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.You need to login to perform this action.
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