Surface Area Problem I am trying to figure out what is going on in this Surface Area problem:

As I attempted to illustrate above, It seems like the formula has been applied incorrectly. Where x has been placed should be f(x) which in this problem is (x^(1/3) + 2). 
Is something going on that changes the input of f(x) to x that I am not seeing here? Thanks!
 A: No, it is incorrect.
It also happens to be example 2 on this page of Paul's Online Math Notes. He explains as best as anyone could hope.
A: The formula you have provided assumes that the surface is being produced by rotating around the x-axis.  In that case, f(x) is the radius of rotation.  It looks like you might instead be creating the surface by rotating around the y-axis, in which case the radius of rotation would be x, as in the solution.  You can think of finding the surface area as a bunch of little slanty sticks each carving out a thin strip of surface as they're carried around the axis of rotation.  Each stick (represented by the stuff in the square root) traces out a strip of circumference 2$\pi$r, where r depends on the set-up of the rotation.
Hope that helps from a conceptual standpoint.
A: 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.)
