We can get a somewhat different point of view by looking at arclength, surface area, and other problems for curves given parametrically.
Let the curve be given by $x=x(t)$, $\:y=y(t)$. Then the surface area obtained when we rotate the chunk of the curve from $t=a$ to $t=b$ around the $x$-axis is $\int_{t=a}^b 2\pi y\sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}\,dt.$ For rotation about the $y$-axis, we have a very similar expression for surface area: $\int_{t=a}^b 2\pi x\sqrt{\left(\frac{dx}{dt}\right)^2 +\left(\frac{dy}{dt}\right)^2}\,dt.$
In our case, we have $y=\sqrt[3]{x}+2$. Let's choose a nice parametrization, something that will simplify the calculations. A sensible choice is $x=t^3$, in which case $y=t+2$. Since $x$ goes from $1$ to $8$, $t$ will go from $1$ to $2$.
Calculate. We have $\frac{dx}{dt}=3t^2$ and $\frac{dy}{dt}=1$. For rotation about the $y$-axis, we get surface area $\int_1^2 2\pi t^3\sqrt{1+9t^4}\,dt.$ This integral can be calculated by a simple substitution. But this was not your problem.
For rotation about the $x$-axis, we get surface area $\int_1^2 2\pi (t+2)\sqrt{1+9t^4}\,dt.$ This is a nightmarish integral. The hardest part, actually the marginally easier $\int\sqrt{1+x^4}\,dx$, has been discussed on this site. It may not be worth looking up.
Whoever was writing out solutions for your book got lucky. If (s)he had not made a mistake, (s)he would have had an awful integral to evaluate. (It would start out looking even worse without the parametric approach.)