Double and Triple Integrals

⚡ Key Takeaways

Double and triple integrals generalize the single-variable definite integral to functions of two and three variables, computing accumulated quantities (volume, mass, charge, average value) over regions of the plane and space. Key result: under mild conditions (continuity, bounded region), a multiple integral over a region equals an iterated integral computed one variable at a time (Fubini's theorem). Where it applies: physics (mass, center of mass, moments of inertia, flux setups), probability (joint distributions), engineering, and any context requiring accumulation over 2D/3D domains. Who/when: roots in Newton & Leibniz (1660s–80s); rigorous formulation by Cauchy, Riemann (1854), and Fubini (1907).

💡 The Core Insight

A multiple integral is a single number — the limit of summing $f(\text{point})\times(\text{cell area or volume})$ over an ever-finer grid. The genius of the iterated-integral method is dimensional reduction: integrate out one variable while holding the others fixed, collapsing an $n$ -dimensional problem into a chain of one-dimensional problems you already know how to solve. The entire art lies in two things: correctly describing the region as inequalities, and choosing a coordinate system (Cartesian, polar, cylindrical, spherical) that makes the region and integrand simple.

DefinitionDouble Integral via Riemann Sums

Let $R=[a,b]\times[c,d]$ be a rectangle and $f:R\to\mathbb{R}$ bounded. Partition $[a,b]$ into subintervals $x_{i-1}\le x_i$ and $[c,d]$ into $y_{j-1}\le y_j$ , forming subrectangles $R_{ij}$ of area $\Delta A_{ij}=\Delta x_i\,\Delta y_j$ . Choose sample points $(x_i^*,y_j^*)\in R_{ij}$ . The double integral is

1

\quad \iint_R f(x,y)\,dA = \lim_{\|P\|\to 0}\sum_{i=1}^m\sum_{j=1}^n f(x_i^*,y_j^*)\,\Delta A_{ij}

Derivation hint: mimic the 1D definite integral — replace intervals of length $\Delta x$ by cells of area $\Delta A$ ; $\|P\|$ is the mesh, the largest cell diameter. The limit, when it exists independent of partition and sample choice, defines integrability. For a general bounded region $R$ , extend $f$ by zero to a bounding rectangle $R^*\supseteq R$ and integrate the extension; the integral exists if $\partial R$ has zero area (measure zero).

DefinitionTriple Integral

For a bounded solid $E\subset\mathbb{R}^3$ and bounded $f:E\to\mathbb{R}$ , partition a bounding box into subboxes of volume $\Delta V_{ijk}=\Delta x_i\,\Delta y_j\,\Delta z_k$ and define

2

\quad \iiint_E f(x,y,z)\,dV = \lim_{\|P\|\to 0}\sum_{i,j,k} f(x_i^*,y_j^*,z_k^*)\,\Delta V_{ijk}

Derivation hint: identical construction in one more dimension; cells become boxes, area becomes volume. Integrability again requires $\partial E$ to have zero volume.

TheoremFubini's Theorem

If $f$ is continuous on a rectangle $R=[a,b]\times[c,d]$ (or more generally Riemann integrable there), then the double integral equals either iterated integral:

3

\quad \iint_R f\,dA = \int_a^b\!\!\left(\int_c^d f(x,y)\,dy\right)dx = \int_c^d\!\!\left(\int_a^b f(x,y)\,dx\right)dy

Derivation hint: both iterated integrals equal the common limit of the Riemann sums; continuity lets you exchange limits and refine the grid in either order. Lebesgue's version requires $\iint |f|\,dA<\infty$ . For a Type I region $R=\{(x,y): a\le x\le b,\ g_1(x)\le y\le g_2(x)\}$ :

4

\quad \iint_R f\,dA = \int_a^b\!\!\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx

Derivation hint: for each fixed $x$ , the inner integral sweeps $y$ across the vertical slice of $R$ ; the outer sum stacks slices. For a Type II region $R=\{(x,y): c\le y\le d,\ h_1(y)\le x\le h_2(y)\}$ :

5

\quad \iint_R f\,dA = \int_c^d\!\!\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy

Derivation hint: swap the roles of $x$ and $y$ ; integrate along horizontal slices first.

TheoremChange of Variables / Jacobian

Let $T:(u,v)\mapsto(x(u,v),y(u,v))$ be a $C^1$ bijection from region $S$ onto $R$ . Then

6

\quad \iint_R f(x,y)\,dA = \iint_S f\bigl(x(u,v),y(u,v)\bigr)\,\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv

where the Jacobian determinant is

7

\quad \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{vmatrix} = x_u y_v - x_v y_u

Derivation hint: a small cell $du\,dv$ maps to a parallelogram spanned by the columns of the Jacobian matrix; its area is the absolute value of the determinant — the local area-scaling factor of $T$ .

DefinitionPolar, Cylindrical, Spherical Volume Elements

The most-used coordinate transformations and their Jacobian factors:

8

\quad dA = r\,dr\,d\theta \quad(\text{polar}),\qquad dV = r\,dz\,dr\,d\theta\quad(\text{cylindrical})

9

\quad dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\quad(\text{spherical}),\quad x=\rho\sin\phi\cos\theta,\ y=\rho\sin\phi\sin\theta,\ z=\rho\cos\phi

Derivation hint: compute the Jacobian of $(x,y)=(r\cos\theta,r\sin\theta)$ to get $r$ ; for spherical, $\det = \rho^2\sin\phi$ after expanding the $3\times3$ Jacobian. Geometrically, a spherical cell has dimensions $d\rho\times\rho\,d\phi\times\rho\sin\phi\,d\theta$ .

ProofFubini for a Continuous Function on a Rectangle

Goal: show $\iint_R f\,dA = \int_a^b\int_c^d f\,dy\,dx$ . 1. Define the slice function. For fixed $x$ , let $A(x)=\int_c^d f(x,y)\,dy$ . Since $f$ is continuous on a compact rectangle, it is uniformly continuous, so $A(x)$ is well-defined and continuous in $x$ . 2. Partition. Take partitions $\{x_i\}$ of $[a,b]$ and $\{y_j\}$ of $[c,d]$ . On each subrectangle $R_{ij}$ , by continuity and compactness $f$ attains a min $m_{ij}$ and max $M_{ij}$ . 3. Bound the slice integral. For $x\in[x_{i-1},x_i]$ :

\sum_j m_{ij}\,\Delta y_j \le \int_c^d f(x,y)\,dy = A(x) \le \sum_j M_{ij}\,\Delta y_j.

4. Integrate over $x$ . Multiplying by $\Delta x_i$ and summing over $i$ :

\sum_{i,j} m_{ij}\,\Delta A_{ij} \le \int_a^b A(x)\,dx \le \sum_{i,j} M_{ij}\,\Delta A_{ij}.

5. Squeeze. The left and right sums are the lower and upper Riemann sums for $\iint_R f\,dA$ . As $\|P\|\to0$ both converge to the double integral (since $f$ is integrable). Hence $\int_a^b A(x)\,dx = \iint_R f\,dA$ . The same argument with roles reversed gives the other order. $\blacksquare$

∎

GF

Guido Fubini

1879–1943

An Italian mathematician of immense breadth, Fubini proved in 1907 the theorem now bearing his name, settling precisely when a multiple (Lebesgue) integral may be computed as an iterated integral. His result is the rigorous engine behind every textbook reduction of a double integral to two single integrations. Fubini also made deep contributions to differential geometry (Fubini–Study metric on complex projective space) and the calculus of variations. A Jewish professor in fascist Italy, he was expelled from his post in 1938 and emigrated to the United States, teaching at Princeton until his death.

🧠 Intuition

Think of the double integral as stacking cardboard slices. To find the volume under a surface, slice the solid with a knife parallel to one axis. Each slice is a thin sheet whose cross-sectional area is a single integral $A(x)=\int f(x,y)\,dy$ ; the volume is $\int A(x)\,dx$ — you sum the areas times their thickness. This is exactly the iterated integral, and it is why the answer cannot depend on slicing direction (Fubini). For change of variables, picture warping graph paper. A nice grid of $du\,dv$ squares gets stretched and sheared into curvy cells. The Jacobian $|\partial(x,y)/\partial(u,v)|$ is the local "stretch factor" telling you how much area each tiny square gains or loses. In polar coordinates the factor is $r$ : cells far from the origin are wider (the arcs $r\,d\theta$ are longer), so you weight them more — that lone $r$ in $r\,dr\,d\theta$ is not arbitrary, it is geometry.

Figure

Type I region: vertical slice for the inner integral

ExampleVolume Under a Paraboloid over a Disk (Polar)

Given: Find the volume of the solid under $z = 4 - x^2 - y^2$ above the disk $x^2+y^2\le 4$ .

Step 1 — Switch to polar. The region is $0\le r\le 2$ , $0\le\theta\le2\pi$ , and $x^2+y^2=r^2$ , with $dA=r\,dr\,d\theta$ :

V = \int_0^{2\pi}\!\!\int_0^2 (4-r^2)\,r\,dr\,d\theta.

Step 2 — Inner integral in $r$ .

\int_0^2 (4r - r^3)\,dr = \left[2r^2 - \tfrac{r^4}{4}\right]_0^2 = 8 - 4 = 4.

Step 3 — Outer integral in $\theta$ .

V = \int_0^{2\pi} 4\,d\theta = 8\pi.

\boxed{V = 8\pi \approx 25.13\ \text{cubic units}}

(Check: the paraboloid meets $z=0$ exactly at the disk boundary $r=2$ , so the region is consistent.)

ExampleMass and Center of Mass of a Solid (Triple Integral)

Given: A solid hemisphere $x^2+y^2+z^2\le a^2$ , $z\ge0$ , has density $\delta(x,y,z)=k z$ (kg/m³, $k$ constant). Find its mass and the $z$ -coordinate of its center of mass with $a=2$ m, $k=3$ kg/m⁴.

Step 1 — Spherical coordinates. Region: $0\le\rho\le a$ , $0\le\phi\le\pi/2$ , $0\le\theta\le2\pi$ ; $z=\rho\cos\phi$ , $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ .

Step 2 — Mass.

M = \int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^a (k\rho\cos\phi)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

Separate: $\displaystyle\int_0^a\rho^3 d\rho=\frac{a^4}{4}$ , $\displaystyle\int_0^{\pi/2}\sin\phi\cos\phi\,d\phi=\frac12$ , $\displaystyle\int_0^{2\pi}d\theta=2\pi$ .

M = k\cdot\frac{a^4}{4}\cdot\frac12\cdot2\pi = \frac{k\pi a^4}{4}.

With $a=2,\ k=3$ : $M = \dfrac{3\pi(16)}{4}=12\pi\approx 37.70$ kg.

Step 3 — Numerator for $\bar z$ . $\displaystyle M_{xy}=\iiint z\,\delta\,dV = k\int z^2\,dV$ with $z^2=\rho^2\cos^2\phi$ :

M_{xy}= k\int_0^{2\pi}\!\!\int_0^{\pi/2}\!\!\int_0^a \rho^4\cos^2\phi\sin\phi\,d\rho\,d\phi\,d\theta = k\cdot\frac{a^5}{5}\cdot\frac13\cdot2\pi = \frac{2k\pi a^5}{15}.

Step 4 — Center of mass.

\bar z = \frac{M_{xy}}{M} = \frac{2k\pi a^5/15}{k\pi a^4/4} = \frac{8a}{15}.

With $a=2$ : $\bar z = \dfrac{16}{15}\approx 1.067$ m.

\boxed{M = 12\pi\approx 37.70\ \text{kg},\qquad \bar z = \tfrac{8a}{15} = \tfrac{16}{15}\approx 1.07\ \text{m}}

⚠️ Common mistake

Order of integration is NOT free for the bounds. When you swap $dx\,dy$ to $dy\,dx$ , you must completely re-describe the region, not just flip the differentials. The limits change because Type I and Type II descriptions use different functions. A common fatal error is writing $\int_0^1\int_0^x f\,dy\,dx$ and swapping to $\int_0^1\int_0^x f\,dx\,dy$ — the new outer variable $y$ cannot have $x$ in its bounds.

The Jacobian uses absolute value. Equation [6] has $|\partial(x,y)/\partial(u,v)|$ — area can never be negative. Forgetting the absolute value, or forgetting the Jacobian factor entirely (especially the $r$ in polar or $\rho^2\sin\phi$ in spherical), is the single most frequent computational mistake.

$dA \ne dx\,dy$ in curved coordinates. Students often write $\iint f\,r\,d\theta$ forgetting the $dr$ , or use $dA = dr\,d\theta$ without the $r$ .

CounterexampleWhen Order of Integration Changes the Answer

Fubini requires integrability of $|f|$ . Consider

f(x,y) = \frac{x^2-y^2}{(x^2+y^2)^2} \quad\text{on } (0,1]\times(0,1].

Computing the iterated integrals in the two orders gives

\int_0^1\!\!\int_0^1 f\,dy\,dx = \frac{\pi}{4}, \qquad \int_0^1\!\!\int_0^1 f\,dx\,dy = -\frac{\pi}{4}.

The two orders disagree! The resolution: $\iint |f|\,dA = \infty$ , so $f$ is not absolutely integrable near the origin, and Fubini's hypotheses fail. Continuity on the open region is not enough when the function blows up. For all bounded continuous integrands on bounded regions, no such pathology occurs.

connection

Double and triple integrals sit at the heart of multivariable calculus and connect outward in every direction. They are the substrate for Vector Calculus: Green's theorem turns a double integral over a region into a line integral around its boundary, while the divergence theorem links a triple integral to a surface flux integral. The change-of-variables formula [6] is the multivariable child of $u$ -substitution from Single-Variable Integration, with the Jacobian playing the role of $du/dx$ . Through Measure Theory and the Lebesgue Integral, the Riemann construction of [1]–[2] is generalized to a setting where Fubini's theorem reaches its sharpest form (Fubini–Tonelli). In Probability Theory, joint densities are integrated over regions to obtain probabilities and expectations, making double/triple integrals the computational backbone of continuous random vectors.

🔬 Where this appears

⏳ History

- 1660s–1680s — Newton and Leibniz develop the integral as accumulation; Leibniz's

\int

notation and the idea of summing infinitesimal areas lay the groundwork.

- 1730s — Euler computes volumes and uses repeated integration freely, treating iterated integrals as obviously equal to the "true" volume.

- 1769 — Euler systematically introduces double integrals and discusses changing the order of integration for explicit functions.

- 1773 — Lagrange uses triple integrals to study the attraction of ellipsoids, an early sophisticated physical application.

- 1820s — Cauchy provides the first rigorous definition of the definite integral via sums, paving the way for careful multivariable treatment.

- 1841 — Jacobi publishes his theory of functional determinants, giving the Jacobian its name and its central role in change of variables.

- 1854 — Riemann's Habilitationsschrift defines the Riemann integral precisely, clarifying integrability conditions that extend to multiple integrals.

- 1902 — Lebesgue introduces measure-theoretic integration, dramatically widening the class of integrable functions.

- 1907 — Fubini proves his theorem for Lebesgue integrals; Tonelli (1909) supplies the companion result for nonnegative measurable functions, completing the modern theory of iterated integration.

open-problem

The clean picture of [1]–[6] hides genuine depth. Improper multiple integrals over unbounded regions or with unbounded integrands require care: unlike the 1D case, there is no canonical "principal value" and absolute convergence is essential for order-swapping (see the counterexample). Highly irregular regions — fractal boundaries with positive-measure frontier — are not Riemann integrable; only Lebesgue/measure-theoretic tools handle them, and deciding integrability touches descriptive set theory. Numerical evaluation in high dimensions suffers the curse of dimensionality: grid-based quadrature needs $N^d$ points, so for $d\gtrsim 4$ deterministic methods become infeasible and Monte Carlo / quasi-Monte Carlo integration takes over — yet sharp error bounds for these stochastic methods on general regions remain an active research frontier. Finally, the change-of-variables theorem under merely Lipschitz (non- $C^1$ ) or non-injective maps, and its sharpest hypotheses, connect to ongoing work in geometric measure theory.

Formal Statement

Why It Holds / Derivation

Building Intuition

Worked Examples

Common Misconceptions

Connections

Real-World Applications

Historical Development

Subtleties & Limits

See Also