Answer
$$
x=t \sin t, \quad y=t \cos t, \quad 0 \leq t \leq \pi / 2
$$
The surface area, which is obtained by rotating the given curve about the x-axis, is
$$
S=\int 2 \pi y d s=\int_{0}^{\pi / 2} 2 \pi t \cos t \sqrt{t^{2}+1} d t \approx 4.7394
$$
Work Step by Step
$$
x=t \sin t, \quad y=t \cos t, \quad 0 \leq t \leq \pi / 2
$$
then
$$
d x / d t=t \cos t+\sin t , \quad d y / d t =-t \sin t+\cos t,
$$
so
$$
(d x / d t)^{2}+(d y / d t)^{2}=t^{2} \cos ^{2} t+2 t \sin t \cos t+\sin ^{2} t+t^{2} \sin ^{2} t-2 t \sin t \cos t+\cos ^{2} t \\ \quad \quad \quad \quad\quad\quad=t^{2}\left(\cos ^{2} t+\sin ^{2} t\right)+\sin ^{2} t+\cos ^{2} t=t^{2}+1
$$
$$
\begin{aligned}\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2} &=t^{2} \cos ^{2} t+2 t \sin t \cos t+\sin ^{2} t+t+\\
& \quad \quad +^{2} \sin ^{2} t-2 t \sin t \cos t+\cos ^{2} t \\
&=t^{2}\left(\cos ^{2} t+\sin ^{2} t\right)+\sin ^{2} t+\cos ^{2} t \\
&=t^{2}+1
\end{aligned}
$$
Thus the surface area, which is obtained by rotating the given curve about the x-axis, is
$$
S=\int 2 \pi y d s=\int_{0}^{\pi / 2} 2 \pi t \cos t \sqrt{t^{2}+1} d t \approx 4.7394
$$
