Question

The area enclosed by the curve $x=a{\cos }^{3}t,y=b{\sin }^{3}t,0\le t\le 2\pi$ is:

• A. $\frac{3\pi ab}{8}$
• B. $\frac{3\pi ab}{4}$
• C. $\frac{3ab}{8}$
• D. none of these

Answer 1

The correct option is A $\frac{3\pi ab}{8}$

The area $A$ of the region is given by

$A=\frac{1}{2}{\int }_0^{2\pi }(x\frac{dy}{dt}-y\frac{dx}{dt})dt$

$=\frac{1}{2}{\int }_0^1\{a{\cos }^{3}t\left(3b{\sin }^{2}t\cos t\right)-b{\sin }^{3}t(-3a{\cos }^{2}t\sin t)\}dt$

$=\frac{3ab}{2}{\int }_0^{2\pi }{\cos }^{2}t{\sin }^{2}tdt=\frac{3ab}{2}\times 4{\int }_0^{2\pi }{\cos }^{2}t{\sin }^{2}tdt$

$=6ab\times \frac{1.1}{4.2}\times \frac{\pi }{2}=\frac{3ab\pi }{8}$