Change of Variables Theorem — Definition, Formula & Examples

The Change of Variables Theorem states that when you switch from one set of integration variables to another, the integral's value is preserved as long as you account for how the transformation stretches or compresses space, captured by the Jacobian determinant.

Let $T: \mathbb{R}^n \to \mathbb{R}^n$ be a continuously differentiable, injective transformation mapping a region $S$ onto a region $R$ . If $f$ is integrable over $R$ , then $\int_R f(\mathbf{x})\,d\mathbf{x} = \int_S f(T(\mathbf{u}))\,\left|\det\left(\frac{\partial T}{\partial \mathbf{u}}\right)\right|\,d\mathbf{u}$ , where $\frac{\partial T}{\partial \mathbf{u}}$ is the Jacobian matrix of $T$ .

Key Formula

\int_R f(\mathbf{x})\,d\mathbf{x} = \int_S f\bigl(T(\mathbf{u})\bigr)\,\left|\det J_T(\mathbf{u})\right|\,d\mathbf{u}

Where:

$R$ = Original region of integration in the x-coordinate system
$S$ = Transformed region of integration in the u-coordinate system
$T$ = The coordinate transformation mapping u to x
$J_T$ = Jacobian matrix of the transformation T
$|\det J_T|$ = Absolute value of the Jacobian determinant, the local scaling factor

How It Works

You choose a new set of variables that simplifies the region of integration or the integrand. Define a transformation

T

that maps the new variables to the old ones. Compute the Jacobian matrix of

T

and take the absolute value of its determinant — this factor corrects for the distortion the transformation introduces. Replace the integrand, the differential, and the bounds, then evaluate the simpler integral. In single-variable calculus, this reduces to

u

-substitution, where the Jacobian is simply

|du/dx|

Worked Example

Problem: Evaluate

\iint_R (x^2 + y^2)\,dA

where

R

is the disk

x^2 + y^2 \le 4

, using polar coordinates.

Step 1: Define the transformation: Use

x = r\cos\theta

y = r\sin\theta

. The disk becomes

0 \le r \le 2

0 \le \theta \le 2\pi

Step 2: Compute the Jacobian: The Jacobian determinant for polar coordinates is

r

\left|\det J_T\right| = \left|\det \begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{pmatrix}\right| = r

Step 3: Rewrite and evaluate: Replace

x^2 + y^2

with

r^2

and

dA

with

r\,dr\,d\theta

\int_0^{2\pi}\int_0^{2} r^2 \cdot r\,dr\,d\theta = \int_0^{2\pi}\int_0^{2} r^3\,dr\,d\theta = \int_0^{2\pi} 4\,d\theta = 8\pi

Answer:

8\pi

Why It Matters

This theorem is essential in multivariable calculus courses whenever a region has circular, spherical, or other non-rectangular symmetry. Physicists use it routinely when converting between Cartesian, polar, cylindrical, and spherical coordinates to compute quantities like mass, charge, and flux.

Common Mistakes

Mistake: Forgetting the absolute value of the Jacobian determinant when substituting.

Correction: The Jacobian determinant can be negative (reflecting an orientation reversal), but the scaling factor for area or volume must be positive. Always take the absolute value

|\det J_T|