Lecture 9: The Change of Variables Formula
KSM3E04: Analysis on Manifolds — Fall 2025 | September 10, 3:00–4:30 PM
In this lecture, we prove one of the most important theorems in multivariable integration: the Change of Variables Formula. This formula gives us a correct way to evaluate integrals after changing coordinates (for example, from Cartesian coordinates $(x,y)$ to polar coordinates $(r,\theta)$). The proof connects ideas from linear algebra, calculus, and measure theory. We will build the proof step-by-step, starting from simple functions and moving to the general case.
1. The Change of Variables Theorem
In single-variable calculus, when we use substitution $x = \phi(t)$ to solve an integral $\int_{a}^{b} f(x) \, dx$, the integral becomes $\int_{\phi^{-1}(a)}^{\phi^{-1}(b)} f(\phi(t)) \phi'(t) \, dt$. The term $\phi'(t)$ is a scaling factor. It corrects for how the change from $x$ to $t$ stretches or shrinks the interval of integration. The Change of Variables Formula is the version of this for multiple dimensions. In this case, the scaling factor is the absolute value of the determinant of the Jacobian matrix.
Theorem: Change of Variables
Let $\Omega$ and $\Omega'$ be open subsets of $\mathbf{R}^n$, and let $\phi: \Omega \to \Omega'$ be a $C^1$-diffeomorphism. For any Lebesgue-integrable function $f: \Omega' \to \mathbf{R}$, the following formula holds:
Here, $m$ denotes the Lebesgue measure and $D\phi(\mathbf{x})$ is the Jacobian matrix of $\phi$ at the point $\mathbf{x}$.
2. Proof of the Change of Variables Theorem
The proof is structured in four main steps. We begin by proving the formula for the simplest case—affine transformations—and progressively build toward the general result for any $C^1$-diffeomorphism using tools from measure theory.
Step 1: The Formula for Affine Transformations
The Idea: First, we prove the formula for the simplest functions: affine maps. These are functions that stretch, rotate, reflect, and shift space uniformly. This means the scaling factor for volume is the same at every point.
The Rigorous Goal: An affine map has the form $\phi(\mathbf{x}) = M\mathbf{x} + \mathbf{b}$, where $M$ is an invertible matrix. Its Jacobian is the constant matrix $M$. We will prove that for any measurable set $A \subset \Omega$, the measure of its image is scaled by a constant factor: $m(\phi(A)) = |\det(M)|m(A)$. This is exactly what the theorem states, since $|\det(D\phi(\mathbf{x}))| = |\det(M)|$.
Click to expand the proof for Step 1
We prove this by considering the parts of an affine map separately.
- Case 1: Translation ($\phi(\mathbf{x}) = \mathbf{x} + \mathbf{b}$): The Lebesgue measure is defined to be translation-invariant. Shifting a set does not change its volume. Therefore, $m(\phi(A)) = m(A + \mathbf{b}) = m(A)$. The Jacobian matrix is the identity matrix $I$, so $\det(D\phi) = 1$. The formula holds: $m(A) = |1| \cdot m(A)$.
- Case 2: Linear Transformation ($\phi(\mathbf{x}) = M\mathbf{x}$): From linear algebra, any invertible matrix $M$ can be written as a product of elementary matrices, $M = E_k E_{k-1} \dots E_1$. We only need to prove the formula for these elementary matrices.
- Permutation (swaps two coordinates): Swapping axes does not change the volume of a set, so $m(E(A)) = m(A)$. The determinant is $\det(E) = -1$. The formula holds: $m(E(A)) = |-1| \cdot m(A) = m(A)$.
- Scaling (multiplies one coordinate by $c \neq 0$): This stretches the set along one axis by a factor of $|c|$, which scales the volume by $|c|$. So, $m(E(A)) = |c|m(A)$. The determinant is $\det(E) = c$. The formula holds: $m(E(A)) = |c|m(A)$.
- Shear (adds a multiple of one coordinate to another): A shear "tilts" a set. By Fubini's theorem, integrating slice-by-slice, we can show that the volume does not change. So, $m(E(A)) = m(A)$. The determinant is $\det(E) = 1$. The formula holds: $m(E(A)) = |1| \cdot m(A) = m(A)$.
- Case 3: General Affine Map ($\phi(\mathbf{x}) = M\mathbf{x} + \mathbf{b}$): This is a composition of a linear map $L(\mathbf{x}) = M\mathbf{x}$ and a translation $T(\mathbf{y}) = \mathbf{y} + \mathbf{b}$. Using the previous cases: $$m(\phi(A)) = m(T(L(A))) = m(L(A)) = |\det(M)|m(A)$$ The formula is proven for all affine transformations.
Step 2: Local Approximation by Linear Maps
The Idea: A general smooth function is not linear. However, if we zoom in very close to a point, it looks almost linear. This step proves that the "infinitesimal" scaling factor for volume at any point $\mathbf{x}$ is exactly the absolute value of the Jacobian determinant at that point, $|\det(D\phi(\mathbf{x}))|$.
The Rigorous Goal: We will prove that for almost every $\mathbf{x} \in \Omega$, the ratio of the volume of the image of a small ball to the volume of the ball itself approaches the Jacobian determinant as the ball shrinks to zero:
$$\lim_{r \to 0} \frac{m(\phi(B(\mathbf{x},r)))}{m(B(\mathbf{x},r))} = |\det(D\phi(\mathbf{x}))|$$
Click to expand the proof for Step 2
Let $\mathbf{x}$ be a point in $\Omega$. The affine approximation of $\phi$ near $\mathbf{x}$ is given by $\phi(\mathbf{x}+\mathbf{h}) \approx \phi(\mathbf{x}) + D\phi(\mathbf{x})\mathbf{h}$. To isolate the non-linear part of $\phi$, we define an auxiliary function $\psi$: $$\psi(\mathbf{h}) = (D\phi(\mathbf{x}))^{-1}(\phi(\mathbf{x}+\mathbf{h}) - \phi(\mathbf{x}))$$ By construction, $\psi(\mathbf{0}) = \mathbf{0}$ and its derivative is $D\psi(\mathbf{0}) = I$. The fact that the derivative is the identity matrix implies that $\psi$ preserves volume locally at the origin, which we prove in the following lemma.
Lemma: Local Volume Preservation
Statement: If $\psi: \mathbf{R}^n \to \mathbf{R}^n$ is differentiable at $\mathbf{0}$ with $\psi(\mathbf{0})=\mathbf{0}$ and $D\psi(\mathbf{0})=I$, then $\lim_{r\to 0} \frac{m(\psi(B(\mathbf{0},r)))}{m(B(\mathbf{0},r))} = 1$.
Click to expand the proof of the lemma
Proof: Define the error term $E(\mathbf{x}) := \psi(\mathbf{x}) - \mathbf{x}$. Since $D\psi(\mathbf{0}) = I$, we know $E(\mathbf{x}) = o(\lVert \mathbf{x}\rVert)$. This means for every $\varepsilon > 0$, there exists an $r_0 > 0$ such that if $\lVert\mathbf{x}\rVert \le r_0$, then $\lVert E(\mathbf{x})\rVert \le \varepsilon\lVert\mathbf{x}\rVert$.
By the triangle inequality, for $\lVert\mathbf{x}\rVert \le r_0$: $$\lVert\psi(\mathbf{x})\rVert = \lVert\mathbf{x} + E(\mathbf{x})\rVert \le \lVert\mathbf{x}\rVert + \lVert E(\mathbf{x})\rVert \le \lVert\mathbf{x}\rVert + \varepsilon\lVert\mathbf{x}\rVert = (1+\varepsilon)\lVert\mathbf{x}\rVert$$ $$\lVert\mathbf{x}\rVert = \lVert\psi(\mathbf{x}) - E(\mathbf{x})\rVert \le \lVert\psi(\mathbf{x})\rVert + \lVert E(\mathbf{x})\rVert \le \lVert\psi(\mathbf{x})\rVert + \varepsilon\lVert\mathbf{x}\rVert \implies (1-\varepsilon)\lVert\mathbf{x}\rVert \le \lVert\psi(\mathbf{x})\rVert$$ So we have the two-sided bound $(1-\varepsilon)\lVert\mathbf{x}\rVert \le \lVert\psi(\mathbf{x})\rVert \le (1+\varepsilon)\lVert\mathbf{x}\rVert$.
This shows that for a ball $B(\mathbf{0}, r)$ with $r \le r_0$, its image $\psi(B(\mathbf{0},r))$ contains the ball $B(\mathbf{0}, (1-\varepsilon)r)$ and is contained within the ball $B(\mathbf{0}, (1+\varepsilon)r)$. Comparing their measures and letting $\omega_n = m(B(\mathbf{0},1))$: $$m(B(\mathbf{0}, (1-\varepsilon)r)) \le m(\psi(B(\mathbf{0},r))) \le m(B(\mathbf{0}, (1+\varepsilon)r))$$ $$\omega_n ((1-\varepsilon)r)^n \le m(\psi(B(\mathbf{0},r))) \le \omega_n ((1+\varepsilon)r)^n$$ Dividing by $m(B(\mathbf{0},r)) = \omega_n r^n$, we get: $$(1-\varepsilon)^n \le \frac{m(\psi(B(\mathbf{0},r)))}{m(B(\mathbf{0},r))} \le (1+\varepsilon)^n$$ Since $\varepsilon > 0$ can be made arbitrarily small by choosing $r$ small enough, letting $r \to 0$ forces the ratio to approach 1 by the Squeeze Theorem. This completes the proof of the lemma.
Now, we rearrange the definition of $\psi$ to express $\phi$ in terms of $\psi$: $$\phi(\mathbf{x}+\mathbf{h}) = \phi(\mathbf{x}) + D\phi(\mathbf{x})\psi(\mathbf{h})$$ This shows that the image of a small ball $B(\mathbf{x}, r)$ under $\phi$ is just a translation of the image of $\psi(B(\mathbf{0},r))$ under the linear map $D\phi(\mathbf{x})$. Using the properties from Step 1 (translation invariance and scaling for linear maps): $$m(\phi(B(\mathbf{x},r))) = m( D\phi(\mathbf{x})[\psi(B(\mathbf{0},r))] ) = |\det(D\phi(\mathbf{x}))| \cdot m(\psi(B(\mathbf{0},r)))$$ Now we can evaluate our limit: $$\lim_{r \to 0} \frac{m(\phi(B(\mathbf{x},r)))}{m(B(\mathbf{x},r))} = \lim_{r \to 0} \frac{|\det(D\phi(\mathbf{x}))| \cdot m(\psi(B(\mathbf{0},r)))}{m(B(\mathbf{0},r))}$$ Since $m(B(\mathbf{x},r)) = m(B(\mathbf{0},r))$, this simplifies to: $$= |\det(D\phi(\mathbf{x}))| \cdot \lim_{r \to 0} \frac{m(\psi(B(\mathbf{0},r)))}{m(B(\mathbf{0},r))} = |\det(D\phi(\mathbf{x}))| \cdot 1$$ This completes the proof of the local scaling property.
Step 3: The Radon–Nikodym Theorem
The Idea: We now know the correct scaling factor at every single point. The challenge is to combine all these local factors to get a global formula for the volume of any set. We use a powerful machine from measure theory, the Radon–Nikodym theorem. This tool allows us to define a new measure for the volume in the output space and prove that its "density" (or rate of change) with respect to the input volume is exactly the Jacobian determinant we found in Step 2.
The Rigorous Goal: We define a new measure $\mu$ on the domain $\Omega$ by letting $\mu(A) = m(\phi(A))$. We will show that $\mu$ is absolutely continuous with respect to $m$ (written $\mu \ll m$). Then, the Radon-Nikodym theorem guarantees there is a density function $g$ such that $\mu(A) = \int_A g \, dm$. Finally, we will use the Lebesgue Differentiation Theorem to prove that this density is exactly $g(\mathbf{x}) = |\det(D\phi(\mathbf{x}))|$.
Click to expand the proof for Step 3
We begin by defining a new measure $\mu$ on the domain $\Omega$ for any measurable set $A \subset \Omega$: $$\mu(A) = m(\phi(A))$$
Theorem: Radon–Nikodym (see Tao, 245B Notes 1, Thm 2 & Cor 1)
Let $\mu$ and $m$ be $\sigma$-finite measures. If $\mu$ is absolutely continuous with respect to $m$ ($\mu \ll m$), meaning that for every set $A$, $m(A) = 0$ implies $\mu(A) = 0$, then there exists a non-negative measurable function $g$ such that for any measurable set $A$:
The function $g$ is called the Radon–Nikodym derivative, written $g = \frac{d\mu}{dm}$.
To apply this theorem, we must check that $\mu \ll m$. If $m(A)=0$, we need to show $\mu(A) = m(\phi(A))=0$. Since $\phi$ is a $C^1$ function, it is locally Lipschitz. A key property of Lipschitz functions is that they map sets of measure zero to sets of measure zero. Thus, the condition holds and the Radon-Nikodym theorem applies.
The theorem guarantees there is a density function $g$ such that $\mu(A) = \int_A g(\mathbf{x}) \, dm(\mathbf{x})$. To find out what $g$ is, we use the Lebesgue Differentiation Theorem, which states that we can recover the density by differentiating the measure:
This is exactly the limit we calculated in Step 2! We have therefore identified the density function as $g(\mathbf{x}) = |\det(D\phi(\mathbf{x}))|$ almost everywhere. This proves the formula for sets: $$m(\phi(A)) = \int_A |\det(D\phi(\mathbf{x}))| \, dm(\mathbf{x})$$
Step 4: Extension to Arbitrary Functions
The Idea: So far, we have a formula for how the volume of a set changes. The final step is to extend this result from sets to integrals of any function. We do this by building up from the simplest functions (which are 1 on a set and 0 elsewhere) to more complex ones.
The Rigorous Goal: We use a standard four-step process. We prove the formula for (1) indicator functions, then extend it by linearity to (2) simple functions, then use the Monotone Convergence Theorem for (3) non-negative measurable functions, and finally for (4) all integrable functions by splitting them into their positive and negative parts.
Click to expand the proof for Step 4
- Indicator Functions: Let $f = \chi_B$ be the indicator function for a measurable set $B \subset \Omega'$. Let $A = \phi^{-1}(B)$. The left-hand side (LHS) of the theorem is $\int_{\Omega'} \chi_B(\mathbf{y}) \, dm(\mathbf{y}) = m(B) = m(\phi(A))$. The right-hand side (RHS) is $\int_{\Omega} (\chi_B \circ \phi)(\mathbf{x}) |\det(D\phi)| \, dm(\mathbf{x}) = \int_{\Omega} \chi_A(\mathbf{x}) |\det(D\phi)| \, dm(\mathbf{x}) = \int_A |\det(D\phi)| \, dm(\mathbf{x})$. By Step 3, we know $m(\phi(A)) = \int_A |\det(D\phi)| \, dm(\mathbf{x})$, so the formula holds.
- Simple Functions: A simple function is a finite sum $s(\mathbf{y}) = \sum_{i=1}^k c_i \chi_{B_i}(\mathbf{y})$, where $c_i \ge 0$. By the linearity of the integral, the formula holds for $s$ because it holds for each $\chi_{B_i}$.
- Non-negative Measurable Functions: For any non-negative measurable function $f$, there exists a sequence of simple functions $s_n$ such that $0 \le s_n(\mathbf{y}) \le s_{n+1}(\mathbf{y})$ and $\lim_{n \to \infty} s_n(\mathbf{y}) = f(\mathbf{y})$. We know the formula holds for each $s_n$: $$\int_{\Omega'} s_n(\mathbf{y}) \, dm(\mathbf{y}) = \int_{\Omega} (s_n \circ \phi)(\mathbf{x}) |\det(D\phi)| \, dm(\mathbf{x})$$ By applying the Monotone Convergence Theorem to both sides as $n \to \infty$, we get the formula for $f$.
- General Integrable Functions: Any integrable function $f$ can be written as $f = f^+ - f^-$, where $f^+ = \max(f,0)$ and $f^- = \max(-f,0)$ are non-negative. We apply the formula to $f^+$ and $f^-$ separately and subtract the results. Linearity of the integral gives the final result for all integrable functions.
4. Problems for Practice
Exercise 1: Computational Mastery via Clever Transformations
The true power of the change of variables formula lies in identifying non-obvious transformations that dramatically simplify a problem. The following integrals test this skill.
- Region Transformation: Let $D$ be the region in the first quadrant bounded by the curves $xy=1$, $xy=3$, $x^2-y^2=1$, and $x^2-y^2=4$. Evaluate the integral: $$ I = \iint_{D} (x^4 - y^4) \,dx\,dy $$ (Hint: Consider the transformation $u = x^2 - y^2$ and $v = xy$.)
- Gaussian Integral with a Quadratic Form: Prove the identity: $$ J = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(ax^2 + 2bxy + cy^2)} \,dx\,dy = \frac{\pi}{\sqrt{ac-b^2}} $$ where $a > 0$ and $ac-b^2 > 0$. (Hint: Complete the square to identify a linear transformation that diagonalizes the quadratic form $ax^2 + 2bxy + cy^2$.)
- Dirichlet's Integral: Evaluate the triple integral $$ K = \iiint_V x^{l-1} y^{m-1} z^{n-1} \,dx\,dy\,dz $$ where $V$ is the region in the first octant defined by $x+y+z \le 1$, and $l,m,n > 0$. The result should be expressed in terms of the Gamma function. (Hint: Consider the transformation $x+y+z=u$, $y+z=uv$, $z=uvw$.)
Exercise 2: Volume of an n-dimensional Ellipsoid
Let $E$ be the solid ellipsoid in $\mathbb{R}^n$ defined by $\frac{x_1^2}{a_1^2} + \frac{x_2^2}{a_2^2} + \dots + \frac{x_n^2}{a_n^2} \le 1$, where $a_i > 0$. The volume of the unit ball $B_1(0)$ in $\mathbb{R}^n$ is $V_n = \frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}$.
- Find the Transformation: Construct a linear transformation $\phi: \mathbb{R}^n \to \mathbb{R}^n$ that maps the unit ball $B_1(0)$ precisely onto the ellipsoid $E$.
- Calculate the Volume: Using the Change of Variables formula, prove that the volume of the ellipsoid $E$ is $m(E) = (a_1 a_2 \cdots a_n) V_n$.
Exercise 3: The Coarea Formula (A Glimpse of Geometric Measure Theory)
The coarea formula is a profound generalization of Fubini's theorem. This problem asks you to prove a special case. Let $f: \mathbb{R}^n \to \mathbb{R}$ be a smooth function with $\nabla f \neq 0$ a.e. Prove that for any integrable function $g$:
Here $d\sigma$ is the $(n-1)$-dimensional surface measure on the level set $\{y \mid f(y)=t\}$. (Hint: Consider a change of coordinates $\Phi(x_1, \dots, x_n) = (x_1, \dots, x_{n-1}, f(x_1, \dots, x_n))$. Calculate its Jacobian.)
Exercise 4: Generalization for Non-Injective Maps
Suppose $\phi: \Omega \to \Omega'$ is a $C^1$ proper map with $\det(D\phi(x)) \neq 0$. Assume for almost every $y \in \Omega'$, the preimage $\phi^{-1}(y)$ consists of exactly $N$ distinct points (an $N$-sheeted covering map). Prove that the correct change of variables formula is:
(Hint: Use the Inverse Function Theorem to argue that $\Omega$ can be partitioned into $N$ disjoint open sets $V_i$ where $\phi$ is a diffeomorphism on each.)
Exercise 5: The Blaschke-Petkantschin Formula
Integral geometry is filled with beautiful change of variables formulas. Any three non-collinear points $x_1, x_2, x_3 \in \mathbb{R}^2$ define a line $L$ (through $x_1, x_2$), coordinates $s_1, s_2$ along $L$, and a signed height $h_3$ of $x_3$ from $L$. This defines a map $\Phi(x_1, x_2, x_3) = (L, s_1, s_2, h_3)$. Prove that the volume elements are related by:
Note that $|s_2 - s_1||h_3|$ is twice the area of the triangle $\triangle(x_1, x_2, x_3)$.