Multiple integrals and change of variables

A height must be multiplied by an area

For f(x,y)=x²+y² on the unit square, a small rectangular patch contributes approximately f(x*,y*) Δx Δy to the volume under the nonnegative graph. The height alone is not a volume; it must be weighted by the patch’s area. Summing over a fine partition leads to a double integral. The exact-coordinate examples in this article use dimensionless rescaled coordinates; physical units are restored when applying the formulas.

If the square is divided into n-by-n equal cells and the height is sampled at each midpoint, the sum is Sₙ=2/3−1/(6n²). Thus n=1 gives 1/2, n=2 gives 5/8, and n=4 gives 21/32. The sums approach 2/3. Here the finite formula can be checked exactly; in general, agreement of a few meshes illustrates convergence rather than establishing it.

Preparation is the single-variable integral and substitution lessons, functions/domains 130, derivatives and Jacobians 131, and determinants 125. A double integral accumulates over a two-dimensional region of inputs; the graph of the integrand is not itself that input region. The same distinction extends to three inputs and triple integrals.

The limiting sum is independent of sample choices when integrability holds

On a rectangle R=[a,b]×[c,d], a Riemann sum has the form ΣᵢΣⱼ f(ξᵢⱼ,ηᵢⱼ) Δxᵢ Δyⱼ, with a sample point in each cell. The double integral ∬R f dA is the common limit as the largest cell side tends to zero, when the limit exists independently of allowed sample choices. A continuous function on a compact rectangle is Riemann integrable.

Linearity and additivity express accumulation: ∬(αf+βg)=α∬f+β∬g, and integrals over disjoint regular pieces add. Pieces sharing only boundaries of area zero can also be combined without counting a positive-area overlap twice. For a nonnegative height, the integral is the volume between the graph and the input plane; for a signed height it is the volume above minus the corresponding volume below.

Area is the special case ∬R 1 dA. If m≤f≤M throughout a region of area A, then mA≤∬f≤MA. This gives a useful check on magnitude and units. A bounded integrand alone is not a general integrability guarantee on every imaginable region; the elementary examples here use compact regions with piecewise smooth boundaries and integrands continuous on them.

Fubini turns a two-dimensional accumulation into slices

For f continuous on a compact rectangle, Fubini’s theorem gives ∬R f dA=∫ₐᵇ[∫cᵈ f(x,y) dy] dx=∫cᵈ[∫ₐᵇ f(x,y) dx] dy. In the first iterated integral, hold x fixed while integrating in y; the resulting function of x is then integrated. In the second, the roles are reversed. Read the differential beside the innermost integral first.

For f=x²+y² on [0,1]², integrating in y first gives x²+1/3; integrating that in x gives 2/3. Reversing the order gives the same value. The theorem explains why both calculations represent the same double integral; computing both is a useful independent check, not the source of the theorem.

For nonrectangular regions, bounds can depend on an outer variable. The function produced by an inner integral then contains the moving endpoints as well as the integrand. Do not turn an integral over a triangle into a product of two independent one-dimensional integrals unless the integrand and region actually permit that separation.

Bounds describe the actual feasible inputs

A vertically sliced region has the form D={(x,y): a≤x≤b, α(x)≤y≤β(x)}. Under the stated continuity and regular-boundary assumptions, ∬D f dA=∫ₐᵇ∫α(x)^β(x) f(x,y) dy dx. A horizontal description instead uses c≤y≤d and γ(y)≤x≤δ(y), leading to the corresponding dx dy order. These are descriptions of the same set, not interchangeable strings of symbols.

Take D={(x,y):0≤x≤1,0≤y≤x}. With f=x+2y, the vertical calculation is ∫₀¹[x y+y²]ᵧ₌₀ˣ dx=∫₀¹2x² dx=2/3. Reversing order requires 0≤y≤1 and y≤x≤1. Then ∫₀¹∫ᵧ¹(x+2y) dx dy also gives 2/3. The bounds y≤x≤1 are obtained from the triangle, not by exchanging the old endpoint labels.

For the curved region 0≤x≤1 and x²≤y≤x, fixing y gives y≤x≤√y with 0≤y≤1. Its area is ∫₀¹(x−x²) dx=1/6, equivalently ∫₀¹(√y−y) dy=1/6. The restrictions x≥0 and 0≤y≤1 matter when solving the inequalities; square roots do not erase the original domain.

A single pair of bounds may not cover every slice

Let D be the L-shaped union [0,2]×[0,1] together with [0,1]×[1,2]. Vertical slices have height 2 for 0≤x≤1 and height 1 for 1≤x≤2. Thus an integral over D can be written as ∫₀¹∫₀² f dy dx+∫₁²∫₀¹ f dy dx. The shared boundary x=1 does not add area.

The horizontal description also splits: ∫₀¹∫₀² f dx dy+∫₁²∫₀¹ f dx dy. For f=x+y, the two rectangular contributions are 3 and 2, giving total 5. Writing every vertical slice as 0≤y≤2 for 0≤x≤2 would include the missing top-right square and answer a different question.

Holes, disconnected slices and changes in which boundary is uppermost often require such splits. A sketch helps identify them, but the final bounds should stand on their own as inequalities. Before evaluating an integral, test a representative slice near each change point and check whether every included input is intended.

The same integral can accumulate a density instead of a height

For a planar lamina with nonnegative area density ρ(x,y), its mass is M=∬Dρ dA. When 0<M<∞ and the required moments are finite, its centre of mass is x̄=(1/M)∬D xρ dA and ȳ=(1/M)∬D yρ dA. The integrals with x and y weights are first moments; the coordinate associated with each numerator should be kept explicit rather than inferred from an axis label.

On the triangle 0≤y≤x≤1, let ρ=1+x in the rescaled example. Then M=∫₀¹(1+x)x dx=5/6. The x-weighted moment is ∫₀¹(x²+x³) dx=7/12; the y-weighted moment is ∫₀¹(1+x)x²/2 dx=7/24. Division gives (x̄,ȳ)=(7/10,7/20), inside the triangle. Uniform density would give its geometric centroid instead; a varying density shifts the mass centre.

For L>0 and ρ₀>0, take a physical triangle with coordinates x=L u and y=L v, and density ρ₀(1+u). The same calculation yields mass (5/6)ρ₀L² and centre (7L/10,7L/20). If ρ₀ is in kilograms per square metre and L in metres, mass is in kilograms and centre coordinates in metres. A density is not an extra geometric Jacobian; they play different roles.

The average of a scalar field over a finite positive-area region is (1/area(D))∬D f dA. Probability uses a related normalisation: a nonnegative density integrates to 1 over its full domain. Neither a centre-of-mass formula nor an average is meaningful with a zero normalising denominator.

The Jacobian determinant scales small coordinate cells

Let (x,y)=Φ(u,v). Near a point where Φ is differentiable, a small parameter displacement is mapped approximately by its derivative matrix. The two sides of a tiny parameter rectangle become the two derivative-column vectors multiplied by Δu and Δv. Their parallelogram area is |det DΦ| Δu Δv. Determinants 125 and local linearisation 131 therefore explain the factor used in integration.

The sign of det DΦ records orientation, while its absolute value records unsigned area scaling. Mass and ordinary volume do not become negative because a coordinate map reverses orientation. Signed orientation reappears in line/surface integrals and differential forms; it is a different choice of object.

In n dimensions, the same principle gives the volume factor |det DΦ| for square derivative matrices. For a surface described by two parameters in three-dimensional space, the derivative is rectangular and has no determinant; its area factor is the norm of a cross product, treated in lesson 135. Do not put an arbitrary rectangular Jacobian into the multiple-volume formula.

A coordinate substitution changes three things

A safe classical change-of-variables setting is a C¹ one-to-one map Φ on an open neighbourhood of a compact regular parameter region U, with det DΦ nonzero there, and a continuous integrand f on its image D=Φ(U). Then ∬D f(x,y) dx dy=∬U f(Φ(u,v)) |det DΦ(u,v)| du dv. More general integrable versions exist, but their hypotheses must be supplied rather than inferred from this convenient setting.

Change the function, the region and the volume weight together. If the substitution is given as (u,v)=Ψ(x,y), invert it to obtain Φ for this formula. Where the maps are C¹ inverses, DΨ(Φ(u,v))=(DΦ(u,v))⁻¹, so the determinants are reciprocals at corresponding points. An inverse determinant at an unrelated input is not the required weight.

For Φ(u,v)=(2u+v,u+3v) on the unit parameter square, DΦ=[[2,1],[1,3]] has determinant 5. The image is a parallelogram of area 5. The integrand x+y becomes 3u+4v, and its integral is 5∫₀¹∫₀¹(3u+4v) du dv=35/2. Replacing Φ by (u+2v,3u+v) reverses the determinant sign to −5; the area factor remains 5 and the corresponding x+y integral is still 35/2.

Use the transformed domain to simplify a sloping boundary

Consider Φ(u,v)=(u,uv) with 1≤u≤2 and 0≤v≤1. Its image is D={(x,y):1≤x≤2,0≤y≤x}. The derivative is [[1,0],[v,u]], so det DΦ=u. Since u>0, the inverse is u=x and v=y/x on an open neighbourhood of this compact region; there is no singular origin to hide in the substitution.

The area is ∫₁²∫₀¹u dv du=3/2. For f(x,y)=y/x, the transformed integrand is v and the full weight is uv. Thus ∬D(y/x) dA=∫₁²∫₀¹uv dv du=3/4. Directly, ∫₁²∫₀ˣ(y/x) dy dx gives the same value.

Leaving out the factor u would give 1/2 instead of 3/4. A visually rectangular parameter domain does not mean its cells have equal area in the image. Here the higher-u strips are wider in physical y even though every parameter strip spans the same v interval.

Polar area has a radial weight and a seam

In polar coordinates, x=r cosθ and y=r sinθ. With derivative columns ordered as r,θ, the determinant is r, so dA=r dr dθ for r>0. A small angular change sweeps an arc of length approximately rΔθ, which explains why the same angle covers a larger area farther from the origin.

For the unit disk and f=x²+y²=r², integration gives ∫₀²π∫₀¹r³ dr dθ=π/2. For f=1, the same radial weight gives disk area π. Omitting r would instead compute the area of a rectangle in parameter coordinates, not the area of the disk.

The full closed parameter rectangle 0≤r≤R,0≤θ≤2π is not one-to-one: the seam is repeated and every angle describes the origin when r=0. Apply the smooth one-to-one formula on cut regions with r>0 and an open angular interval, then account for the omitted seam and centre as area-zero sets, or use corresponding limits. For continuous bounded integrands this does not change the disk integral.

If θ runs from 0 to 4π, almost every noncentral disk point is covered twice, and the ordinary integral with the polar weight counts it twice. A many-to-one map requires a multiplicity-aware statement or an appropriate one-to-one decomposition, not a silent use of the injective theorem. For singular integrands, removing the centre also requires a genuine convergence check.

In three dimensions, slice a solid into area-weighted layers

A triple integral ∭E f dV accumulates over a three-dimensional input region. Its Riemann sums use products Δx Δy Δz. For a continuous function on a compact box, repeated Fubini permits any coordinate order. A nonbox solid needs bounds describing all three coordinates in a consistent nesting.

For the tetrahedron E={x,y,z≥0:x+y+z≤1}, one order is ∫₀¹∫₀¹⁻ˣ∫₀¹⁻ˣ⁻ʸ1 dz dy dx. Integrating in z gives the height 1−x−y over the triangular projection x,y≥0,x+y≤1. The resulting volume is 1/6. If z is the outer coordinate instead, a valid order is ∫₀¹∫₀¹⁻ᶻ∫₀¹⁻ᶻ⁻ˣ1 dy dx dz, with the same value.

The average of a field over a finite positive-volume solid divides its triple integral by the solid’s volume. A volume density in kilograms per cubic metre produces mass after multiplication by dV. The extra integration coordinate changes the geometric weight and its units; a double integral over a projection alone is not the solid’s volume unless its integrand includes the appropriate height.

State the angles before using a memorised factor

Cylindrical coordinates use x=r cosθ, y=r sinθ and z=z. Their unsigned volume factor is r, so dV=r dr dθ dz in a compatible iterated order. Bounds must describe the actual solid: a cylinder, a cone and a ball have different z ranges even if their x,y projections look similar.

Use spherical coordinates x=ρ sinφ cosθ, y=ρ sinφ sinθ, z=ρ cosφ, where ρ is distance from the origin, φ is the colatitude measured from the positive z axis, and θ is azimuth in the x,y plane. With 0<φ<π, the unsigned volume factor is ρ²sinφ. Other books may swap angle names or measure latitude from the equatorial plane, which changes the written factor; the coordinate definitions decide the formula.

For a ball of radius R, the integral of 1 is ∫₀²π∫₀π∫₀ᴿρ²sinφ dρ dφ dθ=(2π)(2)(R³/3)=4πR³/3. The polar axis, origin and azimuthal seam are exceptional parameter sets; they do not invalidate the volume formula when treated by charts, zero-volume boundaries and suitable limits. Their presence does prevent calling the whole closed parameter box a one-to-one nonsingular map.

For a constant physical density ρ₀ in the ball, the moment of inertia about the z axis is ∭ρ₀(x²+y²) dV. Since x²+y²=ρ²sin²φ, separation gives ρ₀(R⁵/5)(4/3)(2π)=8πρ₀R⁵/15. Its units are mass times length squared. The density symbol ρ₀ is a constant here; the coordinate ρ is a radial distance.

Existing iterated improper values do not guarantee one double integral

Let F(x,y)=(x²−y²)/(x²+y²)² away from the origin in the unit square, with an arbitrary assigned value at the origin. For each y>0, integrating first in x gives −1/(1+y²), since F=−∂ₓ[x/(x²+y²)]. The subsequent improper integral in y is −π/4. For each x>0, integrating first in y gives 1/(1+x²), since F=∂ᵧ[y/(x²+y²)]. The subsequent improper integral in x is π/4.

The two calculations disagree because the required integrability hypothesis fails, not because Fubini has two answers. Near the origin, polar substitution gives F=cos(2θ)/r², and |F|dA=|cos(2θ)| dr dθ/r. On an angular interval where |cos(2θ)| is bounded below by a positive constant, the radial integral diverges logarithmically. Thus F is not absolutely integrable on the square.

The inner slice at a zero outer coordinate is itself divergent; the displayed iterated improper values mean performing the inner integral for positive outer coordinates and then taking the outer endpoint limit. Assigning a value at one point cannot repair the divergent mass of |F| in neighbourhoods. A joint truncation or a principal-value convention is an additional definition and must not be substituted silently for an ordinary double integral.

Two broader theorems explain which exchanges are safe

In the measure-theoretic setting on Euclidean spaces, Tonelli’s theorem applies to nonnegative measurable functions: integrating in either order agrees with the full integral, allowing the common value +∞. Fubini’s theorem applies to an absolutely integrable measurable function, meaning ∫|f|<∞; the slice functions are absolutely integrable for almost every outer input, and the resulting iterated integrals are finite and agree. These are different hypotheses, not a rule that any two existing repeated integrals may be exchanged.

The corresponding product-space results use σ-finite measures, meaning that each space is covered by countably many measurable sets of finite measure. Lebesgue measure on finite-dimensional Euclidean space has that property because bounded boxes provide such a cover. Measurability and almost-everywhere qualification belong to these more general statements. The earlier continuous compact-region examples fit the elementary theory without needing this additional machinery.

A useful practical route for a signed integrand is to establish absolute integrability before changing order. For a nonnegative integrand, an infinite answer may be the correct accumulation rather than a failure of order independence. In both cases, an infinity-minus-infinity cancellation is not an ordinary signed integral.

Keep the geometry, the weight and the theorem together

Before computing, identify the input region, what is accumulated, and its units. After choosing an order or coordinate map, write the transformed inequalities, evaluate the integrand at that map, calculate the correct absolute determinant, and check whether inputs are counted once. Test the constant integrand 1 when a known area or volume provides an independent comparison.

The explorer should compare finite Riemann sums with a proved limiting value, show the same triangle by vertical and horizontal slices, and display how a small parameter cell changes under an affine or nonlinear map. A polar-cell comparison should retain equal radial and angular increments while making the radial area factor visible. Those pictures explain a relationship; their scales and declared angle conventions must be consistent.

For an improper or unbounded integrand, finish by stating the convergence argument and the order or truncation used. A numerical grid near a singularity can conceal divergence or create an apparent cancellation. The theorem’s hypotheses, not a stable-looking decimal, determine whether the computation represents the intended integral.

Worked examples

Try it yourself

1. Why does a midpoint height need a factor ΔxΔy in a double Riemann sum?

   Show the answer and reasoning

   The factor is the cell’s area.

   A height contributes a small volume only after multiplication by its horizontal input area. The number of sample points alone is not the weight.

2. Does a signed double integral always equal geometric volume?

   Show the answer and reasoning

   No.

   A nonnegative height gives volume above the input plane. A signed integrand subtracts contributions below that plane.

3. What does ∬D1 dA measure?

   Show the answer and reasoning

   The area of D, when the integral is defined.

   A unit density or unit height weights every small input patch by its area.

4. Reverse the order for 0≤x≤1,0≤y≤x.

   Show the answer and reasoning

   Use 0≤y≤1 and y≤x≤1.

   For a fixed y, the triangle starts at the diagonal x=y and ends at x=1. Merely exchanging differentials would not update the bounds.

5. For 0≤x≤1,x²≤y≤x, what x interval is allowed at a fixed y∈[0,1]?

   Show the answer and reasoning

   y≤x≤√y.

   The upper inequality y≤x gives the lower x bound. Since x≥0, x²≤y gives the upper bound √y.

6. What is the area of the L-shaped region in the article?

   Show the answer and reasoning

   3.

   The 2-by-1 lower rectangle has area 2 and the 1-by-1 upper-left square has area 1. Their common boundary contributes no area.

7. Why can the two first moments not be used directly as centre coordinates?

   Show the answer and reasoning

   They must be divided by the finite positive mass.

   Moments have mass-times-length units; division by mass gives length. A zero denominator does not define a mass centre.

8. A one-to-one planar coordinate map has determinant −5. What weight is used for ordinary area integration?

   Show the answer and reasoning

   5.

   The negative sign records reversed orientation. The absolute determinant supplies unsigned area scaling.

9. Which three objects change in a coordinate substitution?

   Show the answer and reasoning

   The integrand, the region and the volume weight.

   Use f∘Φ, the actual parameter domain and |det DΦ| together. Changing only the variables written in the function is incomplete.

10. If Ψ=Φ⁻¹, is det DΨ evaluated at a corresponding point equal to det DΦ?

    Show the answer and reasoning

    It is the reciprocal where the derivatives are invertible.

    DΨ(Φ(u))=(DΦ(u))⁻¹. The forward substitution formula uses the absolute determinant of the forward map.

11. For x=u,y=uv with u>0, what is the area factor?

    Show the answer and reasoning

    u.

    The derivative matrix is [[1,0],[v,u]], whose determinant is u. Parameter cells widen in the image as u increases.

12. What factor must accompany dr dθ in ordinary planar polar integration?

    Show the answer and reasoning

    r for r≥0.

    A fixed angular increment has arc length proportional to radius. Without r, the integral measures parameter area rather than plane area.

13. Does θ∈[0,4π] cover a noncentral disk point once under the usual polar map?

    Show the answer and reasoning

    It covers almost every such point twice.

    The ordinary polar-weight integral then counts multiplicity. The one-to-one change-of-variables theorem cannot be invoked without addressing that repetition.

14. In the stated spherical convention, what is φ and what is the volume factor?

    Show the answer and reasoning

    φ is colatitude from the positive z axis; the factor is ρ²sinφ.

    The coordinate map fixes the angle meaning. A latitude convention uses a different written trigonometric factor.

15. What units result from integrating kilograms per cubic metre against dV in cubic metres?

    Show the answer and reasoning

    Kilograms.

    The volume weight converts the density into a mass accumulation.

16. Do two finite iterated improper integrals automatically agree?

    Show the answer and reasoning

    No.

    The displayed rational function gives −π/4 and π/4. Absolute integrability or another applicable theorem is required before exchanging the orders.

17. Must a nonnegative measurable function have a finite integral before Tonelli permits either order?

    Show the answer and reasoning

    No.

    The common value may be +∞ under the stated product-measure conditions. This is different from an undefined infinity-minus-infinity signed cancellation.

18. Can redefining F at the origin repair the nonabsolute integrability of the rational counterexample?

    Show the answer and reasoning

    No.

    The divergence comes from neighbourhoods of the origin, not the value of a single point. The relevant |F| accumulation remains infinite.

What to carry forward

• A multiple integral combines a field or density with an area or volume weight; preserve the interpretation and units.
• Continuous compact-region examples admit elementary Fubini reductions. Bounds must describe the same region after an order change.
• Split regions when their slices change or have gaps; do not integrate over an accidental bounding rectangle.
• Mass centres and averages require the correct finite, positive normalising quantity.
• A coordinate change transforms the function, the domain and the absolute Jacobian determinant together.
• Polar, cylindrical and spherical weights follow from the stated coordinate maps. Seams, singular sets and multiplicity need explicit treatment.
• For signed unbounded functions, absolute integrability is a safe order-exchange condition. Existing iterated improper values alone are insufficient.
• Tonelli permits nonnegative measurable accumulation with a possible infinite value; Fubini’s absolutely integrable setting gives finite almost-everywhere slices.
• Exact finite sums, slice comparisons and numerical grids support understanding; analytic hypotheses and convergence arguments establish the intended integral.