When using the shell method to compute volume, why not use exact circumference? I'm going through MIT's online calculus course, and often we're asked to calculate the volume of a 2d object that is rotated around an axis to create a 3d solid.  
Given the x-y plane, there's usually two ways to set the problem up: either integrate over $dx$ or $dy$.  Depending on how the problem is set up, one way will be called the "disk" method (summing up a bunch of cylinders, each with a tiny height), and the other will be called the "shell" method (summing up a bunch of tubes, each with a tiny thickness).
For an example problem, you can check the 42min mark of the lecture video, but I'll describe it here.  We're taking the line $y = e^x$ over the interval $x = [0,1]$, meaning it's also over the interval $y = [1, e]$, and rotating that around the y-axis to create a bowl shape.  The problem asks to find the volume of that bowl.
I got the correct answer by using the shell method, and setting the problem up like so:
$$\int_0^1 2 \pi x (e^1 - e^x) dx$$
where $(e^1 - e^x)$ is the height of each "tube" (or "shell"), dx is the thickness of each tube, and $2 \pi x$ is the circumference of each tube.
But I had the thought that $2 \pi x$ is NOT the actual, intuitive circumference of each tube, $2 \pi (x + \frac{dx}{2})$ is because it's actually the midpoint of each rectangle that's rotating around the y-axis to generate the circumference.  First, am I correct?  And if so, why are we able to just disregard this term?
 A: You can almost use the $dx$ in the integral as a variable that is added and multiplied, but you have to be careful when you do so.
One way to look at the underlying reasoning is like this: What we're really doing is taking a sum of the form
$$
S=\sum_{i=1}^n 2\pi \left(x_i+\frac{\Delta x}{2}\right) h(x_i)\Delta x
$$
(where $h(x)$ is the height of each shell) and taking a limit in which $\Delta x\to 0$ and $n\to\infty$.  We can write this sum as
$$
S=\sum_{i=1}^n 2\pi x_i h(x_i)\Delta x
+ \Delta x\sum_{i=1}^n \pi h(x_i)\Delta x
$$
Taking the limit as $\Delta x\to 0$, these sums become
$$
S=\int_0^1 2\pi \,x \,h(x)\,dx
+ 0\cdot \int_0^1 \pi\, h(x)\,dx
$$
Or, in other words: 
$$
S=\int_0^1 2\pi \,x \,h(x)\,dx
$$

An even less formal way to think about it is to note that these shells are "really thin", so that using the approximation $(x+\frac{\Delta x}{2})\approx x$ doesn't change our result "noticeably".
A: First note that speaking of $dx$ as a nonzero, infinitesimal thickness is problematic. We must speak of a limit process here.
If we consider tubes of positive thickness $\Delta x$, then we only get a rough approximation of the volume: Each tube has volume $(\pi(x+\Delta x)^2-\pi x^2)\cdot f(\xi)$ where we choose $\xi\in[x,x+\Delta x]$ to our liking (e.g. $\xi = x$ or $\xi=x+\frac12\Delta x$ or so that $f(\xi)$ becomes maximal or becomes minimal). Simplifying, the tube volume becomes $$(2\pi x\Delta x +\pi(\Delta x)^2)f(\xi)=2\pi x\Delta x f(\xi) +\pi(\Delta x)^2f(\xi)$$
When we sum over these volumes and then take the limit as we let $\Delta x\to 0$ (and the number $n$ of summands $\to\infty$, the following happens:


*

*For continuous $f$, the distinction between $f(\xi)$ and $f(x)$ becomes negligible

*For bounded $f$, the contribution of the summands $\pi(\Delta x)^2f(\xi)$ becomes negligible because we have $n$ summands of size $\sim \frac 1{n^2}$

*The sum turns into an integral $\int_a^b 2\pi xf(x)\,\mathrm dx$


In principle the same argumentation takes place with the disk method and it already takes places when you show that the area under a function graph is given by an integral.
A: Expand what you have: $$\int_0^1 2\pi (x + \frac{dx}{2}) (e^1 - e^x) dx $$
$$\int_0^1 2\pi x (e^1 - e^x) dx + \int_0^1 \pi dx (e^1 - e^x) dx $$
$$\int_0^1 2\pi x (e^1 - e^x) dx + \int_0^1 (\pi (e^1 - e^x) dx) dx $$
This second term is very interesting, we have two $dx$s. I'm not entirely sure what this means formally (I can't seem to write it as a Riemann sum). But it appears that you're integrating a function where every value is infinitesimal, and since you're integrating over a finite interval, the result should also be infinitesimal?
It reminds me of the ring theory approach to calculus, where you include an $\epsilon$ such that $\epsilon^2 = 0$. The $dx$ behaves like this $\epsilon$, in that when there's just one, it's infinitesimal, but still important, but squaring it makes it negligible.
