Question

The area enclosed by the curves $x=a{{\cos }^{3}}t,\text{ }y=b{{\sin }^{3}}t$ is
A. $\dfrac{\pi ab}{4}$
B. $\dfrac{3\pi ab}{4}$
C. $\dfrac{3\pi ab}{8}$
D. None of these

Answer 1

Hint: We know that the area bounded by a curve is given by the formula
\[Area=\dfrac{1}{2}\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{\left( x\dfrac{dy}{dt}-y\dfrac{dx}{dt} \right)dt}\]
Now, consider $x=a{{\cos }^{3}}t$ and $y=b{{\sin }^{3}}t$ separately and differentiate both equations with respect to $t$. Then, put the values in the formula and integrate using the formula $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{m}}t}.{{\cos }^{n}}t=\dfrac{\left( m-1 \right)\left( m-3 \right)...1\cdot \left( n-1 \right)\left( n-3 \right)...1}{\left( m+n \right)\left( m+n-2 \right)...1}\times \dfrac{\pi }{2}$ with even $m,n$ to obtain the desired result.

Complete step-by-step answer:
We have given equations $x=a{{\cos }^{3}}t$ and $y=b{{\sin }^{3}}t$
Now, first we will find $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$.
$x=a{{\cos }^{3}}t$
Now, differentiate the equation with respect to $t$, we get
$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( a{{\cos }^{3}}t \right)$
Taking out the constant term, we get
\[\Rightarrow \dfrac{dx}{dt}=a\dfrac{d}{dt}\left( {{\cos }^{3}}t \right)\]
\[\Rightarrow \dfrac{dx}{dt}=a3{{\cos }^{2}}t\dfrac{d}{dt}\left( \cos t \right)\]
Now, we know that derivative of $\cos t=-\sin t$
So, when we solve further, we get
\[\begin{align}
& \Rightarrow \dfrac{dx}{dt}=a3{{\cos }^{2}}t\left( -\sin t \right) \\
& \Rightarrow \dfrac{dx}{dt}=-3a{{\cos }^{2}}t\sin t \\
\end{align}\]
Now, we will solve the equation $y=b{{\sin }^{3}}t$
Differentiating the equation with respect to $t$, we get
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( b{{\sin }^{3}}t \right)$
Taking out the constant term, we get
$\begin{align}
& \Rightarrow \dfrac{dy}{dt}=b\dfrac{d}{dt}\left( {{\sin }^{3}}t \right) \\
& \Rightarrow \dfrac{dy}{dt}=b3{{\sin }^{2}}t\dfrac{d}{dt}\left( \sin t \right) \\
\end{align}$
Now, we know that derivative of $\sin t=\cos t$
So, when we solve further, we get
$\begin{align}
& \Rightarrow \dfrac{dy}{dt}=b3{{\sin }^{2}}t\left( \cos t \right) \\
& \Rightarrow \dfrac{dy}{dt}=3b{{\sin }^{2}}t\cos t \\
\end{align}$
Now, we will find the value of \[x\dfrac{dy}{dt}-y\dfrac{dx}{dt}\]
\[\begin{align}
& \Rightarrow x\dfrac{dy}{dt}-y\dfrac{dx}{dt}=a{{\cos }^{3}}t\left( 3b{{\sin }^{2}}t\cos t \right)-b{{\sin }^{3}}t\left( -3a{{\cos }^{2}}t\sin t \right) \\
& \Rightarrow x\dfrac{dy}{dt}-y\dfrac{dx}{dt}=3ab{{\cos }^{4}}t{{\sin }^{2}}t+3ab{{\cos }^{2}}t{{\sin }^{4}}t \\
\end{align}\]
Now, taking common terms out we get
\[\Rightarrow x\dfrac{dy}{dt}-y\dfrac{dx}{dt}=3ab{{\cos }^{2}}t{{\sin }^{2}}t\left( {{\cos }^{2}}t+{{\sin }^{2}}t \right)\]
We know that \[{{\cos }^{2}}x+{{\sin }^{2}}x=1\]
So, we get
\[\Rightarrow x\dfrac{dy}{dt}-y\dfrac{dx}{dt}=3ab{{\cos }^{2}}t{{\sin }^{2}}t\]
Now, the required area will be
\[Area=\dfrac{1}{2}\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{\left( x\dfrac{dy}{dt}-y\dfrac{dx}{dt} \right)dt}\]
Put the values in the formula, we get
\[Area=\dfrac{1}{2}\int\limits_{{{t}_{1}}}^{{{t}_{2}}}{\left( 3ab{{\cos }^{2}}t{{\sin }^{2}}t \right)dt}\]
Now, to obtain a closed curve the value of $t$ varies from $0$ to $2\pi $

\[Area=\dfrac{1}{2}\int\limits_{0}^{2\pi }{\left( 3ab{{\cos }^{2}}t{{\sin }^{2}}t \right)dt}\]
Taking out the constant term, we get
\[Area=\dfrac{3ab}{2}\int\limits_{0}^{2\pi }{\left( {{\cos }^{2}}t{{\sin }^{2}}t \right)dt}\]
We can split the limits of integral into 4 equal parts with limits \[\left[ 0,\dfrac{\pi }{2} \right],\left[ \dfrac{\pi }{2},\pi \right],\left[ \pi ,\dfrac{3\pi }{2} \right],\left[ \dfrac{3\pi }{2},2\pi \right]\] which enclose equal area traced in the first, second, third and fourth quadrant respectively. So we have,
\[Area=\dfrac{3ab}{2}\times 4\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\cos }^{2}}t{{\sin }^{2}}t \right)dt}\]
Now, we know that integration formula that
\[\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{m}}t}.{{\cos }^{n}}t=\left\{ \begin{matrix}
   \dfrac{\left( m-1 \right)\left( m-3 \right)...1\cdot \left( n-1 \right)\left( n-3 \right)...1}{\left( m+n \right)\left( m+n-2 \right)...1}\times \dfrac{\pi }{2} & \text{if }m,n\text{ even} \\
   \dfrac{\left( m-1 \right)\left( m-3 \right)....\left( 2\text{ or 1} \right)\left( n-1 \right)\left( n-3 \right)...\left( 2\text{ or 1} \right)}{\left( m+n \right)\left( m+n-2 \right)...\left( 2\text{ or 1} \right)} & \text{otherwise} \\
\end{matrix} \right.\]
We use the formula for the even condition with that is $m=n=2$in this problem.
\[\begin{align}
  & Area=\dfrac{3ab}{2}\times 4\times \dfrac{\left( 2-1 \right)\left( 2-1 \right)}{\left( 2+2 \right)\left( 2+2-2 \right)}\times \dfrac{\pi }{2} \\
 & Area=\dfrac{3ab}{2}\times 4\times \dfrac{1\times 1}{4\times 2}\times \dfrac{\pi }{2} \\
 & Area=\dfrac{3ab\pi }{8} \\
\end{align}\]
So, option C is the correct answer.

Note: Definite integral gives the area bounded by a curve. Always consider a closed curve to find the area. The formula $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{m}}x}.{{\cos }^{n}}x=\dfrac{\left( m-1 \right)\left( m-3 \right).........\left( n-1 \right)\left( n-3 \right)...}{\left( m+n \right)\left( m+n-2 \right)...}$ used is known as Wallis’s formula for definite integrals of powers of sine and cosine functions. We can find point of intersection of the curve with axes using $x=a{{\cos }^{3}}t=0,y=b{{\sin }^{3}}t=0$ and obtain the limits of for each quadrant.