Moment of inertia related question Sorry for the undetailed title, but when the moment of inertia is calculated in a solid cylinder, the volume of a "sheet" of the cylinder is calculated. I've only seen the volume as $$length*thickness*height = 2\pi r*dr*height$$, but I disagree with that. Shouldn't only the inner length of the "sheet" be $2\pi r$ and the outer part $2\pi (r+dr)$? That way the volume would be different.
 A: Your visual image is correct, but it doesn't change the volume meaningfully.  Thinking informally, the difference between $2\pi r$ and $2\pi(r+dr)$ is infinitesimally small, and has a vanishingly small effect.  It's similar to how 5 and 5+0.000000000000000000000000001 are nearly the same and become increasingly close to the same as I add more zeros.
In fact, the difference you mention is the difference between simple Riemann sum integration and trapezoidal integration.  That tiny difference matters when you are approximating the volume of the cylinder with a small number of layers, but becomes vanishingly small when the number of layers approaches infinity.
A: Since ${\rm d}r \ll r$ it means that $(r + {\rm d}r) = r$
What is the area of the cross section
$$ \pi (r+{\rm d}r)^2 - \pi r^2 = \pi (2 r + {\rm d}r) {\rm d}r = 2 \pi r {\rm d}r $$
A: Suppose we take the volume of the cylinder with radius $r+dr$:
$$ V_\text{outer} = \pi(r+dr)^2h $$
and subtract the volume of the cylinder with radius $r$:
$$ V_\text{inner} = \pi r^2h $$
Then the volume we are left with is the volume of the shell:
$$\begin{align}
 V_\text{shell} &= V_\text{outer} - V_\text{inner} \\
                &= \pi(r+dr)^2h - \pi r^2 h \\
                &= \pi h(2rdr + dr^2) \\
                &= \pi r^2 h\left(2\frac{dr}{r} + \left(\frac{dr}{r}\right)^2\right)
\end{align}$$
But $dr \ll r$ so the fraction $dr/r \ll 1$ and that means that $(dr/r)^2 \ll dr/r$. That means we can approximate the volume of the shell by:
$$  V_\text{shell} \approx 2\pi r h dr $$
because $(dr/r)^2$ is so much smaller than $dr/r$ that it can be ignored. And in the limit of $dr\rightarrow 0$ the approximation becomes exact and we get:
$$  dV= 2\pi r h dr $$
