Find the surface area of the solid generated by the revolution of the asteroid ${{x}^{{2}/{3}\;}}+{{y}^{{2}/{3}\;}}={{a}^{{2}/{3}\;}}$ about the $x$ axis.

Hint: Here we have to calculate the surface area of the solid generated by the revolution of the asteroid ${{x}^{{2}/{3}\;}}+{{y}^{{2}/{3}\;}}={{a}^{{2}/{3}\;}}$ about the x-axis. We will calculate the surface area by using integration. We will substitute the value $x=a{{\cos }^{3}}t$ and $y=a{{\sin }^{3}}t$ , and we will find the appropriate limits for integration.
The result which we will get after integration will be the required surface area of the solid generated.
Complete step-by-step answer:
The given asteroid is ${{x}^{{2}/{3}\;}}+{{y}^{{2}/{3}\;}}={{a}^{{2}/{3}\;}}$. First we will substitute the value of $x$ as $aco{{s}^{3}}t$ and $y$ as $a{{\sin }^{3}}t$i.e.
$x=a{{\cos }^{3}}t$
$y=a{{\sin }^{3}}t$
The given asteroid is symmetrical about the $x$ axis.
Now, we will calculate $\dfrac{dx}{dt}$ by differentiating with respect to.
$\Rightarrow \dfrac{dx}{dt}=\frac{d\left( a{{\cos }^{3}}t \right)}{dt}$
On differentiating, we get
$\Rightarrow \dfrac{dx}{dt}=-3a{{\cos }^{2}}t\sin t$
Similarly, we will calculate $\dfrac{dy}{dt}$ by differentiating with respect to.
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d\left( a{{\sin }^{3}}t \right)}{dt}$
On differentiating, we get
$\Rightarrow \dfrac{dy}{dt}=3a{{\sin }^{2}}t\cos t$
Now, we will calculate $\dfrac{ds}{dt}$, where $\dfrac{ds}{dt}=\sqrt{{{\left( \dfrac{dx}{dt} \right)}^{2}}+{{\left( \dfrac{dy}{dt} \right)}^{2}}}$
We will put the value of $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ here
Thus, $\dfrac{ds}{dt}=\sqrt{{{\left( -3a{{\cos }^{2}}t\sin t \right)}^{2}}+{{\left( 3a{{\sin }^{2}}t\cos t \right)}^{2}}}$
We will calculate the squares of the terms.
$\dfrac{ds}{dt}=\sqrt{9{{a}^{2}}{{\cos }^{4}}t{{\sin }^{2}}t+9{{a}^{2}}{{\sin }^{4}}t{{\cos }^{2}}t}$
On further simplification, we get
$\Rightarrow \dfrac{ds}{dt}=3a\sin t\cos t\sqrt{{{\cos }^{2}}t+{{\sin }^{2}}t}$
We know from trigonometric identities,
${{\cos }^{2}}t+{{\sin }^{2}}t=1$
Thus, $\dfrac{ds}{dt}$becomes;
$\dfrac{ds}{dt}=3a\sin t\cos t$
We will find the limits of $t$ using the limits of $x$ and $y$.
As the value of $x$ is varying from $-a$ to $a$.
We will find the limit of $t$ using $x=a{{\cos }^{3}}t$.
Therefore, value of $t$ when $x$ is $a$ is
$\begin{align}
  & a=a{{\cos }^{3}}t \\
 & t=0 \\
\end{align}$
Therefore, value of $t$ when $x$ is $-a$ is
$\begin{align}
  & -a=a{{\cos }^{3}}t \\
 & t=\pi \\
\end{align}$
Thus, $t$ is varying from $0$ to $\pi $.
The surface area of the solid generated $=\int\limits_{0}^{\pi }{2\pi x\dfrac{ds}{dt}.dt}$
On simplifying the integration, we get
The surface area of the solid generated $=2\int\limits_{0}^{\dfrac{\pi }{2}}{2\pi x\dfrac{ds}{dt}.dt}$
On putting the value of$\dfrac{ds}{dt}$ and $x$, we get
The surface area of the solid generated$=2\int\limits_{0}^{\dfrac{\pi }{2}}{2\pi a.{{\cos }^{3}}t.3a\sin t\cos tdt}$
Taking constants out of integration, we get
The surface area of the solid generated $=12\pi {{a}^{2}}\int\limits_{0}^{\dfrac{\pi }{2}}{\sin t{{\cos }^{4}}tdt}$
Let $\cos t$ be $v$,so $-a\sin t$ will become $dv$. The limits become $\cos \left( 0 \right)=1$ to $\cos \dfrac{\pi }{2}=0$.
Therefore,
The surface area of the solid generated $=12\pi {{a}^{2}}\int\limits_{1}^{0}{-{{v}^{4}}}dv$
On integration, we get
The surface area of the solid generated
$\begin{align}
  & =-12\pi {{a}^{2}}\left( \dfrac{{{v}^{5}}}{5} \right)_{1}^{0} \\
 & =-12\pi {{a}^{2}}\left( \dfrac{0-1}{5} \right) \\
 & =\dfrac{12}{5}\pi {{a}^{2}} \\
\end{align}$
Thus, the required surface area is $\dfrac{{12}}{5}\pi {a^2}$

Note: Here we have used the substitution method of integration. The limit will be changed after substituting the value and it will depend on the substituted value. We will first put the upper limit and then the lower limit in the integration. We should be careful while substituting the value.