# How do you find the volume bounded by y=ln(x) and the lines y=0, x=2 revolved about the y-axis?

Aug 7, 2017

For the solution by cylindrical shells, see below.

#### Explanation:

Here is a picture of the region and a representative slice taken parallel to the axis of rotation. The slice is taken at some value of $x$ and has thickness $\mathrm{dx}$. So our functions will need to be functions of $x$

Revolving about the $y$ axis will result in a cylindrical shell.

The volume of this representative shell is

$2 \pi r h \text{ thickness}$

The radius is shown as a dashed black line in the picture and has length $r = x$

The height of the shell will be the great $y$ value minus the lesser $y$ value. Since the lesser $y$ value is $0$, we have $h = \ln x$
As already mentioned, the thickness is $\mathrm{dx}$

The representative volume is
$2 \pi x \ln x \mathrm{dx}$

$x$ varies from $1$ to $2$, so the solid has volume

$V = {\int}_{1}^{2} 2 \pi x \ln x \mathrm{dx} = 2 \pi {\int}_{1}^{2} x \ln x \mathrm{dx}$

Use integration by parts to get

$= 2 \pi \left(2 \ln 2 - \frac{3}{4}\right)$

Aug 7, 2017

For the solution by washers see below.

#### Explanation:

Here is a picture of the region and a representative slice taken perpendicular to the axis of rotation. The slice is taken at some value of $y$ and has thickness $\mathrm{dy}$. So our functions will need to be functions of $y$

The curve $y = \ln x$ can also be represented by $x = {e}^{y}$

Revolving about the $y$ axis will result in a washer.

The volume of this representative washer is

$\pi {R}^{2} \cdot \text{ thickness" - pir^2 " thickness" = pi(R^2-r^2) *" thickness}$

Where $R$ is the greater radius -- shown as a dashed line. And $r$ is the lesser radius -- shown as a dotted line.

In this case $R = \text{the "x" value on the right} = 2$
and $R = \text{the "x" value on the left} = {e}^{y}$

The representative volume is
$\pi \left({2}^{2} - {\left({e}^{y}\right)}^{2}\right) \mathrm{dy} = \pi \left(4 - {e}^{2 y}\right) \mathrm{dy}$

$x$ varies from $1$ to $2$,

so $y$ varies from $0$ to $\ln 2$

And the volume is

$V = {\int}_{0}^{\ln 2} \pi \left(4 - {e}^{2 y}\right) \mathrm{dy} = \pi {\int}_{0}^{\ln 2} \left(4 - {e}^{2 y}\right) \mathrm{dy}$

Use integration to get

$= \pi \left[\left(4 \ln 2 - \frac{1}{2} {e}^{2 \ln 2}\right) - \left(- \frac{1}{2} {e}^{0}\right)\right] = \pi \left(4 \ln 2 - \frac{3}{2}\right)$