Change of Variables and the Jacobian

The Change of Variables Theorem is the engine behind evaluating almost every nontrivial multiple integral encountered in physics and engineering: moments of inertia, centers of mass, electric/gravitational potentials, and fluid flow computations routinely switch to polar, cylindrical, or spherical coordinates to exploit symmetry, exactly using the Jacobian factors $r$ and $\rho^2\sin\phi$ derived above. In probability and statistics, the same theorem underlies the transformation-of-random-variables formula: if $Y = g(X)$ for a diffeomorphism $g$, the density of $Y$ is obtained from the density of $X$ by multiplying by $|\det Dg^{-1}|$, the key computational tool for deriving distributions like the chi-squared, Rayleigh, or multivariate Gaussian marginals. In numerical analysis and computer graphics, Jacobians measure mesh distortion in finite-element methods and control texture-mapping and rendering transformations. In differential geometry and the theory that follows this article — Surface Integrals and Parametrized Surfaces and the Divergence Theorem — the Jacobian idea generalizes to the "area element" $|\partial_u \mathbf{r} \times \partial_v \mathbf{r}|$ of a parametrized surface and to the volume-flux computations that make Stokes-type theorems work. In cryptography and coding theory, Jacobian determinants of algebraic curves (a more sophisticated relative of the concept) underlie elliptic-curve cryptosystems.

Ask the tutor: Can you walk through why the Jacobian for cylindrical coordinates is just $r$ (not $r^2$ or something involving $z$), using the same "small box distortion" reasoning as for spherical coordinates?

Think first in one variable. If $g:[a,b]\to\mathbb{R}$ is a smooth increasing bijection, an infinitesimal interval $[u, u+du]$ near a point $u$ is stretched by $g$ into an interval of length approximately $g'(u)\,du$. Summing (integrating) these stretched pieces is exactly the substitution rule. The derivative $g'(u)$ is a 1-by-1 "Jacobian," and its absolute value is the local length-scaling factor. Now go to $\mathbb{R}^n$. A linear map $T:\mathbb{R}^n \to \mathbb{R}^n$ sends the unit cube to a parallelepiped whose volume is exactly $|\det T|$ — this is a basic fact of linear algebra, proved below as a lemma. A general $C^1$ map $g$ is, near a point $u_0$, well approximated by the affine map

by definition of differentiability. So a tiny cube of volume $du$ near $u_0$ gets mapped to a region whose volume is approximately $|\det Dg(u_0)|\,du$. The Change of Variables Theorem is the rigorous, global version of "add up all these tiny distortions." For polar coordinates $g(r,\theta) = (r\cos\theta, r\sin\theta)$, a small rectangle $[r, r+dr]\times[\theta,\theta+d\theta]$ in $(r,\theta)$-space maps to a small "curved rectangle" (an annular sector) in $(x,y)$-space with radial thickness $dr$ and arc length $r\,d\theta$, hence area approximately $r\,dr\,d\theta$ — matching $J_g = r$ exactly.

Example 1 (an asymmetric planar region and a genuinely nonlinear substitution). Compute

where $R$ is the parallelogram with vertices $(0,0), (2,0), (3,1), (1,1)$ — i.e., $R = \{(x,y): 0 \le y \le 1,\ y \le x - 2y \cdot 0 ,\ \}$; more precisely, $R$ is bounded by $y=0$, $y=1$, $x - y =$ between $0$ and $2$. Let's set this up via the natural substitution suggested by the integrand: let

Check the vertices: $(0,0)\mapsto(u,v)=(0,0)$; $(2,0)\mapsto(2,2)$; $(3,1)\mapsto(4,2)$; $(1,1)\mapsto(2,0)$. So in $(u,v)$-coordinates the region is the rectangle $[0,4]\times[0,2]$... let's double check by finding which edges map to which. Edge from $(0,0)$ to $(2,0)$: $y=0$, $x\in[0,2]$, so $u=x, v=x$, giving $u=v$ from $(0,0)$ to $(2,2)$. Edge from $(2,0)$ to $(3,1)$: this is $x - 2y = 2$ i.e. $x = 2+2y$... check: at $y=0$, $x=2$ ✓; at $y=1$, $x=4\ne 3$. Let's recompute: the segment from $(2,0)$ to $(3,1)$ has direction $(1,1)$, so $x - y$ is constant $=2$ along it. Indeed $v = x-y = 2$ at both endpoints. Good — so this edge maps to $v=2$, $u$ from $2$ to $4$. Edge from $(3,1)$ to $(1,1)$: $y=1$, $x$ from $3$ to $1$, so $u = x+1 \in[2,4]$, $v = x - 1\in[0,2]$, with $u - v = 2$ constant — this is the segment $u-v=2$. Edge from $(1,1)$ to $(0,0)$: direction $(1,1)$ so $v=x-y$ constant $=0$, $u$ from $0$ to $2$. So $R$ in $(u,v)$-coordinates is the parallelogram with vertices $(0,0),(2,2),(4,2),(2,0)$ bounded by $v=0$ (below-ish), $v=2$, $u - v = 0$, and $u-v=2$. This is exactly the image of the square $S = \{(v,w): 0\le v \le 2,\ 0 \le w \le 2\}$ under $u = v + w,\ v=v$... Let's instead directly change variables a second time to make Fubini trivial: set $w = u - v$ so that $R$ corresponds exactly to $\{(v,w): 0\le v\le 2, 0 \le w \le 2\}$ with $u = v+w$. But it is cleaner to just directly integrate over the parallelogram in $(u,v)$ using Fubini with $v$ from $0$ to $2$ and, for fixed $v$, $u$ ranging over $[v, v+2]$ (checking: at $v=0$, $u\in[0,2]$ ✓ matches vertex range; at $v=2$, $u \in [2,4]$ ✓). Now compute the Jacobian of $(x,y)\mapsto(u,v)$: $u=x+y,\ v=x-y$, so

We want $\partial(x,y)/\partial(u,v)$, the reciprocal Jacobian (valid since the map is a linear bijection): $\det \dfrac{\partial(x,y)}{\partial(u,v)} = 1/(-2) = -\tfrac12$, so $|J| = \tfrac12$. (Directly: $x = \tfrac{u+v}{2}, y=\tfrac{u-v}{2}$, Jacobian matrix $\begin{pmatrix}1/2 & 1/2\\ 1/2 & -1/2\end{pmatrix}$, determinant $-1/2$.) By the Change of Variables Theorem,

Inner integral: $\int_v^{v+2} \sin^2 u\, du = \int_v^{v+2} \frac{1-\cos 2u}{2}\,du = 1 - \frac{\sin(2v+4) - \sin(2v)}{4}$. So

The first piece: $\frac12 \cdot \frac{8}{3} = \frac{4}{3}$. For the second, use $\int v^2 \sin(av+b)\,dv = -\frac{v^2}{a}\cos(av+b) + \frac{2v}{a^2}\sin(av+b) + \frac{2}{a^3}\cos(av+b) + C$ (standard integration by parts twice), with $a=2$. For $\sin(2v+4)$ ($b=4$), antiderivative $F_1(v) = -\frac{v^2}{2}\cos(2v+4) + \frac{v}{2}\sin(2v+4) + \frac14 \cos(2v+4)$. For $\sin(2v)$ ($b=0$), antiderivative $F_2(v) = -\frac{v^2}{2}\cos(2v) + \frac{v}{2}\sin(2v) + \frac14\cos(2v)$. Numerically (radians): $\cos 4 \approx -0.653644$, $\sin 4\approx -0.756802$, $\cos 8 \approx -0.145500$, $\sin 8 \approx 0.989358$, $\cos 0 = 1,\ \sin 0 = 0$. $F_1(2) = -2\cos 8 + \sin 8 + \tfrac14\cos 8 = -2(-0.1455) + 0.989358 + 0.25(-0.1455) = 0.2910 + 0.989358 - 0.036375 = 1.243983$. $F_1(0) = 0 + 0 + \tfrac14\cos4 = 0.25(-0.653644) = -0.163411$. $\int_0^2 v^2\sin(2v+4)\,dv = F_1(2)-F_1(0) = 1.243983 + 0.163411 = 1.407394$. $F_2(2) = -2\cos4 + \sin4 + \tfrac14\cos4 = -2(-0.653644) + (-0.756802) + 0.25(-0.653644) = 1.307288 - 0.756802 -0.163411 = 0.387075$. $F_2(0) = 0+0+\tfrac14(1) = 0.25$. $\int_0^2 v^2 \sin(2v)\,dv = F_2(2) - F_2(0) = 0.387075 - 0.25 = 0.137075$. So $\int_0^2 v^2(\sin(2v+4)-\sin(2v))\,dv \approx 1.407394 - 0.137075 = 1.270319$. Then $I \approx \frac{4}{3} - \frac18(1.270319) = 1.333333 - 0.158790 = 1.174544$. So $I \approx 1.1745$, exactly

— the point of the example is the mechanism (linear change of variables turning a skew parallelogram into a rectangle, with constant Jacobian $\tfrac12$), not the closed form of the trigonometric antiderivative; the numerical value $I\approx 1.1745$ is the answer.

Example 2 (spherical coordinates on an asymmetric solid: mass of a "truncated wedge"). A solid $E$ occupies the spherical wedge $1 \le \rho \le 3$, $0\le \phi \le \pi/3$ (colatitude from the north pole), $0\le\theta\le \pi/2$ (a quarter turn in azimuth), with mass density $\delta(x,y,z) = z = \rho\cos\phi$. Find the total mass

Recall the spherical coordinate map

with Jacobian determinant $J_g = \rho^2 \sin\phi$ (derived in the Intuition/Scope discussion and standard; we recompute it here for completeness).

Expanding along the bottom row: $\det = \cos\phi \cdot M_{31} - (-\rho\sin\phi)\cdot M_{32} + 0$, where

So

Good, confirmed: $J_g = \rho^2\sin\phi \ge 0$ on $0\le\phi\le\pi$, so $|J_g| = \rho^2\sin\phi$. Now apply the theorem:

using Fubini's Theorem to split the integral since the integrand factors and the region is a box in $(\rho,\phi,\theta)$. Compute each factor:

So

So $M = \dfrac{15\pi}{4} \approx 11.781$.