The Change-of-Variables Theorem for Multiple Integrals

Why a Change of Variables Needs a Scale Factor

A substitution can make an integral easier by replacing a complicated region or integrand with a simpler one. But changing coordinates also changes the sizes of the small pieces being added together. A small rectangle in parameter coordinates may become a parallelogram or a curved patch in the original coordinates, so the integrand alone does not account for the same amount of area or volume.

The Jacobian determinant supplies the local scale factor. The change-of-variables theorem combines that factor with the transformed integrand and gives precise conditions under which the rewritten integral has the same value.

Central principle: If a smooth coordinate map is one-to-one and valid on the region being integrated over, then the integral in the original coordinates equals the integral in the parameter coordinates with the Jacobian determinant’s magnitude included.

The Theorem in Two and Three Dimensions

Let \(E\) be a bounded region in \(\mathbb{R}^n\), where \(n=2\) or \(n=3\), whose boundary has \(n\)-dimensional volume zero. Suppose \(E\) lies in an open set \(U\), and \(T:U\to V\) is a one-to-one, onto map with continuous first derivatives and a continuously differentiable inverse. Assume the determinant of its derivative matrix is nonzero on a neighborhood of \(E\). If \(f\) is continuous on \(T(E)\), the change-of-variables theorem states:

\int_{T(E)} f(\mathbf{x})\,d\mathbf{x} = \int_E f(T(\mathbf{u}))\, \left|\det DT(\mathbf{u})\right|\,d\mathbf{u}.

Here \(\mathbf{u}\) denotes the parameter coordinates and \(\mathbf{x}=T(\mathbf{u})\) the original coordinates. In two dimensions, write \(\mathbf{u}=(u,v)\) and \(\mathbf{x}=(x,y)\); then \(DT\) is a \(2\times2\) Jacobian matrix, and the differentials are \(du\,dv\) and \(dx\,dy\). In three dimensions, \(\mathbf{u}=(u,v,w)\) and \(\mathbf{x}=(x,y,z)\), and \(DT\) is a \(3\times3\) matrix.

\iint_{T(E)} f(x,y)\,dx\,dy = \iint_E f(x(u,v),y(u,v)) \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv,

and, in three dimensions,

\iiint_{T(E)} f(x,y,z)\,dx\,dy\,dz = \iiint_E f(x(u,v,w),y(u,v,w),z(u,v,w)) \left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|\,du\,dv\,dw.

The notation \(\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\), for example, means the absolute value of the determinant of the matrix of first partial derivatives. The hypotheses matter: the map must be sufficiently smooth, one-to-one on the region, and nonsingular there. If a map folds the region onto itself or has a singularity in the patch, this version of the theorem cannot simply be applied to the whole region.

Why the Jacobian Appears

At a point \(\mathbf{a}\), the derivative matrix \(DT(\mathbf{a})\) gives the best linear approximation to \(T\) near that point. For a small displacement \(\mathbf{h}\), differentiability says

T(\mathbf{a}+\mathbf{h}) = T(\mathbf{a})+DT(\mathbf{a})\mathbf{h} +o(\|\mathbf{h}\|).

The final term becomes negligible compared with \(\|\mathbf{h}\|\) as \(\mathbf{h}\) shrinks. Thus a sufficiently small box near \(\mathbf{a}\) is mapped approximately as the linear map \(DT(\mathbf{a})\). A linear map scales area or volume by the magnitude of its determinant. Therefore, a small parameter-space piece of volume \(\Delta V\) near \(\mathbf{a}\) maps to a piece with volume approximately \(\left|\det DT(\mathbf{a})\right|\Delta V\).

This local observation explains the integral formula. Partition \(E\) into small boxes. On a box near \(\mathbf{a}\), continuity makes \(f(T(\mathbf{u}))\) close to \(f(T(\mathbf{a}))\), while the linear approximation makes the image volume close to \(\left|\det DT(\mathbf{a})\right|\Delta V\). The contribution from that box is therefore approximately

f(T(\mathbf{a}))\left|\det DT(\mathbf{a})\right|\Delta V.

Adding these approximations over the partition gives Riemann sums for the integral on the right-hand side of the theorem. As the boxes become smaller, the approximation errors tend to zero. Because the map is one-to-one, distinct box interiors do not overlap in the image; their boundaries have zero volume and do not affect the integral. The sums therefore converge to the integral over \(T(E)\), establishing the stated formula.

Why this is a theorem, not just a substitution rule: The transformed integrand \(f(T(\mathbf{u}))\) gives the function values, while \(\left|\det DT(\mathbf{u})\right|\) corrects for the local change in area or volume. The smoothness, nonsingularity, and one-to-one hypotheses ensure that this local accounting applies consistently across the region.

Worked Examples

Worked Example: A Linear Map of a Square

Let \(E=[0,1]\times[0,1]\), let \(T(u,v)=(u+v,u-v)\), and integrate \(f(x,y)=x+y\) over \(T(E)\).

First compute the derivative matrix and determinant:

DT(u,v)= \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}, \qquad \det DT=(1)(-1)-(1)(1)=-2, \qquad |\det DT|=2.

The map is invertible: \(u=(x+y)/2\) and \(v=(x-y)/2\). Its determinant is nonzero, so it satisfies the theorem’s conditions on this square. Substitute \(x=u+v\) and \(y=u-v\), giving \(x+y=2u\). Thus

\begin{aligned} \iint_{T(E)}(x+y)\,dx\,dy &=\int_0^1\int_0^1 2u\cdot 2\,dv\,du\\ &=\int_0^1\int_0^1 4u\,dv\,du\\ &=\int_0^1 4u(1)\,du\\ &=\left[2u^2\right]_0^1=2. \end{aligned}

Check by describing the image directly. The inverse relations give \(0\le x+y\le2\) and \(0\le x-y\le2\). For \(0\le x\le1\), the image has \(-x\le y\le x\); for \(1\le x\le2\), it has \(x-2\le y\le2-x\). Hence

\begin{aligned} \iint_{T(E)}(x+y)\,dx\,dy &=\int_0^1\int_{-x}^{x}(x+y)\,dy\,dx +\int_1^2\int_{x-2}^{2-x}(x+y)\,dy\,dx\\ &=\int_0^1 2x^2\,dx+\int_1^2(4x-2x^2)\,dx\\ &=\frac{2}{3}+\frac{4}{3}=2. \end{aligned}

The direct calculation agrees with the transformed integral.

Worked Example: A Nonlinear Map on a Rectangle

Let \(E=[1,2]\times[0,1]\) and \(T(u,v)=(u,uv)\). Find the integral of \(f(x,y)=x\) over \(T(E)\).

The derivative matrix, determinant, and inverse are

DT(u,v)= \begin{pmatrix} 1&0\\ v&u \end{pmatrix}, \qquad \det DT=u, \qquad u=x,\quad v=\frac{y}{x}.

Throughout \(E\), \(u\ge1\), so the determinant is nonzero, and the displayed inverse shows the map is one-to-one. Since \(x=u\), the transformed integrand is \(u\), and the Jacobian magnitude is \(u\). Therefore

\begin{aligned} \iint_{T(E)}x\,dx\,dy &=\int_1^2\int_0^1 u\cdot u\,dv\,du\\ &=\int_1^2\int_0^1 u^2\,dv\,du\\ &=\int_1^2 u^2(1)\,du\\ &=\left[\frac{u^3}{3}\right]_1^2 =\frac{8-1}{3}=\frac{7}{3}. \end{aligned}

To check, the image is \(1\le x\le2\) and \(0\le y\le x\), since \(x=u\) and \(y=uv\) with \(0\le v\le1\). Integrating in these original coordinates gives

\int_1^2\int_0^x x\,dy\,dx = \int_1^2 x(x)\,dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{7}{3}.

Worked Example: A Three-Dimensional Scaling

Let \(E=[0,1]^3\), let \(T(u,v,w)=(2u,v,3w)\), and integrate \(f(x,y,z)=x+y+z\) over \(T(E)\).

The derivative matrix is diagonal, so

DT= \begin{pmatrix} 2&0&0\\ 0&1&0\\ 0&0&3 \end{pmatrix}, \qquad \det DT=2\cdot1\cdot3=6.

The map is one-to-one, with inverse \(u=x/2,\ v=y,\ w=z/3\). The transformed integrand is \(2u+v+3w\), so

\begin{aligned} \iiint_{T(E)}(x+y+z)\,dx\,dy\,dz &=6\int_0^1\int_0^1\int_0^1(2u+v+3w)\,dw\,dv\,du\\ &=6\left(2\cdot\frac12+\frac12+3\cdot\frac12\right)\\ &=6(1+\tfrac12+\tfrac32)=18. \end{aligned}

As a check, the image is the box \(0\le x\le2,\ 0\le y\le1,\ 0\le z\le3\). Integrating the three terms separately gives

\begin{aligned} \iiint_{T(E)}x\,dV&=\left(\int_0^2x\,dx\right)(1)(3)=2\cdot3=6,\\ \iiint_{T(E)}y\,dV&=(2)\left(\int_0^1y\,dy\right)(3)=2\cdot\tfrac12\cdot3=3,\\ \iiint_{T(E)}z\,dV&=(2)(1)\left(\int_0^3z\,dz\right)=2\cdot\tfrac92=9. \end{aligned}

Their sum is \(6+3+9=18\), as obtained by the theorem.

A Common Error: Omitting the Jacobian

In the nonlinear example, replacing \(x\) by \(u\) is not enough. A tempting but incorrect calculation is

\int_1^2\int_0^1 u\,dv\,du = \int_1^2 u\,du = \frac{3}{2}.

This is wrong because it treats each small \(du\,dv\) rectangle as if it had the same area after mapping. But the map \(T(u,v)=(u,uv)\) scales local area by \(u\); at larger values of \(u\), the same parameter rectangle produces a larger image patch. The corrected integrand includes that factor:

\int_1^2\int_0^1 u\cdot u\,dv\,du = \frac{7}{3}.

The check in the original coordinates confirms \(7/3\), not \(3/2\). In every application, transform the function and multiply by the Jacobian determinant’s magnitude before integrating over the parameter region.

Using the Theorem Reliably

The theorem is useful when a coordinate map makes either the region or the integrand simpler. Before applying it, verify the conditions and write down the pieces of the formula explicitly:

Identify the parameter region \(E\) and its image \(T(E)\).
Check that the map is smooth and one-to-one on the region, with a nonzero Jacobian determinant there.
Substitute the coordinate formulas into the integrand to obtain \(f(T(\mathbf{u}))\).
Compute \(\det DT(\mathbf{u})\), take its magnitude, and include it in the transformed integrand.
Integrate over \(E\), then check the result when a direct calculation is practical.

These requirements prevent two distinct failures: using a map that does not provide valid coordinates over the whole region, and using valid coordinates but assigning the wrong area or volume to their small pieces. The theorem handles both issues through its hypotheses and its Jacobian factor.

Check Your Understanding

Answer each question by identifying the relevant condition or calculation.

State the change-of-variables formula for a double integral and identify what the Jacobian factor measures.
For \(T(u,v)=(u,uv)\), why is the map one-to-one on \(1\le u\le2,\ 0\le v\le1\)?
What can go wrong if the determinant of \(DT\) vanishes at points in the parameter region?
For \(T(u,v,w)=(2u,v,3w)\), what is the volume-scaling factor?
In a transformed integral, which two changes must be made to the original integrand and area or volume element?