Question

The area bounded by the curve $x=a{\cos }^{3}t,y=a{\sin }^{3}t$ is

• A. $\frac{3\pi a^2}{8}$
• B. $\frac{3\pi a^2}{16}$
• C. $\frac{3\pi a^2}{32}$
• D. $3\pi a^2$

Answer 1

The correct option is A $\frac{3\pi a^2}{8}$

Eliminating $t$, we have
$x^{2/3}+y^{2/3}=a^{2/3}$
$x=0\Rightarrow y=\pm a$
$y=0\Rightarrow x=\pm a$

Symmetric about both the axis

Required area
$A=4\underset{0}{\overset{a}{\int }}ydx$
Using $y=a{\sin }^{3}t,x=a{\cos }^{3}t$, we get
$\Rightarrow dx=-3a{\cos }^{2}t\sin tdt$
Therefore,
$A=4\underset{\pi /2}{\overset{0}{\int }}a{\sin }^{3}t(-3a{\cos }^{2}t\sin t)dt\Rightarrow A=12a^2\underset{0}{\overset{\pi /2}{\int }}{\sin }^{4}t\cdot {\cos }^{2}tdt$
Putting $t\to \frac{\pi }{2}-t$, we get
$\Rightarrow A=12a^2\underset{0}{\overset{\pi /2}{\int }}{\cos }^{4}t\cdot {\sin }^{2}tdt$
Adding both, we get
$\Rightarrow A=6a^2\underset{0}{\overset{\pi /2}{\int }}({\cos }^{4}t\cdot {\sin }^{2}t+{\sin }^{4}t\cdot {\cos }^{2}t)dt\Rightarrow A=6a^2\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}t{\cos }^{2}tdt\Rightarrow A=\frac{3a^2}{2}\underset{0}{\overset{\pi /2}{\int }}{\sin }^{2}2tdt\Rightarrow A=\frac{3a^2}{2}\underset{0}{\overset{\pi /4}{\int }}2{\sin }^{2}2tdt\Rightarrow A=\frac{3a^2}{2}\underset{0}{\overset{\pi /4}{\int }}(1-\cos 4t)dt\Rightarrow A=\frac{3a^2}{2}{[t-\frac{\sin 4t}{4}]}_0^{\pi /4}\therefore A=\frac{3\pi a^2}{8}$