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