question_answer

The area bounded by the curve \[x=a{{\cos }^{3}}t,y=a\,\,{{\sin }^{3}}t\]is

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 :

\[x=a{{\cos }^{3}}t,\,\,\,y=a{{\sin }^{3}}t\,\,\,\Rightarrow \,\,\,\,{{x}^{2/3}}+{{y}^{2/3}}={{a}^{2/3}}\]

\[A=4\int\limits_{0}^{\pi /2}{y\frac{dx}{dt}dt=4\int\limits_{0}^{\pi /2}{3{{a}^{2}}{{\sin }^{3}}t{{\cos }^{2}}t(-\sin )dt}}\]

\[=\left| -12{{a}^{2}}\int\limits_{0}^{\pi /2}{{{\sin }^{4}}t{{\cos }^{2}}t\,dt} \right|=\left| -12{{a}^{2}}\frac{3,1,1}{6,4,2}\times \frac{\pi }{2} \right|\]

\[=\frac{3}{8}\pi {{a}^{2}}\,\,sq.\,\,units\]