How to visualise the curve $y=\ln x$ rotating around its $y$-axis

Question

Let the area limited by the curve $y=\ln x$, the line $x=e$ and the $x$-axis rotate around the $y$-axis. Decide the volume of the resulting rotational body.

First thing, I drew the graph:

But then I got stuck on how to imagine/visualise the curve rotating around its $y$-axis. How should I think when visualising the function $lnx$ rotating around its $y$-axis?

EDIT: In the comment section it was written that "if you rotate the rectangle $[0,e]×[0,1]$, it makes a cylinder which volume is easy to compute", but I am not very familiar with the notation for $[0,e]×[0,1]$.

Answer 1

Find the intersection of $e^y$ and $e$ and that is $y = 1$

Inner Radius $= e^{y}$ and Outer Radius $= e$

Then the area of the surface rotating $y = \ln x$ around y axis

$$V = \int_{0}^{1} \pi (e^2-e^{2y})dy = \pi(e^2y - \pi\frac{e^{2y}}{2})|_{0}^{1} = (\pi e^2 - 0) - (\frac{\pi e^2}{2} -1) = \frac{\pi (e^2+1)}{2}$$

Answer 2

take a look here with wolfy for $$f=log(\sqrt{x^2+y^2}$$ http://m.wolframalpha.com/input/?i=plot+log%28%28x%5E2%2By%5E2%29%5E.5%29

Answer 3

Here I (the OP) will add my solution to the problem.

Think of it like follows. If the integral $\int_0^1 dy \;\pi (e)^2$ represents the area of

And the integral $\int_0^1 dy \;\pi (e^y)^2$ represents the area of

Then the area of

is computed like this:

$$\int_0^1 dy \;\pi (e)^2-\int_0^1 dy \;\pi (e^y)^2=\int_0^1 dy \;\pi(e^2-e^{2y})$$

And this integral gives you the answer $\frac{\pi (e^2+1)}{2} \text{v.u.}$