From a Box to a Shifted Solid
A rectangular box is easy to describe with three independent interval bounds. A shifted, tilted box may be harder to describe directly in \(x,y,z\)-coordinates: its faces need not be parallel to the coordinate planes. An affine substitution handles both features at once. It translates a box and then applies an invertible linear map, producing a parallelepiped.
The translation identifies where the solid is located. The linear map controls the directions and lengths of its edges. The determinant of that map gives the volume scale, while the inverse substitution expresses an integrand in the simpler box coordinates.
Write a rectangular parameter box as \(B=[a_1,b_1]\times[a_2,b_2]\times[a_3,b_3]\). Its affine image is the set of all points \(\mathbf{x}=\mathbf{b}+\mathbf{A}\mathbf{u}\) with \(\mathbf{u}\in B\). When \(\mathbf{A}\) is invertible, this image is a parallelepiped. The translation does not change the shape: it moves every point by the same vector.
Edges, Invertibility, and Volume
The columns of \(\mathbf{A}\) describe the transformed coordinate directions. If these columns are \(\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\), then a change in \(u\) moves \(\mathbf{x}\) in the direction \(\mathbf{a}_1\), a change in \(v\) moves it in the direction \(\mathbf{a}_2\), and a change in \(w\) moves it in the direction \(\mathbf{a}_3\). Thus the box's edge directions become the columns of \(\mathbf{A}\).
If the parameter-box side lengths are \(L_1=b_1-a_1\), \(L_2=b_2-a_2\), and \(L_3=b_3-a_3\), then its transformed edge vectors are \(L_1\mathbf{a}_1\), \(L_2\mathbf{a}_2\), and \(L_3\mathbf{a}_3\). Their determinant gives the signed volume. Ordinary volume is nonnegative, so its value is the absolute value of that determinant.
This is the affine case of the change-of-variables theorem. Its Jacobian matrix is constant: differentiating \(\mathbf{x}=\mathbf{b}+\mathbf{A}\mathbf{u}\) with respect to \((u,v,w)\) gives \(\mathbf{A}\). The translation contributes no derivative, and the volume factor is therefore \(\lvert\det(\mathbf{A})\rvert\). Invertibility ensures that each point in the image comes from exactly one parameter point; this matters because the formula uses the image once, not with extra multiplicity.
Why the Determinant Gives the Volume Scale
For a small parameter box with side lengths \(\Delta u,\Delta v,\Delta w\), the transformed edge vectors are \(\Delta u\,\mathbf{a}_1,\Delta v\,\mathbf{a}_2,\Delta w\,\mathbf{a}_3\). The parallelepiped they span has signed volume equal to the determinant of the matrix with those vectors as columns:
Taking absolute values gives the ordinary volume. The same scale factor applies throughout the box because \(\mathbf{A}\) is constant. Adding the volumes of small pieces gives the factor \(\lvert\det(\mathbf{A})\rvert\) in a triple integral. This also shows why a translation changes neither local volumes nor the Jacobian.
Worked Example: Volume of a Shifted Parallelepiped
Worked Example: Volume of a Shifted Parallelepiped
Let \(B=[-1,1]\times[0,2]\times[1,3]\), and map it by \(\mathbf{x}=\mathbf{b}+\mathbf{A}\mathbf{u}\), where \(\mathbf{b}=(4,-2,1)^\top\) and \(\mathbf{A}=\begin{pmatrix}1&1&0\\0&2&1\\0&0&3\end{pmatrix}\). Find the volume of the image.
The box has side lengths \(2\), \(2\), and \(2\), so its volume is \(2\cdot2\cdot2=8\). The matrix is upper triangular, so its determinant is the product of its diagonal entries: \(\det(\mathbf{A})=1\cdot2\cdot3=6\). Therefore the image volume is \(\lvert6\rvert\cdot8=48\).
The columns of \(\mathbf{A}\) are \((1,0,0)^\top\), \((1,2,0)^\top\), and \((0,1,3)^\top\). Multiplying them by the respective side lengths shows that the image edges have vectors \((2,0,0)^\top\), \((2,4,0)^\top\), and \((0,2,6)^\top\). Their determinant is \(48\), confirming the volume directly. The translation by \((4,-2,1)^\top\) changes the location, not the edge vectors or volume.
Transforming a Polynomial Integrand
For an integral, transform both the region and the function. If \(f\) is given in \(x,y,z\), substitute the affine expressions for \(x,y,z\) to obtain \(f(\mathbf{b}+\mathbf{A}\mathbf{u})\). Then integrate over the original rectangular box and include the absolute determinant. For polynomial integrands, the transformed integrand remains a polynomial, which can often be integrated one variable at a time.
Specify the intervals for \(u,v,w\) that parameterize the solid.
Identify the translation vector and the matrix, and check that the determinant is nonzero.
Substitute into the integrand and multiply by \(\lvert\det(\mathbf{A})\rvert\).
Use the independent interval bounds and verify the result where possible.
Worked Example: An Integral of a Linear Polynomial
Worked Example: An Integral of a Linear Polynomial
Map the unit box \(B=[0,1]^3\) by \(x=2+u+v,\ y=-1+v+w,\ z=3+2w\). Evaluate \(\iiint_R(x+y+z)\,dV\), where \(R\) is the image of \(B\).
The matrix and translation vector are \(\mathbf{A}=\begin{pmatrix}1&1&0\\0&1&1\\0&0&2\end{pmatrix}\) and \(\mathbf{b}=(2,-1,3)^\top\). Thus \(\det(\mathbf{A})=1\cdot1\cdot2=2\), so the volume factor is \(2\). Substituting into the integrand gives \(x+y+z=(2+u+v)+(-1+v+w)+(3+2w)=4+u+2v+3w\).
Therefore, \(\iiint_R(x+y+z)\,dV=2\int_0^1\int_0^1\int_0^1(4+u+2v+3w)\,dw\,dv\,du\). The integral over the unit box is \(4+\frac12+2\cdot\frac12+3\cdot\frac12=7\), so the requested value is \(2\cdot7=14\).
As a check, the center of the parameter box is \((\frac12,\frac12,\frac12)\). Its image is \((3,0,4)\), whose coordinates sum to \(7\). A linear function's average over a box equals its value at the center, so the integral is \(7\) times the image volume \(2\), again giving \(14\).
Worked Example: A Quadratic Polynomial Integral
Worked Example: A Quadratic Polynomial Integral
Let \(R\) be the image of \(B=[0,1]^3\) under \(x=1+u+v,\ y=2+v,\ z=-1+u+2w\). Evaluate \(\iiint_R\big((x-y+1)^2+(z+1)^2\big)\,dV\).
The matrix is \(\mathbf{A}=\begin{pmatrix}1&1&0\\0&1&0\\1&0&2\end{pmatrix}\). Expanding along the first row gives \(\det(\mathbf{A})=1(2)-1(0)=2\), so the volume factor is \(2\). Under the substitution, \(x-y+1=(1+u+v)-(2+v)+1=u\), and \(z+1=u+2w\). The transformed integrand is \(u^2+(u+2w)^2\).
Expanding the square gives \(u^2+(u+2w)^2=2u^2+4uw+4w^2\). On the unit box, \(\iiint_Bu^2\,dV=\frac13\), \(\iiint_Buw\,dV=(\frac12)(\frac12)=\frac14\), and \(\iiint_Bw^2\,dV=\frac13\). Hence the parameter-box integral is \(2(\frac13)+4(\frac14)+4(\frac13)=\frac23+1+\frac43=3\). Multiplying by the Jacobian factor gives \(2\cdot3=6\).
For an arithmetic check, the two squared terms have integrals \(\iiint_Bu^2\,dV=\frac13\) and \(\iiint_B(u+2w)^2\,dV=\frac13+4(\frac14)+4(\frac13)=\frac83\). Their sum is \(3\), and the determinant factor again gives \(6\).
A Common Mistake: Keeping the Sign of the Determinant
The determinant records orientation as well as volume scaling. It can be negative when the map reverses orientation, but a geometric volume and the measure factor in a triple integral cannot be negative. For example, consider the map \(x=u,\ y=v,\ z=-2w\) from the unit box. Its matrix has determinant \(-2\). Using \(-2\) as a volume would give the nonsensical result that the image has volume \(-2\). The correction is to use the absolute value: the image volume is \(\lvert-2\rvert\cdot1=2\).
When This Method Applies
This method applies when the target solid is the image of a rectangular box under one invertible affine map. Check the determinant before using the inverse substitution: if \(\det(\mathbf{A})=0\), the map collapses at least one direction and does not give a one-to-one change of three-dimensional coordinates. Also make sure the parameter bounds describe the entire box whose image is the region of integration.
Affine substitutions are useful because they convert a solid with tilted faces into independent rectangular bounds. They also make the geometric effects transparent: the columns set the edge directions, the translation sets the location, and the absolute determinant scales both volume and integrals. For polynomial integrands, the resulting box integral can often be evaluated using elementary one-variable antiderivatives.
Check Your Understanding
Use the affine-map and change-of-variables formulas to answer these questions.
- What condition on \(\mathbf{A}\) ensures that \(\mathbf{x}=\mathbf{b}+\mathbf{A}\mathbf{u}\) is invertible?
- Why does the translation vector not affect the volume of the image?
- A box has side lengths \(1,2,3\), and the affine map has \(\lvert\det(\mathbf{A})\rvert=4\). What is the image volume?
- For an integral over an affine image, where does the factor \(\lvert\det(\mathbf{A})\rvert\) enter?
- If \(\det(\mathbf{A})=-3\) and the parameter box has volume \(5\), what is the image volume, and why?