Answer
$$
y^{2}=x+1 \quad(\text { for } 0 \leq x \leq 3)
$$
The exact area of the surface obtained by rotating the given curve about the x-axis is
$$
S=\int 2 \pi y d s=\int_{0}^{3} 2 \pi y \sqrt{1+\left(y^{\prime}\right)^{2}} d x=\frac{\pi}{6}(17 \sqrt{17}-5 \sqrt{5})
$$
Work Step by Step
$$
y^{2}=x+1 \quad(\text { for } 0 \leq x \leq 3)
$$
The curve
$$
y^{2}=x+1 \Rightarrow y=\sqrt{x+1} \quad(\text { for } 0 \leq x \leq 3 \text { and } 1 \leq y \leq 2)
$$
We have
$$
y^{\prime}=1 /(2 \sqrt{x+1})
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\
&=\sqrt{1+\{1 /(2 \sqrt{x+1})\}^{2}} dx
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the x-axis
$$
\begin{aligned}
S &=\int 2 \pi y d s\\
&=\int_{0}^{3} 2 \pi y \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\
&=2 \pi \int_{0}^{3} \sqrt{x+1} \sqrt{1+\frac{1}{4(x+1)}} d x\\
&=2 \pi \int_{0}^{3} \sqrt{x+1+\frac{1}{4}} d x \\
&=2 \pi \int_{0}^{3} \sqrt{x+\frac{5}{4}} d x\\
&=2 \pi \int_{5 / 4}^{17 / 4} \sqrt{u} d u \quad\left[\begin{array}{c}
u=x+\frac{5}{4}, \\
d u=d x
\end{array}\right] \\
&=2 \pi\left[\frac{2}{3} u^{3 / 2}\right]_{5 / 4}^{17 / 4}\\
&=2 \pi \cdot \frac{2}{3}\left(\frac{17^{3 / 2}}{8}-\frac{5^{3 / 2}}{8}\right)\\
&=\frac{\pi}{6}(17 \sqrt{17}-5 \sqrt{5})
\end{aligned}
$$
The exact area of the surface obtained by rotating the given curve about the x-axis is
$$
S=\int 2 \pi y d s=\int_{0}^{3} 2 \pi y \sqrt{1+\left(y^{\prime}\right)^{2}} d x=\frac{\pi}{6}(17 \sqrt{17}-5 \sqrt{5})
$$
