Volume problem (begginer integration) Not sure if this is correct, can stack exchange help me confirm or deny my work?
determine the volume of revolution by revolving the region bounded by $y=2$, $y=\ln(x)$, $y=0$ and $x=0$ around the y axis
This region is illustrated in the attached photo 
it is clear that $y=\ln(x)$ and $y=2$ intersect at at $(e^2,x)$ given that
$\ln(x)=2$, $e^2=x$. The lines in respect to $y$ are $x=0, x=e^y$
I believe the best option is to integrate with respect to y as a series of circles, where a circle is given by the eqn: $A=\pi r^{2}$ 
so the volume:
$$\begin{align}
V&=\int_{0}^{2}\pi[(e^{y})^{2}-(0)^{2}]\:dy\\
&=\pi\int_{0}^{2}[e^{2y}-0]\:dy\\
&=\pi[\int_{0}^{2}e^{2y}\:dy]\\
&=\pi\big[[\frac{1}{2}e^{2y}]_{0}^{2}\big]\\
&=\pi[\frac{1}{2}e^{4}-\frac{1}{2}]\\
&=1/2\pi[e^{4}-1]\end{align}$$
My questions: 


*

*because it is a definite integral can I just ignore the integral of $0$ (which should be a constant)

*$y=\ln(x)$ crosses into the first quadrant at $x=1/2$ will this change how i set up the integral

*am i missing elements, ignoring important information?
Thanks
p.s feel free to fix my LaTex
 A: So to answer your questions in order:


*

*I am not too sure what you mean by ignoring the integral of $0$ because there are no instances where you are integrating $0$ in this problem. Please elaborate and I'll edit this answer with a response.

*So first off, $y=\ln{x}$ crosses the $x$-axis at $x=1$. Now to answer your question, it will not change how you set up the integral. This is due to the fact that you are interpreting the graph as a function of $y$. Because of this, the only values that matter when setting up your integrals (namely the lower and upper limits of integration) are $y$ values. You need only determine these two values when setting up your integral. The $x$ value where the graph crosses into quadrant I is irrelevant as that value will be computed and taken into consideration when you integrate the function.

*It seems that at least on a conceptual level, you understand what is going on. You got the most critical information

A: The volume you want to calculate is bounded by $x=0$, $x=\mathrm{e}^2$ and $y=0$, $y=ln(x)$. Then you revolve this volume around the $x\text{-axis}$, starting from $0$ to $2\pi$.
You can express this volume as 
$$\int_0^{2\pi} (\int_0^{ln(x)} (\int_0^{\mathrm{e}^2} dx)dy)dz$$
