A) \[\frac{3\pi {{a}^{2}}}{8}\]
B) \[\frac{3\pi {{a}^{2}}}{16}\]
C) \[\frac{3\pi {{a}^{2}}}{32}\]
D) \[3\pi {{a}^{2}}\]
Correct Answer: A
Solution :
\[x=a{{\cos }^{3}}t,\,\,\,y=a{{\sin }^{3}}t\,\,\,\Rightarrow \,\,\,\,{{x}^{2/3}}+{{y}^{2/3}}={{a}^{2/3}}\] |
\[A=4\int\limits_{0}^{\pi /2}{y\frac{dx}{dt}dt=4\int\limits_{0}^{\pi /2}{3{{a}^{2}}{{\sin }^{3}}t{{\cos }^{2}}t(-\sin )dt}}\] |
\[=\left| -12{{a}^{2}}\int\limits_{0}^{\pi /2}{{{\sin }^{4}}t{{\cos }^{2}}t\,dt} \right|=\left| -12{{a}^{2}}\frac{3,1,1}{6,4,2}\times \frac{\pi }{2} \right|\] |
\[=\frac{3}{8}\pi {{a}^{2}}\,\,sq.\,\,units\] |
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