Composing Coordinate Changes in Stages

Why Compose Coordinate Changes?

Some regions and integrands are simpler after more than one change of coordinates. A first map may reshape a convenient parameter region into an intermediate region; a second may take that intermediate region to the region of integration. Keeping the two steps separate can make the geometry and the algebra easier to track.

The key calculation is governed by the chain rule. The Jacobian matrix of a composite map is a product of Jacobian matrices, evaluated at the appropriate points. Consequently, its determinant is the product of the two determinants. For a scalar triple integral, the corresponding volume factors are their absolute values.

Central rule: If $\mathbf{H}$ maps parameters $(r,s,t)$ to intermediate coordinates $(u,v,w)$, and $\mathbf{G}$ maps $(u,v,w)$ to $(x,y,z)$, then the composite map is $\mathbf{G}\circ\mathbf{H}$. Its Jacobian determinant is $\det D(\mathbf{G}\circ\mathbf{H})=\det D\mathbf{G}(\mathbf{H}(r,s,t))\det D\mathbf{H}(r,s,t)$. For a scalar volume integral, use the absolute value of this product.

Set Up the Two Maps

Let $Q$ be a region in parameter space. Suppose $\mathbf{H}:Q\to U$ maps it to an intermediate region $U$, and $\mathbf{G}:U\to D$ maps that region to the target region $D$. Write the maps in components as

\mathbf{H}(r,s,t) = \bigl(u(r,s,t),v(r,s,t),w(r,s,t)\bigr), \qquad \mathbf{G}(u,v,w) = \bigl(x(u,v,w),y(u,v,w),z(u,v,w)\bigr).

The final coordinates are obtained by substitution: $(x,y,z)=\mathbf{G}(\mathbf{H}(r,s,t))$. In particular, every entry of $D\mathbf{G}$ in the composite determinant is evaluated at $\mathbf{H}(r,s,t)$, not at $(r,s,t)$.

Conditions for the staged formula: Assume the maps are continuously differentiable and one-to-one on the regions used, with nonsingular Jacobian matrices in their interiors. Assume also that the regions have boundaries for which the change-of-variables theorem applies. Then, for an integrable function $f$, $$ \iiint_D f(x,y,z)\,dx\,dy\,dz = \iiint_Q f\bigl(\mathbf{G}(\mathbf{H}(r,s,t))\bigr) \left|\det D\mathbf{G}(\mathbf{H}(r,s,t))\right| \left|\det D\mathbf{H}(r,s,t)\right| \,dr\,ds\,dt. $$

The same integral can be calculated in two successive substitutions: first from $D$ to $U$, then from $U$ to $Q$. Each step transforms both the integrand and the volume element. The absolute determinant is required because the volume element measures nonnegative volume, whether a map preserves or reverses orientation.

Why the Jacobian Factors Multiply

The matrix chain rule gives the essential identity. The derivative of the composite map is the matrix product of the derivative of the outer map and the derivative of the inner map:

D(\mathbf{G}\circ\mathbf{H})(r,s,t) = D\mathbf{G}\bigl(\mathbf{H}(r,s,t)\bigr) D\mathbf{H}(r,s,t).

For square matrices, the determinant of a product is the product of the determinants. Taking determinants therefore gives the composite formula. Taking absolute values gives the volume factor in a scalar integral.

Differentiate the inner map.
$D\mathbf{H}(r,s,t)$ converts small changes in the parameters to small changes in the intermediate coordinates.

Differentiate the outer map at the intermediate point.
$D\mathbf{G}(\mathbf{H}(r,s,t))$ converts those intermediate changes to changes in $(x,y,z)$.

Take determinants.
The determinant of the matrix product is the product of the determinants, so the total local volume factor agrees with the composite map.

This also justifies the staged integral formula. Applying the change-of-variables theorem for $\mathbf{G}$ contributes $\left|\det D\mathbf{G}\right|$. Applying it for $\mathbf{H}$ contributes $\left|\det D\mathbf{H}\right|$. Their product equals $\left|\det D(\mathbf{G}\circ\mathbf{H})\right|$, because $|ab|=|a||b|$. Thus either route gives the same transformed integral under the stated conditions.

Worked Examples

Worked Example: A Variable Jacobian Factor in the Outer Map

Let $Q=[0,1]\times[0,1]\times[0,1]$, and define $\mathbf{H}(r,s,t)=(2r,3s,t)$, followed by $\mathbf{G}(u,v,w)=(u,v,(1+u)w)$. Find the volume of the image $D=\mathbf{G}(\mathbf{H}(Q))$ by transforming the integral of $1$.

The Jacobian matrices and determinants are

D\mathbf{H} = \begin{pmatrix} 2&0&0\\ 0&3&0\\ 0&0&1 \end{pmatrix}, \qquad \det D\mathbf{H}=6, $$ $$ D\mathbf{G}(u,v,w) = \begin{pmatrix} 1&0&0\\ 0&1&0\\ w&0&1+u \end{pmatrix}, \qquad \det D\mathbf{G}(u,v,w)=1+u.

Since $u=2r$ after the first map, the outer factor becomes $1+2r$. The staged change of variables gives

\begin{aligned} \iiint_D 1\,dx\,dy\,dz &=\iiint_{H(Q)}(1+u)\,du\,dv\,dw\\ &=\int_0^1\int_0^1\int_0^1 6(1+2r)\,dt\,ds\,dr\\ &=6\int_0^1(1+2r)\,dr\\ &=6\left[r+r^2\right]_0^1\\ &=6(2-0)=12. \end{aligned}

Now check the composite determinant directly. The composite map is $(x,y,z)=(2r,3s,(1+2r)t)$, with Jacobian matrix

D(\mathbf{G}\circ\mathbf{H}) = \begin{pmatrix} 2&0&0\\ 0&3&0\\ 2t&0&1+2r \end{pmatrix}, \qquad \det D(\mathbf{G}\circ\mathbf{H})=6(1+2r).

This is exactly $(1+2r)\cdot 6$, the product of the staged determinants. As a geometric check, $x$ ranges from $0$ to $2$, $y$ from $0$ to $3$, and $z$ from $0$ to $1+x$. The volume is $3\int_0^2(1+x)\,dx=3(4)=12$, agreeing with the transformed integral.

Worked Example: A Nonlinear Inner Map and a Reversing Outer Map

Let $Q=[1,2]\times[0,1]\times[0,1]$. Set $\mathbf{H}(r,s,t)=(r^2,s,t)$, and then $\mathbf{G}(u,v,w)=(u+v,u-v,3w)$. Evaluate $\iiint_D x\,dx\,dy\,dz$, where $D=\mathbf{G}(\mathbf{H}(Q))$.

The inner determinant is $2r$, which is positive on $Q$. The outer Jacobian matrix has determinant $-6$, so its absolute volume factor is $6$. Under the composite map, $x=r^2+s$. Therefore, transforming in stages gives

\begin{aligned} \iiint_D x\,dx\,dy\,dz &=\iiint_{H(Q)}(u+v)\,6\,du\,dv\,dw\\ &=\int_1^2\int_0^1\int_0^1 (r^2+s)\,6(2r)\,dt\,ds\,dr\\ &=\int_1^2 12r\left(r^2+\frac12\right)\,dr\\ &=\int_1^2(12r^3+6r)\,dr\\ &=\left[3r^4+3r^2\right]_1^2\\ &=(48+12)-(3+3)=54. \end{aligned}

For the composite check, the map is $(x,y,z)=(r^2+s,r^2-s,3t)$, so

D(\mathbf{G}\circ\mathbf{H}) = \begin{pmatrix} 2r&1&0\\ 2r&-1&0\\ 0&0&3 \end{pmatrix}, \qquad \det D(\mathbf{G}\circ\mathbf{H})=-12r.

The staged determinants give $(-6)(2r)=-12r$, as required. Their absolute product is $12r$, matching the factor in the integral. The antiderivative check is $\frac{d}{dr}(3r^4+3r^2)=12r^3+6r$, the integrand after integrating over $s$ and $t$. Its endpoint difference is $54$.

Worked Example: Tracking Orientation Through Both Stages

Let $Q=[0,1]^3$, with $\mathbf{H}(r,s,t)=(r+s,r-s,2t)$ and $\mathbf{G}(u,v,w)=(2u,v,w+u)$. Evaluate $\iiint_D(x+y+z)\,dx\,dy\,dz$.

The inner Jacobian matrix and determinant are

D\mathbf{H} = \begin{pmatrix} 1&1&0\\ 1&-1&0\\ 0&0&2 \end{pmatrix}, \qquad \det D\mathbf{H}=-4.

The outer Jacobian matrix has determinant $2$. The integrand becomes $3u+v+w$ after expressing $x+y+z$ in intermediate coordinates. Thus the staged formula, using absolute determinants, is

\begin{aligned} \iiint_D(x+y+z)\,dx\,dy\,dz &=\iiint_{H(Q)}(3u+v+w)\,2\,du\,dv\,dw\\ &=\int_0^1\int_0^1\int_0^1 8(4r+2s+2t)\,dt\,ds\,dr\\ &=8\left(4\cdot\frac12+2\cdot\frac12+2\cdot\frac12\right)\\ &=8(2+1+1)=32. \end{aligned}

Indeed, the composite coordinates are $x=2r+2s$, $y=r-s$, and $z=r+s+2t$, so the transformed integrand is $4r+2s+2t$. Multiplying the Jacobian matrices gives

D(\mathbf{G}\circ\mathbf{H}) = \begin{pmatrix} 2&0&0\\ 0&1&0\\ 1&0&1 \end{pmatrix} \begin{pmatrix} 1&1&0\\ 1&-1&0\\ 0&0&2 \end{pmatrix} = \begin{pmatrix} 2&2&0\\ 1&-1&0\\ 1&1&2 \end{pmatrix}, \qquad \det D(\mathbf{G}\circ\mathbf{H})=-8.

The product of the staged determinants is $2(-4)=-8$. Its absolute value is $8$, which is the factor used above. The integral of $4r+2s+2t$ over the unit cube is $4$, so the final result is $8\cdot4=32$.

A Common Sign Error

In the last example, using the signed composite determinant directly in the scalar integral would give $-8$ as the volume factor and hence the incorrect value $-32$. The map reverses orientation, as the negative determinant records, but a scalar triple integral uses ordinary, nonnegative volume. The correction is to use $\left|\det D(\mathbf{G}\circ\mathbf{H})\right|=8$, or equivalently $\left|\det D\mathbf{G}\right|\left|\det D\mathbf{H}\right|=2\cdot4=8$.

Practical check: Compare the composite determinant with the product of the staged determinants at the same parameter point. Then take the absolute value for a scalar volume integral. A mismatch often signals that a derivative matrix was evaluated at the wrong coordinates, the order of matrix multiplication was reversed, or a sign was incorrectly carried into the volume factor.

When Staging Is Useful

A staged substitution is useful when each map has a clear purpose: one may simplify a curved boundary, while another may scale or shear the resulting coordinates. It can also make a complicated determinant easier to compute as a product of smaller determinants. The intermediate region must still be tracked correctly, and the outer Jacobian must be evaluated at the intermediate coordinates produced by the inner map.

The method does not remove the need to check the hypotheses of change of variables. If a map is not one-to-one on the region, the formula cannot be applied as though it were a one-to-one coordinate patch; multiplicity must be handled appropriately. Likewise, points where a Jacobian determinant vanishes require attention to the valid regions of the map. Under the regularity and injectivity conditions stated above, however, staged and composite calculations agree exactly.

Check Your Understanding

For each question, keep track of where the outer Jacobian is evaluated and whether the integral uses a signed or absolute determinant.

If $\mathbf{H}:Q\to U$ and $\mathbf{G}:U\to D$, write the chain-rule formula for $D(\mathbf{G}\circ\mathbf{H})$.
Why is $D\mathbf{G}$ evaluated at $\mathbf{H}(r,s,t)$ in the composite formula?
If $\det D\mathbf{G}=-3$ and $\det D\mathbf{H}=4$ at corresponding points, what is the composite determinant, and what volume factor belongs in a scalar integral?
What injectivity condition is needed to apply the stated one-to-one change-of-variables formula in stages?
For $\mathbf{H}(r,s,t)=(2r,s,t)$ and $\mathbf{G}(u,v,w)=(u,v,3w)$, compute each determinant and verify their product against the determinant of the composite map.