Definite Integration Question 142
Question: The area bounded by the curve $ y^{2}(2a-x)=x^{3} $ and the line $ x=2a $ is
Options:
A) $ 3\pi a^{2} $ sq. unit
B) $ \frac{3\pi a^{2}}{2} $ sq. unit
C) $ \frac{3\pi a^{2}}{4} $ sq. unit
D) $ \frac{6\pi a^{2}}{5} $ sq. unit
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Let the equation of curve $ y^{2}(2a-x)=x^{3}…(i) $ and equation of line $ x=2a…(ii) $ The given curve is symmetrical about x-axis and passes through origin.
From (i) we have, $ y^{2}=\frac{x^{3}}{2a-x} $
But $ \frac{x^{3}}{2a-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\le x\le 2a $ .
$ \therefore $ Area bounded by the curve and line $ =\int\limits_0^{2a}{ydx=\int\limits_0^{2a}{\frac{{x^{3/2}}}{\sqrt{2a}-x}dx}} $ Put $ x=2a{{\sin }^{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}[ \frac{3}{4}.\frac{1}{2}.\frac{\pi }{2} ]} $ $ =\frac{3\pi a^{2}}{2} $ sq. unit
BETA
