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.  
 A: 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 $2\pi$. 
A: $\int_0^2\pi x^2(\frac14x^2)dx=\frac\pi4\left[\frac{x^5}5\right]_0^2=\frac{8\pi}5$
