Answer
$$
y=\cos \left(\frac{1}{2} x\right) \quad(\text { for } 0 \leq x \leq \pi)
$$
The exact area of the surface obtained by rotating the given curve about the x-axis is
$$
S=\int 2 \pi y d s=\int_{0}^{3} 2 \pi y \sqrt{1+\left(y^{\prime}\right)^{2}} d x=\pi \sqrt{5}+4 \pi \ln \left(\frac{1+\sqrt{5}}{2}\right)
$$
Work Step by Step
The curve
$$
y=\cos \left(\frac{1}{2} x\right) \quad(\text { for } 0 \leq x \leq \pi)
$$
We have
$$
y^{\prime}=-\frac{1}{2} \sin \left(\frac{1}{2} x\right)
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\
&=\sqrt{1+\{-\frac{1}{2} \sin \left(\frac{1}{2} x\right)\}^{2}} dx
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the x-axis
$$
\begin{aligned}
S &=\int 2 \pi y d s\\
&=\int_{0}^{3} 2 \pi y \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\
&=2 \pi \int_{0}^{\pi} \cos \left(\frac{1}{2} x\right) \sqrt{1+\frac{1}{4} \sin ^{2}\left(\frac{1}{2} x\right)} d x \\
&=2 \pi \int_{0}^{1} \sqrt{1+\frac{1}{4} u^{2}}(2 d u) \quad\left[\begin{array}{c}
u=\sin \left(\frac{1}{2} x\right) \\
d u=\frac{1}{2} \cos \left(\frac{1}{2} x\right) d x
\end{array}\right] \\
&=2 \pi \int_{0}^{1} \sqrt{4+u^{2}} d u\\
& \quad\quad\quad\quad\quad\left[\begin{array}{c}
\text{ If we look at the Table of Integrals we see that} , \\ \text{ the closest entry is number 21 with a=2 }
\end{array}\right] \\
&\stackrel{21}{=} 2 \pi\left[\frac{u}{2} \sqrt{4+u^{2}}+2 \ln \left(u+\sqrt{4+u^{2}}\right)\right]_{0}^{1} \\
&=2 \pi\left[\left(\frac{1}{2} \sqrt{5}+2 \ln (1+\sqrt{5})\right)-(0+2 \ln 2)\right] \\
&=\pi \sqrt{5}+4 \pi \ln \left(\frac{1+\sqrt{5}}{2}\right)
\end{aligned}
$$
The exact area of the surface obtained by rotating the given curve about the x-axis is
$$
S=\int 2 \pi y d s=\int_{0}^{3} 2 \pi y \sqrt{1+\left(y^{\prime}\right)^{2}} d x=\pi \sqrt{5}+4 \pi \ln \left(\frac{1+\sqrt{5}}{2}\right)
$$
