Definite Integration Question 431
Question: The area enclosed between the curve $ y^{2}(2a-x)=x^{3} $ and the line x = 2 above the x-axis is
Options:
A) $ \pi a^{2} $ sq. units
B) $ \frac{3\pi a^{2}}{2} $ sq. units
C) $ 2\pi a^{2} $ sq. units
D) $ 3\pi a^{2} $ sq. units
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The required area $ A=\int_0^{2a}{\sqrt{\frac{x^{3}}{2a-x}}dx} $ Put $ x=2a{{\sin }^{2}}\theta $
$ \Rightarrow dx=2a2\sin \theta \cos \theta d\theta $
$ \Rightarrow A=8a^{2}{{\int_0^{\pi }{( \frac{1-\cos 2\theta }{2} )}}^{2}}d\theta $
$ =2a^{2}\int_0^{\pi }{(1-2cos2\theta +cos^{2}2\theta )d\theta } $
$ =2a^{2}\int_0^{\pi }{( 1-2\cos \theta +\frac{1+\cos 4\theta }{2} )d\theta =\frac{3\pi a^{2}}{2}} $
BETA
