Question

The area of the curve $x=a{\cos }^{3}t$,$y=b{\sin }^{3}t$ in sq. units is :

• A. $\frac{3\pi ab}{4}$
• B. $\frac{3\pi ab}{8}$
• C. $\frac{\pi ab}{4}$
• D. $\frac{\pi ab}{8}$

Answer 1

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

required area $=\left|4{\int }_0^{a}ydx\right|$

$=\left|4{\int }_0^{\frac{\pi }{2}}y\frac{dx}{dt}dt\right|$

$=|4{\int }_{\frac{\pi }{2}}^{0}b{\sin }^{3}t\cdot 3a{\cos }^{2}t(-\sin t\left)dt\right|$

$=12ab{\int }_0^{\frac{\pi }{2}}{\sin }^{4}t\cdot {\cos }^{2}tdt$

$=12ab\frac{(4-1)(4-3)(2-1)}{6\times 4\times 2}\frac{\pi }{2}$ [using reduction formula]

$=\frac{3ab\pi }{8}$