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Q.
The area bounded by the curve $y^{2}\left(\right.2a-x\left.\right)=x^{3}$ and the line $x=2a$ is
NTA AbhyasNTA Abhyas 2022
Solution:
Let the equation of curve $y^{2}\left(\right.2a-x\left.\right)=x^{3}$
and equation of line $x=2a$ The given curve is symmetrical about $x$ -axis and passes through origin.
From
(i) we have, $y^{2}=\frac{x^{3}}{2 a - x}$
But $\frac{x^{3}}{2 a - x} < 0$ for $x>2a$ and $x < 0$ So, curve does not lie in the portion $x>2a$ and $x < 0,$ therefore curve lies in $0\leq x\leq 2a$ .
$\therefore $ Area bounded by the curve and line
$=\int\limits _{0}^{2 a}ydx=\int\limits _{0}^{2 a}\frac{x^{3 / 2}}{\sqrt{2 a} - x}dx$
Put $x=2asin^{2}\theta $ and $dx=4a\,sin\theta \,cos\theta d\theta $
$\therefore I=\int\limits _{0}^{\pi / 2}8a^{2}sin^{4}\theta d\theta =8a^{2}\left[\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi }{2}\right]$
using $\int\limits _{0}^{\pi / 2}sin^{m}xcos^{n}xdx$
$= \frac{\Gamma \frac{m + 1}{2} \Gamma \frac{n + 1}{2}}{2 \Gamma \frac{m + n + 2}{2}}=\frac{3 \pi a^{2}}{2}$ sq. unit