Justification of $dV = r dz dr d \theta$

Question

Context: I'm taking calc based physics, and we are supposed to be able to integrate moment of inertia for a cylinder. I referenced a mit vid, and though I have no education on multiple integrals, I got all but one thing. https://www.youtube.com/watch?v=iYFogDTPlRo

My progress is:

1) $\int r^2 dm$

2) $ \int r^2 \delta dV $ because $ \delta = \frac{dm}{dV} $ -> $dm = \delta dV$ where $\delta$ is density

3) $\delta \int_0^{2\pi} \int_0^r \int_0^h r^2 * r dzdrd\theta$ <- I dont fully understand this step.

Can someone explain why $dV = r dz dr d\theta$. Can it be justified in a similar manner as step 2, perhaps in multiple parts even? I get the idea I believe, but can't justify it.

EDIT: Thank you everyone. I don't have enough rep to upvote, but the answers given by David, Hamed, B. Pasternak, and root were all helpful. I have a pretty good understanding now.

Answer 1

Here is an intuitive (though not rigorous) geometrical argument.

Consider the volume element you get if you start at a point $(r,\theta,z)$ and increase $r$ by $dr$ and increase $\theta$ by $d\theta$ and increase $z$ by $dz$. You get a "curved box" where one side is a line with length $dr$, one is a line with length $dz$ and one is an arc of a circle with radius $r$, angle $d\theta$ and therefore length $r\,d\theta$. If this were an actual rectangular box (not curved), its volume would be $$dV=r\,dr\,d\theta\,dz\ ;$$ but as $dr$, $d\theta$ and $dz$ become vanishingly small, this will become a better and better approximation to the volume of the "curved box".

Another point worth considering: the "obvious" answer $dV=dr\,d\theta\,dz$ cannot be right as it is dimensionally incorrect: the right hand side is an area, not a volume.

Answer 2

Other than David's geometrical method which is nicer, here is an algebraic method. (Remember $r$ is the radial length in the x-y plane) $$ (x,y,z)=(r\cos\theta, r\sin\theta, z) $$ Now $dx dy dz = Jdrd\theta dz$, where $J$ is the Jacobian and is defined as (this works for any change of coordinates) $$ J=\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z}\\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z} \end{vmatrix}= \begin{vmatrix} \cos\theta & -r\sin\theta & 0\\ \sin\theta & r\cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix}=r(\cos^2\theta + \sin^2 \theta) = r $$ Hence $dV=rdrd\theta dz$ (by the way $|\bullet|$ above means the determinant).

Answer 3

For more-dimensional change of coordinates (in this case: cartesian $\rightarrow$ cylindric) one uses, that (under certain circumstances you can find in every textbook) substitution is given by

$$ \int_{g(D)} f(\mathbf{y}) d\mathbf{y} = \int_D (f \circ g)(x) \left| \det \left( \frac{dg}{d\mathbf{x}}(\mathbf{x}) \right) \right| d\mathbf{x}$$

This is the substitution rule in $\Bbb R^n$. $ \frac{dg}{d\mathbf{x}}(\mathbf{x})$ is the Jacobi matrix, given by

$$J = \frac{dg}{d\mathbf{x}}(\mathbf{x}) = \begin{pmatrix} \nabla g^1(\mathbf{x}) \\ \vdots \\ \nabla g^m (\mathbf{x}) \end{pmatrix}$$

So the entry $(J_{kj})$ is given by $\frac{\partial g^k(\mathbf{x})}{\partial x_j}$, where $g: \Bbb R^n \to \Bbb R^m$ is in your case the mapping cartesian $\rightarrow$ cylindric coordinates and $g^k(\mathbf{x})$ is the $k^{\text{th}}$ coordinate of $g$, which is in your case

$$g: (x,y,z) \rightarrow (r \cos(\varphi), r \sin(\varphi), z) $$

E. g. the first entry of $J$ would be $ J_{11}= \frac{\partial g^1(\mathbf{x})}{\partial x_1} = \cos(\varphi)$, since $(x_1,x_2,x_3) = (r,\varphi,z)$. Summarizing:

$$d\mathbf{x} \equiv dxdydz = \left| \det \left( \frac{dg}{d(r,\varphi,z)}(r,\varphi,z) \right) \right| d(r,\varphi,z) = r drd\varphi dz$$