# 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?

Jun 7, 2018

$V = \setminus \frac{14 \setminus \pi}{15}$ (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.
$y = \setminus \frac{1}{x}$ and $y = 2$ intersect for $x = 2$ and $y = {x}^{2}$ and $y = \setminus \frac{1}{x}$ intersect for $x = 1$.

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

As a result, the volume of the resulting solid of revolution is
$V = \setminus {\int}_{0}^{1} A \left(x\right) \mathrm{dx} = \setminus {\int}_{0}^{\setminus \frac{1}{2}} A \left(x\right) \mathrm{dx} + \setminus {\int}_{\setminus} {\frac{1}{2}}^{1} A \left(x\right) \mathrm{dx}$ and
$\setminus {\int}_{0}^{\setminus} \frac{1}{2} A \left(x\right) \mathrm{dx} = \setminus {\int}_{0}^{\setminus} \frac{1}{2} \setminus \pi \left(4 - {x}^{4}\right) \mathrm{dx} = \setminus \pi {\left[4 x - \setminus \frac{{x}^{5}}{5}\right]}_{0}^{\setminus \frac{1}{2}} = \setminus \pi \left(2 - \setminus \frac{1}{160}\right)$.

On other hand, $\setminus {\int}_{\setminus} {\frac{1}{2}}^{1} A \left(x\right) \mathrm{dx} = \setminus {\int}_{\setminus} {\frac{1}{2}}^{1} \setminus \pi \left(\setminus \frac{1}{{x}^{2}} - {x}^{4}\right) = \setminus \pi {\left[- \setminus \frac{1}{x} - \setminus \frac{{x}^{5}}{5}\right]}_{\setminus} {\frac{1}{2}}^{1} = \setminus \pi \left(- 1 - \setminus \frac{1}{5} + 2 + \setminus \frac{1}{160}\right)$.

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

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