Change of Variables & the Jacobian Determinant

Let $g: [\alpha, \beta] \to [a, b]$ be a $C^1$ bijection with $g'(t) \neq 0$ on $(\alpha, \beta)$. Then for any continuous $f: [a, b] \to \mathbb{R}$:

$\int_a^b f(x)\,dx = \int_\alpha^\beta f(g(t))\,|g'(t)|\,dt.$

The factor $|g'(t)|$ adjusts for how $g$ stretches or compresses the interval. When $g$ is increasing, $|g'(t)| = g'(t)$; when decreasing, $|g'(t)| = -g'(t)$, which also reverses the limits to compensate.

In 1D, you can track orientation via limit ordering. In $\mathbb{R}^n$, there are no “limits” to reverse — the absolute value of the Jacobian determinant is the only way to ensure positivity of the volume element. The absolute value is the conceptual bridge from 1D to $n$D.

Compute $\int_0^1 \sqrt{1 - x^2}\,dx$ via $x = \sin\theta$. We have $g(\theta) = \sin\theta$, $g'(\theta) = \cos\theta$, limits $[0, \pi/2]$:

$\int_0^1 \sqrt{1 - x^2}\,dx = \int_0^{\pi/2} \sqrt{1 - \sin^2\theta} \cdot \cos\theta\,d\theta = \int_0^{\pi/2} \cos^2\theta\,d\theta = \frac{\pi}{4}.$

This is one quarter of the unit disk area — a preview of polar coordinates.

Compute $\int_0^\infty e^{-x^2/2}\,dx$ via $u = x^2/2$. We get $x = \sqrt{2u}$, $dx = du/\sqrt{2u}$:

$\int_0^\infty e^{-x^2/2}\,dx = \frac{1}{\sqrt{2}} \int_0^\infty u^{-1/2} e^{-u}\,du = \frac{1}{\sqrt{2}} \,\Gamma(1/2) = \sqrt{\pi/2}.$

This connects to the Gamma function (Topic 8) and previews the Gaussian integral computation later in this topic.

A coordinate transformation (or change of variables) on an open set $D^* \subseteq \mathbb{R}^2$ is a $C^1$ function $\varphi: D^* \to D \subseteq \mathbb{R}^2$ that is a diffeomorphism: bijective, $C^1$, and with $C^1$ inverse $\varphi^{-1}: D \to D^*$.

The IFT (Topic 12) guarantees a local diffeomorphism wherever $\det J_\varphi \neq 0$. For the change of variables formula, we need a global diffeomorphism on $D^*$ (or at least injectivity, with $\det J_\varphi = 0$ only on a set of measure zero). Polar coordinates fail to be a global diffeomorphism on $\{r > 0\}$ because of the $2\pi$-periodicity in $\theta$, but they are a diffeomorphism on any domain that doesn’t wrap all the way around.

Let $\varphi: D^* \to D$ be a $C^1$ diffeomorphism between open subsets of $\mathbb{R}^2$, and let $f: D \to \mathbb{R}$ be continuous. Then:

$\iint_D f(x, y)\,dA = \iint_{D^*} f(\varphi(u, v))\,|\det J_\varphi(u, v)|\,du\,dv.$

The Jacobian determinant $|\det J_\varphi|$ converts the area element $du\,dv$ in the $(u,v)$-coordinate system to the area element $dA$ in the $(x,y)$-coordinate system.

For $\varphi(u, v) = (au + bv,\, cu + dv)$ with $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $\det A \neq 0$: $\det J_\varphi = \det A$ (constant). The formula reduces to $\iint_D f(x,y)\,dA = |\det A| \iint_{D^*} f(Au)\,du\,dv$. Linear maps scale all areas by the same factor.

For $\varphi(u,v) = (u + v, v)$: $\det J_\varphi = 1$. The integral is unchanged: $\iint_D f\,dA = \iint_{D^*} f(u+v, v)\,du\,dv$. Area-preserving transformations have a unit Jacobian determinant.

Often we start with an integral in $(x, y)$ and want to transform to $(u, v)$. If $\varphi: D^* \to D$ is the transformation $(u, v) \mapsto (x, y)$, then we substitute $x = \varphi_1(u,v)$, $y = \varphi_2(u,v)$, $dA = |\det J_\varphi|\,du\,dv$, and change the region from $D$ to $D^* = \varphi^{-1}(D)$. The conceptual direction is: “I want simpler limits, so I find a $\varphi$ that maps a simple $D^*$ to the complicated $D$.”

The polar coordinate transformation is $\varphi(r, \theta) = (r\cos\theta, r\sin\theta)$, mapping from $D^* = \{(r, \theta) : r > 0,\, \theta \in (0, 2\pi)\}$ to $D = \mathbb{R}^2 \setminus \{x \ge 0, y = 0\}$ (the plane minus the positive $x$-axis). The Jacobian is:

$J_\varphi(r, \theta) = \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix}, \qquad |\det J_\varphi| = r.$

The area element is $dA = r\,dr\,d\theta$.

At $r = 0$, $\det J_\varphi = 0$, and $\varphi$ collapses all $\theta$ values to the origin — it is not injective. The origin is a single point (measure zero), so excluding it does not affect the integral. This is typical: changes of variables are allowed to fail on sets of measure zero.

$\iint_{x^2+y^2 \le 1} dA = \int_0^{2\pi} \int_0^1 r\,dr\,d\theta = 2\pi \cdot \frac{1}{2} = \pi.$

Compare with the Cartesian computation from Topic 13, Example 7: $\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} dy\,dx = \int_{-1}^1 2\sqrt{1-x^2}\,dx = \pi$. Same answer, but polar reduces the integral to a product of two elementary 1D integrals.

$\iint_D (x^2 + y^2)\,dA$ over the annulus $1 \le x^2+y^2 \le 4$: in polar, $x^2 + y^2 = r^2$:

$\int_0^{2\pi} \int_1^2 r^2 \cdot r\,dr\,d\theta = 2\pi \cdot \frac{r^4}{4}\Big|_1^2 = 2\pi \cdot \frac{15}{4} = \frac{15\pi}{2}.$

The circular symmetry of both the region and the integrand makes polar coordinates the natural choice.

From Topic 13, §7: the volume between $z = x^2 + y^2$ and $z = 4$ over the disk $x^2 + y^2 \le 4$. In polar:

$V = \int_0^{2\pi} \int_0^2 (4 - r^2)\,r\,dr\,d\theta = 2\pi \int_0^2 (4r - r^3)\,dr = 2\pi \left[2r^2 - \frac{r^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi.$

$\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}.$

Step 1: Square the integral. Let $I = \int_{-\infty}^{\infty} e^{-x^2}\,dx$. Then:

$I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right)\left(\int_{-\infty}^{\infty} e^{-y^2}\,dy\right) = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA.$

The first equality uses Fubini (Topic 13, Theorem 1) to convert the product of two 1D integrals into a double integral. This requires justifying that $e^{-(x^2+y^2)}$ is integrable over $\mathbb{R}^2$ — it suffices to note that $\int_0^R \int_0^{2\pi} e^{-r^2} r\,d\theta\,dr = 2\pi \int_0^R r e^{-r^2}\,dr = \pi(1 - e^{-R^2}) \to \pi$ as $R \to \infty$, so the improper integral converges.

Step 2: Switch to polar coordinates. Using $x^2 + y^2 = r^2$ and $dA = r\,dr\,d\theta$:

$I^2 = \int_0^{2\pi}\int_0^{\infty} e^{-r^2}\,r\,dr\,d\theta.$

The inner integral is elementary: $\int_0^{\infty} r e^{-r^2}\,dr = \left[-\frac{1}{2}e^{-r^2}\right]_0^\infty = \frac{1}{2}$.

Step 3: Evaluate. $I^2 = 2\pi \cdot \frac{1}{2} = \pi$. Since $I > 0$, we conclude $I = \sqrt{\pi}$.

The function $e^{-x^2}$ has no elementary antiderivative (this is a theorem, not a failure of technique). The 1D integral is genuinely intractable without the 2D “trick.” The change-of-variables formula transforms a hard 1D problem into an easy 2D one by exploiting radial symmetry.

The PDF of $\mathcal{N}(\mu, \sigma^2)$ is $\frac{1}{\sigma\sqrt{2\pi}} e^{-(x-\mu)^2/(2\sigma^2)}$. Verify normalization using the substitution $u = (x - \mu)/(\sigma\sqrt{2})$ and $\int_{-\infty}^{\infty} e^{-u^2}\,du = \sqrt{\pi}$:

$\int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}} e^{-(x-\mu)^2/(2\sigma^2)}\,dx = \frac{1}{\sigma\sqrt{2\pi}} \cdot \sigma\sqrt{2} \cdot \sqrt{\pi} = 1.$

This is why the $\sqrt{2\pi}$ appears in the Gaussian density — it is the Gaussian integral.

The cylindrical coordinate transformation is $\varphi(r, \theta, z) = (r\cos\theta, r\sin\theta, z)$, with $r > 0$, $\theta \in (0, 2\pi)$, $z \in \mathbb{R}$. The Jacobian is:

$J_\varphi = \begin{pmatrix} \cos\theta & -r\sin\theta & 0 \\ \sin\theta & r\cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}, \qquad |\det J_\varphi| = r.$

The volume element is $dV = r\,dr\,d\theta\,dz$.

The spherical coordinate transformation is $\varphi(\rho, \theta, \phi) = (\rho\sin\phi\cos\theta,\, \rho\sin\phi\sin\theta,\, \rho\cos\phi)$, with $\rho > 0$, $\theta \in (0, 2\pi)$, $\phi \in (0, \pi)$. The Jacobian matrix is:

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

Computing $|\det J_\varphi|$ by cofactor expansion along the third row:

$\det J_\varphi = \cos\phi \cdot (-\rho^2\sin\phi\cos\phi\sin^2\theta - \rho^2\sin\phi\cos\phi\cos^2\theta) + (-\rho\sin\phi)(\rho\sin^2\phi\cos^2\theta + \rho\sin^2\phi\sin^2\theta)$

$= -\rho^2\sin\phi\cos^2\phi - \rho^2\sin^3\phi = -\rho^2\sin\phi(\cos^2\phi + \sin^2\phi) = -\rho^2\sin\phi.$

Since $\sin\phi \ge 0$ for $\phi \in [0, \pi]$: $|\det J_\varphi| = \rho^2\sin\phi$. The volume element is $dV = \rho^2\sin\phi\,d\rho\,d\theta\,d\phi$.

$V = \iiint_{x^2+y^2+z^2 \le 1} dV = \int_0^{2\pi}\int_0^{\pi}\int_0^1 \rho^2 \sin\phi\,d\rho\,d\phi\,d\theta = 2\pi \cdot 2 \cdot \frac{1}{3} = \frac{4\pi}{3}.$

Three decoupled 1D integrals — the spherical symmetry of the ball makes the computation trivial.

For a uniform solid sphere of mass $M$ and radius $R$ with density $\rho_{\text{mass}} = \frac{3M}{4\pi R^3}$, the moment of inertia about the $z$-axis is $I = \iiint \rho_{\text{mass}} (x^2 + y^2)\,dV$. In spherical coordinates, $x^2 + y^2 = \rho^2 \sin^2\phi$:

$I = \frac{3M}{4\pi R^3} \int_0^{2\pi}\int_0^{\pi}\int_0^R \rho^2 \sin^2\phi \cdot \rho^2 \sin\phi\,d\rho\,d\phi\,d\theta.$

Evaluating each factor: $\int_0^{2\pi}d\theta = 2\pi$, $\int_0^{\pi}\sin^3\phi\,d\phi = \frac{4}{3}$, $\int_0^R \rho^4\,d\rho = \frac{R^5}{5}$. So $I = \frac{3M}{4\pi R^3} \cdot 2\pi \cdot \frac{4}{3} \cdot \frac{R^5}{5} = \frac{2}{5}MR^2$.

• Polar: circular symmetry in 2D ($x^2 + y^2$ appears, region is a disk or annulus).
• Cylindrical: cylindrical symmetry in 3D (the region or integrand is symmetric about the $z$-axis).
• Spherical: spherical symmetry ($x^2 + y^2 + z^2$ appears, region is a ball or spherical shell).
• If none of these symmetries are present, a custom coordinate transformation may still simplify the integral.

A function $\varphi: D^* \to D$ between open subsets of $\mathbb{R}^n$ is a $C^1$ diffeomorphism if: (i) $\varphi$ is bijective, (ii) $\varphi$ is $C^1$ (continuously differentiable), and (iii) $\varphi^{-1}: D \to D^*$ is $C^1$. By the Inverse Function Theorem (Topic 12, Theorem 1), conditions (ii) and (iii) are equivalent to: $\varphi$ is $C^1$ and $\det J_\varphi(u) \neq 0$ for all $u \in D^*$.

Let $\varphi: D^* \to D$ be a $C^1$ diffeomorphism between open subsets of $\mathbb{R}^n$. Let $f: D \to \mathbb{R}$ be continuous and integrable over $D$. Then:

$\int_D f(x)\,dx = \int_{D^*} f(\varphi(u))\,|\det J_\varphi(u)|\,du.$

This reduces to the 1D substitution rule (Theorem 1) when $n = 1$, and to the 2D formula (Theorem 2) when $n = 2$.

We follow Spivak’s approach (Calculus on Manifolds, Theorem 3-13).

Step 1: The linear case. If $\varphi(u) = Au + b$ is an affine map with $A$ invertible, then $D = A(D^*) + b$, $J_\varphi = A$, and the formula follows from the definition of the Riemann integral: the Riemann sum over $D$ with partition $P$ corresponds to the Riemann sum over $D^*$ with partition $A^{-1}(P - b)$, with each cell volume scaled by $|\det A|$.

Step 2: Local validity. For a general $C^1$ diffeomorphism and any $u_0 \in D^*$, the linear approximation $\varphi(u) \approx \varphi(u_0) + J_\varphi(u_0)(u - u_0)$ is accurate on a small ball $B_\epsilon(u_0)$. Because $J_\varphi$ is continuous and $\det J_\varphi \neq 0$, on a sufficiently small ball:

$\left|\frac{|\det J_\varphi(u)|}{|\det J_\varphi(u_0)|} - 1\right| < \delta$

for any prescribed $\delta > 0$. The integral formula holds locally up to an error controlled by $\delta$.

Step 3: Partition of unity. Cover $D^*$ with a finite collection of balls $\{B_k\}$ (using compactness of $\overline{D^*}$ if $D^*$ is bounded; the general case uses exhaustion by compact subsets — see Topic 3 for compactness). Choose a subordinate partition of unity $\{\psi_k\}$: smooth functions $\psi_k \ge 0$ with $\text{supp}(\psi_k) \subset B_k$ and $\sum_k \psi_k = 1$ on $D^*$. Then:

$\int_D f\,dx = \sum_k \int_D (\psi_k \circ \varphi^{-1}) \cdot f\,dx = \sum_k \int_{B_k} (\psi_k \cdot f \circ \varphi)\,|\det J_\varphi|\,du = \int_{D^*} f(\varphi(u))\,|\det J_\varphi(u)|\,du.$

Each step uses: (i) the partition of unity decomposes $f$ into locally supported pieces, (ii) the linear case applies locally on each $B_k$ (up to the error in Step 2), (iii) summing recovers the full integral. The rigorous details involve showing the error terms from Step 2 sum to zero — this uses the uniform continuity of $J_\varphi$ on compact subsets.

The theorem extends to $\varphi$ that fails to be injective or has $\det J_\varphi = 0$ on a set of measure zero (e.g., the origin for polar coordinates, or the $z$-axis for cylindrical coordinates). The Lebesgue version of the change-of-variables theorem handles this rigorously — see Sigma-Algebras & Measures and The Lebesgue Integral (coming soon).

Let $\varphi(u, v) = (au\cos v, bu\sin v)$ with $a, b > 0$. The Jacobian determinant is $|\det J_\varphi| = abu$. The area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1$:

$\int_0^{2\pi}\int_0^1 ab\,u\,du\,dv = 2\pi \cdot ab \cdot \frac{1}{2} = \pi ab.$

This generalizes the disk area $\pi r^2$ to $\pi ab$.

$\varphi(u, v) = \left(\frac{u^2 - v^2}{2},\, uv\right)$ has $|\det J_\varphi| = u^2 + v^2$. Useful for problems with parabolic symmetry (e.g., electrostatics near a parabolic reflector).

If $\varphi_1: D_1^* \to D_1$ and $\varphi_2: D_1 \to D$ are $C^1$ diffeomorphisms, then $\varphi = \varphi_2 \circ \varphi_1: D_1^* \to D$ is a $C^1$ diffeomorphism with:

$\det J_\varphi(u) = \det J_{\varphi_2}(\varphi_1(u)) \cdot \det J_{\varphi_1}(u).$

This follows from the chain rule for Jacobians (Topic 10, Theorem 2) and the multiplicativity of determinants. Geometrically: volume distortions compose multiplicatively, as established in Topic 10, Proposition 2.

To integrate over the region $x^2 + xy + y^2 \le 1$ (a rotated ellipse), first rotate by $\pi/4$ to diagonalize the quadratic form, then scale to a unit disk. The composed Jacobian determinant is the product of the individual determinants.

Let $Z$ be a random vector with density $p_Z$ and let $X = \varphi(Z)$ where $\varphi$ is a $C^1$ diffeomorphism. The density of $X$ is:

$p_X(x) = p_Z(\varphi^{-1}(x)) \cdot |\det J_{\varphi^{-1}}(x)| = \frac{p_Z(\varphi^{-1}(x))}{|\det J_\varphi(\varphi^{-1}(x))|}.$

This follows directly from the change of variables theorem applied to $P(X \in A) = \int_A p_X(x)\,dx = \int_{\varphi^{-1}(A)} p_Z(z)\,dz = \int_A p_Z(\varphi^{-1}(x)) |\det J_{\varphi^{-1}}(x)|\,dx$ for all measurable $A$.

If $Z \sim \mathcal{N}(0, 1)$ and $X = e^Z$, then $\varphi(z) = e^z$, $\varphi^{-1}(x) = \ln x$, $|(\varphi^{-1})'(x)| = 1/x$. So:

$p_X(x) = \frac{1}{\sqrt{2\pi}} e^{-(\ln x)^2/2} \cdot \frac{1}{x} = \frac{1}{x\sqrt{2\pi}} e^{-(\ln x)^2/2} \quad \text{for } x > 0.$

This is the log-normal density.

A normalizing flow is a composition of $K$ diffeomorphisms: $x = f_K \circ f_{K-1} \circ \cdots \circ f_1(z)$. By the composition rule (Proposition 2):

$\log p_X(x) = \log p_Z(z) - \sum_{k=1}^K \log |\det J_{f_k}(z_{k-1})|$

where $z_0 = z$ and $z_k = f_k(z_{k-1})$. The entire architecture of flow-based generative models (RealNVP, Glow, Neural Spline Flows) is designed to make each $|\det J_{f_k}|$ cheap to compute — typically $O(n)$ instead of $O(n^3)$ — by using triangular Jacobians (coupling layers, autoregressive transforms). This is the change-of-variables formula doing heavy lifting in generative modeling.

The RealNVP coupling layer splits $z = (z_a, z_b)$ and defines $x_a = z_a$, $x_b = z_b \odot \exp(s(z_a)) + t(z_a)$. The Jacobian is lower-triangular with diagonal entries $1$ (for $x_a$) and $\exp(s_i(z_a))$ (for $x_b$). So $\log|\det J| = \sum_i s_i(z_a)$ — a sum, not a determinant. This is $O(n)$ and trivially differentiable. The IFT (Topic 12) guarantees invertibility since $\exp(s_i) > 0$ everywhere.

In variational autoencoders, we need $\nabla_{\mu, \sigma} \mathbb{E}_{q_\phi(z)}[f(z)]$ where $q_\phi(z) = \mathcal{N}(\mu, \sigma^2)$. The expectation depends on $\mu, \sigma$ through the distribution, so we can’t just differentiate under the integral sign. The trick: write $z = \mu + \sigma\epsilon$ with $\epsilon \sim \mathcal{N}(0, 1)$. This is a change of variables from $\epsilon$ to $z$. Now $\mathbb{E}_{q_\phi(z)}[f(z)] = \mathbb{E}_{\epsilon \sim \mathcal{N}(0,1)}[f(\mu + \sigma\epsilon)]$, and the gradient moves inside: $\nabla_{\mu, \sigma} \mathbb{E}[f(\mu + \sigma\epsilon)]$ is just a standard derivative of a deterministic function of $\mu, \sigma$ evaluated at a random $\epsilon$. The change-of-variables formula separates the randomness ($\epsilon$) from the parameters ($\mu, \sigma$).