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Jacobians and General Substitutions · Tutorial 691 of 999

Three-Variable Jacobians by Determinant Expansion

Compute three-variable Jacobians carefully, check them with determinant-preserving row operations, and use their absolute values to transform triple integrals.

Advanced 10 min read

What You'll Learn

  • Arrange the partial derivatives of a three-variable map in its Jacobian matrix.
  • Expand a three-by-three determinant along a chosen row or column.
  • Use row replacements to simplify a Jacobian determinant without changing its value.
  • Check a determinant by comparing expansion and row-operation calculations.
  • Identify conditions needed to apply the change-of-variables formula.
  • Set up and evaluate triple integrals over parameterized regions.

Why Three-Variable Jacobians Need Care

A substitution in a triple integral replaces the original coordinates \(x,y,z\) with parameters \(u,v,w\). The substitution changes not only the formula for the integrand but also the volume of small regions. The factor that accounts for this local volume change is the absolute value of a three-variable Jacobian determinant.

In two dimensions, a Jacobian determinant has four entries. In three dimensions, it has nine, and sign errors in a cofactor expansion can be easy to miss. A reliable calculation starts by arranging the partial derivatives consistently, then uses expansion or determinant-preserving row operations, and finally checks the result by an independent calculation when possible.

The Jacobian Matrix and Determinant

Let a differentiable coordinate map \(T\) be given by \(x=x(u,v,w)\), \(y=y(u,v,w)\), and \(z=z(u,v,w)\). Its Jacobian matrix has one row for each output coordinate and one column for each parameter:

$$ \mathbf{J}_T(u,v,w) = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w}\\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{pmatrix}. $$

The Jacobian determinant, also called the determinant of the derivative matrix, is denoted \(\det(\mathbf{J}_T)\), or sometimes \(\frac{\partial(x,y,z)}{\partial(u,v,w)}\). The order of the coordinates matters: rows correspond to \(x,y,z\), and columns correspond to \(u,v,w\).

Definition: For a differentiable map \(T(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))\), its Jacobian determinant is \(\det(\mathbf{J}_T)\), where \(\mathbf{J}_T\) is the matrix of first partial derivatives arranged as above.

For a matrix with entries arranged as \(\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}\), expansion along the first row gives

$$ \det \begin{pmatrix} a&b&c\\ d&e&f\\ g&h&i \end{pmatrix} = a(ei-fh)-b(di-fg)+c(dh-eg). $$

Each entry is multiplied by the determinant of the \(2\times2\) matrix left after deleting its row and column. The signs across the first row are \(+,-,+\). Expansion along another row or column is valid as well, provided its signs are handled correctly.

Why Cofactor Expansion Works

The determinant measures oriented volume scaling: a linear map sends a small box to a parallelepiped, and the determinant gives its signed volume factor. The cofactor formula is a way of calculating that factor from the matrix entries.

For a \(3\times3\) matrix with rows \(\mathbf{r}_1,\mathbf{r}_2,\mathbf{r}_3\), its determinant can be viewed as the scalar triple product \(\mathbf{r}_1\cdot(\mathbf{r}_2\times\mathbf{r}_3)\). The cross product \(\mathbf{r}_2\times\mathbf{r}_3\) has components \(ei-fh,\ fg-di,\ dh-eg\). Taking its dot product with \((a,b,c)\) gives \(a(ei-fh)+b(fg-di)+c(dh-eg)\), which is exactly \(a(ei-fh)-b(di-fg)+c(dh-eg)\). Thus the alternating signs in cofactor expansion follow directly from the scalar triple product formula.

Row operations can make a determinant easier to calculate. Replacing one row by that row plus a multiple of another row leaves the determinant unchanged. Swapping two rows reverses its sign, and multiplying one row by a constant multiplies the determinant by that constant. In particular, row replacements can create zeros without changing the determinant.

Useful rule: A row replacement of the form \(R_j\leftarrow R_j+cR_k\), with \(j\ne k\), preserves the determinant. Track row swaps and row scaling explicitly; they change the determinant.

Worked Example: A Constant Three-Variable Jacobian

Worked Example: A Constant Three-Variable Jacobian

Consider the map \(x=u+v\), \(y=v+w\), and \(z=u+2w\). Compute its Jacobian determinant by cofactor expansion, then verify it using a row replacement.

The partial derivatives form the matrix

$$ \mathbf{J}_T= \begin{pmatrix} 1&1&0\\ 0&1&1\\ 1&0&2 \end{pmatrix}. $$

Expand along the first row. The first minor has determinant \(1\cdot2-1\cdot0=2\). The second minor has determinant \(0\cdot2-1\cdot1=-1\). The third term is zero. Therefore

$$ \det(\mathbf{J}_T) =1(2)-1(-1)+0 =2+1 =3. $$

For a check, replace the third row by the third row minus the first row. This row replacement preserves the determinant:

$$ \begin{pmatrix} 1&1&0\\ 0&1&1\\ 1&0&2 \end{pmatrix} \longrightarrow \begin{pmatrix} 1&1&0\\ 0&1&1\\ 0&-1&2 \end{pmatrix}. $$

Expanding the resulting matrix along its first column gives \(1\) times the determinant of \(\begin{pmatrix}1&1\\-1&2\end{pmatrix}\), which is \(1\cdot2-1(-1)=3\). The two calculations agree.

Worked Example: A Nonlinear Map and Its Jacobian

Worked Example: A Nonlinear Map and Its Jacobian

Let \(x=u+v\), \(y=u-v\), and \(z=2w+uv\). Find the Jacobian determinant and check it using a row replacement.

Differentiate each coordinate with respect to each parameter:

$$ \mathbf{J}_T(u,v,w)= \begin{pmatrix} 1&1&0\\ 1&-1&0\\ v&u&2 \end{pmatrix}. $$

Expansion along the first row gives

$$ \begin{aligned} \det(\mathbf{J}_T) &=1\bigl((-1)(2)-0\cdot u\bigr) -1\bigl(1\cdot2-0\cdot v\bigr) +0\\ &=-2-2\\ &=-4. \end{aligned} $$

To check, replace row two by row two minus row one, and row three by row three minus \(v\) times row one. Both operations preserve the determinant. The resulting matrix is

$$ \begin{pmatrix} 1&1&0\\ 0&-2&0\\ 0&u-v&2 \end{pmatrix}. $$

Expansion down the first column gives the determinant of the lower-right \(2\times2\) matrix: \((-2)(2)-0(u-v)=-4\). Thus \(\det(\mathbf{J}_T)=-4\) at every parameter point. The map reverses orientation, while its local volume scale is \(|-4|=4\).

Using a Jacobian in a Triple Integral

Suppose \(T\) maps a parameter region \(U\) one-to-one onto a region \(E\), is continuously differentiable, and has nonzero Jacobian determinant in the interior of \(U\). Under the usual change-of-variables conditions on the regions and integrand, the triple integral over \(E\) can be written over \(U\): substitute the coordinate formulas into the integrand and multiply by \(|\det(\mathbf{J}_T)|\). The absolute value is used because ordinary volume is nonnegative, even when the map reverses orientation.

$$ \iiint_E f(x,y,z)\,dV = \iiint_U f\bigl(x(u,v,w),y(u,v,w),z(u,v,w)\bigr) \left|\det(\mathbf{J}_T(u,v,w))\right|\,du\,dv\,dw. $$

Before using this formula, identify the parameter region \(U\), verify that the map is one-to-one on the region being used, compute the Jacobian, and determine the transformed integrand and bounds. The nonzero determinant condition ensures local invertibility in the interior; it does not by itself guarantee that the map is globally one-to-one, so that must be checked separately.

Worked Example: Transforming a Triple Integral

Worked Example: Transforming a Triple Integral

Use \(x=u+v\), \(y=u-v\), and \(z=2w+uv\) to evaluate the integral of \(x+y+z\) over the image of \(U=[0,1]\times[0,1]\times[0,1]\).

The inverse relations for \(u\) and \(v\) follow from \(x+y=2u\) and \(x-y=2v\): \(u=(x+y)/2\) and \(v=(x-y)/2\). Then \(w=(z-uv)/2\). Thus each image point determines at most one parameter triple. The determinant computed above is \(-4\), so its absolute value is \(4\).

Substitute the coordinate formulas into the integrand:

$$ x+y+z=(u+v)+(u-v)+(2w+uv)=2u+2w+uv. $$

The change-of-variables formula gives

$$ \iiint_E(x+y+z)\,dV = \int_0^1\int_0^1\int_0^1 (2u+2w+uv)\,4\,dw\,dv\,du. $$

Evaluate the integral by separating the three terms. On the unit cube, \(\int_0^1\int_0^1\int_0^1 2u\,dw\,dv\,du=\int_0^1 2u\,du=1\). Similarly, the integral of \(2w\) is \(\int_0^1 2w\,dw=1\). The integral of \(uv\) is \((\int_0^1u\,du)(\int_0^1v\,dv)(\int_0^1dw)=(1/2)(1/2)(1)=1/4\). Hence

$$ \iiint_E(x+y+z)\,dV =4\left(1+1+\frac14\right) =4\cdot\frac94 =9. $$

The result is checked by substituting the transformed integrand and the absolute Jacobian into the formula: the parameter integral equals \(4(9/4)=9\). Omitting the absolute value would instead produce \(-9\), which cannot represent the integral of this positive integrand over the image region.

Worked Example: Cofactor Expansion and an Integral Setup

Worked Example: Cofactor Expansion and an Integral Setup

Consider the linear map \(x=u+v+w\), \(y=u+2v+3w\), and \(z=2u-v+w\), and let \(U=[0,1]^3\). Set up and evaluate the integral of \(x\) over the image \(E=T(U)\).

The Jacobian matrix is constant:

$$ \mathbf{J}_T= \begin{pmatrix} 1&1&1\\ 1&2&3\\ 2&-1&1 \end{pmatrix}. $$

Expanding along the first row gives

$$ \begin{aligned} \det(\mathbf{J}_T) &=1\bigl(2\cdot1-3(-1)\bigr) -1\bigl(1\cdot1-3\cdot2\bigr) +1\bigl(1(-1)-2\cdot2\bigr)\\ &=1(5)-1(-5)+1(-5)\\ &=5+5-5\\ &=5. \end{aligned} $$

The determinant is nonzero, and a nonsingular linear map is one-to-one, so the change-of-variables formula applies. Since \(x=u+v+w\), the transformed integral is

$$ \iiint_E x\,dV = \int_0^1\int_0^1\int_0^1 (u+v+w)\,5\,dw\,dv\,du. $$

Each of \(u,v,w\) has integral \(1/2\) over the unit cube, with the other two variables contributing factors of \(1\). Therefore

$$ \iiint_E x\,dV =5\left(\frac12+\frac12+\frac12\right) =\frac{15}{2}. $$

As a check, the substituted integrand has integral \(3/2\) over \(U\), and the constant volume factor is \(5\), giving \(5(3/2)=15/2\).

A Common Sign Error

A frequent mistake is to forget the alternating signs when expanding along a row. For the matrix in the nonlinear example, the first-row expansion is not the sum of the two minor products. Its second term must be subtracted:

$$ 1(-2)+1(2) \quad\text{is incorrect, whereas}\quad 1(-2)-1(2)=-4. $$

The incorrect calculation gives \(0\), not \(-4\). A practical correction is to write the sign pattern \(+,-,+\) above the chosen row before evaluating its minors. Then check the result with a row replacement or a different expansion. Also distinguish the signed determinant from the volume factor: a negative determinant is valid, but the factor in a triple-integral volume element is its absolute value.

A Reliable Calculation Sequence

1
Build the matrix.
Place the partial derivatives of \(x,y,z\) in rows and derivatives with respect to \(u,v,w\) in columns.
2
Choose a calculation.
Expand along a row or column with convenient zeros, or use determinant-preserving row replacements to create zeros first.
3
Check the determinant.
Recalculate by another expansion or row-reduction path, carefully accounting for swaps and scaling.
4
Set up the transformed integral.
Check the map and its region, substitute into the integrand, and multiply by the absolute Jacobian determinant.

Why the Calculation Matters

A correct Jacobian is essential because it measures how parameter-space volume changes under the coordinate map. Even if the transformed bounds and integrand are otherwise correct, a missing factor or incorrect sign handling changes the integral. Writing the matrix in a fixed coordinate order and checking the determinant independently makes the setup more dependable.

Check Your Understanding

Use these questions to test both determinant calculations and integral setup.

  1. For a map \(T(u,v,w)=(x,y,z)\), which coordinate is represented by each row and each column of \(\mathbf{J}_T\)?
  2. What sign pattern is used when expanding a \(3\times3\) determinant along its first row?
  3. Why does replacing one row by that row plus a multiple of another row preserve a determinant?
  4. Why does a triple-integral substitution use the absolute value of the Jacobian determinant?
  5. What must be checked about a coordinate map in addition to computing a nonzero Jacobian determinant?

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