Definite Integration Question 258
Question: The area of the curve $ xy^{2}=a^{2}(a-x) $ bounded by y-axis is
[RPET 1996]
Options:
A) $ \pi a^{2} $
B) $ 2\pi a^{2} $
C) $ 3\pi a^{2} $
D) $ 4\pi a^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
Since the curve is symmetrical about x-axis, therefore Required area $ A=2\int_0^{a}{a\sqrt{\frac{a-x}{x}}dx} $
Put $ x=a{{\sin }^{2}}\theta $
$ \Rightarrow dx=2a\sin \theta .\cos \theta d\theta $
$ A=2\int_0^{\pi /2}{a\sqrt{\frac{a{{\cos }^{2}}\theta }{a{{\sin }^{2}}\theta }}}a\sin 2\theta d\theta $
$ =2a^{2}\int_0^{\pi /.2}{\frac{\cos \theta }{\sin \theta }2\sin \theta \cos \theta d\theta } $
$ A=4a^{2}\int_0^{\pi /2}{{{\cos }^{2}}\theta d\theta } $
therefore $ A=4a^{2}.\frac{1}{2}.\frac{\pi }{2}=\pi a^{2} $ .
BETA
