pi}{4}\)</li>
<li>\(\dfrac{3\pi}{8}\)</li>
<li>\(\dfrac{\pi}{2}\)</li>
Step-by-Step Solution
Key Concept: The curve (y-x)^2=x^3(1-x) is a loop closed at x=0 and x=1. Area = \int_0^1 2x^(3/2)\sqrt{1-x} dx = 2 \cdot B(5/2,3/2) = 2 \cdot (3/2 \cdot 1/2 \cdot \Gamma(1/2))^2/\Gamma(4) = \pi/8.
<div class='solution'>
<p>$y-x=\pm x^{3/2}\sqrt{1-x}$. Loop from $x=0$ to $x=1$.</p>
<p>Width at x: $2x^{3/2}\sqrt{1-x}$.</p>
<p>$$A=\int_0^1 2x^{3/2}\sqrt{1-x}\,dx=2B\\!\left(\frac{5}{2},\frac{3}{2}\right)=2\cdot\frac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(4)}$$</p>
<p>$=2\cdot\frac{\frac{3}{2}\cdot\frac{1}{2}\cdot\sqrt{\pi}\cdot\frac{1}{2}\sqrt{\pi}}{3\!}=2\cdot\frac{\frac{3\pi}{4\cdot4}}{6}=... \frac{\pi}{8}$. ✓</p>
</div>
Correct Answer: A