Question

The area enclosed by the curves $x=a{\sin }^{3}t$ and $y=a{\cos }^{3}t$ is equal to

• A. $12a^2\underset{0}{\overset{\pi /2}{\int }}{\cos }^{4}t{\sin }^{2}tdt$
• B. $12a^2\underset{0}{\overset{\pi /2}{\int }}{\cos }^{2}t{\sin }^{4}tdt$
• C. $2\underset{-a}{\overset{a}{\int }}{(a^{\frac{2}{3}}-x^{\frac{2}{3}})}^{\frac{3}{2}}dx$
• D. $4\underset{0}{\overset{a}{\int }}{(a^{\frac{2}{3}}-x^{\frac{2}{3}})}^{\frac{3}{2}}dx$

Answer 1

Answer: (A), (C), (D)

$4\underset{0}{\overset{a}{\int }}{(a^{\frac{2}{3}}-x^{\frac{2}{3}})}^{\frac{3}{2}}dx$

Eliminating $t$, we have
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$
$\Rightarrow y={(a^{2/3}-x^{2/3})}^{3/2}$
From the figure,
$A=2\underset{-a}{\overset{a}{\int }}{(a^{2/3}-x^{2/3})}^{3/2}dx$
$=4\underset{0}{\overset{a}{\int }}{(a^{2/3}-x^{2/3})}^{3/2}dx$
$=4\underset{0}{\overset{a}{\int }}ydx$
$=4a^2\underset{0}{\overset{\pi /2}{\int }}3{\cos }^{3}t{\sin }^{2}t\cos tdt$