Definite Integration Question 484
Question: Area enclosed between the curve $ y^{2}(2a-x)=x^{3} $ and line $ x=2a $ above x-axis is
[MP PET 2001]
Options:
A) $ \pi a^{2} $
B) $ \frac{3\pi a^{2}}{2} $
C) $ 2\pi a^{2} $
D) $ 3\pi a^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Curve $ y^{2}(2a-x)=x^{3} $ is symmetrical about x-axis and passes through origin. Also $ \frac{x^{3}}{2a-x}<0 $ for $ x>2a $ or $ x<0 $ . So curve does not lie in $ x>2a $ and $ x<0, $ curve lies wholly on $ 0\le x\le 2a $ .
$ \therefore $ Area $ =\int_0^{2a}{\frac{{x^{3/2}}}{\sqrt{2a-x}}dx} $
$ =\int_0^{\pi /2}{8a^{2}{{\sin }^{4}}\theta d\theta } $ , (Put $ x=2a{{\sin }^{2}}\theta ) $
$ =8a^{2}[ \frac{3}{4}.\frac{1}{2}.\frac{\pi }{2} ] $
$ =\frac{3\pi a^{2}}{2} $ , (Applying Gamma function).
BETA
