Calculus Teacher being Tricky (Integrals) So our calc teacher is being tricky and sending us off to fend for ourselves in the world of mathematics.
Here's the question:
We need to set up an integral to find the volume of the solid formed by the area bounded by the functions y=ln(x), x=3, and the x-axis revolved around the line x=3.
The problem?
Most of us put something similar to $\pi \int_0^{\ln 3} (e^y - 3)^2\ dy$, which is believed to be correct.
He claims it can also be written as $\pi \int_0^{e^3} (3 - e^y)^2\ dy$, but hasn't given us any idea as to why.
Many thanks to anyone who can help us out.
EDIT:
Multiple choice options given:


*

*$\pi \int_0^{e^3} (3 - e^y)^2\ dy$

*$\pi \int_0^{e^3} (9 - e^{2y})\ dy$

*$\pi \int_1^{3} (\ln_e(x))^2\ dy$

*None of These <- Not this answer!

*$\pi \int_1^{3} (9 - (\ln_e(x))^2)\ dy$

 A: From my calculations, it seems that your integral is correct.  The other one is not.
The volume is indeed given by $$V=\pi \int_0^{\ln 3} (e^y-3)^2 dy=5.929$$ (We go from 0 to $\ln 3$, and the radius of the circle at height $y$ is $3-e^y$)
Alternatively, integrating the $x$-coordinate we find the volume is $$V=2\pi \int_{1}^{3} (3-x)\ln x dx=5.929$$ (Using the so called "shells" method)
Hope that helps, sometimes people just make typos.
A: Let $f$ be some function. I'll demonstrate how to calculate the volume $V$ of the solid obtained by revolving the region under the graph $f \ge 0$ on some interval $I$ around the horizontal axis. 
Let $P = \{ t_0, \cdots, t_n \}$ be any partition of $I$. Let $m_i = \inf \{ f(x) : t_{i-1} \le x \le t_i \}$ and $M_i = \sup \{ f(x) : t_{i-1} \le x \le t_i \}$, be as always. 
It's easy to see, that $\pi m_i^2(t_i - t_{i-1}) $ is the volume of some cylinder that lies inside $V$ on $[t_{i-1}, t_i]$. Similarly, $\pi M_i^2(t_i - t_{i-1}) $ is some cylinder that contains a part of $V$.
We know that as we sum over the partition of $I$, the lower sum will be a lower bound for the volume of $V$ and the upper sum an upper bound. So we have:
$$\sum_{i=1}^n \pi m_i^2 (t_i - t_{i-1}) \le volume(V) \le \sum_{i=1}^n \pi M_i^2 (t_i - t_{i-1})$$
These sums are the lower and upper sums for $f^2$ on $I$, so it follows that:
$$volume(V) = \pi \int_a^b f(x)^2 dx$$
Now, we need to adapt your region to this enviroment. Here we need to rotate the region about a line that is parallel to the $y$-axis. So we can shift back the region $3$ units towards the $y$-axis. Now our region is just the area under $\ln(x+3)$ on the interval $[-2,0]$.
As we are rotating about the $y$-axis, we must express the function in terms of $y$, i.e. $x=e^y - 3$, on the interval from $0$ to $\ln(0+3) = \ln 3$, so indeed, plugging this into our formula, yields:
$$volume (V) = \pi \int_0^{\ln 3} (e^y - 3)^2\ dy$$
so yes, your first integral is correct.
