Question

The whole area of the curves $x=a{\cos }^{3}t,y=b{\sin }^{3}t$ is given by?

• A. $\frac{3}{8}\pi ab$
• B. $\frac{5}{8}\pi ab$
• C. $\frac{1}{8}\pi ab$
• D. None of these

Answer 1

Answer: (A)

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

$x=a{\cos }^3\theta$

$y=b{\sin }^3\theta$

To get the area,

$A={\int }_0^{2x}xdy$

$={\int }_0^{2\pi }a{\cos }^3\theta 3b{\sin }^2\theta \cos \theta d\theta$

$=3ab{\int }_0^{2\pi }{\sin }^2\theta {\cos }^4\theta d\theta ⟶\left(1\right)$

Substituting $\theta \to \theta +\pi /2$ in $1$,

$A=3ab{\int }_0^{2\pi }{\cos }^2\theta {\sin }^4\theta d\theta ⟶\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$

$2A=3ab{\int }_0^{2\pi }{\cos }^2\theta {\sin }^2\theta d\theta ⟶\left(3\right)$

$\{{\sin }^2\theta +{\cos }^2\theta =1\}$

Multiplying $\left(3\right)$ by $\left(4\right)$

$8A=3ab{\int }_0^{2\pi }{\sin }^{2}2\theta d\theta ⟶\left(4\right)$

Substituting $\theta \to \theta +\pi /4$ in $\left(4\right)$

$8A=3ab{\int }_0^{2\pi }{\cos }^{2}2\theta d\theta ⟶\left(5\right)$

Adding $\left(4\right)$ and $\left(5\right)$

$16A=3ab{\int }_0^{2\pi }1d\theta$

$=6ab\pi$

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