A Three-Dimensional Substitution Problem Workshop

From a Substitution to a Complete Integral

A three-dimensional substitution is useful when it turns a complicated solid or integrand into a more manageable one. But a correct solution requires more than replacing $x,y,z$ in the integrand. The map must be inverted, the region must be described in the new variables, and the volume element must be multiplied by the absolute value of the Jacobian determinant.

This workshop develops those steps through three complete problems. The examples use coupled variables: the new coordinates depend on more than one original coordinate, or the original coordinates depend on more than one new coordinate. Such substitutions are not products of three independent one-variable changes.

Change-of-variables formula: Suppose $\mathbf{r}(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))$ is one-to-one on a region $S$, has continuously differentiable components, and has nonzero Jacobian determinant in the interior of $S$. If $R=\mathbf{r}(S)$ and $f$ is integrable on $R$, then $$ \iiint_R f(x,y,z)\,dV = \iiint_S f\bigl(x(u,v,w),y(u,v,w),z(u,v,w)\bigr) \left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|\,du\,dv\,dw. $$

The absolute value is essential: volume is nonnegative whether the map preserves or reverses orientation. In each problem, also check that the proposed inverse is valid on the stated region. A determinant formula by itself does not establish that the map describes the region one-to-one.

Worked Example: A Coupled Linear Map

Worked Example: Mass in a Parallelepiped

Let $R$ be the image of the unit cube $0\leq u,v,w\leq 1$ under $u=x+y,\ v=y+z,\ w=z+x$. Find the volume of $R$ and its mass for density $\rho(x,y,z)=x+y+z$.

Step 1: Invert the map. From $u=x+y$, $v=y+z$, and $w=z+x$, adding the first and third equations and subtracting the second gives $2x=u-v+w$. The other coordinates follow in the same way: $$ x=\frac{u-v+w}{2},\qquad y=\frac{u+v-w}{2},\qquad z=\frac{-u+v+w}{2}. $$ Substitution verifies the inverse: $x+y=u$, $y+z=v$, and $z+x=w$. Thus the cube gives a one-to-one parametrization of $R$.

Step 2: Compute the Jacobian. The forward map has derivative matrix $$ \frac{\partial(u,v,w)}{\partial(x,y,z)} = \begin{pmatrix} 1&1&0\\ 0&1&1\\ 1&0&1 \end{pmatrix}, \qquad \det\frac{\partial(u,v,w)}{\partial(x,y,z)}=2. $$ Therefore $\left|\partial(x,y,z)/\partial(u,v,w)\right|=1/2$. The transformed bounds are precisely $0\leq u,v,w\leq1$.

Step 3: Find the volume. Set $f=1$ in the change-of-variables formula: $$ \operatorname{Vol}(R) =\int_0^1\int_0^1\int_0^1\frac12\,dw\,dv\,du =\frac12(1)(1)(1)=\frac12. $$

Step 4: Find the mass. Adding the inverse-coordinate formulas gives $x+y+z=(u+v+w)/2$. Including the Jacobian factor, the transformed mass integrand is $(u+v+w)/4$. Hence $$ m=\int_0^1\int_0^1\int_0^1\frac{u+v+w}{4}\,dw\,dv\,du =\frac14\left(\frac12+\frac12+\frac12\right) =\frac38. $$ Each half is the integral of one coordinate over the unit cube; for example, $\int_0^1\int_0^1\int_0^1u\,dw\,dv\,du=1/2$.

Check: The inverse gives $x+y+z=(u+v+w)/2$, which is nonnegative throughout the cube, as the stated density requires. The mass computation includes the volume factor $1/2$, and the volume result is the same cube volume multiplied by that factor.

A Nonlinear Map with Dependent Bounds

The next example illustrates a useful feature of triangular substitutions. They can untangle bounds in stages: first one variable is shifted, then a second variable is shifted using the first two. The Jacobian may be especially simple even though the map is nonlinear.

Worked Example: Mass in a Curved Solid

Let $R$ be described by $0\leq x\leq1$, $x^2\leq y\leq x^2+1$, and $x(y-x^2)\leq z\leq x(y-x^2)+1$. Find its mass when $\rho(x,y,z)=1+x+y+z$.

Step 1: Define and invert the substitution. Choose $$ u=x,\qquad v=y-x^2,\qquad w=z-x(y-x^2). $$ Solving in order gives $x=u$, $y=v+u^2$, and $z=w+uv$. Thus the forward map is $$ x=u,\qquad y=v+u^2,\qquad z=w+uv. $$ The two formulas are inverse: $y-x^2=v$, and $z-x(y-x^2)=w$.

Step 2: Transform the bounds. The $x$-bounds give $0\leq u\leq1$. The $y$-bounds give $0\leq y-x^2=v\leq1$. The $z$-bounds give $0\leq z-x(y-x^2)=w\leq1$. Therefore the transformed region is the unit cube.

Step 3: Calculate the Jacobian. Differentiating the forward map gives $$ \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{pmatrix} 1&0&0\\ 2u&1&0\\ v&u&1 \end{pmatrix}, \qquad \det\frac{\partial(x,y,z)}{\partial(u,v,w)}=1. $$ The transformed density is $$ 1+x+y+z=1+u+(v+u^2)+(w+uv) =1+u+v+u^2+w+uv. $$

Step 4: Integrate. With Jacobian factor $1$, integrate each term over the unit cube: $$ \begin{aligned} m &=\int_0^1\int_0^1\int_0^1 (1+u+v+u^2+w+uv)\,dw\,dv\,du\\ &=1+\frac12+\frac12+\frac13+\frac12+\frac14 =\frac{37}{12}. \end{aligned} $$ The individual terms use the unit-cube integrals of $1,u,v,u^2,w,$ and $uv$, respectively.

Check: The inverse substitutions recover $u=x$, $v=y-x^2$, and $w=z-x(y-x^2)$, so the transformed bounds correspond exactly to the original inequalities. The density is positive on $R$, since $x,y,z$ are nonnegative there, consistent with the positive mass $37/12$.

Why the Triangular Substitution Works

For the preceding example, the integral equality can be justified directly by three ordinary one-variable substitutions. This makes clear how the transformed bounds and the Jacobian fit together. Let $g$ be any continuous function on $R$. In the original iterated integral, fix $x,y$ and set $w=z-x(y-x^2)$. Then $dw=dz$, and the $z$-limits become $0$ and $1$. Next set $v=y-x^2$, so $dv=dy$ and the $y$-limits become $0$ and $1$. Finally set $u=x$, with $du=dx$. The result is $$ \begin{aligned} \iiint_R g(x,y,z)\,dV &=\int_0^1\int_{x^2}^{x^2+1} \int_{x(y-x^2)}^{x(y-x^2)+1}g(x,y,z)\,dz\,dy\,dx\\ &=\int_0^1\int_0^1\int_0^1 g(u,v+u^2,w+uv)\,dw\,dv\,du. \end{aligned} $$ Here the Jacobian is $1$, so this is exactly the change-of-variables formula for this map. The direct substitutions explain why the shifted limits become cube limits; the determinant calculation gives the same volume factor in the general formula.

Worked Example: A Variable Jacobian on a Positive Region

Worked Example: Volume and Mass in a Scaled Solid

Use the map $x=uv,\ y=uw,\ z=u$, with $1\leq u\leq2$ and $0\leq v,w\leq1$. Find the volume of its image and its mass for density $\rho(x,y,z)=x+y+z$.

Step 1: Invert the map and identify the image. Since $z=u\geq1$, division by $z$ is valid. The inverse is $$ u=z,\qquad v=\frac{x}{z},\qquad w=\frac{y}{z}. $$ Substitution gives $uv=z(x/z)=x$, $uw=z(y/z)=y$, and $u=z$. The image region is therefore $1\leq z\leq2,\ 0\leq x\leq z,\ 0\leq y\leq z$. These bounds also show the map is one-to-one.

Step 2: Calculate the Jacobian. The derivative matrix and determinant are $$ \frac{\partial(x,y,z)}{\partial(u,v,w)} = \begin{pmatrix} v&u&0\\ w&0&u\\ 1&0&0 \end{pmatrix}, \qquad \det\frac{\partial(x,y,z)}{\partial(u,v,w)}=u^2. $$ Since $u\geq1$, the absolute Jacobian factor is $u^2$.

Step 3: Find the volume. $$ \operatorname{Vol}(R) =\int_1^2\int_0^1\int_0^1u^2\,dw\,dv\,du =\int_1^2u^2\,du =\left[\frac{u^3}{3}\right]_1^2 =\frac{8-1}{3}=\frac73. $$

Step 4: Find the mass. The density becomes $uv+uw+u=u(v+w+1)$. After multiplying by $u^2$, the integrand is $u^3(v+w+1)$. Thus $$ \begin{aligned} m &=\int_1^2\int_0^1\int_0^1u^3(v+w+1)\,dw\,dv\,du\\ &=\left(\int_1^2u^3\,du\right) \left(\int_0^1\int_0^1(v+w+1)\,dw\,dv\right)\\ &=\frac{16-1}{4}\left(\frac12+\frac12+1\right) =\frac{15}{4}\cdot2 =\frac{15}{2}. \end{aligned} $$

Check: In the original region, each cross-section at height $z$ is a square of side length $z$, so its area is $z^2$. Therefore the volume is $\int_1^2 z^2\,dz=7/3$, agreeing with the transformed calculation. At fixed $z$, integrating $x+y+z$ over $0\leq x,y\leq z$ gives $z^3/2+z^3/2+z^3=2z^3$. Then $\int_1^2 2z^3\,dz=15/2$, independently confirming the mass.

A Pitfall: Dropping a Variable Jacobian Factor

In the last example, the Jacobian is $u^2$, not a constant. If it is mistakenly omitted, the volume calculation becomes $$ \int_1^2\int_0^1\int_0^1 1\,dw\,dv\,du=1, $$ which is incorrect. The map stretches a small volume near larger $u$ more than one near smaller $u$; the factor $u^2$ accounts for that change in scale. Including it gives $\int_1^2u^2\,du=7/3$, in agreement with the direct cross-section check.

For a complete substitution problem, keep the roles of the calculations distinct: the inverse helps establish how the regions correspond, the transformed inequalities determine the new limits, and the absolute Jacobian converts volume. The density or integrand is then rewritten using the inverse map. Losing any one of these steps can produce an integral over the wrong region or with the wrong volume weighting.

Practical checklist: Write down the forward map and its inverse; verify them by substitution; translate every bound; calculate the Jacobian determinant and take its absolute value; transform the integrand; then check the result using a geometric or algebraic calculation when possible.

Check Your Understanding

For each question, identify the transformation, region, or volume factor that must be used.

For the map $u=x+y,\ v=y+z,\ w=z+x$, what is the absolute value of $\partial(x,y,z)/\partial(u,v,w)$?
Under the map $x=u,\ y=v+u^2,\ z=w+uv$, what original-coordinate expressions give $v$ and $w$?
For $x=uv,\ y=uw,\ z=u$, why is it valid to write $v=x/z$ on the stated region?
What volume factor must accompany integration in $u,v,w$ for the map $x=uv,\ y=uw,\ z=u$?
For a substitution whose Jacobian determinant is negative throughout the region, which Jacobian factor belongs in the volume integral?