# Volume of revolution with integration by parts

You will find that this is an almost-duplicate post from another poster, but I need more explanation than what was given.

Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve $y=e^x$ and the line $x=ln4$, rotated about the line $x=ln4$.

How to do this using the shell method?

Also, can shell method be used for all volumes of revolution?

Volume of a solid involving integration by parts.

• You should link to that original post. – Mike Pierce Dec 14 '14 at 0:53
• Mainly, why is the shell's height "x" or "ln4-x" for any shell method? – Eric Johnson Dec 14 '14 at 1:01

From the picture we can see a general set up for the shell method. What you do is to take shells and imagine them as a wall. We are revolving $f(x)$ (bounded by $x=a$ and $x=b$) around a line $x=L$. In this case $L<a$. So next step is to sum all these shells to obtain the volume of the solid. This is why we take the integral. We are actually taking the limit
$$\lim_{\Delta x\to0}\sum\limits_{x=a}^b(\text{volume of shells of radius x})=\lim_{\Delta x\to0}\sum\limits_{x=a}^b(\text{height})(\text{width})(\text{base})\\=\lim_{\Delta x\to0}\sum\limits_{x=a}^bf(x)2\pi(x-L)\Delta x=\int_a^b2\pi(x-L)f(x)dx.$$
• If you are revolving around some $y=y_0$ then express the radius as a function of $y$, and the curve as a function of $y$ as well. If you are revolving around some $x=x_0$ do the analogous thing.
• Identify the radius. This can be a bit tricky. In the link you provided it happens that $f(x)$ is at the left of the axis that you will be revolving around, this is why the radius is not $x-\ln 5$ but rather $\ln 5 -x$. The radius is the distance from the axis of rotation and some $x$ between $a$ and $b$ (or $y$ between $a$ and $b$). Of course the radius must be a function of the variable you are integrating with respect to.