A) \[\frac{\pi ab}{4}\]
B) \[\frac{\pi ab}{32}\]
C) \[\frac{3\pi ab}{32}\]
D) \[\frac{5\pi ab}{32}\]
Correct Answer: C
Solution :
[c] \[y=0\], when \[t=0\] and then \[x=a\] So desire area \[A=\int\limits_{0}^{a}{ydx=\int\limits_{\pi /2}^{0}{b{{\sin }^{3}}t(-3a\,co{{s}^{2}}t\,\,sin\,\,tdt)}}\] \[A=3ab\int\limits_{0}^{\pi /2}{{{\sin }^{4}}t\,\,{{\cos }^{2}}t\,\,dt=3ab\int\limits_{0}^{\pi /2}{{{\cos }^{4}}t\,\,{{\sin }^{2}}tdt}}\] \[\therefore 2A=3ab\int\limits_{0}^{\frac{\pi }{2}}{{{\cos }^{2}}t\,{{\sin }^{2}}tdt=3ab\cdot \frac{\pi }{16}\Rightarrow A=\frac{3\pi ab}{32}}\]
You need to login to perform this action.
You will be redirected in
3 sec
