Multiple Integrals

Why this concept exists

A definite integral

∫ₐᵇ g(x) dx

is best understood as an accumulation process: you slice the x-interval into tiny widths Δx, evaluate g on each slice, and add up g(xᵢ)Δx. In the limit as Δx → 0, that sum becomes a single number: the signed area under g.

In many real problems, “one coordinate isn’t enough.” You might want:

• •The volume under a surface z = f(x,y) above some region D in the xy-plane
• •The total mass of a thin plate with varying density ρ(x,y)
• •The amount of heat in a 3D object with density ρ(x,y,z)

In each case, you’re adding up contributions over a region in 2D or 3D, not just an interval.

The big idea (extension of Riemann sums)

A double integral is the limit of sums over tiny rectangles (or more general shapes) in a region D ⊂ ℝ².

Partition D into small pieces with area ΔAᵢ. Pick a sample point (xᵢ, yᵢ) in each piece. Then form a sum:

∑ f(xᵢ, yᵢ) ΔAᵢ

If the limit exists as the partition is refined (max ΔAᵢ → 0), we define

∬_D f(x,y) dA = lim ∑ f(xᵢ, yᵢ) ΔAᵢ.

Similarly, a triple integral adds over tiny boxes in a region E ⊂ ℝ³:

∭_E f(x,y,z) dV = lim ∑ f(xᵢ, yᵢ, zᵢ) ΔVᵢ.

What “dA” and “dV” really mean

The symbols dA and dV are reminders of the “infinitesimal piece” you’re summing.

• •In Cartesian coordinates, dA is written as dx dy (or dy dx).
• •In 3D Cartesian coordinates, dV is written as dx dy dz (in some order).

It is tempting to treat dx dy as algebraic factors—and in computations it often behaves that way—but conceptually it stands for the limiting area of a tiny patch.

Signed accumulation (not always geometric volume)

In one variable, ∫ can be negative if the function is below the axis. The same is true here:

• •∬_D f(x,y) dA is signed: positive where f > 0, negative where f < 0.
• •If you want physical volume under z = f above D, you typically assume f ≥ 0.

A key sanity check

If f(x,y) = 1, then

⎧ ∬_D 1 dA = Area(D)

⎨

⎩ ∭_E 1 dV = Volume(E)

This is a powerful check: if your limits describe the wrong region, your “area/volume of 1” will come out wrong.

Two ways to think about the output

A multiple integral returns one number. You can interpret it as:

1) Geometric accumulation: volume under a surface (2D region) or hypervolume (higher dimensions)

2) Total quantity: if f is a density (mass/area, mass/volume, probability density, etc.), the integral gives the total amount.

In many applications, the main challenge is not the integration itself—it’s correctly describing the region D or E and choosing a convenient differential element.

Why limits are the real problem

In single-variable integrals, the interval [a,b] is straightforward. In multiple integrals, the region can be:

• •A rectangle: easy
• •A triangle: piecewise bounds
• •A disk: curved boundary
• •A region between curves: requires careful reading

Your goal is to translate a geometric description of D (or E) into limits of integration.

Type I and Type II regions in ℝ²

Many 2D regions can be described in two common ways.

Type I (x-simple):

D = { (x,y) : a ≤ x ≤ b, g₁(x) ≤ y ≤ g₂(x) }

Then

∬_D f(x,y) dA = ∫ₓ=aᵇ ∫ᵧ=g₁(x)^{g₂(x)} f(x,y) dy dx.

Type II (y-simple):

D = { (x,y) : c ≤ y ≤ d, h₁(y) ≤ x ≤ h₂(y) }

Then

∬_D f(x,y) dA = ∫ᵧ=cᵈ ∫ₓ=h₁(y)^{h₂(y)} f(x,y) dx dy.

Same region, different descriptions. The choice can make the integral easy or miserable.

Reading boundaries from a picture (the method)

A reliable workflow:

1) Sketch D (even roughly).

2) Decide an order (dy dx or dx dy).

3) For the inner variable, draw a vertical or horizontal line through the region.

4) Find where that line enters and exits the region: those are the inner bounds.

5) The outer bounds are the range over which that line actually intersects D.

Example of why order matters

Suppose D is the region between y = x² and y = 2x for x ≥ 0. Intersections:

x² = 2x

x² − 2x = 0

x(x − 2) = 0 ⇒ x = 0, 2

If you integrate with dy first (vertical slices):

0 ≤ x ≤ 2,

x² ≤ y ≤ 2x.

So

∬_D f dA = ∫₀² ∫_{x²}^{2x} f(x,y) dy dx.

If you reverse order (horizontal slices), you must solve for x in terms of y:

y = x² ⇒ x = √y (since x ≥ 0)

y = 2x ⇒ x = y/2

But the left/right boundary switches depending on y. You’d need to split into ranges where y/2 ≤ √y (and check where they meet).

That’s the typical pattern: reversing order can require piecewise integrals.

Differential area element as geometry

In Cartesian coordinates:

• •A tiny rectangle has side lengths dx and dy
• •Its area is approximately dx·dy

So dA = dx dy.

This becomes especially meaningful later in other coordinate systems (polar, cylindrical, spherical), where the “tiny patch” is not a rectangle. Even in Cartesian coordinates, it helps to remember that dA represents shape and scale of the local patch.

A table: choosing an order

In ℝ³: describing solids with bounds

For triple integrals, the same ideas extend, but you now have three variables and nested bounds.

A common pattern is a “z-simple” solid:

E = { (x,y,z) : (x,y) ∈ D, z₁(x,y) ≤ z ≤ z₂(x,y) }

Then

∭_E f(x,y,z) dV = ∬_D

\left( ∫_{z=z₁(x,y)}^{z₂(x,y)} f(x,y,z) dz \right) dA.

In Cartesian coordinates, dV = dx dy dz (in some order). Geometrically, you’re summing tiny boxes with volume dx·dy·dz.

A key mental model: “stacking”

If E is between two surfaces z = z₂(x,y) (top) and z = z₁(x,y) (bottom), then for each (x,y) in the base region D you have a vertical segment of height z₂ − z₁. You can think:

Volume(E) = ∬_D (z₂(x,y) − z₁(x,y)) dA

because that’s exactly ∭_E 1 dV with dz integrated out first:

∭_E 1 dV

= ∬_D \left( ∫_{z₁}^{z₂} 1 dz \right) dA

= ∬_D (z₂ − z₁) dA.

That identity is one of the most useful bridges between 2D and 3D integration.

Why iterated integrals work

The double integral is defined via a 2D limiting process, but in practice we compute it using iterated integrals (one integral inside another). Under mild conditions (e.g., f continuous on a nice region), Fubini’s Theorem tells us we can compute

∬_D f(x,y) dA

by integrating in one variable and then the other, provided the bounds correctly describe D.

So computation becomes familiar: it’s just repeated single-variable integration.

Anatomy of an iterated integral

Take

∬_D f(x,y) dA = ∫ₓ=aᵇ ∫ᵧ=g₁(x)^{g₂(x)} f(x,y) dy dx.

Read it as:

1) For a fixed x, integrate f(x,y) with respect to y from y = g₁(x) to y = g₂(x).

2) The result is a function of x.

3) Integrate that result from x = a to x = b.

The inner integral treats the “outer variable” as a constant.

Worked algebra pattern (keep constants straight)

If f(x,y) = x y² + 3x, then when integrating with respect to y:

∫ (x y² + 3x) dy

= x ∫ y² dy + 3x ∫ 1 dy

= x \left( y³/3 \right) + 3x(y)

= (x/3) y³ + 3xy.

The key move is recognizing x is constant during dy integration.

Double integrals as volume under a surface

If f(x,y) ≥ 0 and D is a region in the plane, then

Volume under z = f(x,y) above D = ∬_D f(x,y) dA.

Why? Because above each small patch ΔA, the “column” has base area ΔA and height f(xᵢ,yᵢ), so volume contribution is approximately f(xᵢ,yᵢ)ΔA.

Triple integrals as accumulation in 3D

In 3D, if ρ(x,y,z) is a density (mass per unit volume), then total mass is

Mass = ∭_E ρ(x,y,z) dV.

Again, each tiny box has mass approximately ρ(xᵢ,yᵢ,zᵢ)ΔV.

Changing order: not just algebra, geometry

Changing dx dy to dy dx is not a symbolic swap. It means:

• •You are slicing the region differently.
• •The inner bounds must represent the “enter/exit” points of the slice.

This is where many mistakes happen: the integrand stays the same, but the bounds must be rewritten to describe the same region.

A practical test: integrate 1

If you’re unsure whether your bounds match the region, compute ∬_D 1 dA.

• •If D is a triangle of area 3, you must get 3.
• •If D is a disk of radius 2, you must get 4π.

This is the fastest debugging tool for region setup.

Common region templates (Cartesian)

1) Rectangle: a ≤ x ≤ b, c ≤ y ≤ d

∬_D f dA = ∫ₐᵇ ∫_c^d f(x,y) dy dx

2) Between curves y = g₁(x) and y = g₂(x):

∬_D f dA = ∫ₐᵇ ∫_{g₁(x)}^{g₂(x)} f(x,y) dy dx

3) Between surfaces z = z₁(x,y) and z = z₂(x,y):

∭_E f dV = ∬_D ∫_{z₁(x,y)}^{z₂(x,y)} f(x,y,z) dz dA

Differential element as bookkeeping

In an iterated integral, the differential at the end is a reminder of the order:

• •dx dy means “integrate in x first (inner), then y (outer)” only if written as ∫∫ … dx dy.
• •dy dx means “integrate in y first, then x.”

So the pair (bounds, differential order) must match.

A helpful habit: when you write

∫ₓ=aᵇ ∫ᵧ=g₁(x)^{g₂(x)} f(x,y) dy dx

say out loud: “dy then dx.” This prevents accidental mismatches like writing y-bounds that depend on y (nonsense) or forgetting which variable is currently being integrated.

1) Area and volume as special cases

As noted earlier:

Area(D) = ∬_D 1 dA

Volume(E) = ∭_E 1 dV

And for a solid under z = f(x,y) above base D (with f ≥ 0):

Volume = ∬_D f(x,y) dA.

This is the “volume under surfaces” story that motivates double integrals.

2) Mass with variable density

If a lamina (thin plate) occupies D and has density ρ(x,y) (mass per unit area), then

Mass = ∬_D ρ(x,y) dA.

If a 3D object occupies E with density ρ(x,y,z) (mass per unit volume), then

Mass = ∭_E ρ(x,y,z) dV.

These formulas are identical in structure to area/volume, but the integrand now encodes material variation.

3) Averages in 2D and 3D

In 1D, the average value of g on [a,b] is (1/(b−a))∫ₐᵇ g(x) dx.

In 2D, the average value of f on region D is

f_avg = (1/Area(D)) ∬_D f(x,y) dA.

In 3D, the average value of f on region E is

f_avg = (1/Volume(E)) ∭_E f(x,y,z) dV.

This is a common use in physics (average temperature, average density) and probability (expected values with densities).

4) Probability (continuous random variables)

If (X,Y) has joint density p(x,y) over D, then

P((X,Y) ∈ D) = ∬_D p(x,y) dA.

And for expectations, you integrate the quantity of interest times density:

E[g(X,Y)] = ∬ g(x,y) p(x,y) dA.

Even if you haven’t studied probability yet, it’s helpful to recognize that the same accumulation logic powers it.

5) Where this node leads next (coordinate systems)

Many regions (disks, cylinders, spheres) are awkward in Cartesian bounds but simple in polar/cylindrical/spherical coordinates. In those systems, the differential element changes (e.g., dA = r dr dθ), which is fundamentally the same “infinitesimal patch” idea.

So mastering regions + iterated integrals in Cartesian coordinates sets you up to understand Jacobian factors later: the scaling needed when your tiny patch is a wedge or shell rather than a rectangle.

6) Connection to vector calculus

Multiple integrals are also the foundation for:

• •Line integrals and surface integrals (integrating along curves or over surfaces)
• •Divergence and curl theorems (turning complicated flux integrals into volume integrals)

Those topics reuse the same mental pattern: define a region, identify the correct differential element, and accumulate.