This note unpacks how tiny changes transform across coordinates using the Jacobian matrix--from displacement to area to volume, in Cartesian, cylindrical, and spherical systems.

Sources:

1. Fawwaz Ulaby. (2020). Chapter 3: Vector Analysis. Fundamentals of applied electromagnetics (8rd ed., pp. 130–163). Pearson.

Local transformations and Differential Elements

Notation

Symbol	Type	Explanation
$x,y,z$	$\in \mathbb{R}$	Cartesian coordinates
$u,v,w$	$\in \mathbb{R}$	General curvilinear coordinates
$r,\varphi ,z$	$\in \mathbb{R}$	Cylindrical coordinates
$R,\theta ,\varphi$	$\in \mathbb{R}$	Spherical coordinates
$\mathbf{r}$	$\in {\mathbb{R}}^3$	Position vector
$\hat{\mathbf{x}},\hat{\mathbf{y}},\hat{\mathbf{z}}$	$\in {\mathbb{R}}^3$	Cartesian unit basis vectors
$\hat{\mathbf{r}},\hat{\mathit{\varphi }},\hat{\mathbf{z}}$	$\in {\mathbb{R}}^3$	Cylindrical unit basis vectors
$\hat{\mathbf{R}},\hat{\mathit{\theta }},\hat{\mathit{\varphi }}$	$\in {\mathbb{R}}^3$	Spherical unit basis vectors
${\mathbf{e}}_u,{\mathbf{e}}_v,{\mathbf{e}}_w$	$\in {\mathbb{R}}^3$	Base vectors in the transformed coordinate system
$d\mathbf{l}$	$\in {\mathbb{R}}^3$	Differential displacement vector
$d{\mathbf{l}}_u,d{\mathbf{l}}_v,d{\mathbf{l}}_w$	$\in {\mathbb{R}}^3$	Differential displacement vectors along each coordinate
$d\mathbf{s}$	$\in {\mathbb{R}}^3$	Differential area vector
$d{\mathbf{s}}_u,d{\mathbf{s}}_v,d{\mathbf{s}}_w$	$\in {\mathbb{R}}^3$	Differential area vectors along each coordinate
$dV$	$\in \mathbb{R}$	Differential volume element
$J$	$\in {\mathbb{R}}^{3\times 3}$	Jacobian matrix of the coordinate transformation
$\det J$	$\in \mathbb{R}$	Determinant of the Jacobian matrix, used for volume scaling
${\mathbf{e}}_u\cdot ({\mathbf{e}}_v\times {\mathbf{e}}_w)$	$\in \mathbb{R}$	Scalar triple product of base vectors, equal to $\det J$
$\mathrm{\nabla }$	Operator	Gradient operator

Coordinate transformations and local linearization

Consider a general coordinate transformation:

$$x=f_1(u,v,w),y=f_2(u,v,w),z=f_3(u,v,w)$$

For function $f_1(u,v,w)$, the total increment is:

$$\mathrm{\Delta }x\triangleq f_1(u+du,v+dv,w+dw)-f_1(u,v,w)$$

Expanding using the Taylor series:

$$\mathrm{\Delta }x=\frac{\partial x}{\partial u}du+\frac{\partial x}{\partial v}dv+\frac{\partial x}{\partial w}dw+O(d{u}^2,d{v}^2,d{w}^2)$$

Here, $\partial x$ means $\partial f_1$.

Defining the total differential as only the linear terms: $$dx\triangleq \mathrm{\nabla }x\cdot d\mathbf{u}$$

where:

$$\mathrm{\nabla }x=(\frac{\partial x}{\partial u},\frac{\partial x}{\partial v},\frac{\partial x}{\partial w}),d\mathbf{u}=\left[\begin{array}{l}du\\ dv\\ dw\end{array}\right]$$

yields:

$$\mathrm{\Delta }x=dx+\frac{\partial x}{\partial w}dw+O(d{u}^2,d{v}^2,d{w}^2)$$

When $du,dv,dw$ are small, higher-order terms can be neglected, leading to $$\mathrm{\Delta }x\approx dx.$$

Therefore, $dx$ is a linear approximation of $\mathrm{\Delta }x$, and we shall use $dx$ to perform integration.

Since the same linear approximation applies to $x,y,z$, we can similarly approximate $\mathrm{\Delta }y$ and $\mathrm{\Delta }z$ using $dy$ and $dz$. Thus, we extend this to all three coordinates:

$$\left[\begin{array}{l}dx\\ dy\\ dz\end{array}\right]=\left[\begin{array}{l}\mathrm{\nabla }x\\ \mathrm{\nabla }y\\ \mathrm{\nabla }z\end{array}\right]d\mathbf{u}$$

Jacobian matrix

Writing in matrix form:

$$\left[\begin{array}{l}dx\\ dy\\ dz\end{array}\right]=J\left[\begin{array}{l}du\\ dv\\ dw\end{array}\right]$$

where $$J=\left[\begin{array}{l}\mathrm{\nabla }x\\ \mathrm{\nabla }y\\ \mathrm{\nabla }z\end{array}\right]=\left[\begin{array}{lll}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}& \frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}& \frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u}& \frac{\partial z}{\partial v}& \frac{\partial z}{\partial w}\end{array}\right].$$ $J$ is often referred to as the Jacobian matrix.

Transforming the differential displacement vector

Consider a position vector $\mathbf{r}$ in Cartesian coordinates, expressed as:

$$\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}.$$

A small displacement in space is described by the differential displacement vector:

$$d\mathbf{l}=\hat{\mathbf{x}}dx+\hat{\mathbf{y}}dy+\hat{\mathbf{z}}dz$$

If coordinates $x,y,z$ are functions of a new coordinate system $(u,v,w)$, we express this transformation as:

$$\left[\begin{array}{l}dx\\ dy\\ dz\end{array}\right]=J\left[\begin{array}{l}du\\ dv\\ dw\end{array}\right]$$

Substituting this into the expression for $d\mathbf{l}$, we obtain:

$$d\mathbf{l}=\left[\begin{array}{lll}\hat{\mathbf{x}}& \hat{\mathbf{y}}& \hat{\mathbf{z}}\end{array}\right]J\left[\begin{array}{l}du\\ dv\\ dw\end{array}\right].$$

Note that

$$\left[\begin{array}{lll}\hat{\mathbf{x}}& \hat{\mathbf{y}}& \hat{\mathbf{z}}\end{array}\right]J=\left[\begin{array}{lll}\frac{\partial \mathbf{r}}{\partial u}& \frac{\partial \mathbf{r}}{\partial v}& \frac{\partial \mathbf{r}}{\partial w}\end{array}\right].$$ Thus,

$$ d =

= $$

where the base vectors (not necessarily unit) of the new coordinate system are given by:

$${\mathbf{e}}_i=\frac{\partial \mathbf{r}}{\partial u^i}$$ where $u^i=u,v,w$.

This shows that the Jacobian not only transforms differentials but also provides the (coordinates of the) base vectors (not necessarily unit) in the transformed coordinate system.

To cylindrical coordinates $(r,\varphi ,z)$

Transforming to cylindrical coordinates, $$x=r\cos \varphi ,y=r\sin \varphi ,z=z$$

The position vector becomes

$$\mathbf{r}=r\cos \varphi \hat{\mathbf{x}}+r\sin \varphi \hat{\mathbf{y}}+z\hat{\mathbf{z}}$$

The Jacobian matrix is

$$J=\left[\begin{array}{ccc}\cos \varphi & -r\sin \varphi & 0\\ \sin \varphi & r\cos \varphi & 0\\ 0& 0& 1\end{array}\right]$$

The base vectors are the columns of $J$ :

$$\begin{array}{c}{\mathbf{e}}_r=\cos \varphi \hat{\mathbf{x}}+\sin \varphi \hat{\mathbf{y}}\\ {\mathbf{e}}_{\varphi }=-r\sin \varphi \hat{\mathbf{x}}+r\cos \varphi \hat{\mathbf{y}}\\ {\mathbf{e}}_z=\hat{\mathbf{z}}\end{array}$$ Normalize them, obtain base unit vectors:

$$\begin{array}{c}\hat{\mathbf{r}}=\cos \varphi \hat{\mathbf{x}}+\sin \varphi \hat{\mathbf{y}}\\ \hat{\mathit{\varphi }}=-\sin \varphi \hat{\mathbf{x}}+\cos \varphi \hat{\mathbf{y}},\\ \hat{\mathbf{z}}={\mathbf{e}}_z.\end{array}$$

Since ${\mathbf{e}}_{\varphi }$ has a magnitude of $r$, we write

$${\mathbf{e}}_{\varphi }=r\hat{\mathit{\varphi }}$$

Thus,

$${\mathbf{e}}_r=\hat{\mathbf{r}},{\mathbf{e}}_{\varphi }=r\hat{\mathit{\varphi }},{\mathbf{e}}_z=\hat{\mathbf{z}}$$ Using the transformation formula,

$$d\mathbf{l}=\left[\begin{array}{lll}{\mathbf{e}}_r& {\mathbf{e}}_{\varphi }& {\mathbf{e}}_z\end{array}\right]\left[\begin{array}{l}dr\\ d\varphi \\ dz\end{array}\right]$$

Substituting the base vectors,

$$d\mathbf{l}=\left[\begin{array}{lll}\hat{\mathbf{r}}& r\hat{\mathit{\varphi }}& \hat{\mathbf{z}}\end{array}\right]\left[\begin{array}{l}dr\\ d\varphi \\ dz\end{array}\right].$$

or explicitly,

$$d\mathbf{l}=\hat{\mathbf{r}}dr+r\hat{\mathit{\varphi }}d\varphi +\hat{\mathbf{z}}dz$$

To spherical coordinates $(R,\theta ,\varphi )$

Transforming to spherical coordinates,

$$x=R\sin \theta \cos \varphi ,y=R\sin \theta \sin \varphi ,z=R\cos \theta$$

The position vector is

$$\mathbf{r}=R\sin \theta \cos \varphi \hat{\mathbf{x}}+R\sin \theta \sin \varphi \hat{\mathbf{y}}+R\cos \theta \hat{\mathbf{z}}$$

The Jacobian matrix is

$$J=\left[\begin{array}{ccc}\sin \theta \cos \varphi & R\cos \theta \cos \varphi & -R\sin \theta \sin \varphi \\ \sin \theta \sin \varphi & R\cos \theta \sin \varphi & R\sin \theta \cos \varphi \\ \cos \theta & -R\sin \theta & 0\end{array}\right].$$

The coordinates of the base vectors are the columns of $J$:

$$\begin{array}{c}{\mathbf{e}}_R=\sin \theta \cos \varphi \hat{\mathbf{x}}+\sin \theta \sin \varphi \hat{\mathbf{y}}+\cos \theta \hat{\mathbf{z}}\\ {\mathbf{e}}_{\theta }=R\cos \theta \cos \varphi \hat{\mathbf{x}}+R\cos \theta \sin \varphi \hat{\mathbf{y}}-R\sin \theta \hat{\mathbf{z}}\\ {\mathbf{e}}_{\varphi }=-R\sin \theta \sin \varphi \hat{\mathbf{x}}+R\sin \theta \cos \varphi \hat{\mathbf{y}}\end{array}$$

To obtain base unit vectors, we normalize: $$\begin{array}{c}\hat{\mathbf{R}}=\sin \theta \cos \varphi \hat{\mathbf{x}}+\sin \theta \sin \varphi \hat{\mathbf{y}}+\cos \theta \hat{\mathbf{z}}\\ \hat{\mathit{\theta }}=\cos \theta \cos \varphi \hat{\mathbf{x}}+\cos \theta \sin \varphi \hat{\mathbf{y}}-\sin \theta \hat{\mathbf{z}}\\ \hat{\mathit{\varphi }}=-\sin \varphi \hat{\mathbf{x}}+\cos \varphi \hat{\mathbf{y}}\end{array}$$

Since ${\mathbf{e}}_{\theta }$ has a magnitude of $R$ and ${\mathbf{e}}_{\varphi }$ has a magnitude of $R\sin \theta$, we write

$${\mathbf{e}}_{\theta }=R\hat{\mathit{\theta }},{\mathbf{e}}_{\varphi }=R\sin \theta \hat{\mathit{\varphi }}$$

Thus,

$${\mathbf{e}}_R=\hat{\mathbf{R}},{\mathbf{e}}_{\theta }=R\hat{\mathit{\theta }},{\mathbf{e}}_{\varphi }=R\sin \theta \hat{\mathit{\varphi }}$$

Using the transformation formula,

$$d\mathbf{l}=\left[\begin{array}{lll}\hat{\mathbf{R}}& R\hat{\mathit{\theta }}& R\sin \theta \hat{\mathit{\varphi }}\end{array}\right]\left[\begin{array}{c}dR\\ d\theta \\ d\varphi \end{array}\right]$$

or explicitly,

$$d\mathbf{l}=\hat{\mathbf{R}}dR+R\hat{\mathit{\theta }}d\theta +R\sin \theta \hat{\varphi }d\varphi$$

Transforming the differential area vector

The differential area vector $d\mathbf{s}$ is a vector perpendicular to an infinitesimal surface element with magnitude equal to its area. For a surface parameterized by coordinates $(u,v)$, the differential displacement vectors along these coordinate directions are:

$$d{\mathbf{l}}_u={\mathbf{e}}_{u}du,d{\mathbf{l}}_v={\mathbf{e}}_{v}dv$$

The differential area vector is then given by their cross product:

$$d\mathbf{s}=d{\mathbf{l}}_u\times d{\mathbf{l}}_v=\left({\mathbf{e}}_{u}du\right)\times \left({\mathbf{e}}_{v}dv\right)$$

Its magnitude represents the infinitesimal surface area:

$$ds=|{\mathbf{e}}_u\times {\mathbf{e}}_v|dudv$$

where $|\cdot |$ means vector magnitude.

In Cartesian coordinates, the differential displacement vectors along each coordinate direction are: $$d{\mathbf{l}}_x=\hat{\mathbf{x}}dx,d{\mathbf{l}}_y=\hat{\mathbf{y}}dy,d{\mathbf{l}}_z=\hat{\mathbf{z}}dz$$

Thus, the differential area vectors are:

$$\begin{array}{rl}& d{\mathbf{s}}_x=d{\mathbf{l}}_y\times d{\mathbf{l}}_z=\hat{\mathbf{x}}dydz,\\ & d{\mathbf{s}}_y=d{\mathbf{l}}_z\times d{\mathbf{l}}_x=\hat{\mathbf{y}}dxdz,\\ & d{\mathbf{s}}_z=d{\mathbf{l}}_x\times d{\mathbf{l}}_y=\hat{\mathbf{z}}dxdy.\end{array}$$

To cylindrical coordinates $(r,\varphi ,z)$

The differential displacement vectors in cylindrical coordinates are:

$$d{\mathbf{l}}_r=\hat{\mathbf{r}}dr,d{\mathbf{l}}_{\varphi }=r\hat{\mathit{\varphi }}d\varphi ,d{\mathbf{l}}_z=\hat{\mathbf{z}}dz$$

Taking cross products to obtain the differential area vectors:

$$\begin{array}{c}d{\mathbf{s}}_r=d{\mathbf{l}}_{\varphi }\times d{\mathbf{l}}_z=(r\hat{\mathit{\varphi }}d\varphi )\times (\hat{\mathbf{z}}dz)=\hat{\mathbf{r}}rd\varphi dz\\ d{\mathbf{s}}_{\varphi }=d{\mathbf{l}}_z\times d{\mathbf{l}}_r=(\hat{\mathbf{z}}dz)\times (\hat{\mathbf{r}}dr)=\hat{\mathit{\varphi }}drdz\\ d{\mathbf{s}}_z=d{\mathbf{l}}_r\times d{\mathbf{l}}_{\varphi }=(\hat{\mathbf{r}}dr)\times (r\hat{\mathit{\varphi }}d\varphi )=\hat{\mathbf{z}}rdrd\varphi \end{array}$$

To spherical coordinates $(R,\theta ,\varphi )$

The differential displacement vectors in spherical coordinates are: $$d{\mathbf{l}}_R=\hat{\mathbf{R}}dR,d{\mathbf{l}}_{\theta }=R\hat{\theta }d\theta ,d{\mathbf{l}}_{\varphi }=R\sin \theta \hat{\varphi }d\varphi$$

Taking cross products to obtain the differential area vectors:

$$\begin{array}{rl}& d{\mathbf{s}}_R=d{\mathbf{l}}_{\theta }\times d{\mathbf{l}}_{\varphi }=(R\hat{\mathit{\theta }}d\theta )\times (R\sin \theta \hat{\mathit{\varphi }}d\varphi )=\hat{\mathbf{R}}{R}^2\sin \theta d\theta d\varphi \\ & d{\mathbf{s}}_{\theta }=d{\mathbf{l}}_{\varphi }\times d{\mathbf{l}}_R=(R\sin \theta \hat{\mathit{\varphi }}d\varphi )\times (\hat{\mathbf{R}}dR)=\hat{\mathit{\theta }}R\sin \theta dRd\varphi \\ & d{\mathbf{s}}_{\varphi }=d{\mathbf{l}}_R\times d{\mathbf{l}}_{\theta }=(\hat{\mathbf{R}}dR)\times (R\hat{\mathit{\theta }}d\theta )=\hat{\mathit{\varphi }}RdRd\theta \end{array}$$

Transforming the differential volume element

The differential volume element represents an infinitesimal volume in space. It is computed as the scalar triple product of threedifferential displacement vectors that span the volume element.

For a coordinate system $(u,v,w)$, the differential displacement vectors along each coordinate direction are:

$$d{\mathbf{l}}_u={\mathbf{e}}_{u}du,d{\mathbf{l}}_v={\mathbf{e}}_{v}dv,d{\mathbf{l}}_w={\mathbf{e}}_{w}dw$$

The differential volume element is then given by:

$$dV=|d{\mathbf{l}}_u\cdot (d{\mathbf{l}}_v\times d{\mathbf{l}}_w)|$$

where $|\cdot |$ means absolute value.

Expanding: $$dV=|{\mathbf{e}}_u\cdot ({\mathbf{e}}_v\times {\mathbf{e}}_w)|dudvdw=|\det J|dudvdw$$

The later equation holds because the definition of cross product.

In Cartesian coordinates, the position vector is:

$$\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}+z\hat{\mathbf{z}}$$

The Jacobian matrix is:

$$J=\left[\begin{array}{lll}\frac{\partial x}{\partial x}& \frac{\partial x}{\partial y}& \frac{\partial x}{\partial z}\\ \frac{\partial y}{\partial x}& \frac{\partial y}{\partial y}& \frac{\partial y}{\partial z}\\ \frac{\partial z}{\partial x}& \frac{\partial z}{\partial y}& \frac{\partial z}{\partial z}\end{array}\right]=\left[\begin{array}{lll}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$$

Since $\det J=1$, the differential volume element is:

$$dV=|\det J|dxdydz=dxdydz$$

To cylindrical coordinates $(r,\varphi ,z)$

From the differential displacement vectors: $$d{\mathbf{l}}_r=\hat{\mathbf{r}}dr,d{\mathbf{l}}_{\varphi }=r\hat{\mathit{\varphi }}d\varphi ,d{\mathbf{l}}_z=\hat{\mathbf{z}}dz$$

The differential volume element is given by: $$dV=|d{\mathbf{l}}_r\cdot (d{\mathbf{l}}_{\varphi }\times d{\mathbf{l}}_z)|$$

Since $d{\mathbf{l}}_{\varphi }\times d{\mathbf{l}}_z$ gives a differential area element $d{\mathbf{s}}_r=\hat{\mathbf{r}}rd\varphi dz$, we compute:

$$dV=|dr\cdot (rd\varphi dz)|=rdrd\varphi dz$$ as $r$ is non-negative by definition.

To spherical coordinates $(R,\theta ,\varphi )$

From the differential displacement vectors:

$$d{\mathbf{l}}_R=\hat{\mathbf{R}}dR,d{\mathbf{l}}_{\theta }=R\hat{\mathit{\theta }}d\theta ,d{\mathbf{l}}_{\varphi }=R\sin \theta \hat{\mathit{\varphi }}d\varphi$$

The differential volume element is given by:

$$dV=|d{\mathbf{l}}_R\cdot (d{\mathbf{l}}_{\theta }\times d{\mathbf{l}}_{\varphi })|$$

Since $d{\mathbf{l}}_{\theta }\times d{\mathbf{l}}_{\varphi }$ gives a differential area element $d{\mathbf{s}}_R=\hat{\mathbf{R}}{R}^2\sin \theta d\theta d\varphi$, we compute:

$$dV=|dR\cdot (R^2\sin \theta d\theta d\varphi )|=R^2\sin \theta dRd\theta d\varphi$$