Question

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?

Answer 1

`V=\frac{14\pi}{15}` (where the unit is `u^3` if `u` is the unit of length)

Explanation:

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.