d3x - Cartesian to Cylindrical Coordinates

Question

Given is $d^3x = dxdydz$ and I need to convert it to cylindrical coordinates (given through: $x = r\cos\varphi$ and $y = r\sin\varphi$).

The expected result is: $(dz)(dr)(r)(d\varphi)$ and I cannot seem to get it right.

This is what I am doing:

$dz = dz$
$dy = \frac{dy}{d\varphi} d\varphi = r \cos\varphi d\varphi = \frac{dy}{dr} = \sin\varphi dr$
$dx = \frac{dx}{d\varphi} d\varphi = - r \sin\varphi d\varphi = \frac{dx}{dr} dr = \cos\varphi dr$

But I can't see how to create the expected solution from this.

Answer 1

The colored area under the picture is the unit area in polar coordinates.

$\Delta A = (\pi.(r+ \Delta r)^2 - \pi(r)^2). \frac{\Delta \theta}{2\pi}$

$\Delta A = \frac{(2r\Delta r + {\Delta r}^2).\Delta\theta}{2}$

Since we assume $(\Delta r)^2$ is negligable because of the square, we have

$\Delta A = r\Delta r\Delta\theta$

The $dz$ contribution is trivial.

Answer 2

The Jacobian of $(r,\phi,z)\mapsto (r\cos\phi,r\sin\phi,z)$ is $$ \left|\matrix{ \cos\phi & \sin\phi & 0\cr -r\sin\phi & r\cos\phi & 0\cr 0 & 0 & 1 }\right| = r, $$ so the volume element is $r\,dr\,d\phi\,dz$.