The three-dimensional version of plane polar coordinates.
\(z\) is the distance from the reference plane \(z=0\), perpendicular to the line \(\rho=0\).
A point is the intersection of three mutually perpendicular surfaces: a circular
cylinder (\(\rho=\text{constant}\)), a plane (\(\phi=\text{constant}\)) and a plane
(\(z=\text{constant}\)). On the \(z\) axis one plane suffices.
Each unit vector \(\av{\rho}\), \(\av{\phi}\), \(\av{z}\) points toward increasing values of its
coordinate, normal to the surface on which that coordinate is constant (as \(\av{x}\) is normal to
\(x=\text{constant}\)).
At \(P(\rho_1,\phi_1,z_1)\), \(\av{\rho}\) points radially outward, normal to the cylinder
\(\rho=\rho_1\), and lies in the planes \(\phi=\phi_1\) and \(z=z_1\). \(\av{\phi}\) is normal to the
plane \(\phi=\phi_1\), points toward increasing \(\phi\), lies in \(z=z_1\) and is tangent
to the cylinder \(\rho=\rho_1\) (Fig. 1.8(b)).
Rectangular unit vectors are constant, but \(\av{\rho}\) and \(\av{\phi}\) vary with \(\phi\),
so they are not constants when differentiating or integrating with respect to \(\phi\).
\(\av{\rho}\times\av{\phi}=\av{z}\).
Volume element: increase \(\rho\), \(\phi\), \(z\) by \(\dd\rho\), \(\dd\phi\), \(\dd z\).
itemize
The cylinders of radius \(\rho\) and \(\rho+\dd\rho\), the radial planes at \(\phi\) and
\(\phi+\dd\phi\), and the horizontal planes at \(z\) and \(z+\dd z\) enclose a small truncated wedge
(Fig. 1.8(c)).
In the limit it becomes a rectangular box with sides \(\dd\rho\), \(\rho\,\dd\phi\) and
\(\dd z\).
\(\dd\rho\) and \(\dd z\) are lengths but \(\dd\phi\) is not; \(\rho\,\dd\phi\) is the length.
The face areas are \(\rho\,\dd\rho\,\dd\phi\), \(\rho\,\dd\phi\,\dd z\) and \(\dd\rho\,\dd z\), and
the volume is \(\rho\,\dd\rho\,\dd\phi\,\dd z\).
itemize
From Fig. 1.9,
\[
x=\rho\cos\phi,\qquad y=\rho\sin\phi,\qquad z=z,
\]
and conversely
\[
\rho=\sqrt{x^2+y^2}\quad(\rho\ge0),\qquad \phi=\tan^{-1}\frac{y}{x},\qquad z=z.
\]
Solution. Using Table 1.1 with \(x=\rho\cos\phi\), \(y=\rho\sin\phi\):
\begin{align*}
B_\rho&=\vect{B}\cdot\av{\rho}=y(\av{x}\cdot\av{\rho})-x(\av{y}\cdot\av{\rho})
=y\cos\phi-x\sin\phi=\rho\sin\phi\cos\phi-\rho\cos\phi\sin\phi=0,\\
B_\phi&=\vect{B}\cdot\av{\phi}=y(\av{x}\cdot\av{\phi})-x(\av{y}\cdot\av{\phi})
=-y\sin\phi-x\cos\phi=-\rho\sin^2\phi-\rho\cos^2\phi=-\rho,
\end{align*}
and \(B_z=z\). Thus \(\vect{B}=-\rho\,\av{\phi}+z\av{z}\).
