Please, help with this integration problem

Question

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!

Answer 1

Rotating about the vertical line $x=1$ is in principal no different than rotating about the $y$-axis, which after all is just the vertical line $x=0$. The radius of a shell when rotating about the $y$-axis is the distance of $x$ from $0$, which is $r=|x-0|=|x|$; this further simplifies to $|x|=x$ if $0\leq x$. The only modification required when rotating about a different vertical line such as $x=1$ is that now the radius of a shell is going to be the distance of $x$ from $1$, which is $r=|x-1|$, which will further simplify to $|x-1|=1-x$ if $x\leq 1$, as is the case for your problem.

Answer 2

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.

Answer 3

Interesting Question indeed.

I assume that you know the concepts of cylindrical shells already.

Let R be the region by functions  g(x)  and  f(x)  on an interval x=[a,b] where 
g(x) > f(x) on that interval. Then the volume of revolution obtained by rotating 
the region about the y-axis using cylindrical shell is given by the formula

V = 2*pi*integral of x(R) dx = 2*pi*integral of x(g(x)-f(x)) dx

[I really dont know how to use the math symbols on here so i will post an image of my work]

Here is an image of the bounded region.

The limits of integration then are a=0, b=1.

If you want to learn about the concepts go here http://www.mhhe.com/math/calc/smithminton2e/cd/folder_structure/text/chap05/section03.htm.