Change of variables and Jacobians

Transform both the region and the area element

A coordinate map $T(u,v)=(x(u,v),y(u,v))$ can turn a complicated region into a simple one. It also stretches small areas. At a regular point, the linear approximation $DT$ sends a small coordinate rectangle to an approximate parallelogram of area $|\det DT|\,du\,dv$. The determinant's absolute value measures unsigned area, while its sign records orientation.

Definition

Jacobian determinant

For a $C^1$ planar map $T(u,v)=(x(u,v),y(u,v))$, its Jacobian determinant is

$$J_T(u,v)=\frac{\partial(x,y)}{\partial(u,v)} =\det\begin{pmatrix}x_u&x_v\\y_u&y_v\end{pmatrix}.$$

Theorem

Change of variables in two dimensions

Let $S\subset\mathbb R^2$ be bounded and Jordan measurable, with compact closure $\overline S$ contained in an open set $U$. Suppose $T:U\to V$ is a $C^1$ diffeomorphism onto an open set $V$, let $D=T(S)$, and let $f$ be continuous on $T(\overline S)$. Then

$$\iint_D f(x,y)\,dx\,dy =\iint_S f(T(u,v))\,|J_T(u,v)|\,du\,dv.$$

For a bounded set, Jordan measurable means that its boundary has area zero. A $C^1$ diffeomorphism is a bijective $C^1$ map with a $C^1$ inverse; consequently its Jacobian is nonzero. These are sufficient hypotheses; a map covering a positive-area part of $D$ multiple times cannot be used in this formula without accounting for multiplicity.

Proof

Geometric reason for the determinant

For small increments $\Delta u,\Delta v$, differentiability approximates the image of a coordinate cell by the parallelogram generated by $T_u\Delta u$ and $T_v\Delta v$. Its area is $|\det(T_u,T_v)|\Delta u\Delta v$. Summing transformed sample values and taking limits gives the formula under hypotheses ensuring that the local approximations and partition errors can be controlled. This is the idea, not a full proof for arbitrary regions.

Polar coordinates

Set $x=r\cos\theta$, $y=r\sin\theta$, with $r\ge0$. Then

$$\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| =\left|\det\begin{pmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{pmatrix}\right|=r.$$

Thus $dA=r\,dr\,d\theta$. For a disk of radius $R>0$ and an integrand continuous on the closed disk, first restrict to $\varepsilon\le r\le R$ and $\delta\le\theta\le2\pi-\delta$, with $0<\varepsilon<R$ and $0<\delta<\pi$. The map is a diffeomorphism on an open neighbourhood of this rectangle. Apply the theorem there, then let $\varepsilon,\delta\to0$: the omitted disk and angular wedge have area tending to zero, and continuity on the closed disk makes the integrand bounded; the corresponding parameter integrand is also bounded, so the omitted integrals tend to zero. This justifies the disk formula despite the origin and seam, rather than treating them as an automatic exception. An angular interval of length $4\pi$ would count almost every point twice.

Worked example

A disk and a radial integrand

For the disk $D:x^2+y^2\le4$,

$$\iint_D(x^2+y^2)\,dA=\int_0^{2\pi}\int_0^2r^2\,r\,dr\,d\theta=8\pi.$$

The first $r^2$ is the transformed integrand; the additional $r$ is the Jacobian. Forgetting it changes the answer and the dimension of the area element.

Worked example

A linear transformation

Let $D$ be the parallelogram $0\le x+y\le2$, $0\le x-y\le1$, and evaluate $\iint_D(x+y)\,dA$. Set $u=x+y$, $v=x-y$, so $x=(u+v)/2$, $y=(u-v)/2$. The new region is $0\le u\le2$, $0\le v\le1$, and $|\partial(x,y)/\partial(u,v)|=1/2$. Therefore

$$\iint_D(x+y)\,dA=\int_0^2\int_0^1u\cdot\frac12\,dv\,du=1.$$

Note that $\partial(u,v)/\partial(x,y)=-2$ is the inverse Jacobian; its absolute reciprocal is the factor needed here.

Comparison and checks

Rectangular coordinates are often best for regions with simple vertical or horizontal bounds. Polar coordinates suit disks, sectors, and radial functions; a linear substitution suits parallelograms and linear combinations. Always (1) describe the new region, (2) express the integrand in new variables, (3) compute the correct-direction Jacobian, (4) check one-to-one coverage, and (5) verify boundary and sign conventions. Negative Jacobian does not make an ordinary scalar area integral negative.

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.

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.

Practice

1. Compute the area of $x^2+y^2\le9$ by polar integration.
2. Evaluate $\iint_D e^{x^2+y^2}\,dA$ for $D:x^2+y^2\le1$.
3. Let $x=2u$, $y=3v$. Find the image of $[0,1]^2$ and compute its area using a Jacobian.
4. Explain why $0\le r\le1$, $0\le\theta\le4\pi$ is not a one-to-one polar parametrization of the unit disk up to boundary exceptions.

Full solutions

1. $\int_0^{2\pi}\int_0^3r\,dr\,d\theta=9\pi$.
2. $\int_0^{2\pi}\int_0^1e^{r^2}r\,dr\,d\theta=\pi(e-1)$.
3. The image is $[0,2]\times[0,3]$, and the Jacobian is $6$. Its area is $\int_0^1\int_0^1 6\,du\,dv=6$.
4. Every nonzero interior point has two representatives with angles separated by $2\pi$, away from a negligible seam. The integral over this parameter range counts its area twice.

Synthesis

A substitution changes coordinates, the integrand, the domain, and the measure together. The Jacobian accounts for local area scaling, while injectivity or explicit multiplicity control prevents overcounting.