Let R be the region bounded by #y = 1/x#, #y = x^2#, #x = 0#, and #y = 2# and revolved about the x axis. How do you find the volume of rotation using: a) the method of cylindrical shells; b) the method of circular disks?

Because we are looking at revolution about an horizontal axis, using cross-section (See videos 1 through 6 here if you need a refresher) is more natural, so let us start with that.
Here is the region we are rotating about the #x#-axis:
.

The first thing to understand is where the curves intersect.
#y=\frac{1}{x}# and #y=2# intersect for #x=2# and #y=x^2# and #y=\frac{1}{x}# intersect for #x=1#.

So the expression for the area #A(x)# of the cross section by #x=c# is going to be different for #0\leq x\leq frac{1}{2}# and for #\frac{1}{2}\leq x\leq 1#.
Namely, for #0\leq x\leq frac{1}{2}#, the cross-section is a washer with outer-radius 2, and inner-radius #x^2# so that the area is #A(x)=\pi (2^2-(x^2)^2)=\pi(4-x^4)#.
On the other hand if #\frac{1}{2}\leq x\leq 1#, then the cross section is a washer with outer radius #\frac{1}{x}# and inner radius #x^2#, so that the area is
#A(x)=\pi((\frac{1}{x})^2-(x^2)^2)=\pi(\frac{1}{x^2}-x^4)#.

As a result, the volume of the resulting solid of revolution is
#V=\int_0^1A(x) dx=\int_0^{\frac{1}{2}}A(x)dx+\int_\frac{1}{2}^1A(x)dx# and
#\int_0^\frac{1}{2}A(x)dx=\int_0^\frac{1}{2}\pi(4-x^4)dx=\pi[4x-\frac{x^5}{5}]_0^{\frac{1}{2}}=\pi(2-\frac{1}{160})#.

On other hand, #\int_\frac{1}{2}^1A(x)dx=\int_\frac{1}{2}^1 \pi(\frac{1}{x^2}-x^4)=\pi [-\frac{1}{x}-\frac{x^5}{5}]_\frac{1}{2}^1=\pi(-1-\frac{1}{5}+2+\frac{1}{160})#.

Hence, altogether, we obtain
#V=\pi(2-\frac{1}{160}+1-\frac{1}{5}+\frac{1}{160})=\pi(3-\frac{1}{5})=\frac{14\pi}{15}#.

Considering the length of the answer, I am leaving the cylindrical shell part to another time.