Answer
$$
y= \int_{0}^{x} \sqrt {\sqrt {t}-1}) dt, \quad \quad 1 \leq x \leq 16
$$
the area of the surface obtained by rotating the given curve about the y-axis is:
$$
\begin{aligned}
S&=\int 2 \pi x d s \\
&=\int_{1}^{16} 2 \pi x \sqrt {\sqrt {x}} dx\\
&=\frac{4088}{9}\pi
\end{aligned}
$$
Work Step by Step
$$
y= \int_{0}^{x} \sqrt {\sqrt {t}-1}) dt, \quad \quad 1 \leq x \leq 16
$$
$\Rightarrow$
$$
y^{\prime} =\frac{dy}{dx}=\sqrt {\sqrt {x}-1})
$$
$\Rightarrow$
$$
\begin{aligned} ds &= \sqrt {1+\left(y^{\prime}\right)^{2}} dx \\
&=\sqrt{1+\left(\sqrt {\sqrt {x}-1})\right)^{2}} dx\\
&=\sqrt {\sqrt {x}}
\end{aligned}
$$
So, the area of the surface obtained by rotating the curve about the y-axis is:
$$
\begin{aligned}
S&=\int 2 \pi x d s \\
&=\int_{1}^{16} 2 \pi x \sqrt {\sqrt {x}} dx\\
&=2 \pi \int_{1}^{16} x ^{\frac{5}{4}} dx\\
&=2 \pi \frac{4}{9} [x^{\frac{9}{4}} ]_{1}^{16} \\
&=\frac{8\pi}{9} (512-1)\\
&=\frac{4088}{9}\pi
\end{aligned}
$$
