Multiple Integrals

A single integral $\int_a^b f(x)\,dx$ accumulates a function along a line — it sweeps left to right and adds up infinitely many thin slices. But the world is not one-dimensional. To find the volume trapped under a curved surface, the mass of a plate whose density varies from point to point, or the total rainfall over a county, you need to accumulate over an area or a solid, not just a segment. Multiple integrals are the tool that extends the idea of "add up infinitely many tiny pieces" from a line to two and three dimensions.

The beautiful news is that you already know how to do them. A double or triple integral is not a new kind of arithmetic — it is a repeated ordinary integral. Master the single integral, learn to hold one variable still while you integrate the other, and multiple integrals become a matter of careful bookkeeping.

Learning Objectives

• Understand a double integral as the limit of a Riemann sum over a 2D region, and a triple integral over a 3D region.
• Evaluate double and triple integrals by converting them to iterated integrals.
• Set up correct limits of integration for both rectangular and non-rectangular regions.
• Interpret multiple integrals physically: volume, area, mass, and average value.
• Reverse the order of integration and know when doing so simplifies a problem.

Quick Answer

A double integral $\iint_R f(x,y)\,dA$ sums the values of $f$ over every point of a two-dimensional region $R$; if $f \geq 0$ it equals the volume under the surface $z = f(x,y)$ above $R$. A triple integral $\iiint_E f(x,y,z)\,dV$ does the same over a solid region $E$. You compute them as iterated integrals — integrate with respect to one variable at a time, treating the others as constants, working from the inside out. For a region where $y$ runs between two curves $g_1(x)$ and $g_2(x)$, the inner limits depend on $x$ and the outer limits are constants. Getting the limits right is the whole game; the integration itself is single-variable calculus.

Where It Came From

The need was old and practical: people wanted volumes and masses of things that are not simple boxes. Archimedes (c. 250 BCE) already computed the volume of a sphere and areas of curved regions using his "method of exhaustion," slicing shapes into pieces and summing. But this was a bespoke craft — each solid demanded a new trick.

The real leverage came after Newton and Leibniz created calculus in the late 1600s. Once a single integral could compute an area, the natural question followed: could you integrate again to get a volume? Through the 1700s, Euler and others treated double integrals essentially as "integrate twice," and used them to attack problems in mechanics — finding the center of mass of a lamina, the moment of inertia of a rotating body, the gravitational attraction of an extended object. These were not abstract exercises: shipbuilders, artillerists, and astronomers all needed the mass properties of irregular shapes.

The concept was made rigorous in the 1800s. Cauchy and later Riemann defined the integral precisely as a limit of sums over ever-finer partitions, which extended cleanly from intervals on a line to rectangles in a plane and boxes in space. Fubini's theorem (Guido Fubini, 1907) finally proved when it is legitimate to compute a double integral as a repeated single integral — the theoretical license for everything we do by hand. So the story is one of steadily generalizing accumulation: from length, to area, to volume, to mass distributed through space.

The Double Integral: Accumulating Over a Region

Imagine a region $R$ in the $xy$-plane and a surface $z = f(x,y)$ floating above it. Chop $R$ into tiny rectangles of area $\Delta A = \Delta x \, \Delta y$. Over each little rectangle, the surface is nearly flat, so the volume of the thin column above it is approximately (height) × (base) $= f(x_i, y_j)\,\Delta A$. Add up all the columns and refine the grid:

$\iint_R f(x,y)\,dA = \lim_{\Delta A \to 0} \sum_{i,j} f(x_i, y_j)\,\Delta A$

When $f(x,y) \geq 0$, this is exactly the volume of the solid between $R$ and the surface. When $f(x,y) = 1$ everywhere, the integral simply returns the area of $R$.

Fubini's theorem lets us evaluate this as an iterated integral. Over a rectangle $R = [a,b] \times [c,d]$:

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

The inner integral integrates over $y$ while treating $x$ as a constant; the outer integral then finishes the job over $x$.

Worked Example: A Double Integral Over a Rectangle

Evaluate $\displaystyle \iint_R (x + 2y)\,dA$ over $R = [0,2] \times [1,3]$.

Do the inner integral first, holding $x$ constant and integrating over $y$ from $1$ to $3$:

$\int_1^3 (x + 2y)\,dy = \Big[\, xy + y^2 \,\Big]_{1}^{3} = (3x + 9) - (x + 1) = 2x + 8$

Now integrate that result over $x$ from $0$ to $2$:

$\int_0^2 (2x + 8)\,dx = \Big[\, x^2 + 8x \,\Big]_0^2 = (4 + 16) - 0 = 20$

So the integral equals $20$. Because $x + 2y \geq 0$ on this region, this is the volume under the plane $z = x + 2y$ over the rectangle.

Non-Rectangular Regions and the Order of Integration

Most interesting regions are not rectangles. When the region is bounded above and below by curves, the inner limits become functions of the outer variable.

If $R$ is described by $a \leq x \leq b$ and $g_1(x) \leq y \leq g_2(x)$ (a "vertically simple" region), then:

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

The golden rule: the inner limits may depend on the outer variable, but the outer limits must be constants. If your final answer still contains a variable, you have made an error.

Worked Example: A Triangular Region

Evaluate $\displaystyle \iint_R 6xy\,dA$ where $R$ is the triangle with vertices $(0,0)$, $(2,0)$, and $(2,4)$.

Sketch it: the region sits under the line $y = 2x$ (which passes through $(0,0)$ and $(2,4)$) and above the $x$-axis, with $x$ running from $0$ to $2$. So for each $x$, $y$ goes from $0$ to $2x$:

$\int_0^2 \int_0^{2x} 6xy\,dy\,dx$

Inner integral over $y$:

$\int_0^{2x} 6xy\,dy = 6x \cdot \Big[\tfrac{y^2}{2}\Big]_0^{2x} = 6x \cdot \frac{(2x)^2}{2} = 6x \cdot 2x^2 = 12x^3$

Outer integral over $x$:

$\int_0^2 12x^3\,dx = \Big[\, 3x^4 \,\Big]_0^2 = 3 \cdot 16 = 48$

The integral equals $48$.

Reversing the Order

The same region can usually be described the other way — sweeping horizontally instead of vertically. For the triangle above, for a fixed $y$ (from $0$ to $4$), $x$ runs from the line $x = y/2$ over to $x = 2$:

$\int_0^4 \int_{y/2}^{2} 6xy\,dx\,dy$

This must give the same answer, $48$ (you can check it). Why bother reversing? Sometimes the inner integral is impossible in one order but easy in the other. The classic case is $\int_0^1 \int_x^1 e^{y^2}\,dy\,dx$: you cannot integrate $e^{y^2}$ with respect to $y$ in closed form, but if you swap to $\int_0^1 \int_0^y e^{y^2}\,dx\,dy$, the inner integral is trivial and the whole thing evaluates to $\tfrac{1}{2}(e - 1)$. Always sketch the region before reversing — the limits are read off the picture, not swapped mechanically.

Triple Integrals: Mass and Volume of Solids

A triple integral extends the idea one more dimension, accumulating over a solid region $E$ in space:

$\iiint_E f(x,y,z)\,dV = \int \int \int f(x,y,z)\,dz\,dy\,dx$

With $f = 1$, the triple integral gives the volume of $E$. With $f(x,y,z) = \rho(x,y,z)$ a density function, it gives the mass of the solid — this is the single most common physical use. The same idea in 2D gives the mass of a flat plate (lamina): $m = \iint_R \rho(x,y)\,dA$.

Worked Example: Mass of a Solid Box

A rectangular block occupies $0 \leq x \leq 2$, $0 \leq y \leq 1$, $0 \leq z \leq 3$, with density $\rho(x,y,z) = z$ (denser toward the top). Find its mass.

$m = \int_0^2 \int_0^1 \int_0^3 z\,dz\,dy\,dx$

Innermost, over $z$:

$\int_0^3 z\,dz = \Big[\tfrac{z^2}{2}\Big]_0^3 = \frac{9}{2}$

Middle, over $y$ (the integrand $9/2$ is constant in $y$):

$\int_0^1 \frac{9}{2}\,dy = \frac{9}{2}$

Outer, over $x$:

$\int_0^2 \frac{9}{2}\,dx = \frac{9}{2} \cdot 2 = 9$

The mass is $9$ units. Notice how a constant integrand just gets multiplied by the length of the interval — a useful shortcut worth recognizing.

Real-World Applications

• Engineering — moments of inertia: Designing a flywheel or a beam requires $\iint r^2 \, \rho \, dA$, the resistance to rotation, computed as a double integral over the cross-section.
• Physics — center of mass: The balance point of an irregular plate or solid is found by integrating position weighted by density, one integral per coordinate.
• Probability — joint distributions: For two continuous random variables, $P((X,Y) \in R) = \iint_R f(x,y)\,dA$, where $f$ is the joint density. The total probability integrates to $1$ over the whole plane.
• Environmental science: Total rainfall, pollutant load, or heat over a mapped region is a double integral of a spatially varying quantity.
• Computer graphics and rendering: Light arriving at a surface is an integral of incoming radiance over all directions (a double integral over a hemisphere) — the heart of realistic rendering.

Common Mistakes

1. Leaving a variable in the final answer. Misconception: the outer limits can depend on a variable too. Why wrong: after the outermost integration there is nothing left to substitute, so any remaining variable signals that a limit was placed in the wrong integral. Correction: the outermost limits must always be pure constants; only inner limits may reference outer variables.

2. Swapping limits mechanically when reversing order. Misconception: to reverse $dy\,dx$ to $dx\,dy$, just switch the two limit pairs. Why wrong: the new limits describe the region a different way and generally look nothing like the old ones. Correction: re-draw the region and read the new limits off the geometry.

3. Forgetting that a constant integrand times a region gives area or volume. Misconception: $\iint_R 1 \, dA$ needs the surface to mean something. Why wrong: integrating the constant $1$ simply totals up $dA$, which is the area of $R$ (and $\iiint_E 1\,dV$ is the volume of $E$). Correction: use this as a sanity check — if you know the area, verify your setup reproduces it.

Comparison and Connections

The single, double, and triple integral are the same idea at three scales of accumulation.

Feature	Single $\int f\,dx$	Double $\iint f\,dA$	Triple $\iiint f\,dV$
Domain	Interval on a line	Region in a plane	Solid in space
$f=1$ gives	Length	Area	Volume
$f \geq 0$ gives	Area under curve	Volume under surface	(4D "hypervolume")
Density use	Mass of a wire	Mass of a plate	Mass of a solid
Evaluated as	One integration	Two iterated	Three iterated

Multiple integrals also connect forward to coordinate changes — polar coordinates for circular regions (where $dA = r\,dr\,d\theta$), cylindrical and spherical coordinates for solids — and to vector calculus theorems (Green's, Stokes', Divergence) that relate integrals over regions to integrals over their boundaries.

Practice Questions

Recall

What does $\iint_R 1\,dA$ compute, and what does $\iiint_E 1\,dV$ compute?

Answer: The area of the region $R$, and the volume of the solid $E$, respectively.

Understanding

Explain why the outer limits of an iterated integral must be constants while the inner limits may be functions.

Guidance: Each integration eliminates one variable. The inner integral runs over a slice at a fixed value of the outer variable, so its bounds can depend on that value. After the last (outer) integration no variables remain, so its bounds cannot depend on anything — they must be numbers.

Application

Evaluate $\displaystyle \int_0^1 \int_0^{2} (4x + 3y^2)\,dx\,dy$.

Answer: Inner over $x$: $\int_0^2 (4x + 3y^2)\,dx = [2x^2 + 3y^2 x]_0^2 = 8 + 6y^2$. Outer over $y$: $\int_0^1 (8 + 6y^2)\,dy = [8y + 2y^3]_0^1 = 8 + 2 = 10$.

Analysis

The integral $\int_0^2 \int_{y/2}^{1} e^{-x^2}\,dx\,dy$ cannot be done in the given order. Reverse the order of integration and evaluate.

Guidance: The region is $0 \leq y \leq 2$, $y/2 \leq x \leq 1$, i.e. $x$ from $0$ to $1$ with $y$ from $0$ to $2x$. Reversed: $\int_0^1 \int_0^{2x} e^{-x^2}\,dy\,dx = \int_0^1 2x\,e^{-x^2}\,dx$. Let $u = x^2$: this equals $[-e^{-x^2}]_0^1 = 1 - e^{-1} \approx 0.632$.

FAQ

Do I always integrate the inside variable first? Yes — iterated integrals are evaluated from the inside out. The differential nearest the integrand (say $dy$) is done first, with all other variables held constant, then you move outward.

When $f$ can be negative, does the double integral still give volume? It gives signed volume: regions where $f < 0$ contribute negatively. If you want the true geometric volume between the surface and the plane, integrate $|f|$, splitting the region where $f$ changes sign.

How do I choose the order of integration? Look at two things: whether the region is easier to describe one way (fewer pieces), and whether the integrand is integrable in that order. If the inner integral is impossible in one order, try the other before anything else.

What is $dA$ versus $dx\,dy$? $dA$ is the abstract "element of area." When you set up an iterated integral in rectangular coordinates it becomes $dx\,dy$ (or $dy\,dx$). In polar coordinates the same $dA$ becomes $r\,dr\,d\theta$ — the extra $r$ accounts for how area elements stretch.

Can a region need to be split into multiple integrals? Yes. If a region cannot be described by a single set of "top and bottom" curves — for example an L-shape or a region whose boundary changes formula — break it into sub-regions, integrate each, and add the results.

Is there a fourth-dimensional integral? Absolutely — you can integrate over any number of variables, and physics regularly uses integrals over four or more dimensions (space plus time, or configuration spaces). The mechanics are identical: nest one more single integral.

Quick Revision

• $\iint_R f\,dA$ = accumulation over a plane region; $\iiint_E f\,dV$ = over a solid.
• $f = 1$: double integral gives area, triple gives volume.
• $f =$ density: gives mass ($m = \iint \rho\,dA$ or $\iiint \rho\,dV$).
• Evaluate as iterated integrals, inside-out, holding other variables constant (Fubini).
• Vertically simple region: $\int_a^b \int_{g_1(x)}^{g_2(x)} f\,dy\,dx$ — outer limits must be constants.
• To reverse order: redraw the region, then read off new limits; never swap mechanically.
• Reversing order can turn an impossible integral (e.g. $e^{y^2}$, $e^{-x^2}$) into an easy one.

Related Topics

Prerequisites

• Integrals
• Techniques of Integration
• Multivariable Calculus

Related Topics

• Applications of Integration
• Calculus branch overview: Calculus

Next Topics

• Change of variables and polar/cylindrical/spherical coordinates
• Vector calculus: Green's, Stokes', and the Divergence Theorem