finding the volume of the solid via disk or washer method the question is: $y = 1/4x^2$, $x = 2$, $y = 0$; about the $y$-axis

I tried to draw it out, but I can't figure this stuff out. The graphing is the hardest part for me because I don't know what to do to graph whatever information is given and then make some sort of reasonable deduction about what to plug in where for either disk or washer method.
The region that is being rotated is the part of the first quadrant that is below the easily drawn parabola $y=\frac{x^2}{4}$ and to the left of the vertical line $x=2$. 

We are rotating about the $y$-axis. Imagine taking a slice of the solid perpendicular to the $y$-axis, "at" $y$. The cross-section is a disk with a circular hole in it. The outer radius of the cross-section is $2$, and the inner radius is $2\sqrt{y}$. So the area of cross-section is $\pi\left(2^2-(2\sqrt{y})^2\right)$. Note that when $x=2$, we have $y=1$. Thus the volume is
$$\int_0^1 \pi\left(4-4y\right)\,dy.$$
Integrate. We get $2\pi$.

Another way: Alternately, we can use the Method of Cylindrical Shells. Take a thin horizontal slice of width "$dx$" at $x$, and rotate it about the $y$-axis. We get a thin cylindrical shell of radius $x$, height $\frac{x^2}{4}$, and thickness $dx$. Thus the volume of the shell is approximately $(2\pi x)(x^2/4)\,dx$. Now "add up" (integrate) from $x=0$ to $x=2$. Our volume is
$$\int_0^2 (2\pi)\left(\frac{x^3}{4}\right)\,dx.$$
Calculate. Again we get $2\pi$.
$\int_0^2\pi x^2(\frac14x^2)dx=\frac\pi4\left[\frac{x^5}5\right]_0^2=\frac{8\pi}5$