Spatial derivatives and multiple integration

A concentration field assigns a scalar to every location. Its gradient describes local spatial change, while an integral combines the field with area or volume to produce total mass. Coordinate changes simplify geometry but require a Jacobian factor so that the physical quantity does not change with the notation.

1 Gradients and local approximation

For a scalar field $c(x,y,z)$, partial derivatives vary one coordinate while holding the others fixed. If $c$ has continuous first partial derivatives near $a$, it is differentiable and

$$c(a+h)=c(a)+\nabla c(a)\cdot h+o(\|h\|).$$

To see this, split $h$ into coordinate steps, apply the one-variable mean value theorem to each, and use continuity of partials to control the difference from the displayed linear map. For a moving path $r(t)$, the chain rule gives $d c(r(t))/dt=\nabla c\cdot r'$. Along a unit direction $u$, the rate is $\nabla c\cdot u$, maximized at $\|\nabla c\|$ in the gradient direction when it is nonzero.

For $c(x,y)=10-x^2-2y^2$, $\nabla c=(-2x,-4y)$. At $(1,1)$, the steepest ascent direction is $(-1,-2)/\sqrt5$ with rate $2\sqrt5$. This statement is local; the quadratic model may become negative outside its meaningful region.

The Hessian records second partial derivatives. For $C^2$ fields it is symmetric. The second-order approximation adds $h^TH_ch/2$, and a negative-definite Hessian at a stationary point proves a strict local maximum. For the quadratic field above the Hessian is $\operatorname{diag}(-2,-4)$, so the origin is its unique global maximizer, with value 10, under the algebraic model.

2 Multiple integrals and mass

A double or triple integral is a limit of cell sums $\sum f(\xi_i)\Delta V_i$, not a bare list of iterated symbols. Continuous functions on compact rectangles are integrable because uniform continuity makes upper and lower sums arbitrarily close. For continuous rectangle integrands, Fubini's theorem permits iterated integration in either order: the iterated value lies between the same upper and lower cell sums, so squeezing identifies it with the multiple integral.

If $\rho$ is mass density, $M=\iiint_D\rho\,dV$. If $c$ is concentration with units mass per volume, the same integral gives stored mass. For $M>0$, the center of mass is

$$\bar x=\frac1M\iiint_Dx\rho\,dV,$$

with analogous formulas for $\bar y,\bar z$. For a uniform rectangular slab, symmetry gives its geometric midpoint. With variable density, a geometric centroid is not generally the center of mass.

3 Changes of variables and Jacobians

Let $U,V\subset\mathbb R^n$ be open, $n\ge1$, and let $\Phi:U\to V$ be a global $C^1$ diffeomorphism. For a compact Jordan set $K\subset U$, put $D=\Phi(K)$ and assume $f$ is bounded and Riemann integrable on $D$. Then

$$\int_Df(x)\,dx=\int_{\Phi^{-1}(D)}f(\Phi(u))|\det D\Phi(u)|\,du.$$

The determinant is the local volume-scaling factor of the derivative. One conceptual proof partitions parameter space into small cells: differentiability approximates each image by a parallelepiped, whose volume is determinant times cell volume; uniform control on compact subregions and Riemann-sum convergence then give the formula. Boundary or coordinate singularities need separate zero-volume or limiting treatment, not blind substitution into an invertibility hypothesis.

For polar coordinates $(x,y)=(r\cos\theta,r\sin\theta)$, the Jacobian is $r$ for $r>0$, so $dA=r\,dr\,d\theta$. The origin and angular seam are excluded from the one-to-one smooth interior map and restored as zero-area sets. In cylindrical coordinates, $dV=r\,dr\,d\theta\,dz$. In spherical coordinates with polar angle $\phi$ from the positive $z$ axis, $dV=r^2\sin\phi\,dr\,d\phi\,d\theta$.

Worked circular layer. On the disk $r\le R$ with $R>0$, let areal density be $\rho_0(1-r^2/R^2)$ with $\rho_0>0$. Then

$$M=\int_0^{2\pi}\int_0^R\rho_0(1-r^2/R^2)r\,dr\,d\theta=\frac{\pi\rho_0R^2}{2}.$$

The average density is $M/(\pi R^2)=\rho_0/2$. Dropping the factor $r$ would give wrong units and the wrong mass.

Complete second-order estimate under exactly $C^2$ hypotheses

Let $f$ be real-valued and $C^2$ on an open neighborhood of $a\in\mathbb R^n$. Choose $r>0$ so the closed ball of radius $r$ lies there. For $\|h\|<r$, put $\phi(t)=f(a+th)$ for $0\le t\le1$. The chain rule gives $\phi''(t)=h^TH_f(a+th)h$. Integration by parts, or the fundamental theorem applied to $(1-t)\phi'(t)$, yields

$$f(a+h)-f(a)-Df(a)h=\int_0^1(1-t)h^TH_f(a+th)h\,dt.$$

Subtract $\tfrac12h^TH_f(a)h$. The remaining term $R(h)$ satisfies

$$|R(h)|\le\frac{\|h\|^2}{2}\sup_{0\le t\le1}\|H_f(a+th)-H_f(a)\|_{\rm op}.$$

Continuity at $a$ makes the supremum tend to zero because $\|th\|\le\|h\|$ for all $t$. Thus $R(h)=o(\|h\|^2)$, with no third derivative required. If $\|H_f\|_{\rm op}\le M$ on the segment, the first displayed identity also gives the first-order error bound $M\|h\|^2/2$.

For completeness, symmetry of the Hessian follows directly from $C^2$. Fix distinct coordinates $i,j$ and form the rectangular difference

$$D(h,k)=f(a+he_i+ke_j)-f(a+he_i)-f(a+ke_j)+f(a).$$

Applying the one-variable fundamental theorem first in coordinate $i$, then $j$, expresses $D(h,k)$ as the iterated integral of $\partial_j\partial_i f$ over that rectangle. Reversing the order of this argument expresses the same difference as the iterated integral of $\partial_i\partial_j f$. Divide each expression by $hk$ for nonzero $h,k$. Continuity makes the respective averages tend to the two mixed partials at $a$, so they are equal. This comparison uses two evaluations of one finite difference, not an unproved interchange of integrals.

Learning goals and prerequisites

Prove that a differentiable coordinate change preserves negligible Jordan boundaries and obtain a uniform two-sided volume estimate for each small cube. The inner-coverage argument supplies the lower bound alongside the outer linearization estimate. The next chapter uses these estimates to prove the integral theorem in every finite dimension.

We use finite-dimensional Euclidean topology, the fundamental theorem along a line segment, determinants, compact uniform continuity and the ordinary Darboux integral on boxes. Volume and integration are developed through finite Jordan covers and Darboux sums.

For estimates use $\|x\|_\infty=\max_j|x_j|$ and its induced matrix norm. A closed cube of side $h$ centered at $c$ has $\|x-c\|_\infty\le h/2$. Its ordinary volume is $h^n$. Changing the norm does not change continuity or differentiability in finite dimension, but this norm makes the cube constants transparent.

1. Jordan volume and a finite-cover boundary test

A bounded set is Jordan null if, for every $\varepsilon>0$, it can be covered by finitely many axis-parallel boxes with total volume below epsilon. One may require the boxes to be cubes of arbitrarily small sides: subdivide each of a finite box cover finely, allowing arbitrarily small total extra volume at its faces. Jordan volume $v(E)$ is the Riemann integral of the indicator of $E$ in a containing box, when that indicator is integrable.

Lemma: bounded Jordan sets have null boundaries, and conversely

A bounded set $E$ is Jordan measurable exactly when $\partial E$ is Jordan null. Moreover, on a sufficiently fine regular cube grid, the total volume of cubes meeting $\partial E$ tends to zero.

Proof from Darboux sums. A cell whose interior meets both $E$ and its complement has indicator oscillation one, and cells not meeting the boundary have oscillation zero. If the boundary is null, compactness of its closure permits a finite open box cover of arbitrarily small total volume. Enlarge a null cover slightly to make it open while keeping its total volume small. A sufficiently fine grid's boundary cells lie in that open cover: otherwise cells of shrinking diameter would supply points arbitrarily close to the compact boundary but outside its neighborhood, a contradiction. Their total volume is then small, proving the Darboux criterion.

Conversely, choose a grid with indicator upper-lower gap below epsilon. Every boundary point not on a grid face lies in a cell meeting both sets, hence one of those oscillating cells. The finitely many grid faces themselves are null: cover each bounded face by a slab of arbitrarily small thickness. The oscillating cells together with thin face slabs cover the whole boundary with arbitrarily small total volume. This proves nullity. The preceding open-cover argument then gives the regular-grid boundary-cell assertion. The argument also establishes the regular-grid boundary-cell estimate used below.

Finite unions, intersections and differences of bounded Jordan sets are Jordan: their boundaries are contained in the union of the original boundaries. Jordan volume and Riemann integration are finitely additive on pieces whose overlaps are null. To see this, refine their box approximations together; outside a cover of the finitely many null boundaries the relevant indicators add exactly, and the remaining error is bounded by the total covering volume. The same argument proves that changing a bounded integrand on a Jordan-null set does not change its integral.

If a function is continuous on a compact Jordan set $K$, its zero extension to a containing box is integrable. In cubes inside $K$, uniform continuity controls its oscillation. Cubes meeting the boundary have arbitrarily small total volume and a common bound on the function controls their contribution. Outside $K$, the extension is zero. This also shows that tagged sums over the grid cubes wholly inside $K$ converge to its integral; the omitted boundary cubes have vanishing total volume.

2. Affine box volume

Lemma: parallelepiped volume

For an invertible real matrix $A$ and an axis-parallel box $Q$,

$$v(AQ+b)=|\det A|\,v(Q).$$

The image of a box is a parallelepiped.

Elementary geometric proof. Gaussian elimination expresses $A$ as a product of coordinate permutations, nonzero coordinate scalings and shears that add a multiple of one coordinate to another. Permutations preserve box-cover volumes. A coordinate scaling multiplies their volumes by the absolute scale, first exactly for boxes and then for finite box approximations. A shear preserves the volume of a parallelepiped: a slice parallel to the changed coordinate remains an interval of the same length, merely translated by a quantity depending on the other coordinates.

The needed slicing principle for polytopes follows directly from Darboux sums. A bounded polytope has boundary in finitely many hyperplanes, each null. For a bounded hyperplane piece, solve for one coordinate with nonzero coefficient and subdivide the remaining coordinates into cubes of side $h$; its height variation is $O(h)$ on each column, so covering volume is $O(h)$ overall. Slice lengths of a polytope are piecewise affine on finitely many pieces, hence Riemann integrable. Integrating the cellwise lower and upper indicator sums bounds their slice-length integral between the same sums; the boundary-null criterion makes their gap vanish. This proves volume equals the integral of the slice lengths for these polytopes. Therefore a shear's unchanged slice lengths imply unchanged volume.

Apply these elementary operations successively to $Q$. Every intermediate image is a parallelepiped, so the slicing justification continues to apply. The volume factors are one for permutations and shears, and the absolute diagonal factors for scalings. Their product is $|\det A|$ by determinant algebra. Translation preserves the finite box approximations. This proves the lemma.

The absolute value is compulsory for ordinary volume. A coordinate reflection changes determinant sign but cannot turn a nonnegative volume into a negative one.

3. A local Lipschitz map sends a compact null set to a null set

Lemma: finite-cover null preservation

Let $N$ be a compact Jordan-null subset of an open set $U$, and let $F:U\to\mathbb R^n$ be $C^1$. Then $F(N)$ is Jordan null.

Proof. Choose a compact neighborhood of $N$ inside $U$, and bound $\|DF\|_\infty$ there by some $L>0$. Cover $N$ by finitely many sufficiently small cubes in that neighborhood, with total volume as small as desired. On each cube, segments stay in the neighborhood. The fundamental theorem along a segment gives $\|F(x)-F(y)\|_\infty\le L\|x-y\|_\infty$. A cube of side $h$ centered at $c$ consequently has its image in the cube centered at $F(c)$ of side $Lh$. Those image-enclosing cubes have total volume at most $L^n$ times the original cover total. Make that total below $\varepsilon/L^n$ to obtain the required finite image cover. If desired enlarge $L$ to at least one. The cube-by-cube estimate keeps every segment in the open domain.

The same conclusion holds for any bounded subset of such a compact null set, because its image lies in the null image cover. It also applies to the inverse of a $C^1$ diffeomorphism on compact subsets of its target.

Let now $T:U\to V$ be a $C^1$ diffeomorphism between open subsets of $\mathbb R^n$. If $K\subset U$ is compact and Jordan measurable, then $T(K)$ is compact and Jordan measurable. Indeed a homeomorphism of open neighborhoods maps the boundary to the boundary:

$$\partial T(K)=T(\partial K).$$

For justification, $T$ maps the interior of $K$ to the interior of $T(K)$ in both directions because it and its inverse are open; compactness makes both sets closed. Subtracting interiors proves the boundary identity. Apply null preservation to $\partial K$ and then the Jordan boundary criterion.

This is also valid for bounded Jordan subsets whose closure lies in $U$, by applying the same local boundary argument. It will justify pieces cut by a finite grid in the target.

4. Uniform cube distortion: prove both inclusions

Fix compact $K\subset U$. Choose a compact neighborhood $K_*$ of $K$ lying inside $U$. On it, both $DT$ and its inverse matrices are bounded, because their entries are continuous and every determinant is nonzero. Write $\kappa$ for a bound on $\|DT(x)^{-1}\|_\infty$ and $J_*$ for a bound on $J(x)=|\det DT(x)|$.

Uniform continuity gives a modulus $\omega(h)\to0$ bounding $\|DT(x)-DT(y)\|_\infty$ when $x,y\in K_*$ and $\|x-y\|_\infty\le h$. Consider any sufficiently small closed grid cube $Q$ of side $h$ meeting $K$; choose the size so that all of $Q$ lies in $K_*$. Let $c$ be its center and $A=DT(c)$. The segment fundamental theorem gives

$$T(x)=T(c)+A(x-c)+R(x),\qquad \|R(x)\|_\infty\le\omega(h)h/2\quad(x\in Q).$$

Set $\delta(h)=\kappa\omega(h)/2$ and normalize the image by

$$H(x)=c+A^{-1}[T(x)-T(c)].$$

Then $H(c)=c$ and $\|H(x)-x\|_\infty\le\delta(h)h$ throughout $Q$. Let $Q^+$ and $Q^-$ be concentric cubes of side $h(1+2\delta)$ and $h(1-2\delta)$, taking $h$ small enough that $\delta<1/4$.

Outer inclusion. The coordinate error bound immediately gives $H(Q)\subseteq Q^+$.

Inner inclusion. A boundary point $x\in\partial Q$ has some coordinate at distance $h/2$ from the center. Its image under $H$ therefore has that coordinate at distance at least $h/2-\delta h$. Hence $H(\partial Q)$ misses the open cube with the same interior as $Q^-$. Since $H$ is a homeomorphism on an open neighborhood, $\partial H(Q)=H(\partial Q)$, and $c$ is an interior point of $H(Q)$.

If a point of the open inner cube were outside the closed set $H(Q)$, the segment joining it to $c$, wholly in that inner cube, would have to hit $\partial H(Q)$ on leaving the set. That contradicts the just-proved boundary exclusion. Therefore the open inner cube lies in $H(Q)$, and closedness gives $Q^-\subseteq H(Q)$. This is the elementary first-exit property of a segment from an interior point of a closed set.

Undo the affine normalization. The linear-volume lemma and set inclusion give the two-sided estimate

$$J(c)(1-2\delta(h))^n h^n \le v(T(Q))\le J(c)(1+2\delta(h))^n h^n.$$

Thus there is a single $\beta(h)\to0$, independent of the selected cube near $K$, with

$$|v(T(Q))-J(c)v(Q)|\le\beta(h)v(Q).$$

For example take $\beta(h)=J_*\max((1+2\delta)^n-1,\ 1-(1-2\delta)^n)$. Also $v(T(Q))\le C v(Q)$ with fixed $C=J_*(3/2)^n$ for these small cubes. The affine-volume lemma converts the two enclosures into the stated volume bounds.

The uniform relative error allows the cell errors to be summed over a refining grid; the same constants also control boundary cubes.

Worked estimate

In dimension two, if $\delta\le1/100$ and the center Jacobian is $J(c)$, the image area lies between $0.98^2J(c)h^2$ and $1.02^2J(c)h^2$. The relative bounds remain useful as the number of grid cells grows on the order of $h^{-2}$. Uniform derivative continuity forces delta to zero as $h$ decreases.

Graded proof tasks

1. Null cover. Reproduce the Lipschitz null-image proof and explain why the derivative bound need only hold on small cover cubes, not on every segment joining arbitrary points of $U$.
2. Boundary. Prove $\partial T(K)=T(\partial K)$ under the stated compact/homeomorphism hypotheses. Which facts would need rechecking if the set reached the boundary of $U$?
3. Inner coverage. Explain why the outer image enclosure alone cannot establish the stated volume asymptotic. Reproduce the boundary-and-segment argument for the inner enclosure.
4. Uniformity. Identify where compactness supplies common derivative, inverse-matrix and continuity bounds. Explain why a separate little-o statement for every unrelated cell center is not the summed estimate needed later.

Full solutions

1. A finite cube cover of total volume $\eta$ has image enclosed by cubes of total volume at most $L^n\eta$, so choose $\eta<\varepsilon/L^n$. Each small cube and its internal segments stay in a fixed compact neighborhood inside $U$, where the derivative bound is uniform. A distant connecting segment might leave a nonconvex domain; the proof never needs it.
2. Compactness makes $K$ and $T(K)$ closed in the ambient spaces. A homeomorphism between open neighborhoods carries interior to interior in both directions. Thus $T(K\setminus\operatorname{int}K)=T(K)\setminus\operatorname{int}T(K)$, giving the boundary identity. At a domain boundary there may be no open neighborhood for that argument, no uniform derivative bound and no defined image of missing boundary points. The theorem deliberately keeps compact closure inside $U$.
3. An upper enclosure only bounds image volume above and does not show the map covers most of the affine cell. Here the normalized image boundary lies outside the open inner cube, while its center is inside the image. A segment to any allegedly missing inner point would leave the closed image and hence meet its boundary within the inner cube, impossible. Taking closure includes the inner cube's faces. This provides the necessary lower bound before computing the asymptotic.
4. Continuity of $DT$ and its inverse matrices on $K_*$ gives finite common bounds; uniform continuity of $DT$ gives one modulus for all small cubes near $K$. Consequently every volume error is at most the same $\beta(h)$ times cell volume, whose sum stays bounded. Pointwise little-o constants or scales depending independently on cell centers would not by themselves provide a common vanishing bound for the whole grid.

Summary and terms

A Jordan-null set has arbitrarily small finite box covers. A cell-distortion estimate bounds image volume above and below by nearby determinant-scaled volume. Uniformity lets the relative cell errors be summed safely. A diffeomorphism supplies neighborhood homeomorphism and invertible derivatives, not just a formal determinant. These finite-cover and finite-cell facts establish the geometric foundation of the next chapter's Riemann theorem.

Learning goals and precise scope

Prove a finite-dimensional $C^1$ change-of-variables theorem using the preceding chapter's finite Jordan covers and uniform two-sided cube distortion. Then extend the continuous case to bounded Riemann integrands by finite Darboux step brackets. The theorem below specifies the proper Jordan/Riemann setting and its coverage conditions.

Throughout, $n$ is a positive finite integer. A function is Riemann integrable over a bounded Jordan set if its zero extension to an enclosing box is Riemann integrable there; different enclosing boxes give the same integral. Null boundaries may be assigned arbitrary bounded values. The preceding chapter supplies the Jordan boundary and null-image lemmas and the uniform small-cube estimate.

1. The theorem and every hypothesis

Theorem: proper $C^1$ Jordan/Riemann substitution

Let $U,V\subseteq\mathbb R^n$ be open and let $T:U\to V$ be a $C^1$ diffeomorphism: it is a bijection, $T$ and its inverse are continuously differentiable, and in particular $\det DT(x)\ne0$ for every $x\in U$. Let $K\subset U$ be compact and Jordan measurable. Let $f:T(K)\to\mathbb R$ be bounded and Riemann integrable over $T(K)$. Then $T(K)$ is Jordan measurable, the function $x\mapsto f(T(x))|\det DT(x)|$ is Riemann integrable over $K$, and

$$\int_{T(K)}f(y)\,dy =\int_K f(T(x))J(x)\,dx, \qquad J(x)=|\det DT(x)|.$$

The empty or Jordan-null case is allowed. The pointwise determinant magnitude covers both orientations and disconnected domains. A bounded Jordan region $D$ with $\overline D\subset U$ is covered as well: replace it by its compact closure and assign bounded boundary values, since the boundary and its image are null.

Compact containment is substantive. It supplies uniform derivative and inverse-matrix bounds and keeps all small boundary cells inside the map's open domain. Scope. This is a proper bounded Riemann theorem. Unbounded or noncompact limits need convergence estimates; many-to-one maps need coverage accounting. General Lebesgue and lower-dimensional surface-area formulas are separate results.

2. Prepare one uniform grid, including the boundary layer

If $K$ is empty, both integrals are zero, so assume it is nonempty. As proved in the foundations chapter, $E=T(K)$ is compact and Jordan measurable: its boundary is the null image of $\partial K$.

Choose a compact neighborhood $K_*$ of $K$ inside $U$. For all sufficiently small regular grid cubes meeting $K$, the previous chapter supplies common constants $L,C$ and a function $\beta(h)\to0$ such that, for a cube $Q$ of side $h$ and center $c$,

$$\|T(x)-T(z)\|_\infty\le L\|x-z\|_\infty\quad(x,z\in Q),$$ $$|v(T(Q))-J(c)h^n|\le\beta(h)h^n,\qquad v(T(Q))\le Ch^n.$$

The derivative bound is taken along segments inside each cube.

Let $\mathcal I_h$ be the finite collection of grid cubes wholly contained in $K$. Let $\mathcal B_h$ be the cubes meeting $K$ but not wholly contained in it. Every cube in the second class meets $\partial K$, and

$$b_h:=\sum_{Q\in\mathcal B_h}v(Q)\longrightarrow0$$

by the Jordan boundary criterion. Cubes may share faces, but faces are null, as are their images under $T$. Global injectivity therefore makes distinct interior-cell images overlap only on null images of faces. Finite additivity applies to their volumes and integrals.

The part of $E$ not covered by the interior-cell images lies in the images of boundary cubes. Its volume is at most $Cb_h$, because $v(T(Q))\le Cv(Q)$ and the cover is finite. It is itself Jordan measurable: subtract a finite Jordan union from the Jordan set $E$. This gives the target boundary-layer error bound.

3. Prove the theorem first for continuous f

Suppose $f$ is continuous on compact $E$. It is bounded by $M$ and uniformly continuous; write $\omega_f(\eta)\to0$ for a common modulus on $E$. For every interior cube $Q$, the target integral over $T(Q)$ differs from $f(T(c))v(T(Q))$ by at most $\omega_f(Lh)v(T(Q))$. The uncovered target boundary layer contributes an error at most $MCb_h$.

Since the interior-cell images have disjoint interiors up to null sets, their total volume is at most $v(E)$. Thus

$$\left|\int_Ef-\sum_{Q\in\mathcal I_h}f(T(c))v(T(Q))\right| \le\omega_f(Lh)v(E)+MCb_h\longrightarrow0.$$

Now replace each image volume by $J(c)h^n$. For a fixed box $B$ containing $K$, the total additional error is at most

$$M\beta(h)\sum_{Q\in\mathcal I_h}h^n\le M\beta(h)v(B)\longrightarrow0.$$

The resulting sum is the interior-cell tagged sum for $f(T(x))J(x)$ on $K$. That function is continuous on compact $K$, and the preceding Jordan/Darboux lemma proves these sums tend to its integral: uniform continuity controls interior cells and $b_h\to0$ controls omitted boundary cells. Therefore the two integrals are equal for continuous $f$.

The common relative cell bound makes the total volume-replacement error vanish, while the separate boundary estimate controls the omitted cells.

4. Extract the volume identity for Jordan pieces

Taking $f\equiv1$ in the proved continuous case gives

$$v(T(S))=\int_SJ(x)\,dx$$

for every compact Jordan set $S\subset U$. It also holds for bounded Jordan $S$ whose closure lies in $U$: replace it by $\overline S$. The added subset lies in $\partial S$, whose image is null, so neither volume nor integral changes.

We need this identity for pieces obtained by cutting the target with a finite grid. Let $C$ be one grid box and $S=K\cap T^{-1}(C)$. Its boundary is contained in

$$\partial K\ \cup\ T^{-1}(\partial C\cap E).$$

The first set is null. The second is null because the target grid face portion is a bounded subset of a compact null set inside $V$, and the inverse map is $C^1$ on a neighborhood of compact $E$. The finite-cover null-image lemma applies to that inverse. Consequently $S$ is Jordan measurable and has compact closure in $U$.

Choose grid cells with a consistent half-open boundary assignment if an exactly disjoint partition is desired. The resulting pieces still have boundaries in the same finite null collection. Their image is precisely $E\cap C$, with the corresponding assignment. Thus the volume identity proves, by finite linearity, the substitution formula for every finite step function on the target grid.

The continuous theorem's constant-function consequence has thus established substitution for finite target step functions.

5. Bounded Riemann integrands through finite step brackets

Take a target box containing $E$ and let $f_0$ be the zero extension of $f$. Its Riemann integrability means that for every $\varepsilon>0$ there is a finite grid whose lower and upper step functions $\ell,u$ satisfy

$$\ell\le f_0\le u,\qquad \int(u-\ell)<\varepsilon.$$

On a cell, use the infimum and supremum of $f_0$ over its closure, assigning face values consistently. The Darboux criterion gives exactly the displayed gap. Restricting the nonnegative difference to $E$ cannot increase its integral.

The source functions $p(x)=\ell(T(x))J(x)$ and $q(x)=u(T(x))J(x)$ on $K$, extended by zero outside, are Riemann integrable. Each is a finite sum of continuous $J$ times indicators of the Jordan preimage pieces just established. Such a bounded product is integrable by the Darboux product estimate, or directly by uniform continuity off those null piece boundaries. Since $J\ge0$,

$$p(x)\le f(T(x))J(x)\le q(x),$$

and the step-function identity gives

$$\int_K(q-p)=\int_E(u-\ell)<\varepsilon.$$

A bounded function with integrable lower and upper brackets of arbitrarily small integral gap is Riemann integrable. Indeed its lower integral is at least the lower bracket's integral and its upper integral is at most the upper bracket's integral, so their difference is below every positive epsilon.

This proves the transformed integrand is integrable. Its integral and the original $\int_Ef$ both lie between the same numbers $\int_E\ell=\int_Kp$ and $\int_Eu=\int_Kq$, whose gap is below epsilon. Their difference is therefore zero. The theorem, including the bounded Riemann case, is proved.

The finite Darboux brackets and the Jordan volume identity complete the bounded Riemann extension.

6. Worked nonlinear map, including a discontinuous integrand

Let $T(u,v)=(u,v+u^2)$ on $\mathbb R^2$. Its inverse is $(x,y)\mapsto(x,y-x^2)$ and its determinant is one. The unit square maps to

$$E=\{0\le x\le1,\quad x^2\le y\le x^2+1\}.$$

For continuous $f(x,y)=y$, the theorem gives

$$\iint_Ey\,dA=\int_0^1\int_0^1(v+u^2)\,dv\,du=\frac56.$$

The map bends the top and bottom boundaries but preserves area, since the integral of one remains one.

For the bounded indicator $f(x,y)=1$ when $y\ge x^2+1/2$ and zero below, the discontinuity inside $E$ lies on one smooth graph. Its bounded zero extension is Riemann integrable: cover that graph and $\partial E$ by finite thin cells of arbitrarily small total area; away from them it is constant. The transformed function is the step $1_{\{v\ge1/2\}}$, so its integral is $1/2$. The example applies the bounded Riemann extension to a jump across a smooth graph.

7. Polar coordinates as a limiting corollary

On $U=(0,\infty)\times(-\pi,\pi)$, polar $T(r,\theta)=(r\cos\theta,r\sin\theta)$ is a diffeomorphism onto the plane with the nonpositive horizontal ray removed, and $J=r$. An explicit smooth inverse on that slit plane is $r=\sqrt{x^2+y^2}$ and $\theta=2\arctan[y/(\sqrt{x^2+y^2}+x)]$: its denominator is strictly positive there, and the half-angle identities verify both compositions. For $0<\varepsilon<R$ and $0<\delta<\pi$, the compact rectangle $[\varepsilon,R]\times[-\pi+\delta,\pi-\delta]$ satisfies the theorem exactly.

For a bounded continuous integrand on a closed disk, pass from these truncated sectors to the disk only after controlling the omissions. A missing angular wedge of width $2\delta$ in an annulus can be covered using another nonsingular polar branch; its area is $\delta(R^2-\varepsilon^2)$ by the theorem with integrand one. The disk of radius epsilon lies in a square of area $4\varepsilon^2$. Multiplying these bounds by a bound on the integrand makes both omitted contributions tend to zero. The circle and radial seams have null area by finite smooth-curve covers. This proves the usual disk polar formula for such bounded integrands and, with integrand one, gives disk area $\pi R^2$.

Truncation keeps the map regular and single-valued; the displayed bounds recover the origin and angular seam in the limit.

8. Nonzero Jacobian does not imply once-only coverage

For $F(x,y)=(x^2-y^2,2xy)$ on the punctured plane, $\det DF=4(x^2+y^2)>0$. It is locally invertible everywhere, but $F(x,y)=F(-x,-y)$. The annulus $1\le r\le2$ maps twice onto the annulus $1\le\rho\le4$.

The image area is $15\pi$, whereas

$$\iint_{1\le r\le2}|\det DF|\,dA =\int_0^{2\pi}\int_1^2 4r^3\,dr\,d\theta=30\pi.$$

The factor of two records the overlapping coverage. Global injectivity was the condition making distinct image cells disjoint up to null faces. Restricting to injective branches and recording their overlap separates the two traversals from the ordinary image area.

Graded theorem and application tasks

1. Hypothesis audit. Explain why compact containment $K\subset U$ and global injectivity are logically different from a nonzero derivative determinant. Give a boundary-domain example where the proper theorem does not apply without an improper limit.
2. Proof mechanism. Identify exactly where the uniform small-cube relative error and the Jordan boundary-layer estimate enter the continuous proof. Why is neither one dispensable from that argument?
3. Nonlinear shear. Verify the inverse and determinant for $(u,v)\mapsto(u,v+u^2)$ and reproduce both its $y$ integral and the indicated upper-half step integral.
4. Coverage. Verify the determinant and the two-to-one annulus description for the complex-square map. Explain which step of the theorem's proof would overcount if injectivity were silently removed.
5. Bounded Riemann extension. Explain why the finite step brackets have Jordan-integrable pullbacks and why multiplying inequalities by $J$ preserves their direction. State what the proof does not establish for an arbitrary bounded nonintegrable function.
6. Three dimensions. Use $T(u,v,w)=(u,v,w+u^2+v^2)$ on the unit cube to find the image volume and $\iiint_{T(K)}z\,dV$, verifying all map and domain conditions.

Full solutions

1. A nonzero determinant is local. It neither bounds the derivative and inverse matrices near a missing domain boundary nor prevents distinct distant inputs from sharing an image. For example $T(x)=1/x$ on $(0,\infty)$ is a diffeomorphism, but $(0,1]$ is not compactly contained in that domain and its image is unbounded. A substitution there is an improper-integral question requiring truncation and a convergence argument, not a direct use of the proper compact theorem. The complex-square annulus gives the independent global-injectivity failure despite a nonzero determinant everywhere.
2. Uniform relative error controls the replacement of every curved-cell volume by $J(c)h^n$; summing gives at most $M\beta(h)$ times a fixed containing-box volume, tending to zero. The boundary-layer bound controls the target points not covered by wholly interior source cells, by $MCb_h$. Interior linear approximation alone misses that layer, while boundary control alone does not identify the interior volume density. Both estimates are explicit parts of the equality proof.
3. Solving $x=u,y=v+u^2$ gives $u=x,v=y-x^2$, a global smooth inverse. The determinant is one. Over the unit square the transformed $y$ is $v+u^2$, integrating to $1/2+1/3=5/6$. The indicator of $y\ge x^2+1/2$ becomes the indicator of $v\ge1/2$, whose square integral is $1/2$. Boundaries and the separating graph are Jordan null, so bounded assignments on them do not change either value.
4. The derivative matrix is $\begin{pmatrix}2x&-2y\\2y&2x\end{pmatrix}$, determinant $4(x^2+y^2)$. In polar notation it squares radius and doubles angle, producing exactly two antipodal preimages of each image point in the stated annuli. Source determinant integration yields $30\pi$ while image area is $15\pi$. The finite image-cell additivity step would fail: images from distinct source regions overlap in positive area, not merely on null faces.
5. Each preimage grid piece has boundary contained in $\partial K$ plus an inverse image of finitely many target faces. Both are Jordan null by the finite-cover lemma for $T^{-1}$, so each piece is Jordan. Continuous $J$ times its indicator is Riemann integrable. The determinant magnitude is nonnegative, so lower and upper inequalities stay in order. The argument assumes target Riemann integrability to obtain arbitrarily close Darboux brackets; arbitrary bounded functions may have no such brackets, and their pullbacks are not asserted integrable by this theorem.
6. The inverse is $(x,y,z)\mapsto(x,y,z-x^2-y^2)$, smooth on all of $\mathbb R^3$. The triangular derivative has determinant one; the cube is compact Jordan inside that open domain. Its image is the curved slab $0\le x,y\le1$, $x^2+y^2\le z\le x^2+y^2+1$, of volume one. The transformed integrand is $w+u^2+v^2$, whose cube integral is $1/2+1/3+1/3=7/6$. This illustrates the theorem's finite-dimensional form beyond planar area.

Summary

Substitution combines Jordan boundary preservation with uniform two-sided cell volumes. The interior sums converge because their relative errors are uniform, and the boundary layer has vanishing volume. Finite Darboux brackets then extend the continuous formula to bounded Riemann integrands. The nonlinear shear, polar truncation and double-cover examples show how to verify the hypotheses in calculations.

Recovering cylindrical and spherical boundaries by controlled limits

The proper substitution theorem proved here applies in positive finite dimension to a compact Jordan set inside the domain of a global $C^1$ diffeomorphism between open sets, with a bounded Riemann-integrable target function. The local determinant alone does not establish this hypothesis. Here are the limits needed for the displayed coordinate models.

For cylindrical coordinates use $\rho>0$, $-\pi<\theta<\pi$ and bounded $z$ intervals. The inverse has $\rho=\sqrt{x^2+y^2}$, the slit-plane polar inverse, and unchanged $z$; thus the map is a diffeomorphism off the axis and angular seam. On a closed cylinder first take $\varepsilon\le\rho\le R$ and $-\pi+\eta\le\theta\le\pi-\eta$. The omitted axial cylinder lies in a square prism of volume $4\varepsilon^2L$, where $L$ is the height. The omitted angular wedge has volume at most $\eta R^2L$, by the proved planar polar area formula and rectangle Fubini. These vanish, and a bound on the integrand controls the target integral error. The parameter integral has factor $\rho$, so its corresponding omissions also vanish by direct one-variable integration.

2. Derive spherical scale before using radial symmetry

Use coordinates

$$T(r,\phi,\theta)=(r\sin\phi\cos\theta,r\sin\phi\sin\theta,r\cos\phi),$$

where $\phi$ is measured from the positive $z$ axis, $0\le\phi\le\pi$, and $\theta$ is azimuth over one full turn. The derivative vectors have lengths $1,r,r\sin\phi$ and are mutually perpendicular with the corresponding positive orientation on the interior. Their scalar triple product is $r^2\sin\phi$, giving

$$dV=r^2\sin\phi\,dr\,d\phi\,d\theta.$$

To verify the determinant explicitly, the derivative columns in the order $(r,\phi,\theta)$ are

$$T_r=(\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi),$$ $$T_\phi=(r\cos\phi\cos\theta,r\cos\phi\sin\theta,-r\sin\phi),\qquad T_\theta=(-r\sin\phi\sin\theta,r\sin\phi\cos\theta,0).$$

Their triple product is $r^2\sin\phi>0$ when $r>0$ and $0<\phi<\pi$. On the open parameter domain $(0,\infty)\times(0,\pi)\times(-\pi,\pi)$ the map is one-to-one onto its slit image, with smooth inverse $r=\sqrt{x^2+y^2+z^2}$, $\phi=\arccos(z/r)$ and the usual smooth slit-plane azimuth. Thus the proved finite-dimensional change-of-variables theorem applies to every compact truncated parameter box inside that domain.

The omitted target regions also need estimates independent of the formula we are proving. The ball of radius $\varepsilon$ is in a cube of volume $8\varepsilon^3$. In a radius-$R$ ball, the two polar caps with angle at most $\delta<\pi/2$ from the axis lie in the box $|x|,|y|\le R\sin\delta$, $|z|\le R$, of volume $8R^3\sin^2\delta$. An azimuth seam of total angular width $2\eta$, with $0<\eta<\pi/2$, lies in $|x|,|z|\le R$, $|y|\le R\sin\eta$, a box of volume $8R^3\sin\eta$. All three bounds vanish with their cutoff. Multiplying by a bound for the continuous target integrand controls its omitted integral, while the parameter estimates below control the other side. This completes the nonsingular-exhaustion argument without using spherical volume to justify its own formula.

The origin, poles and angular seam are coordinate-boundary exceptions. For bounded continuous integrands on a ball, use nonsingular truncated parameter regions and pass to their limit; the omitted parameter integral bounds tend to zero. For a bounded integrand, the radial factor near zero is bounded by a constant times $\int_0^\varepsilon r^2dr=\varepsilon^3/3$, while a polar cap has factor $\int_0^\varepsilon\sin\phi\,d\phi=1-\cos\varepsilon\to0$ and a shrinking azimuth seam has vanishing angle width. The formula is not a license to ignore an arbitrary unbounded singularity at those locations.

4 Graded practice

E1 Foundation. For $c(x,y,z)=x^2+yz$, find the directional derivative at $(1,2,3)$ along $(2,0,0)$ per unit distance.

E2 Integration. Find total mass and center of mass on the unit square with areal density $\rho=2x$.

E3 Polar coordinates. Evaluate $\iint_{x^2+y^2\le1}(x^2+y^2)dA$.

E4 Cylindrical coordinates. Find the volume of $x^2+y^2\le4$, $0\le z\le3$ using an integral.

5 Full solutions

S1. The gradient is $(2x,z,y)$, so at the point it is $(2,3,2)$. Normalize the specified direction to $(1,0,0)$; the derivative is 2. Dotting with the unnormalized vector would instead give 4, a rate for a different parameter speed.

S2. $M=\int_0^1\int_0^1 2x\,dy\,dx=1$. The $x$ moment is $\int_0^12x^2dx=2/3$ and the $y$ moment is $(\int_0^12x dx)(\int_0^1y dy)=1/2$. Thus center is $(2/3,1/2)$.

S3. $\int_0^{2\pi}\int_0^1r^2r\,dr\,d\theta=2\pi/4=\pi/2$.

S4. $\int_0^{2\pi}\int_0^2\int_0^3r\,dz\,dr\,d\theta=2\pi\cdot2\cdot3=12\pi$, matching base area times height.

Summary

Gradients predict local spatial change and directional response. Multiple integrals convert densities into totals and weighted centers. Coordinate changes preserve the same physical total only when their Jacobian scale factors, domain bounds and singularity qualifications are handled correctly.