The question:
Find the volume of the solid generated by revolving about the y-axis the region bounded by:
$$y = x^2$$ $$x-axis$$ $$x = 2$$
First, I believe we want to do this problem in terms of y. If this is correct, we would then set it up like the following:
$$ V = \int_0^4 \pi [(2)^2 - (y^\frac{1}{2})^2] dy $$
$$ V = \int_0^4 \pi [4 - y] dy $$
$$ V = \pi [4y - \frac{1}{2} y^2 ] $$ from 0 to 4
$$ V = \pi [16 - 8] $$
I feel like I messed up with the problem, the $2^2$ doesn't seem right.
Here's one way of solving this problem:
So you notice that the volume you get by rotating the region around the y-axis sort of resembles a shot glass, or maybe a cylinder with a bullet-shaped object carved out. Either way, it might not be the ideal shape to work with. But you notice that you can "fill the shot glass" or "fit the bullet inside" (whichever analogy you go with) to make a simple cylinder of radius $2$ and height $4$. So now it comes to finding the volume of the cylinder and subtracting the volume of the "drink" or "bullet".
The volume of the cylinder is easy: $\pi r^2 h = \pi \cdot 2^2 \cdot 4 = 16 \pi$.
Now, for the "bullet". You note that translating or rotating an object does not change its volume, so now let's instead consider the object that you get by rotating $y = 4 - x^2$ around the y-axis. By cutting this object into cylindrical "shells" of width dx, we get that each "shell" has the volume of $2\pi y dx = 2\pi (4-x^2) dx$. Integrating this from $0$ to $2$, we get
$\displaystyle \int_0^2 2\pi (4-x^2)dx = 2\pi \left(4x - \frac{x^3}{3}\right) = 2\pi \left(4\cdot2-\frac{2^3}{3}\right) = 2\pi \left(8-\frac{8}{3}\right) = \frac{32\pi}{3}$.
Finally, going back to the original question, we subtract the "bullet" from the cylinder and get $\displaystyle 16\pi - \frac{32\pi}{3} = \frac{16\pi}{3}$.
Of course, it is very possible to do the cylindrical shell method on the original object as well. Each shell has a volume of $2\pi y dx = 2\pi x^2 dx$, and integrating that from $x = 0$ to $x = 2$ gives us $\displaystyle \int_0^2 2\pi x^2 dx = 2\pi \int_0^2 x^2 dx = 2\pi \left(\frac{x^3}{3}\right) = \frac{16\pi}{3}$. And in fact, in this case it was actually easier to calculate it this way.
There's more than one way to go around solving problems like these. But either way, dealing with square roots tends to get a little ugly, so I avoided it by solving it in terms of $x$ instead of $y$. It is, of course, very possible to do it the way you were thinking of it, by solving the integral in terms of $y$. That's the beauty of math - there are multiple ways of solving a problem, but there's only one right answer - but the process can determine which method is the preferred one.