Change of Variables and the Jacobian

Prerequisites

• Derivatives, Differentials, and the Chain Rule — total differentials, partial derivatives
• The Implicit Function Theorem — when a coordinate change is locally invertible

The problem

You want to convert ${\iint }_{D}f(x,y)dxdy$ to polar coordinates. You know $x=r\cos \theta$ and $y=r\sin \theta$. The total differential of $x$ is:

$dx=\cos \theta dr-r\sin \theta d\theta$

Can you substitute this directly into $dxdy$? No — and understanding why not is the point of this note.

Why the naive substitution fails

The symbol $dxdy$ in a double integral is an area element — an infinitesimal piece of area in the plane. It is not the product of two independent length changes. If you multiply $dx$ and $dy$ as if they were numbers, you get cross-terms ($drd\theta$ and $d\theta dr$) with no rule for combining them.

The correct substitution is:

$dxdy=|\det J|drd\theta$

where $J$ is the Jacobian matrix. For polar coordinates $|\det J|=r$, giving the familiar $dxdy=rdrd\theta$. The rest of this note explains where that determinant comes from.

The geometric argument: a rectangle becomes a parallelogram

Take a small rectangle $[r,r+dr]\times [\theta ,\theta +d\theta ]$ in parameter space. What is the area of its image in the $(x,y)$-plane?

For polar coordinates, you can see it directly: the image is a thin wedge with sides $dr$ (radial) and $rd\theta$ (arc length). Area $\approx rdrd\theta$.

For a general transformation $T:(u,v)\mapsto (x(u,v),y(u,v))$, the image of a small rectangle is approximately a parallelogram spanned by two vectors:

$\mathbf{a}=\left(\frac{\partial x}{\partial u},\frac{\partial y}{\partial u}\right)du\mathbf{b}=\left(\frac{\partial x}{\partial v},\frac{\partial y}{\partial v}\right)dv$

These are the images of the two edges of the rectangle under the linear approximation of $T$. The area of a parallelogram in 2D is the absolute value of the cross product of its sides:

$\text{Area}=|a_1{b}_2-a_2{b}_1|dudv=\left|\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right|dudv$

That expression is $|\det J|dudv$, where:

$J=\left(\begin{array}{cc}\partial x/\partial u& \partial x/\partial v\\ \partial y/\partial u& \partial y/\partial v\end{array}\right)$

Verification for polar:

$J=\left(\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right),\det J=r{\cos }^2\theta +r{\sin }^2\theta =r✓$

The change of variables theorem

In words: To integrate over a region $D$ in the $(x,y)$-plane using new coordinates $(u,v)$, you substitute the transformation into $f$, and multiply by the absolute value of the Jacobian determinant — the local area scaling factor. This works as long as the transformation doesn’t collapse any area to zero ($\det J\ne 0$) and doesn’t map two different parameter points to the same output (injective).

Hypothesis: Let $T:G\to D$ be a $C^1$ map, injective on the interior of $G$, with $\det J_T\ne 0$ on $G$.

Thesis: ${\iint }_{D}f(x,y)dxdy={\iint }_{G}f(T(u,v))|\det J_T(u,v)|dudv$

The hypotheses ensure:

• $\det J\ne 0$: the map is locally invertible (Inverse Function Theorem). The parallelogram never collapses — areas scale but don’t vanish.
• Injective: different $(u,v)$ points map to different $(x,y)$ points, so area isn’t double-counted.

When $\det J=0$: the polar origin

At $r=0$, the Jacobian determinant vanishes. Geometrically: every value of $\theta$ maps to the same point $(0,0)$ — the map is not injective, and the “parallelogram” collapses to a point.

In practice this is harmless: the origin is a single point (measure zero) and doesn’t affect the integral. The theorem allows failure of injectivity on a set of measure zero. But it explains why polar coordinates are technically valid only for $r>0$.

The 3D generalization

The same logic extends: a small box $dudvdw$ maps to a parallelepiped whose volume is $|\det J_{3\times 3}|$.

Cylindrical $(r,\theta ,z)$: $\det J=r$, so $dxdydz=rdrd\theta dz$.

Spherical $(\rho ,\varphi ,\theta )$ where $\varphi$ is the polar angle from the $z$-axis:

$\det J={\rho }^2\sin \varphi ,dxdydz={\rho }^2\sin \varphi d\rho d\varphi d\theta$

Geometric reading: a small box in spherical has sides $d\rho$, $\rho d\varphi$, and $\rho \sin \varphi d\theta$. The last factor is the radius of the azimuthal circle at colatitude $\varphi$.

The wedge product: making the naive substitution work

There is a way to multiply differentials and get the Jacobian automatically — the wedge product $\wedge$, which satisfies $dr\wedge dr=0$ and $d\theta \wedge dr=-dr\wedge d\theta$ (antisymmetric).

Compute $dx\wedge dy$:

$dx\wedge dy=(\cos \theta dr-r\sin \theta d\theta )\wedge (\sin \theta dr+r\cos \theta d\theta )$

Expand — all $dr\wedge dr$ and $d\theta \wedge d\theta$ terms vanish:

$=r{\cos }^2\theta (dr\wedge d\theta )+r{\sin }^2\theta (dr\wedge d\theta )=rdr\wedge d\theta$

The determinant falls out of the antisymmetry algebra. This is not a coincidence: the determinant is the wedge product. The change of variables formula says the area 2-form $dx\wedge dy$ transforms by the determinant.

Questions to sit with

1. For a linear map $T(u,v)=(2u+v,u+3v)$, the Jacobian is constant ($\det J=5$). Every rectangle scales by the same factor. What shape is the image of the unit square?

2. In spherical coordinates, $\det J={\rho }^2\sin \varphi$ vanishes on the $z$-axis ($\varphi =0$ or $\pi$) and at the origin ($\rho =0$). What is the geometric failure at each?

3. Why does ordinary multiplication of $dx$ and $dy$ not work, but the wedge product does? What algebraic property of $\wedge$ kills exactly the terms that would give wrong answers?

See also

• The Gradient in Curvilinear Coordinates — scale factors appear there too
• The Implicit Function Theorem — the $\det J\ne 0$ condition
• 1-Forms and the Dual Basis — the wedge product formalism