Question

The surface area of the solid generated by the revolution of the asteroids $x = a{\cos ^3}t$, $y = a{\sin ^3}t$ axis is:
A.\[\dfrac{{\pi {a^2}}}{5}\]
B.\[\dfrac{{3\pi {a^2}}}{5}\]
C.\[\dfrac{{6\pi {a^2}}}{5}\]
D.\[\dfrac{{12\pi {a^2}}}{5}\]

Answer 1

Hint: We calculate the area under the curve using integration. In this question, we are given parametric equations of the curve. We will simplify the expression and find the appropriate limits for integrating the resultant function.

Complete step-by-step answer:
The asteroid is symmetrical about the $x$ axis.
We will calculate $\dfrac{{dx}}{{dt}}$ by differentiating $x$ with respect to $t$.
$\dfrac{{dx}}{{dt}} = - 3a{\cos ^2}t\sin t$
We will calculate $\dfrac{{dy}}{{dt}}$ by differentiating $y$ with respect to $t$.
$\dfrac{{dy}}{{dt}} = 3a{\sin ^2}t\cos t$
Also, $\dfrac{{ds}}{{dt}} = \sqrt {{{\left( {\dfrac{{dx}}{{dt}}} \right)}^2} + {{\left( {\dfrac{{dy}}{{dt}}} \right)}^2}} $
Then,
 $
  \dfrac{{ds}}{{dt}} = \sqrt {9{a^2}{{\cos }^4}t{{\sin }^2}t + 9{a^2}{{\sin }^4}t{{\cos }^2}t} \\
   = 3a\sin t\cos t\sqrt {{{\sin }^2}t + {{\cos }^2}t} \\
   = 3a\sin t\cos t \\
$
The limits for the $t$ will be defined by the limits of $x$ and $y$.
The value of $x$ will be varying from the $ - a$ to $a$.
Thus the limits of $t$ can be found by using the equation $x = a{\cos ^3}t$
The limits of $t$ are
$
  a = a{\cos ^3}t \\
  t = 0 \\
 $
And
$
   - a = a{\cos ^3}t \\
  {\cos ^3}t = - 1 \\
  t = \pi \\
 $
The required surface will be $\int_0^\pi {2\pi x\dfrac{{ds}}{{dt}}.dt} $
The above integration can be simplified as $2\int_0^{\dfrac{\pi }{2}} {2\pi x\dfrac{{ds}}{{dt}}.dt} $
On simplifying the above equation, we get
$
   = 4\pi \int_0^{\dfrac{\pi }{2}} {a{{\cos }^3}t.3a\sin t\cos t.dt} \\
   = 12\pi {a^2}\int_0^{\dfrac{\pi }{2}} {\sin t{{\cos }^4}t.dt} \\
$
Let $\cos t$ be $u$ and $ - \sin t.dt$ be $du$. The limits \[0\] to \[\dfrac{\pi }{2}\]will be changed as \[\cos \left( 0 \right) = 1\] to $\cos \left( {\dfrac{\pi }{2}} \right) = 0$ for solving the integration.
$
   = 12\pi {a^2}\int_1^0 { - {u^4}.du} \\
   = - 12\pi {a^2}{\left[ {\dfrac{{{u^5}}}{5}} \right]_1}^0 \\
   = - 12\pi {a^2}\left( {\dfrac{{0 - 1}}{5}} \right) \\
   = \dfrac{{12}}{5}\pi {a^2} \\
$
Thus the surface area of the asteroid is \[\dfrac{{12}}{5}\pi {a^2}\].
Hence, option D is correct.

Note: In this question, we have integrated using substitution. It is important to take limits according to the substitution. Also, be careful while substituting the limits. We have to put the upper limit first and then followed by the lower limit.