question_answer

The  area  of the  loop  the  curve \[a{{y}^{2}}={{x}^{2}}(a-x)\]is

A) \[\frac{8{{a}^{2}}}{15}sq\,\,unit\]                            

B) \[\frac{4{{a}^{2}}}{15}sq\,\,unit\]

C) \[\frac{2{{a}^{2}}}{15}sq\,\,unit\]            

D)         None of these

Correct Answer: A

Solution :

The area of the loop of the curve\[a{{y}^{2}}={{x}^{2}}(a-x)\]is given by                 \[2\int_{0}^{a}{y\,\,dx}=2\int_{0}^{a}{x\sqrt{\frac{a-x}{a}}dx}\]                 \[=4{{a}^{2}}\int_{0}^{\pi /2}{{{\sin }^{3}}\theta \cdot {{\cos }^{2}}\theta \cdot d\theta }\] [Let\[x=a{{\sin }^{2}}\theta \] \[\Rightarrow \]               \[dx=2a\sin \theta \cdot \cos \theta \cdot d\theta ]\]                 \[=4{{a}^{2}}\cdot \frac{2\cdot 1}{5\cdot 3\cdot 1}=\frac{8{{a}^{2}}}{15}\]sq unit.