Answer
$$
x= t^{2}-t^{3}, \quad y= t+t^{4} , \quad 0 \leq t \leq 1
$$
The surface area of the curve, which is obtained by rotating the given curve about the x-axis, is
$$
S=\int 2 \pi y d s=\int_{0}^{1} 2 \pi\left(t+t^{4}\right) \sqrt{16 t^{6}+9 t^{4}-4 t^{3}+4 t^{2}+1} d t \approx 12.7176
$$
Work Step by Step
$$
x= t^{2}-t^{3}, \quad y= t+t^{4} , \quad 0 \leq t \leq 1
$$
then
$$
d x / d t=2 t-3 t^{2} , \quad d y / d t =1+4 t^{3}
$$
so
$$
\begin{aligned}\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} &=\left(2 t-3 t^{2}\right)^{2}+\left(1+4 t^{3}\right)^{2} \\
&=4 t^{2}-12 t^{3}+9 t^{4}+1+8 t^{3}+16 t^{6},
\end{aligned}
$$
thus the surface area of the curve, which is obtained by rotating the given curve about the x-axis, is
$$
S=\int 2 \pi y d s=\int_{0}^{1} 2 \pi\left(t+t^{4}\right) \sqrt{16 t^{6}+9 t^{4}-4 t^{3}+4 t^{2}+1} d t \approx 12.7176
$$
