Answer
$$
y=\frac{1}{3} x^{3 / 2} \quad(\text { for } 0 \leq x \leq 12)
$$
The exact area of the surface obtained by rotating the given curve about the y-axis is
$$
S=\int 2 \pi x d s=\int_{0}^{3} 2 \pi x \sqrt{1+\left(y^{\prime}\right)^{2}} d x=\frac{3712 \pi}{15}
$$
Work Step by Step
The curve
$$
y=\frac{1}{3} x^{3 / 2} \quad(\text { for } 0 \leq x \leq 12)
$$
We have
$$
y^{\prime}=\frac{1}{2} x^{1 / 2}
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\
&=\sqrt{1+\{\frac{1}{2} x^{1 / 2}\}^{2}} dx\\
&=\sqrt{1+\frac{1}{4} x} dx
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the y-axis
$$
\begin{aligned}
S &=\int 2 \pi x d s\\
&=\int_{0}^{3} 2 \pi x \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\
&=\int_{0}^{12} 2 \pi x \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\
&=2 \pi \int_{0}^{12} x \sqrt{1+\frac{1}{4} x} d x\\
&=2 \pi \int_{0}^{12} x \frac{1}{2} \sqrt{4+x} d x \\
&=\pi \int_{4}^{16}(u-4) \sqrt{u} d u \quad\left[\begin{array}{c}
u=x+4 \\
d u=d x
\end{array}\right] \\
&=\pi \int_{4}^{16}\left(u^{3 / 2}-4 u^{1 / 2}\right) d u\\
&=\pi\left[\frac{2}{3} u^{5 / 2}-\frac{8}{3} u^{3 / 2}\right]_{4}^{16}\\
&=\pi\left[\left(\frac{2}{5} \cdot 1024-\frac{8}{3} \cdot 64\right)-\left(\frac{2}{5} \cdot 32-\frac{8}{3} \cdot 8\right)\right] \\
&=\pi\left(\frac{2}{5} \cdot 992-\frac{8}{3} \cdot 56\right) \\
&=\pi\left(\frac{5952-2240}{15}\right) \\
&=\frac{3712 \pi}{15}
\end{aligned}
$$
The exact area of the surface obtained by rotating the given curve about the y-axis is
$$
S=\int 2 \pi x d s=\int_{0}^{3} 2 \pi x \sqrt{1+\left(y^{\prime}\right)^{2}} d x=\frac{3712 \pi}{15}
$$
