question_answer

The area enclosed between the curve \[{{y}^{2}}(2a-x)={{x}^{3}}\] and the line x = 2 above the x-axis is

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

Correct Answer: B

Solution :

[b] The required area \[A=\int_{0}^{2a}{\sqrt{\frac{{{x}^{3}}}{2a-x}}dx}\] Put \[x=2a{{\sin }^{2}}\theta \] \[\Rightarrow dx=2a\,2\sin \theta \cos \theta d\theta \] \[\Rightarrow A=8{{a}^{2}}{{\int_{0}^{\pi }{\left( \frac{1-\cos 2\theta }{2} \right)}}^{2}}d\theta \] \[=2{{a}^{2}}\int_{0}^{\pi }{(1-2cos2\theta +co{{s}^{2}}2\theta )d\theta }\] \[=2{{a}^{2}}\int_{0}^{\pi }{\left( 1-2\cos \theta +\frac{1+\cos 4\theta }{2} \right)d\theta =\frac{3\pi {{a}^{2}}}{2}}\]