Common Jacobian Errors and Diagnostic Checks

Why Jacobian Checks Matter

A three-dimensional substitution changes more than the variables written in an integrand. It also changes the volume element, the region of integration, and the way the density is evaluated. An error in any one of these parts can produce a plausible-looking integral with the wrong value.

For a map $\Phi(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))$, its derivative matrix records how the original coordinates change as the parameters change. The determinant measures local volume scaling, while its sign records orientation. These are related, but they are not interchangeable: volume uses the absolute value of the determinant.

Change-of-variables formula: Let $\Phi:G\to D$ be a continuously differentiable, one-to-one map between regions in $\mathbb R^3$, with a continuously differentiable inverse and nonzero Jacobian determinant in the interior. For an integrable density $f$, $$ \iiint_D f(x,y,z)\,dV = \iiint_G f\bigl(x(u,v,w),y(u,v,w),z(u,v,w)\bigr) \left|\det D\Phi(u,v,w)\right|\,du\,dv\,dw. $$ Here $G$ is the parameter region and $D=\Phi(G)$ is the corresponding region in the original coordinates. If a map is not one-to-one, the region must be split into suitable one-to-one pieces, with care not to count the same part of $D$ more than once.

Why the Absolute Value Appears

The determinant of a linear map gives a signed volume factor. Its magnitude is the volume scaling factor; its sign distinguishes whether the map preserves or reverses orientation. Reversing orientation cannot make a physical volume negative, so the change-of-variables formula uses the absolute value.

This follows from the geometric meaning of the determinant. A linear map sends a small rectangular box to a parallelepiped. The determinant gives the parallelepiped’s signed volume relative to the original orientation, and its absolute value gives its ordinary, nonnegative volume. For a differentiable map, near a point the map is approximated by its derivative matrix. Thus the local volume factor is $\left|\det D\Phi\right|$; the change-of-variables formula accumulates these local factors over the parameter region.

Worked Example: A Map That Reverses Orientation

Let $G=[0,1]^3$, and define $\Phi(u,v,w)=(1-u,v,w)$. Its image is also the unit cube. Find the integral of the constant density $1$ over that image using the substitution, and diagnose what goes wrong if the determinant is used without an absolute value.

The derivative matrix is

D\Phi= \begin{pmatrix} -1&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}, \qquad \det D\Phi=(-1)(1)(1)=-1.

The correct volume factor is $|-1|=1$. Therefore

\iiint_D 1\,dV = \int_0^1\int_0^1\int_0^1 1\cdot 1\,dw\,dv\,du =1.

Without the absolute value, the same calculation would give $-1$, which cannot be the volume of the unit cube. As a check, $D=[0,1]^3$ has volume $1$, agreeing with the correctly transformed integral.

Check Every Entry Before Expanding the Determinant

The Jacobian matrix is not a collection of guessed scaling factors. Its entry in row $i$, column $j$ is the partial derivative of the $i$-th output coordinate with respect to the $j$-th parameter. Writing the coordinate functions above the matrix is a simple diagnostic: each row must differentiate one output coordinate, and each column must correspond to one parameter.

Worked Example: A Missing Partial Derivative

Consider $\Phi(u,v,w)=(u+v,v+w,w+u)$ on $G=[0,1]^3$. The map is linear and invertible if its determinant is nonzero. Find the correct volume factor, then see what happens if the $v$-derivative of the first coordinate is mistakenly set to zero.

Differentiate each output coordinate with respect to $u,v,w$:

D\Phi= \begin{pmatrix} \frac{\partial (u+v)}{\partial u}&\frac{\partial (u+v)}{\partial v}&\frac{\partial (u+v)}{\partial w}\\ \frac{\partial (v+w)}{\partial u}&\frac{\partial (v+w)}{\partial v}&\frac{\partial (v+w)}{\partial w}\\ \frac{\partial (w+u)}{\partial u}&\frac{\partial (w+u)}{\partial v}&\frac{\partial (w+u)}{\partial w} \end{pmatrix} = \begin{pmatrix} 1&1&0\\ 0&1&1\\ 1&0&1 \end{pmatrix}.

Expanding along the first row gives

\det D\Phi =1(1\cdot1-1\cdot0)-1(0\cdot1-1\cdot1)+0 =1-(-1)=2.

Since the determinant is nonzero, this linear map is invertible, and its volume factor is $2$. The image $D=\Phi(G)$ therefore has volume $2$, as the parameter cube has volume $1$.

If the entry $\partial(u+v)/\partial v=1$ is incorrectly written as $0$, the matrix becomes

\begin{pmatrix} 1&0&0\\ 0&1&1\\ 1&0&1 \end{pmatrix}, \qquad \det=1(1\cdot1-1\cdot0)=1.

This incorrect matrix predicts volume $1$, not $2$. The check is to recompute derivatives directly from the coordinate formulas, rather than rely on memory about which entries “should” be zero.

Bounds Belong to the Region Being Integrated Over

After a substitution, the parameters range over $G$, not automatically over the old numerical bounds for $x,y,z$. If the substitution maps a parameter cube to a different solid, keeping the old bounds may integrate over the wrong region. The reliable procedure is to identify the parameter region first and then express the transformed density and Jacobian on that region.

Worked Example: Using the Wrong Bounds

Use the map $\Phi(u,v,w)=(u+v,u-v,w)$ on $G=[0,1]^3$. Let $D=\Phi(G)$, and find $\iiint_D x\,dV$. Then compare the correct integral with the result from incorrectly using $0\le x,y,z\le1$ as the bounds for the transformed variables.

The derivative matrix and determinant are

D\Phi= \begin{pmatrix} 1&1&0\\ 1&-1&0\\ 0&0&1 \end{pmatrix}, \qquad \det D\Phi=1(-1)-1(1)=-2.

Thus the volume factor is $2$. The original density is $x$, and $x=u+v$, so the correct transformed integral is

\begin{aligned} \iiint_D x\,dV &=\int_0^1\int_0^1\int_0^1 2(u+v)\,dw\,dv\,du\\ &=2\int_0^1\int_0^1 (u+v)\,dv\,du\\ &=2\int_0^1\left(u+\frac12\right)\,du\\ &=2\left(\frac12+\frac12\right)=2. \end{aligned}

The parameter bounds are $0\le u,v,w\le1$. The image is not the unit cube: for example, $x=u+v$ ranges from $0$ to $2$. If one mistakenly treats $x,y,z$ as the parameters and integrates $2x$ over the unit cube, the result is

\int_0^1\int_0^1\int_0^1 2x\,dz\,dy\,dx =2\left(\frac12\right)=1.

This does not match the correct value $2$. The direct check is also consistent: the image has volume $2$, and by symmetry its average $x$-coordinate is $1$, so the integral of $x$ over it is $2$.

Check Whether the Map Covers the Region More Than Once

The one-to-one condition is essential. If different parameter points map to the same point in the original region, integrating over the entire parameter region can count that part of the original region repeatedly. A nonzero determinant alone does not guarantee global one-to-one behavior. When a map has multiple branches, restrict to a single branch or divide the parameter region into pieces whose images do not overlap in their interiors.

Worked Example: A Two-to-One Map

Let $\Phi(u,v,w)=(u^2,v,w)$ on $[-1,1]\times[0,1]\times[0,1]$. The image is the unit cube. Compare the cube’s volume with the integral obtained by using the full parameter region.

The derivative matrix is diagonal, so

D\Phi= \begin{pmatrix} 2u&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}, \qquad \left|\det D\Phi\right|=|2u|.

For every $x$ with $0<x\le1$, both $u=\sqrt{x}$ and $u=-\sqrt{x}$ map to $x=u^2$. Thus the map covers the cube twice, apart from the boundary where the two branches meet. Applying the formula naively over the full parameter region gives

\int_{-1}^1\int_0^1\int_0^1 2|u|\,dw\,dv\,du =\int_{-1}^1 2|u|\,du =2\left(\int_{-1}^0(-u)\,du+\int_0^1u\,du\right) =2\left(\frac12+\frac12\right)=2.

But the image is the unit cube, whose volume is $1$. Restricting to the branch $0\le u\le1$ gives a one-to-one map onto that cube, and

\int_0^1\int_0^1\int_0^1 2u\,dw\,dv\,du =\int_0^1 2u\,du =1.

This agrees with the cube’s volume. The check reveals that the full parameter region counted the image twice.

Compose the Density with the Substitution

The density in the transformed integral must be evaluated at the point $\Phi(u,v,w)$. In other words, it becomes $f(x(u,v,w),y(u,v,w),z(u,v,w))$, not the original formula with parameter letters substituted informally. Keeping the composition explicit helps prevent mixing original coordinates and parameter coordinates.

Worked Example: An Uncomposed Density

Let $\Phi(u,v,w)=(2u,v,w)$ on $G=[0,1]^3$, so $D=[0,2]\times[0,1]\times[0,1]$. For the original density $f(x,y,z)=x$, compute the integral and compare it with the result of forgetting to compose the density.

The matrix and volume factor are

D\Phi= \begin{pmatrix} 2&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}, \qquad |\det D\Phi|=2.

Since $x=2u$, the composed density is $f(\Phi(u,v,w))=2u$. Therefore

\begin{aligned} \iiint_D x\,dV &=\int_0^1\int_0^1\int_0^1 (2u)(2)\,dw\,dv\,du\\ &=\int_0^1 4u\,du =4\left(\frac12\right)=2. \end{aligned}

As a direct check in the original coordinates,

\int_0^1\int_0^1\int_0^2 x\,dx\,dy\,dz =\frac{2^2}{2}=2.

If one incorrectly uses $u$ as the density instead of $2u$, the transformed integral becomes $\int_0^1 2u\,du=1$. The direct calculation in the original region shows why that answer is wrong.

A Compact Diagnostic Routine

When checking a three-dimensional substitution, verify the pieces in an order that makes errors easy to localize:

Identify the map and regions.
Write $\Phi(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))$, name the parameter region $G$, and determine its image $D$.

Differentiate by rows.
Each row of $D\Phi$ differentiates one output coordinate with respect to all three parameters.

Use the volume factor.
Compute the determinant carefully and use its absolute value. A negative determinant is not a negative volume.

Check coverage and bounds.
Confirm that the map is one-to-one on the chosen parameter region, and use that region’s bounds in the transformed integral.

Compose the density.
Replace each original coordinate in $f$ by its expression in the parameters, then multiply by the absolute Jacobian determinant.

Why These Checks Are Useful

Each diagnostic tests a different part of the substitution. Comparing $\det D\Phi$ with a known volume can expose a missing absolute value or a derivative error. Mapping the parameter bounds to the original region can reveal an untransformed region. Checking whether two parameter points can yield the same output detects duplicated coverage. Finally, evaluating the transformed integral independently in the original coordinates, when practical, tests the density composition and the overall setup.

These checks do not replace the change-of-variables conditions. They help establish that the formula has been applied to the correct map and region. In particular, a determinant calculation cannot repair a map that is not one-to-one on the domain being used.

Check Your Understanding

For each question, identify which part of the substitution should be checked and explain why.

If a map has determinant $-3$ throughout the parameter region, what volume factor belongs in the integral?
For $\Phi(u,v,w)=(u+v,v,w)$, which derivative is the entry in row one, column two?
What condition can fail when two distinct parameter points map to the same point in the original region?
If the original density is $f(x,y,z)=x+z$ and $x=2u,\ z=w$, what is the composed density?
Why is it not generally valid to reuse the original-coordinate bounds as parameter bounds?