Chapter 33 — Change of Variables and Jacobians

33.1 The Problem with Hard Regions

You can now set up and evaluate double and triple integrals. But Chapter 32 left you with a quiet frustration: an integral is only as easy as its region and its integrand allow. A circle described in Cartesian coordinates forces you into square roots in your limits. A parallelogram tilted off the axes turns a clean region into a thicket of inequalities. A bell-shaped integrand $e^{-(x^2+y^2)}$ has no elementary antiderivative in $x$ alone, yet collapses to something trivial the moment you switch to polar coordinates.

The fix in single-variable calculus was substitution: when an integral was awkward in $x$, you changed to a new variable $u$ in which it became easy. This chapter is the multivariable version of that idea — and the central new object, the one that makes it all work, is the Jacobian determinant. It is the single most important computation in multivariable integration, and by the end of this chapter you will understand not just how to compute it, but why the area element $r\,dr\,d\theta$ from Chapter 32 carries that factor of $r$.

This is the last chapter of Part VI. It ties together everything from Chapter 29 (functions of several variables) through Chapter 32 (multiple integrals), and it hands the baton to vector calculus in Part VII, where the Jacobian reappears in every coordinate change for line and surface integrals.

The Key Insight. A smooth transformation, viewed up close, looks linear — it acts like its matrix of partial derivatives, the Jacobian matrix. The determinant of that matrix tells you the one thing integration cares about: how much the transformation stretches or shrinks area (in 2D) or volume (in 3D). Change of variables is nothing more than bookkeeping for that local stretching.

33.2 The Single-Variable Recall

Before we climb a dimension, let us be precise about the rule we are generalizing. In single-variable calculus, the substitution rule reads

$$\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du, \qquad u = g(x).$$

The mechanical heart of this is the differential relation

$$du = g'(x)\,dx.$$

Read this carefully. When you change variables from $x$ to $u = g(x)$, a small length $dx$ on the $x$-axis maps to a small length $du = g'(x)\,dx$ on the $u$-axis. The factor $g'(x)$ is the local length-scaling factor of the substitution: near $x$, the map $g$ stretches lengths by $g'(x)$.

When you run the substitution backward — expressing the integral in terms of $u$ — you must account for that stretching. If we write $x = h(u)$ for the inverse map, then $dx = h'(u)\,du$, and $h'(u) = 1/g'(x)$. To keep the value of the integral unchanged, the stretching factor rides along inside the integral.

Geometric Intuition. Imagine the $x$-axis as a stretchy rubber band and $g$ as the act of stretching it onto the $u$-axis. Where $g'(x)$ is large, a tiny interval $dx$ blows up into a large interval $du$; where $g'(x)$ is small, it shrinks. The integral $\int f\,dx$ is a sum of (height $\times$ width) strips. When you switch axes, every width gets rescaled by the local stretch factor, so the factor must appear explicitly to preserve the total. The Jacobian is exactly this stretch factor, promoted to two or three dimensions — where "stretch" means area or volume, not length.

In one dimension, "how much does a small piece scale?" has a one-number answer: $|g'(x)|$. The deep idea of this chapter is that in higher dimensions the answer is still one number — the absolute value of a determinant — because a small square has a single area and a small cube has a single volume, no matter how the transformation tilts and shears them.

Common Pitfall. Some sources sloppily say "the differential transforms as $dx \to du / g'(x)$." That is backward bookkeeping and a frequent source of sign and reciprocal errors. The honest statement is $du = g'(x)\,dx$. When you substitute into an integral written in $u$, you replace $du$ by $g'(x)\,dx$; when you go the other way you replace $dx$ by $du/g'(x)$. Keep track of which variable is "old" and which is "new," and always ask: does the area element belong to the variables I am actually integrating over? The Jacobian factor always multiplies the new differentials.

33.3 The Jacobian Matrix and Determinant

Consider a transformation $T$ that takes a point $(u, v)$ in one plane to a point $(x, y)$ in another:

$$T: (u, v) \mapsto (x, y), \qquad x = x(u, v), \quad y = y(u, v).$$

Think of $(u, v)$ as the "new," convenient coordinates and $(x, y)$ as the "old" coordinates in which the integral was originally posed. The Jacobian matrix of $T$ is the matrix of all four first-order partial derivatives:

$$J_T = \frac{\partial(x, y)}{\partial(u, v)} = \begin{pmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[2ex] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{pmatrix}.$$

This matrix is the best linear approximation to $T$ near a point — it is the multivariable derivative you met in Chapter 30. The notation $\partial(x,y)/\partial(u,v)$ is standard shorthand: the top variables $(x,y)$ are the outputs, the bottom $(u,v)$ the inputs.

The Jacobian determinant (often just "the Jacobian") is its determinant:

$$\det J_T = \frac{\partial x}{\partial u}\,\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\,\frac{\partial y}{\partial u}.$$

This single number measures local area scaling. We will prove this carefully in §33.6, but here is the claim in one sentence: a small rectangle of area $du\,dv$ in the $(u,v)$-plane maps under $T$ to a region of area approximately $|\det J_T|\,du\,dv$ in the $(x,y)$-plane. The absolute value is essential — area cannot be negative, but a determinant can. The sign of $\det J_T$ records orientation (whether $T$ flips the plane like a mirror), which matters for differential forms but not for the size of an area.

The Polar Transformation, Computed

Take the polar transformation $T: (r, \theta) \mapsto (x, y)$ given by $x = r\cos\theta$, $y = r\sin\theta$. Its partial derivatives are

$$\frac{\partial x}{\partial r} = \cos\theta, \quad \frac{\partial x}{\partial \theta} = -r\sin\theta, \qquad \frac{\partial y}{\partial r} = \sin\theta, \quad \frac{\partial y}{\partial \theta} = r\cos\theta.$$

So

$$\det J_T = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = \cos\theta\,(r\cos\theta) - (-r\sin\theta)\,\sin\theta = r\cos^2\theta + r\sin^2\theta = r.$$

There it is: $\det J_T = r$. That is precisely the factor in the area element $dA = r\,dr\,d\theta$ that Chapter 32 asserted and used. We have now derived it. The factor of $r$ is not a memorized rule; it is the local area-scaling factor of the polar map, and it grows with $r$ because a wedge of fixed angular width $d\theta$ spans more arc length, and hence more area, the farther out you go.

Check Your Understanding. A transformation has Jacobian matrix $\begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$ at every point. What happens to the area of a small square under this transformation, and what happens to the integral $\iint_R \,dA$ when you change to these variables?

Answer

$\det J = 3 \cdot 2 - 0 \cdot 0 = 6$, so every small square has its area multiplied by $6$ — the map stretches by $3$ horizontally and $2$ vertically. In the change-of-variables formula the area element becomes $dA = 6\,du\,dv$, so the integral picks up a constant factor of $6$. This is the transformation $x = 3u$, $y = 2v$, which sends the unit square to a $3 \times 2$ rectangle of area $6$.

33.4 The Change of Variables Formula (2D)

We can now state the central theorem of the chapter.

Change of Variables Theorem (double integrals). Let $T: (u,v) \mapsto (x,y)$ be a one-to-one, continuously differentiable transformation mapping a region $S$ in the $(u,v)$-plane onto a region $R$ in the $(x,y)$-plane, with $\det J_T \neq 0$ on the interior of $S$. Then for any continuous $f$, $$\iint_R f(x, y)\,dA = \iint_S f\big(x(u, v),\, y(u, v)\big)\,\big|\det J_T\big|\,du\,dv.$$

Compare this, term for term, with single-variable substitution:

The structural parallel is exact. The factor $|\det J_T|$ is the two-dimensional successor of $g'(x)$, and the absolute value is the only genuinely new wrinkle — it appears because in 2D the "direction" of the scaling lives in the sign of the determinant, which area discards.

The procedure. To use the theorem:

1. Identify or design a useful transformation $T$ — one that makes either the region or the integrand simpler.
2. Compute $\det J_T$ from the partial derivatives.
3. Translate the region: figure out what region $S$ in $(u,v)$-space maps onto $R$.
4. Substitute: replace $f(x,y)$ by $f(x(u,v),y(u,v))$, replace $dA$ by $|\det J_T|\,du\,dv$, and integrate over $S$.

Worked Example 1: A Tilted Parallelogram Becomes a Rectangle

Let $R$ be the parallelogram bounded by the four lines $x + y = 1$, $x + y = 3$, $x - y = 0$, and $x - y = 2$. In Cartesian coordinates the limits are miserable. But the boundaries themselves suggest the substitution: let

$$u = x + y, \qquad v = x - y.$$

The region in $(u,v)$ is then simply $1 \le u \le 3$, $0 \le v \le 2$ — a rectangle. To apply the formula we need $x$ and $y$ as functions of $u$ and $v$: solving the linear system gives $x = (u+v)/2$, $y = (u-v)/2$. The Jacobian matrix is

$$J_T = \begin{pmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{pmatrix} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}, \qquad \det J_T = \left(\tfrac12\right)\left(-\tfrac12\right) - \left(\tfrac12\right)\left(\tfrac12\right) = -\tfrac14 - \tfrac14 = -\tfrac12.$$

So $|\det J_T| = \tfrac12$, and any integral over $R$ becomes

$$\iint_R f(x,y)\,dA = \int_1^3 \int_0^2 f\!\left(\tfrac{u+v}{2}, \tfrac{u-v}{2}\right)\cdot \tfrac12 \,dv\,du.$$

The nightmare parallelogram has become a friendly rectangle, at the cost of a constant factor $\tfrac12$. As a sanity check, set $f \equiv 1$ so the integral is just area. The right side gives $\int_1^3\int_0^2 \tfrac12\,dv\,du = \tfrac12 \cdot 2 \cdot 2 = 2$, so the parallelogram $R$ has area $2$. The relation $|R| = |\det J_T|\cdot|S|$ holds exactly: the $(u,v)$-rectangle $S$ has area $4$, and $\tfrac12 \cdot 4 = 2$. The Jacobian converts the area of $S$ into the area of $R$, just as it should.

Common Pitfall. Notice the subtlety the parenthetical above stumbled into, and learn from it: the direction of your transformation determines which determinant you use. If you define $T:(u,v)\to(x,y)$, then $|\det J_T|$ is the area-scaling factor that multiplies $du\,dv$. If instead you wrote down the inverse map $(x,y)\to(u,v)$ — which is what $u = x+y$, $v=x-y$ literally is — its Jacobian determinant is the reciprocal, $\det \partial(u,v)/\partial(x,y) = -2$, and you would need $1/|-2| = \tfrac12$. Both routes give $\tfrac12$, but only if you invert correctly. When in doubt, always solve for $x,y$ in terms of $u,v$ and differentiate those.

Worked Example 2: An Elliptical Region

Compute $\displaystyle\iint_E \left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) dA$ over the elliptical disk $E:\ \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$.

The ellipse is awkward, but the substitution $x = au$, $y = bv$ turns it into the unit disk $u^2 + v^2 \le 1$. The Jacobian:

$$J_T = \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}, \qquad \det J_T = ab.$$

The integrand $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ becomes simply $u^2 + v^2$. So

$$\iint_E \left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) dA = \iint_{u^2+v^2 \le 1} (u^2 + v^2)\,(ab)\,du\,dv.$$

Now apply a second change of variables — polar in the $(u,v)$ disk, where $u^2 + v^2 = \rho^2$ and the area element is $\rho\,d\rho\,d\theta$ (the polar Jacobian we just derived, here in dummy variables):

$$= ab \int_0^{2\pi}\!\!\int_0^1 \rho^2 \cdot \rho \,d\rho\,d\theta = ab \cdot 2\pi \cdot \frac{1}{4} = \frac{\pi a b}{2}.$$

This example shows two ideas at once: change of variables can be chained (linear stretch, then polar), and a good transformation simultaneously fixes the region and the integrand.

Worked Example 3: Designing a Transformation from the Curves

Let $R$ be the region between $y = x^2$ and $y = 2x^2$, with $1 \le x \le 2$, and compute $\iint_R x\,dA$. The curves $y = cx^2$ for $1 \le c \le 2$ foliate the region, which suggests measuring "which parabola" we are on. Set

$$u = x, \qquad v = \frac{y}{x^2}.$$

Then $v$ ranges over $[1,2]$ (it labels the parabola) and $u$ over $[1,2]$. Solving back, $x = u$, $y = u^2 v$. The Jacobian:

$$J_T = \begin{pmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2uv & u^2 \end{pmatrix}, \qquad \det J_T = (1)(u^2) - (0)(2uv) = u^2.$$

The integrand $x = u$, so

$$\iint_R x\,dA = \int_1^2\!\!\int_1^2 u \cdot u^2 \,dv\,du = \int_1^2 u^3 \,du \int_1^2 dv = \frac{16 - 1}{4}\cdot 1 = \frac{15}{4}.$$

The general skill on display: when the region is bounded by a family of curves, introduce a variable that labels the family. The boundaries become coordinate lines, and the integral becomes a rectangle.

Worked Example 4: Polar Rescues an Impossible Integrand

Compute $\displaystyle\iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA$ over the entire plane. The integrand has no elementary antiderivative in $x$ (this is the Gaussian of Chapter 14's "what FTC does not promise"), so iterated Cartesian integration is a dead end. But $x^2 + y^2 = r^2$ in polar coordinates, and the Jacobian supplies a factor of $r$ that is exactly what makes the radial integral elementary:

$$\iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA = \int_0^{2\pi}\!\!\int_0^\infty e^{-r^2}\,\underbrace{r\,dr}_{\text{Jacobian}}\,d\theta.$$

The inner integral now yields to the substitution $s = r^2$, $ds = 2r\,dr$:

$$\int_0^\infty e^{-r^2}\,r\,dr = \frac{1}{2}\int_0^\infty e^{-s}\,ds = \frac{1}{2}.$$

Therefore the whole integral is $\int_0^{2\pi}\tfrac12\,d\theta = \pi$. This is the famous Gaussian integral: from it, $\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$ follows by taking a square root, because the 2D integral factors as that 1D integral squared. An integral that is impossible in one variable becomes trivial in two — the extra dimension hands you a factor of $r$ from the Jacobian, and that factor is the whole game. This is the single most important payoff of change of variables in all of probability and statistics.

33.5 A Catalog of Common 2D Transformations

A handful of transformations cover most of what you will meet. Memorize their Jacobians; derive them once so you trust them.

Linear / affine. $T(u,v) = (au + bv,\ cu + dv)$. The Jacobian matrix is the constant matrix $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$, so $\det J = ad - bc$ everywhere. A linear map scales every area by the same factor — the determinant of its matrix — which is the geometric meaning of a determinant in the first place. A rotation has $\det J = 1$ (it preserves area), and a uniform scaling by $k$ has $\det J = k^2$.

Polar. $x = r\cos\theta$, $y = r\sin\theta$, with $|\det J| = r$, giving $dA = r\,dr\,d\theta$. Use whenever circles, disks, or rotational symmetry appear, or whenever the integrand depends on $x^2 + y^2$. (You first met polar coordinates as a curve-description tool in Chapter 26; here we see why they carry a factor of $r$ in integration.)

Axis stretch. $x = au$, $y = bv$, with $|\det J| = ab$. Maps the unit disk to an ellipse with semi-axes $a$ and $b$; since the unit disk has area $\pi$, the ellipse has area $ab \cdot \pi = \pi ab$ — the familiar formula, now a one-line consequence of the Jacobian.

Geometric Intuition. Every one of these is the same story: the transformation grabs the little coordinate grid in $(u,v)$-space and redraws it in $(x,y)$-space. Straight grid lines may bend (as in polar, where horizontal lines become circles and vertical lines become rays), and the little grid cells change size. The Jacobian determinant is the area ratio of an output cell to its input cell. For polar, the cells fan out and grow with $r$ — that is the entire content of $dA = r\,dr\,d\theta$ in one picture.

33.6 Why the Jacobian Is the Area Factor

Here is the derivation that justifies everything above. It is worth doing slowly, because it reveals the Jacobian as inevitable rather than arbitrary.

Fix a point $(u_0, v_0)$ and look at a tiny rectangle in $(u,v)$-space with corner there, with sides $du$ along the $u$-direction and $dv$ along the $v$-direction. Where do its sides go under $T$?

The side along $u$ — moving from $(u_0, v_0)$ to $(u_0 + du, v_0)$ — maps to a vector that, to first order, points in the direction the image moves when only $u$ changes. That is the partial-derivative vector

$$\mathbf{T}_u = \left\langle \frac{\partial x}{\partial u},\ \frac{\partial y}{\partial u}\right\rangle du.$$

Likewise the side along $v$ maps to

$$\mathbf{T}_v = \left\langle \frac{\partial x}{\partial v},\ \frac{\partial y}{\partial v}\right\rangle dv.$$

So the image of the tiny rectangle is, to first order, the parallelogram spanned by $\mathbf{T}_u$ and $\mathbf{T}_v$. And the area of a parallelogram spanned by two vectors in the plane is the absolute value of the determinant whose columns are those vectors (the 2D cross product, Chapter 28):

$$\text{area} = \left| \det \begin{pmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{pmatrix}\right| du\,dv = |\det J_T|\,du\,dv.$$

That is it. The Jacobian determinant is the area of the image parallelogram per unit input area. Summing $f \cdot (\text{area})$ over all the little cells — a Riemann sum — and passing to the limit gives the change-of-variables formula. The whole theorem is the statement "a smooth map is locally linear, and a linear map scales area by its determinant."

The Key Insight. This is the multivariable echo of FTC's recurring slogan from Chapter 14 — that calculus is built by approximating a curved, complicated object with a simple linear one and then summing. Here the linear object is the Jacobian matrix, the simple fact is "linear maps scale area by their determinant," and the sum is the integral. Differentiation gives the local linear approximation; integration accumulates it. The two halves of calculus are doing exactly what they always do.

Math Major Sidebar — Why first order suffices. You might worry that the image of the little rectangle is only approximately a parallelogram — its sides are slightly curved and the corner does not quite close. The error in each cell is of order $(du)^2 dv$ and $du\,(dv)^2$: one order smaller than the leading $du\,dv$ term. When you sum over $\sim 1/(du\,dv)$ cells and take the limit, those higher-order errors vanish, exactly as the higher-order terms vanished in the single-variable difference quotient. Making this rigorous (uniform continuity of the partials on a compact region, a careful $\varepsilon$-estimate of the remainder) is the content of the change-of-variables theorem's proof in real analysis. The geometric argument above is honest; it is the limit that needs the care.

33.7 The 3D Jacobian and the Coordinate Systems of Chapter 32

Everything lifts to three dimensions with no new ideas — just a bigger determinant. For a transformation $(u,v,w) \mapsto (x,y,z)$,

$$J_T = \begin{pmatrix} x_u & x_v & x_w \\ y_u & y_v & y_w \\ z_u & z_v & z_w \end{pmatrix}, \qquad dV = |\det J_T|\,du\,dv\,dw.$$

Now $|\det J_T|$ is the volume-scaling factor: a small box $du\,dv\,dw$ maps to a parallelepiped whose volume is $|\det J_T|\,du\,dv\,dw$ (a 3D determinant is the volume of the parallelepiped spanned by its columns — the scalar triple product of Chapter 28). The change-of-variables formula reads

$$\iiint_R f(x,y,z)\,dV = \iiint_S f\big(x(u,v,w),\,y(u,v,w),\,z(u,v,w)\big)\,|\det J_T|\,du\,dv\,dw.$$

This is the chapter's payoff for Chapter 32: it derives the volume elements you were told to use.

Cylindrical Coordinates: $|\det J| = r$

Cylindrical coordinates are $x = r\cos\theta$, $y = r\sin\theta$, $z = z$. The Jacobian matrix is

$$J = \begin{pmatrix} \cos\theta & -r\sin\theta & 0 \\ \sin\theta & r\cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}.$$

Expanding along the bottom row (which is $0,0,1$), the determinant is $1$ times the upper-left $2\times2$ minor — exactly the polar Jacobian:

$$\det J = 1 \cdot \big(\cos\theta\cdot r\cos\theta - (-r\sin\theta)\sin\theta\big) = r.$$

So $dV = r\,dr\,d\theta\,dz$. Cylindrical coordinates are just polar in the $xy$-plane with the $z$-axis left alone, so their Jacobian is the polar Jacobian, untouched.

Spherical Coordinates: $|\det J| = \rho^2 \sin\phi$

Use the standard physics-and-math convention from Chapter 32: $\rho$ is the distance from the origin, $\phi$ the polar angle measured down from the positive $z$-axis ($0 \le \phi \le \pi$), and $\theta$ the azimuthal angle. Then

$$x = \rho\sin\phi\cos\theta, \qquad y = \rho\sin\phi\sin\theta, \qquad z = \rho\cos\phi.$$

The Jacobian matrix has columns of partials with respect to $\rho$, $\phi$, $\theta$:

$$J = \begin{pmatrix} \sin\phi\cos\theta & \rho\cos\phi\cos\theta & -\rho\sin\phi\sin\theta \\ \sin\phi\sin\theta & \rho\cos\phi\sin\theta & \rho\sin\phi\cos\theta \\ \cos\phi & -\rho\sin\phi & 0 \end{pmatrix}.$$

Expanding this determinant and simplifying (we verify it with sympy in §33.11, and you should grind through it by hand once in your life) gives the clean result

$$\det J = \rho^2 \sin\phi, \qquad \text{so} \qquad dV = \rho^2 \sin\phi\,d\rho\,d\phi\,d\theta.$$

Both factors have an intuitive reading. The $\rho^2$ says volume grows like the square of the radius — a fixed solid-angle wedge sweeps out more volume the farther out you go, in two directions at once (a shell of radius $\rho$ has surface area proportional to $\rho^2$). The $\sin\phi$ says that wedges near the poles ($\phi \approx 0$ or $\pi$) are pinched thin, while wedges near the equator ($\phi = \pi/2$, where $\sin\phi = 1$) are fattest. Together they reproduce the spherical volume element you used on faith in Chapter 32.

Warning. Conventions for spherical coordinates differ between disciplines. Many physics texts swap the names of $\phi$ and $\theta$, and some measure the polar angle up from the $xy$-plane instead of down from the $z$-axis. The Jacobian's form — radius-squared times the sine of the polar angle — survives any consistent convention, but the literal expression $\rho^2\sin\phi$ assumes $\phi$ is the angle from the $z$-axis with $0\le\phi\le\pi$. If you measure latitude (from the equator) instead, the factor becomes $\rho^2\cos(\text{latitude})$. Always confirm which angle is which before plugging in.

Check Your Understanding. Without computing a determinant, explain why the spherical volume element must vanish when $\phi = 0$.

Answer

At $\phi = 0$ the point lies on the positive $z$-axis, where the azimuthal angle $\theta$ is undefined — every value of $\theta$ describes the same point. The coordinate system is degenerate there: a whole range of $(\theta)$ collapses to a single point, so the transformation crushes a 3D neighborhood and its local volume-scaling factor must be zero. And indeed $\sin\phi = \sin 0 = 0$. The Jacobian vanishing is the algebraic signature of the coordinates breaking down on the axis.

33.8 The Jacobian, Inverses, and Singular Points

Two structural facts round out the theory.

Inverses multiply to one. If $T:(u,v)\to(x,y)$ is invertible with inverse $T^{-1}:(x,y)\to(u,v)$, then their Jacobian determinants are reciprocals:

$$\det \frac{\partial(x,y)}{\partial(u,v)} \cdot \det \frac{\partial(u,v)}{\partial(x,y)} = 1, \qquad \text{i.e.} \qquad \det J_{T^{-1}} = \frac{1}{\det J_T}.$$

This is the chain rule for Jacobian matrices ($J_{T^{-1}}J_T = I$) combined with the fact that the determinant of a product is the product of determinants. It is the exact multivariable analog of $\frac{dx}{du}\cdot\frac{du}{dx} = 1$ from single-variable calculus. It is also extremely useful in practice: when computing $x,y$ in terms of $u,v$ is messy but the reverse is easy, compute $\det J_{T^{-1}}$ and take the reciprocal.

Singular points and the Inverse Function Theorem. Where $\det J_T = 0$, the transformation locally collapses dimensions — it crushes a 2D neighborhood onto a curve or a point, and is not invertible there. For the polar map, $\det J = r = 0$ at the origin, which is exactly where $\theta$ becomes undefined. Such points are typically a single point or a curve — a set of zero area — so they do not affect the value of an integral. But they are where the geometry breaks, and they deserve respect.

Math Major Sidebar — The Inverse Function Theorem. One of the foundational results of multivariable analysis states: if $T$ is continuously differentiable and $\det J_T(p) \neq 0$ at a point $p$, then $T$ has a continuously differentiable local inverse near $p$, and the Jacobian of that inverse is $J_T^{-1}$. In other words, a nonzero Jacobian determinant is a certificate of local invertibility. This is the higher-dimensional version of "if $f'(a) \neq 0$ then $f$ is locally invertible with $(f^{-1})'(b) = 1/f'(a)$," which you met in Chapter 8. The change-of-variables theorem leans on this: it needs $\det J_T \neq 0$ to guarantee $T$ is one-to-one on the interior, so the map between regions is honest.

33.9 Application: Transforming Probability Densities

Here is where the Jacobian leaves pure geometry and becomes one of the most-used tools in statistics and machine learning. The connection is the Area Under the Normal Curve anchor example we have tracked since Chapter 13 — but now in two or more dimensions.

Suppose random variables $X, Y$ have a joint probability density $p_{XY}(x,y)$, meaning the probability of landing in a region $R$ is $\iint_R p_{XY}\,dA$. Now define new variables $(U,V) = T(X,Y)$. What is the density $p_{UV}$ of the new pair?

A probability is a number — it cannot change just because we renamed the axes. So we demand $\iint_S p_{UV}\,du\,dv = \iint_R p_{XY}\,dA$ for corresponding regions. Applying the change-of-variables formula to the right side forces

$$\boxed{\,p_{UV}(u, v) = p_{XY}\big(x(u,v),\,y(u,v)\big)\,\big|\det J_{T^{-1}}\big|\,}$$

where $T^{-1}:(u,v)\to(x,y)$ expresses the old variables in terms of the new ones. The Jacobian factor is exactly what keeps the total probability equal to $1$ after the change. A density is a thing per unit area, so when area rescales, the density must rescale inversely to compensate — and the Jacobian is the rescaling.

Real-World Application — The Box–Muller transform (data science, simulation). How does a computer generate normally distributed random numbers when it can only produce uniform ones? The Box–Muller method takes two independent uniform variables, maps them through a transformation into polar-like coordinates, and the Jacobian factor of that map turns the flat uniform density into the bell-shaped Gaussian density $\frac{1}{2\pi}e^{-(x^2+y^2)/2}$. Run in reverse, the polar change of variables is also the slick trick for evaluating the otherwise-impossible Gaussian integral $\int_{-\infty}^\infty e^{-x^2}dx = \sqrt\pi$: square it into a 2D integral, switch to polar, and the factor of $r$ from the Jacobian makes the integrand integrable. Every normal random number your simulations have ever used rode in on a Jacobian.

Real-World Application — Normalizing flows (machine learning). A modern class of generative models, normalizing flows (Real NVP, Glow, MAF), builds a complicated probability distribution by pushing a simple one — usually a standard Gaussian — through a chain of learned, invertible transformations $T$. The density of the result is governed by exactly the formula above, generalized to $n$ dimensions: $p(\mathbf{x}) = p_{\text{base}}\!\big(T^{-1}(\mathbf{x})\big)\,\big|\det J_{T^{-1}}\big|$. The whole engineering challenge of these models is designing transformations expressive enough to be interesting yet structured enough that $\det J$ is cheap to compute (often by making $J$ triangular, so its determinant is just the product of its diagonal). The change-of-variables formula you are learning this week is, almost verbatim, the loss function these networks minimize.

33.10 Application: Coordinate Systems in Physics

Physics is a relentless consumer of change of variables, because nature's symmetries rarely line up with Cartesian axes.

• Quantum mechanics. The hydrogen atom is solved in spherical coordinates, because the Coulomb potential depends only on $\rho$. Every normalization integral and expectation value carries the $\rho^2\sin\phi$ Jacobian; without it, the famous orbital probability densities would be wrong.
• Electromagnetism and gravity. Fields with cylindrical symmetry (a charged wire, fluid in a pipe) are handled in cylindrical coordinates, with the factor of $r$; fields with spherical symmetry (a point charge, a planet) in spherical coordinates.
• Statistical mechanics. Liouville's theorem states that Hamiltonian time-evolution preserves volume in phase space — its Jacobian determinant is exactly $1$. This "incompressibility" of phase space is a cornerstone of statistical mechanics, and it is, at bottom, a statement that $\det J = 1$ for the flow.

In every case the Jacobian is the translator that lets the integral be set up in whatever coordinates make the physics simplest.

33.11 Computation: Jacobians in Python

The three-tier pattern — pose analytically, compute by hand, verify by machine — applies perfectly here. SymPy computes Jacobian determinants symbolically, so you can check any hand calculation, including the brutal spherical one.

# Symbolic Jacobian of the polar transformation; should print r.
import sympy as sp

r, theta = sp.symbols('r theta', positive=True)
x = r * sp.cos(theta)
y = r * sp.sin(theta)

J = sp.Matrix([[sp.diff(x, r), sp.diff(x, theta)],
               [sp.diff(y, r), sp.diff(y, theta)]])
det_J = sp.simplify(J.det())
print("polar  |det J| =", sp.simplify(sp.Abs(det_J)))
# Output: polar  |det J| = r

SymPy even has a built-in .jacobian() method, which is the cleanest way to handle the 3D spherical case:

# Symbolic spherical Jacobian via Matrix.jacobian; should print rho**2*sin(phi).
import sympy as sp

rho, phi, theta = sp.symbols('rho phi theta', positive=True)
coords = sp.Matrix([rho * sp.sin(phi) * sp.cos(theta),   # x
                    rho * sp.sin(phi) * sp.sin(theta),   # y
                    rho * sp.cos(phi)])                  # z
J = coords.jacobian([rho, phi, theta])
print("spherical  det J =", sp.simplify(J.det()))
# Output: spherical  det J = rho**2*sin(phi)

Computational Note. coords.jacobian([rho, phi, theta]) builds the full matrix of partials in one call — the rows are the output coordinates $x,y,z$ and the columns are the input variables $\rho,\phi,\theta$, exactly matching the matrix in §33.7. For the numerical Jacobian of a transformation with no closed form, approximate each column by a centered finite difference (Chapter 5's idea, vectorized) and take np.linalg.det. The symbolic route is exact and should always be your first check; the numerical route is what you reach for when the transformation is a learned neural network and no formula exists, as in normalizing flows.

# Numerical Jacobian determinant of polar map at (r, theta) = (1, pi/4); should be ~1.
import numpy as np

def jacobian_numerical(f, point, h=1e-6):
    point = np.asarray(point, dtype=float)
    f0 = f(point)
    J = np.zeros((len(f0), len(point)))
    for i in range(len(point)):
        step = np.zeros_like(point); step[i] = h
        J[:, i] = (f(point + step) - f(point - step)) / (2 * h)
    return J

T = lambda p: np.array([p[0] * np.cos(p[1]), p[0] * np.sin(p[1])])
J = jacobian_numerical(T, [1.0, np.pi / 4])
print("numerical det J =", round(float(np.linalg.det(J)), 6))
# Output: numerical det J = 1.0   (= r at r = 1)

33.12 The Differential Forms Perspective (Optional)

There is a more elegant formulation lurking beneath all of this. The double integral $\iint f\,dx\,dy$ is, properly understood, the integral of a differential 2-form $f\,dx \wedge dy$, where $\wedge$ is an antisymmetric product ($dx \wedge dy = -\,dy \wedge dx$, and $dx \wedge dx = 0$). Under a transformation, the form transforms as

$$dx \wedge dy = \det J_T \,\,du \wedge dv,$$

and the sign of the determinant — which the absolute value discarded — is precisely the antisymmetry of the wedge keeping track of orientation. This perspective unifies change of variables across all dimensions into a single line and underlies the grand statement of Chapter 38, $\int_{\partial M}\omega = \int_M d\omega$, that generalizes the Fundamental Theorem of Calculus to every dimension at once. We mention it now so that when you meet it in Part VII, you recognize the Jacobian as its humble origin.

Historical Note. The determinant of partial derivatives is named for Carl Gustav Jacob Jacobi (1804–1851), who developed its theory systematically in the 1830s and 1840s, though special cases appear earlier in the work of Cauchy and even Lagrange. Jacobi recognized the determinant $\partial(x,y)/\partial(u,v)$ as the natural object governing change of variables and functional dependence, and he proved that two functions are functionally dependent precisely when their Jacobian vanishes identically. The compact $\partial(x,y)/\partial(u,v)$ notation is his legacy.

33.13 Summary of Important Jacobians

Add to Your Modeling Portfolio. Add a coordinate change to your model, using a Jacobian to translate an integral into the system where the symmetry lives. Biology: model a quantity on a roughly spherical cell or organ and integrate it in spherical coordinates, carrying the $\rho^2\sin\phi$ factor; or transform a population density between two measurement scales. Economics: apply a linear change of variables to a two-good utility or cost integral to decorrelate the variables (the same diagonalization idea behind principal components), and account for the constant Jacobian. Physics: compute the moment of inertia or total mass of a cylindrically or spherically symmetric body, deriving the volume element from its Jacobian rather than quoting it. Data Science: transform a 2D probability density (e.g. a correlated Gaussian) through an invertible map and write down the new density using $p_{UV} = p_{XY}\,\lvert\det J_{T^{-1}}\rvert$ — the exact mechanism of a single normalizing-flow layer.

Looking Ahead

You have closed Part VI with the keystone of multivariable integration. Every awkward region and every symmetric integrand now has a strategy: find the coordinates in which it is simple, and let the Jacobian pay the toll for the change.

Part VII opens with Chapter 34: Vector Fields — functions $\mathbf{F}(\mathbf{x})$ that assign a vector to every point, the natural language for fluid flow, electric and gravitational fields, and force. From there, Chapter 35 builds line integrals and Green's Theorem, Chapter 36 surface integrals, and Chapter 37 Stokes' and the Divergence Theorems — each one a higher-dimensional Fundamental Theorem of Calculus, and each one quietly relying on the Jacobian whenever a coordinate change is needed. Chapter 38 unifies them all. The Jacobian you learned here is not a one-chapter tool; it is the change-of-coordinates engine for everything that follows.

Reflection

The Jacobian is one of those rare objects that is at once utterly concrete — a determinant of partial derivatives you can compute in thirty seconds — and quietly profound. It is the single number that captures how a transformation stretches space, and it converts the intimidating problem of "integrate over this horrible region" into the manageable one of "find better coordinates." It retroactively explains the $r$ and the $\rho^2\sin\phi$ you took on faith. It powers the algorithms that generate random numbers and the neural networks that generate images. And it is, when you see it through differential forms, the seed of the deepest unification in calculus. Master it, and you have mastered the multivariable version of the most useful trick in all of integration: when a problem is hard, change your point of view.