Definite Integration Question 169
Question: The area enclosed by the curve $ x=acos^{3}t, $
$ y=bsin^{3}t $ and the positive directions of x-axis and y-axis is
Options:
A) $ \frac{\pi ab}{4} $
B) $ \frac{\pi ab}{32} $
C) $ \frac{3\pi ab}{32} $
D) $ \frac{5\pi ab}{32} $
Show Answer
Answer:
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(-3acos^{2}tsintdt)}} $
$ A=3ab\int\limits_0^{\pi /2}{{{\sin }^{4}}t{{\cos }^{2}}tdt=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}} $
BETA
