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diff --git a/source/know/concept/cylindrical-parabolic-coordinates/index.md b/source/know/concept/cylindrical-parabolic-coordinates/index.md
index 766c9b6..58358dd 100644
--- a/source/know/concept/cylindrical-parabolic-coordinates/index.md
+++ b/source/know/concept/cylindrical-parabolic-coordinates/index.md
@@ -8,82 +8,97 @@ categories:
 layout: "concept"
 ---
 
-**Cylindrical parabolic coordinates** are a coordinate system
-that describes a point in space using three coordinates $$(\sigma, \tau, z)$$.
-The $$z$$-axis is unchanged from the Cartesian system,
-hence it is called a *cylindrical* system.
-In the $$z$$-isoplane, however, confocal parabolas are used.
-These coordinates can be converted to the Cartesian $$(x, y, z)$$ as follows:
+**Cylindrical parabolic coordinates** extend parabolic coordinates $$(\sigma, \tau)$$ to 3D,
+by describing a point in space using the variables $$(\sigma, \tau, z)$$.
+The $$z$$-axis is the same as in the Cartesian system, (hence the name *cylindrical*),
+while the coordinate lines of $$\sigma$$ and $$\tau$$ are confocal parabolas.
+
+[Cartesian coordinates](/know/concept/cartesian-coordinates/) $$(x, y, z)$$
+and this system $$(\sigma, \tau, z)$$ are related by:
 
 $$\begin{aligned}
     \boxed{
-        x = \frac{1}{2} (\tau^2 - \sigma^2 )
-        \qquad
-        y = \sigma \tau
-        \qquad
-        z = z
+        \begin{aligned}
+            x
+            &= \frac{1}{2} (\tau^2 - \sigma^2)
+            \\
+            y
+            &= \sigma \tau
+            \\
+            z
+            &= z
+        \end{aligned}
     }
 \end{aligned}$$
 
-Converting the other way is a bit trickier.
-It can be done by solving the following equations,
-and potentially involves some fiddling with signs:
+Conversely, a point given in $$(x, y, z)$$ can be converted
+to $$(\sigma, \tau, z)$$ using these formulae:
 
 $$\begin{aligned}
-    2 x
-    = \frac{y^2}{\sigma^2} - \sigma^2
-    \qquad \qquad
-    2 x
-    = - \frac{y^2}{\tau^2} + \tau^2
+    \boxed{
+        \begin{aligned}
+            \sigma
+            &= \sgn(x) \sqrt{\sqrt{x^2 + y^2} - x}
+            \\
+            \tau
+            &= \sqrt{\sqrt{x^2 + y^2} + x}
+            \\
+            z
+            &= z
+        \end{aligned}
+    }
 \end{aligned}$$
 
 Cylindrical parabolic coordinates form
 an [orthogonal curvilinear system](/know/concept/orthogonal-curvilinear-coordinates/),
-so we would like to find its scale factors $$h_\sigma$$, $$h_\tau$$ and $$h_z$$.
-The differentials of the Cartesian coordinates are as follows:
+whose **scale factors** $$h_\sigma$$, $$h_\tau$$ and $$h_z$$ we need.
+To get those, we calculate the unnormalized local basis:
 
 $$\begin{aligned}
-    \dd{x} = - \sigma \dd{\sigma} + \tau \dd{\tau}
-    \qquad
-    \dd{y} = \tau \dd{\sigma} + \sigma \dd{\tau}
-    \qquad
-    \dd{z} = \dd{z}
+    h_\sigma \vu{e}_\sigma
+    &= \vu{e}_x \pdv{x}{\sigma} + \vu{e}_y \pdv{y}{\sigma} + \vu{e}_z \pdv{z}{\sigma}
+    \\
+    &= - \vu{e}_x \sigma + \vu{e}_y \tau
+    \\
+    h_\tau \vu{e}_\tau
+    &= \vu{e}_x \pdv{x}{\tau} + \vu{e}_y \pdv{y}{\tau} + \vu{e}_z \pdv{z}{\tau}
+    \\
+    &= \vu{e}_x \tau + \vu{e}_y \sigma
+    \\
+    h_\sigma \vu{e}_\sigma
+    &= \vu{e}_x \pdv{x}{z} + \vu{e}_y \pdv{y}{z} + \vu{e}_z \pdv{z}{z}
+    \\
+    &= \vu{e}_z
 \end{aligned}$$
 
-We calculate the line segment $$\dd{\ell}^2$$,
-skipping many terms thanks to orthogonality:
-
-$$\begin{aligned}
-    \dd{\ell}^2
-    &= (\sigma^2 + \tau^2) \:\dd{\sigma}^2 + (\tau^2 + \sigma^2) \:\dd{\tau}^2 + \dd{z}^2
-\end{aligned}$$
-
-From this, we can directly read off the scale factors $$h_\sigma^2$$, $$h_\tau^2$$ and $$h_z^2$$,
-which turn out to be:
+By normalizing the **local basis vectors**
+$$\vu{e}_\sigma$$, $$\vu{e}_\tau$$ and $$\vu{e}_z$$,
+we arrive at these expressions,
+where we have defined the abbreviation $$\rho$$ for convenience:
 
 $$\begin{aligned}
     \boxed{
-        h_\sigma = \sqrt{\sigma^2 + \tau^2}
-        \qquad
-        h_\tau = \sqrt{\sigma^2 + \tau^2}
-        \qquad
-        h_z = 1
+        \begin{aligned}
+            h_\sigma
+            &= \rho
+            \equiv \sqrt{\sigma^2 + \tau^2}
+            \\
+            h_\tau
+            &= \rho
+            \equiv \sqrt{\sigma^2 + \tau^2}
+            \\
+            h_z
+            &= 1
+        \end{aligned}
     }
-\end{aligned}$$
-
-With these scale factors, we can use
-the general formulae for orthogonal curvilinear coordinates
-to easily to convert things from the Cartesian system.
-The basis vectors are:
-
-$$\begin{aligned}
+    \qquad\qquad
     \boxed{
         \begin{aligned}
             \vu{e}_\sigma
-            &= \frac{- \sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y
+            &= - \vu{e}_x \frac{\sigma}{\rho} + \vu{e}_y \frac{\tau}{\rho}
             \\
             \vu{e}_\tau
-            &= \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y
+            &= \vu{e}_x \frac{\tau}{\rho} + \vu{e}_y \frac{\sigma}{\rho}
             \\
             \vu{e}_z
             &= \vu{e}_z
@@ -91,13 +106,54 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The basic vector operations (gradient, divergence, Laplacian and curl) are given by:
+Thanks to these scale factors, we can easily convert calculus from the Cartesian system
+using the standard formulae for orthogonal curvilinear coordinates.
+
+
+
+## Differential elements
+
+For line integrals,
+the tangent vector element $$\dd{\vb{\ell}}$$ for a curve is as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vb{\ell}}
+        = \vu{e}_\sigma \: \rho \dd{\sigma} + \: \vu{e}_\tau \: \rho \dd{\tau} + \: \vu{e}_z \dd{z}
+    }
+\end{aligned}$$
+
+For surface integrals,
+the normal vector element $$\dd{\vb{S}}$$ for a surface is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vb{S}}
+        = \vu{e}_\sigma \: \rho \dd{\tau} \dd{z} + \: \vu{e}_\tau \: \rho \dd{\sigma} \dd{z} + \: \vu{e}_z \: \rho^2 \dd{\sigma} \dd{\tau}
+    }
+\end{aligned}$$
+
+And for volume integrals,
+the infinitesimal volume $$\dd{V}$$ takes the following form:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{V}
+        = \rho^2 \dd{\sigma} \dd{\tau} \dd{z}
+    }
+\end{aligned}$$
+
+
+
+## Common operations
+
+The basic vector operations (gradient, divergence, curl and Laplacian) are given by:
 
 $$\begin{aligned}
     \boxed{
         \nabla f
-        = \frac{\vu{e}_\sigma}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\sigma}
-        + \frac{\vu{e}_\tau}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\tau}
+        = \vu{e}_\sigma \frac{1}{\rho} \pdv{f}{\sigma}
+        + \vu{e}_\tau \frac{1}{\rho} \pdv{f}{\tau}
         + \vu{e}_z \pdv{f}{z}
     }
 \end{aligned}$$
@@ -105,78 +161,135 @@ $$\begin{aligned}
 $$\begin{aligned}
     \boxed{
         \nabla \cdot \vb{V}
-        = \frac{1}{\sigma^2 + \tau^2}
-        \Big( \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\sigma} + \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\tau} \Big) + \pdv{V_z}{z}
+        = \frac{1}{\rho} \pdv{V_\sigma}{\sigma} + \frac{\sigma V_\sigma}{\rho^3}
+        + \frac{1}{\rho} \pdv{V_\tau}{\tau} + \frac{\tau V_\tau}{\rho^3}
+        + \pdv{V_z}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \quad \vu{e}_\sigma \bigg( \frac{1}{\rho} \pdv{V_z}{\tau} - \pdv{V_\tau}{z} \bigg)
+            \\
+            &\quad\: + \vu{e}_\tau \bigg( \pdv{V_\sigma}{z} - \frac{1}{\rho} \pdv{V_z}{\sigma} \bigg)
+            \\
+            &\quad\: + \vu{e}_z \bigg( \frac{1}{\rho} \pdv{V_\tau}{\sigma} + \frac{\sigma V_\tau}{\rho^3}
+            - \frac{1}{\rho} \pdv{V_\sigma}{\tau} - \frac{\tau V_\sigma}{\rho^3} \bigg)
+        \end{aligned}
     }
 \end{aligned}$$
 
 $$\begin{aligned}
     \boxed{
         \nabla^2 f
-        = \frac{1}{\sigma^2 + \tau^2} \Big( \pdvn{2}{f}{\sigma} + \pdvn{2}{f}{\tau} \Big) + \pdvn{2}{f}{z}
+        = \frac{1}{\rho^2} \pdvn{2}{f}{\sigma} + \frac{1}{\rho^2} \pdvn{2}{f}{\tau} + \pdvn{2}{f}{z}
     }
 \end{aligned}$$
 
+
+
+## Uncommon operations
+
+Uncommon operations include:
+the gradient of a divergence $$\nabla (\nabla \cdot \vb{V})$$,
+the gradient of a vector $$\nabla \vb{V}$$,
+the advection of a vector $$(\vb{U} \cdot \nabla) \vb{V}$$ with respect to $$\vb{U}$$,
+the Laplacian of a vector $$\nabla^2 \vb{V}$$,
+and the divergence of a 2nd-order tensor $$\nabla \cdot \overline{\overline{\vb{T}}}$$:
+
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
-            \nabla \times \vb{V}
-            &= \vu{e}_\sigma \Big( \frac{\vu{e}_1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\tau} - \pdv{V_\tau}{z} \Big)
+            \nabla (\nabla \cdot \vb{V})
+            &= \quad \vu{e}_\sigma \bigg( \frac{1}{\rho^2} \pdvn{2}{V_\sigma}{\sigma} + \frac{1}{\rho^2} \mpdv{V_\tau}{\sigma}{\tau}
+            + \frac{1}{\rho} \mpdv{V_z}{\sigma}{z}
+            \\
+            &\qquad\qquad + \frac{\tau}{\rho^4} \pdv{V_\tau}{\sigma} - \frac{\sigma}{\rho^4} \pdv{V_\tau}{\tau}
+            + \frac{\rho^2 - 3 \sigma^2}{\rho^6} V_\sigma - \frac{3 \sigma \tau V_\tau}{\rho^6} \bigg)
             \\
-            &+ \vu{e}_\tau \Big( \pdv{V_\sigma}{z} - \frac{1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\sigma} \Big)
+            &\quad\: + \vu{e}_\tau \bigg( \frac{1}{\rho^2} \mpdv{V_\sigma}{\tau}{\sigma} + \frac{1}{\rho^2} \pdvn{2}{V_\tau}{\tau}
+            + \frac{1}{\rho} \mpdv{V_z}{\tau}{z}
             \\
-            &+ \frac{\vu{e}_z}{\sigma^2 + \tau^2}
-            \Big( \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\sigma} - \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\tau} \Big)
+            &\qquad\qquad - \frac{\tau}{\rho^4} \pdv{V_\sigma}{\sigma} + \frac{\sigma}{\rho^4} \pdv{V_\sigma}{\tau}
+            - \frac{3 \sigma \tau V_\sigma}{\rho^6} + \frac{\rho^2 - 3 \tau^2}{\rho^6} V_\tau \bigg)
+            \\
+            &\quad\: + \vu{e}_z \bigg( \frac{1}{\rho} \mpdv{V_\sigma}{z}{\sigma} + \frac{1}{\rho} \mpdv{V_\tau}{z}{\tau} + \pdvn{2}{V_z}{z}
+            + \frac{\sigma}{\rho^3} \pdv{V_\sigma}{z} + \frac{\tau}{\rho^3} \pdv{V_\tau}{z} \bigg)
         \end{aligned}
     }
 \end{aligned}$$
 
-The differential element of volume $$\dd{V}$$
-in cylindrical parabolic coordinates is given by:
-
 $$\begin{aligned}
     \boxed{
-        \dd{V} = (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau} \dd{z}
+        \begin{aligned}
+            \nabla \vb{V}
+            &= \quad \vu{e}_\sigma \vu{e}_\sigma \bigg( \frac{1}{\rho} \pdv{V_\sigma}{\sigma} + \frac{\tau V_\tau}{\rho^3} \bigg)
+            + \vu{e}_\sigma \vu{e}_\tau \bigg( \frac{1}{\rho} \pdv{V_\tau}{\sigma} - \frac{\tau V_\sigma}{\rho^3} \bigg)
+            + \vu{e}_\sigma \vu{e}_z \frac{1}{\rho} \pdv{V_z}{\sigma}
+            \\
+            &\quad\: + \vu{e}_\tau \vu{e}_\sigma \bigg( \frac{1}{\rho} \pdv{V_\sigma}{\tau} - \frac{\sigma V_\tau}{\rho^3} \bigg)
+            + \vu{e}_\tau \vu{e}_\tau \bigg( \frac{1}{\rho} \pdv{V_\tau}{\tau} + \frac{\sigma V_\sigma}{\rho^3} \bigg)
+            + \vu{e}_\tau \vu{e}_z \frac{1}{\rho} \pdv{V_z}{\tau}
+            \\
+            &\quad\: + \vu{e}_z \vu{e}_\sigma \pdv{V_\sigma}{z}
+            + \vu{e}_z \vu{e}_\tau \pdv{V_\tau}{z}
+            + \vu{e}_z \vu{e}_z \pdv{V_z}{z}
+        \end{aligned}
     }
 \end{aligned}$$
 
-The differential elements of the isosurfaces are as follows,
-where $$\dd{S_\sigma}$$ is the $$\sigma$$-isosurface, etc.:
-
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
-            \dd{S_\sigma} &= \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
+            (\vb{U} \cdot \nabla) \vb{V}
+            &= \quad \vu{e}_\sigma \bigg( \frac{U_\sigma}{\rho} \pdv{V_\sigma}{\sigma} + \frac{U_\tau}{\rho} \pdv{V_\sigma}{\tau} + U_z \pdv{V_\sigma}{z}
+            + \frac{\tau}{\rho^3} U_\sigma V_\tau - \frac{\sigma}{\rho^3} U_\tau V_\tau \bigg)
             \\
-            \dd{S_\tau} &= \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
+            &\quad\: + \vu{e}_\tau \bigg( \frac{U_\sigma}{\rho} \pdv{V_\tau}{\sigma} + \frac{U_\tau}{\rho} \pdv{V_\tau}{\tau} + U_z \pdv{V_\tau}{z}
+            + \frac{\sigma}{\rho^3} U_\tau V_\sigma - \frac{\tau}{\rho^3} U_\sigma V_\sigma \bigg)
             \\
-            \dd{S_z} &= (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
+            &\quad\: + \vu{e}_z \bigg( \frac{U_\sigma}{\rho} \pdv{V_z}{\sigma} + \frac{U_\tau}{\rho} \pdv{V_z}{\tau} + U_z \pdv{V_z}{z} \bigg)
         \end{aligned}
     }
 \end{aligned}$$
 
-The normal element $$\dd{\vu{S}}$$ of a surface and
-the tangent element $$\dd{\vu{\ell}}$$ of a curve are respectively:
-
 $$\begin{aligned}
     \boxed{
-        \dd{\vu{S}}
-        = \vu{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
-        + \vu{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
-        + \vu{e}_z (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
+        \begin{aligned}
+            \nabla^2 \vb{V}
+            &= \quad \vu{e}_\sigma \bigg( \frac{1}{\rho^2} \pdvn{2}{V_\sigma}{\sigma} + \frac{1}{\rho^2} \pdvn{2}{V_\sigma}{\tau} + \pdvn{2}{V_\sigma}{z}
+            + \frac{2 \tau}{\rho^4} \pdv{V_\tau}{\sigma} - \frac{2 \sigma}{\rho^4} \pdv{V_\tau}{\tau} - \frac{V_\sigma}{\rho^4} \bigg)
+            \\
+            &\quad\: + \vu{e}_\tau \bigg( \frac{1}{\rho^2} \pdvn{2}{V_\tau}{\sigma} + \frac{1}{\rho^2} \pdvn{2}{V_\tau}{\tau} + \pdvn{2}{V_\tau}{z}
+            - \frac{2 \tau}{\rho^4} \pdv{V_\sigma}{\sigma} + \frac{2 \sigma}{\rho^4} \pdv{V_\sigma}{\tau} - \frac{V_\tau}{\rho^4} \bigg)
+            \\
+            &\quad\: + \vu{e}_z \bigg( \frac{1}{\rho^2} \pdvn{2}{V_z}{\sigma} + \frac{1}{\rho^2} \pdvn{2}{V_z}{\tau} + \pdvn{2}{V_z}{z} \bigg)
+        \end{aligned}
     }
 \end{aligned}$$
 
 $$\begin{aligned}
     \boxed{
-        \dd{\vu{\ell}}
-        = \vu{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\sigma}
-        + \vu{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\tau}
-        + \vu{e}_z \dd{z}
+        \begin{aligned}
+            \nabla \cdot \overline{\overline{\mathbf{T}}}
+            &= \vu{e}_\sigma \bigg( \frac{1}{\rho} \pdv{T_{\sigma \sigma}}{\sigma} + \frac{1}{\rho} \pdv{T_{\tau \sigma}}{\tau} + \pdv{T_{z \sigma}}{z}
+            + \frac{\sigma T_{\sigma \sigma}}{\rho^3} + \frac{\tau T_{\sigma \tau}}{\rho^3}
+            + \frac{\tau T_{\tau \sigma}}{\rho^3} - \frac{\sigma T_{\tau \tau}}{\rho^3} \bigg)
+            \\
+            &+ \vu{e}_\tau \bigg( \frac{1}{\rho} \pdv{T_{\sigma \tau}}{\sigma} + \frac{1}{\rho} \pdv{T_{\tau \tau}}{\tau} + \pdv{T_{k \tau}}{z}
+            - \frac{\tau T_{\sigma \sigma}}{\rho^3} + \frac{\sigma T_{\sigma \tau}}{\rho^3}
+            + \frac{\sigma T_{\tau \sigma}}{\rho^3} + \frac{\tau T_{\tau \tau}}{\rho^3} \bigg)
+            \\
+            &+ \vu{e}_z \bigg( \frac{1}{\rho} \pdv{T_{\sigma z}}{\sigma} + \frac{1}{\rho} \pdv{T_{\tau z}}{\tau} + \pdv{T_{zz}}{z}
+            + \frac{\sigma T_{\sigma z}}{\rho^3} + \frac{\tau T_{\tau z}}{\rho^3} \bigg)
+        \end{aligned}
     }
 \end{aligned}$$
 
 
+
 ## References
 1.  M.L. Boas,
     *Mathematical methods in the physical sciences*, 2nd edition,