Answer
$$
x= \sin t, \quad y= \sin 2 t, \quad 0 \leq t \leq \pi / 2
$$
The surface area of the curve, which is obtained by rotating the given curve about the x-axis, is
$$
S=\int 2 \pi y d s=\int_{0}^{\pi / 2} 2 \pi \sin 2 t \sqrt{\cos ^{2} t+4 \cos ^{2} 2 t} d t \approx 8.0285
$$
Work Step by Step
$$
x= \sin t, \quad y= \sin 2 t, \quad 0 \leq t \leq \pi / 2
$$
then
$$
d x / d t=\cos t , \quad d y / d t = 2 \cos 2 t,
$$
so
$$
\begin{aligned}\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} &= \cos ^{2} t+ 4\cos ^{2} 2t
\end{aligned}
$$
Thus the surface area of the curve, which is obtained by rotating the given curve about the x-axis, is
$$
S=\int 2 \pi y d s=\int_{0}^{\pi / 2} 2 \pi \sin 2 t \sqrt{\cos ^{2} t+4 \cos ^{2} 2 t} d t \approx 8.0285
$$
