Please, help with this integration problem Consider the region bounded by the curves $y=e^x$, $y=e^{-x}$, and $x=1$. Use the method of cylindrical shells to find the volume of the solid obtained by rotating this region about the y-axis.
I drew the corresponding graph. I'm confused by the fact that the area is rotating about one of the bounding lines (x=1) that bound the region (I have solved volumes this way, until now), but about the y-axis, which is a previous point to the bounding line. How does this changes the procedure?
Thank you!


I drew the corresponding graph. I'm confused by the fact that the area is rotating about one of the bounding lines (x=1) that bound the region (I have solved volumes this way, until now), but about the y-axis, which is a previous point to the bounding line. How does this changes the procedure?

The $y$-axis is just the line $x = 0$.  To rotate about this line, you need to know the inner radius and the outer radius, which are the horizontal distance to this line.
If you draw a picture, you should be able to see


*

*The inner radius, as a function of $y$, is $- \ln y$ when
$\frac1e \le y \le 1$, and $\ln y$ when $1 \le y \le e$.

*The outer radius, as a function of $y$, is $1$ everywhere.
Then the value of the integral is
$$
\int_\frac{1}{e}^1 \pi \left[ R^2 - r^2 \right] \; dx + 
\int_1^e \pi \left[ R^2 - r^2 \right] \; dx
$$
where $R$ is outer radius and $r$ is inner radius.
