Answer
$$
x= t^{3}, \quad y= t^{2} , \quad 0 \leq t \leq 1
$$
The exact area of the surface obtained by rotating
the given curve about the x-axis is:
$$
S=\int 2 \pi y d s=\int_{0}^{1} 2 \pi t^{2} \sqrt{9 t^{4}+4 t^{2} } d t=\frac{2 \pi}{1215}(247 \sqrt{13}+64).
$$
Work Step by Step
$$
x= t^{3}, \quad y= t^{2} , \quad 0 \leq t \leq 1
$$
then
$$
d x / d t=3 t^{2} , \quad d y / d t =2 t
$$
so
$$
\begin{aligned}\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} &=\left(3 t^{2} \right)^{2}+\left(2 t\right)^{2} \\
&=9 t^{4}+4 t^{2},
\end{aligned}
$$
thus the surface area of the curve, which is obtained by rotating the given curve about the x-axis, is
$$
\begin{aligned}
S&=\int 2 \pi y d s=\int_{0}^{1} 2 \pi t^{2} \sqrt{9 t^{4}+4 t^{2} } d t \\
&=2 \pi\int_{0}^{1} t^{2} \sqrt{t^{2}(9 t^{2}+4 )} d t \\
&\quad\quad \quad\left[\begin{array}{l}{ \text {Let }u=9 t^{2}+4, \quad \text { then } t^{2}=(u-4) / 9} \\ {d u=18 t d t, \quad \text {and so }\quad td t=\frac{1}{18} d u}\end{array}\right]\\
&=2 \pi \int_{4}^{13}\left(\frac{u-4}{9}\right) \sqrt{u}\left(\frac{1}{18} d u\right) \\
&=\frac{2 \pi}{9 \cdot 18} \int_{4}^{13}\left(u^{3 / 2}-4 u^{1 / 2}\right) d u \\
& =\frac{\pi}{81}\left[\frac{2}{5} u^{5 / 2}-\frac{8}{3} u^{3 / 2}\right]_{4}^{13} \\
&=\frac{\pi}{81} \cdot \frac{2}{15}\left[3 u^{5 / 2}-20 u^{3 / 2}\right]_{4}^{13}\\
&=\frac{2 \pi}{1215}\left[\left(3 \cdot 13^{2} \sqrt{13}-20 \cdot 13 \sqrt{13}\right)-(3 \cdot 32-20 \cdot 8)\right]\\
&=\frac{2 \pi}{1215}(247 \sqrt{13}+64).
\end{aligned}
$$
The exact area of the surface obtained by rotating
the given curve about the x-axis is
$$
S=\int 2 \pi y d s=\int_{0}^{1} 2 \pi t^{2} \sqrt{9 t^{4}+4 t^{2} } d t=\frac{2 \pi}{1215}(247 \sqrt{13}+64).
$$
