📐 Concept diagram

06-09 — Double Integrals

Phase: 6 — Calculus III: Multivariable Calculus Subject: 06-09 Prerequisites: Single-variable integration (Phase 4), 06-01 — Functions of Several Variables Next subject: 06-10 — Double Integrals in Polar Coordinates

Learning Objectives

By the end of this subject, you will be able to:

1. Define the double integral as a limit of Riemann sums and interpret it as a volume under a surface
2. Set up and evaluate iterated integrals over rectangular regions using Fubini's theorem
3. Set up and evaluate double integrals over general (non-rectangular) regions as iterated integrals
4. Determine the correct order of integration (dx dy vs. dy dx) for a given region
5. Reverse the order of integration when one order is computationally intractable

Core Content

1. Definition — Riemann Sums for Double Integrals

⚠️ CRITICAL FOUNDATION: The double integral ∬_R f(x,y) dA computes volume under a surface. The skill is setting up bounds correctly — sketching the region, choosing Type I (dy dx) or Type II (dx dy), and knowing when to reverse integration order when one direction is intractable.

The double integral of f(x, y) over a region R is:

$∬_R f(x, y) dA = lim    Σ Σ f(xᵢ*, yⱼ*) ΔA
              Δ→0   i  j
$

Where ΔA = Δx Δy is the area of a small rectangle, and (xᵢ, yⱼ) is a sample point in each sub-rectangle.

Geometric interpretation: If f(x, y) ≥ 0, the double integral represents the volume of the solid under the surface z = f(x, y) and above the region R in the xy-plane.

If f(x, y) = 1, then ∬_R 1 dA = Area(R).

Average value: The average value of f over R is:

$f_avg = (1/Area(R)) ∬_R f(x, y) dA
$

2. Fubini's Theorem — Iterated Integrals

Fubini's Theorem: If f is continuous on the rectangle R = [a, b] × [c, d], then:

$                    b  d                  d  b
∬_R f(x, y) dA = ∫  ∫  f(x, y) dy dx = ∫  ∫  f(x, y) dx dy
                   a  c                  c  a
$

The two iterated integrals give the same result. Choose whichever is easier to compute.

How to read iterated integrals:

$    b  d
∫  ∫  f(x, y) dy dx   means: first integrate w.r.t. y (treating x constant),
   a  c                      then integrate the result w.r.t. x.
$

The inner integral is done first, and its limits can depend on the outer variable (for general regions).

Example 1: Evaluate ∬_R (x² + y²) dA over R = [0, 2] × [0, 1].

    2  1
∫  ∫  (x² + y²) dy dx
   0  0

Inner (w.r.t. y, x constant):
∫₀¹ (x² + y²) dy = [x²y + y³/3]₀¹ = x² + 1/3

Outer:
∫₀² (x² + 1/3) dx = [x³/3 + x/3]₀² = 8/3 + 2/3 = 10/3

Example 2: Same integral, reversed order:

$    1  2
∫  ∫  (x² + y²) dx dy
   0  0

Inner: ∫₀² (x² + y²) dx = [x³/3 + y²x]₀² = 8/3 + 2y²
Outer: ∫₀¹ (8/3 + 2y²) dy = [8y/3 + 2y³/3]₀¹ = 8/3 + 2/3 = 10/3 ✓
$

3. Double Integrals over General Regions

Not all regions are rectangles! We classify regions into two types.

Type I (vertically simple): The region is bounded between two functions of x:

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

The double integral is:

$                    b  g₂(x)
∬_R f(x, y) dA = ∫   ∫     f(x, y) dy dx
                   a  g₁(x)
$

The y-limits are functions of x; the x-limits are constants.

Type II (horizontally simple): The region is bounded between two functions of y:

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

The double integral is:

$                    d  h₂(y)
∬_R f(x, y) dA = ∫   ∫     f(x, y) dx dy
                   c  h₁(y)
$

The x-limits are functions of y; the y-limits are constants.

4. Setting Up Double Integrals — The Procedure

1. Sketch the region R in the xy-plane
2. Choose the order of integration (dy dx = Type I, or dx dy = Type II)
3. Find the bounds:
4. For dy dx: the outer limits on x are the min/max x-values in R. For each x, y goes from the bottom curve to the top curve
5. For dx dy: the outer limits on y are the min/max y-values in R. For each y, x goes from the left curve to the right curve
6. Integrate — inner first, then outer

Example 3: Evaluate ∬_R xy dA, where R is the triangle with vertices (0, 0), (1, 0), (0, 2).

Sketch: R is bounded by x = 0, y = 0, and the line from (1, 0) to (0, 2): y = 2 − 2x.

As Type I (dy dx): x ∈ [0, 1]. For each x, y ∈ [0, 2 − 2x].

$    1  2−2x
∫   ∫     xy dy dx
   0  0

Inner: ∫₀^(2−2x) xy dy = x[y²/2]₀^(2−2x) = x(2−2x)²/2 = x(4 − 8x + 4x²)/2 = 2x − 4x² + 2x³

Outer: ∫₀¹ (2x − 4x² + 2x³) dx = [x² − 4x³/3 + x⁴/2]₀¹ = 1 − 4/3 + 1/2 = 6/6 − 8/6 + 3/6 = 1/6
$

As Type II (dx dy): y ∈ [0, 2]. For each y, x ∈ [0, (2 − y)/2].

$    2  (2−y)/2
∫   ∫       xy dx dy
   0  0

Inner: ∫₀^((2−y)/2) xy dx = y[x²/2]₀^((2−y)/2) = y(2−y)²/8

Outer: 1/8 ∫₀² y(2−y)² dy = 1/8 ∫₀² y(4 − 4y + y²) dy
= 1/8 ∫₀² (4y − 4y² + y³) dy = 1/8[2y² − 4y³/3 + y⁴/4]₀²
= 1/8(8 − 32/3 + 4) = 1/8(12 − 32/3) = 1/8(4/3) = 1/6 ✓
$

5. Changing the Order of Integration

Sometimes the given order leads to an impossible or difficult inner integral. Reversing the order can make it tractable.

Example 4: Evaluate ∫₀¹ ∫_x¹ sin(y²) dy dx.

The inner integral ∫ sin(y²) dy cannot be expressed in elementary functions! Let's reverse the order.

Region (from the limits): x ∈ [0, 1], for each x, y ∈ [x, 1]. This is the triangle above the line y = x, bounded by y = 1, x = 0, and x = 1.

Sketching: R = {(x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y}.

Reversed integral:

$    1  y
∫   ∫   sin(y²) dx dy
   0  0

Inner: ∫₀^y sin(y²) dx = sin(y²)·[x]₀^y = y sin(y²)

Outer: ∫₀¹ y sin(y²) dy
Let u = y², du = 2y dy → y dy = du/2.
When y = 0: u = 0. When y = 1: u = 1.
= ∫₀¹ sin(u) · (du/2) = ½[−cos u]₀¹ = ½(−cos 1 + cos 0) = ½(1 − cos 1)
$

The answer is ½(1 − cos 1). This was impossible in the original order!

6. Area and Volume Calculations

Area of region R: Area = ∬_R 1 dA.

Example 5: Find the area of the region bounded by y = x² and y = 2x − x².

Intersections: x² = 2x − x² → 2x² − 2x = 0 → 2x(x − 1) = 0 → x = 0, 1.

As Type I: x ∈ [0, 1], y ∈ [x², 2x − x²] (since 2x − x² > x² on (0, 1)).

$Area = ∫₀¹ ∫_(x²)^(2x−x²) 1 dy dx = ∫₀¹ [(2x − x²) − x²] dx
     = ∫₀¹ (2x − 2x²) dx = [x² − 2x³/3]₀¹ = 1 − 2/3 = 1/3
$

7. Volume as a Double Integral

Volume under z = f(x, y) above R: V = ∬_R f(x, y) dA (when f ≥ 0).

Example 6: Find the volume under z = x + y and above the triangle R with vertices (0, 0), (2, 0), (0, 1).

R as Type I: x ∈ [0, 2], y ∈ [0, 1 − x/2].

$V = ∫₀² ∫₀^(1−x/2) (x + y) dy dx

Inner: ∫₀^(1−x/2) (x + y) dy = [xy + y²/2]₀^(1−x/2)
    = x(1 − x/2) + (1 − x/2)²/2
    = x − x²/2 + (1 − x + x²/4)/2
    = x − x²/2 + 1/2 − x/2 + x²/8
    = (x − x/2) + (−x²/2 + x²/8) + 1/2
    = x/2 − 3x²/8 + 1/2

Outer: ∫₀² (x/2 − 3x²/8 + 1/2) dx = [x²/4 − x³/8 + x/2]₀²
     = 4/4 − 8/8 + 2/2 = 1 − 1 + 1 = 1
$

Volume = 1 cubic unit.

Key Terms

• Choose the order
• Find the bounds
• Integrate
• Sketch the region

Worked Examples

Worked Example 1: Iterated Integral with ln

Problem: Evaluate ∫₀¹ ∫₀¹ xe^(xy) dy dx.

Solution:

Inner (w.r.t. y): ∫₀¹ x e^(xy) dy = [e^(xy)]₀¹ = e^x − 1
(Since ∂/∂y[e^(xy)] = x e^(xy), so ∫ x e^(xy) dy = e^(xy).)

Outer: ∫₀¹ (e^x − 1) dx = [e^x − x]₀¹ = (e − 1) − (1 − 0) = e − 2

Worked Example 2: General Region (between two curves)

Problem: Evaluate ∬_R x dA where R is the region bounded by y = x² and y = √x.

Solution: Intersections: x² = √x → x⁴ = x → x(x³ − 1) = 0 → x = 0, 1.

As Type I: x ∈ [0, 1], y ∈ [x², √x] (since √x ≥ x² on [0, 1]).

$∬_R x dA = ∫₀¹ ∫_(x²)^(√x) x dy dx

Inner: ∫_(x²)^(√x) x dy = x[√x − x²] = x^(3/2) − x³

Outer: ∫₀¹ (x^(3/2) − x³) dx = [2x^(5/2)/5 − x⁴/4]₀¹ = 2/5 − 1/4 = 8/20 − 5/20 = 3/20
$

Worked Example 3: Reversing Order

Problem: Evaluate ∫₀² ∫_(y/2)^1 e^(x²) dx dy by reversing the order of integration.

Solution: Region: y ∈ [0, 2], x ∈ [y/2, 1]. Sketch: x = y/2 → y = 2x. So y goes from 0 to 2x for each x, and x goes from 0 to 1.

Reversed: x ∈ [0, 1], y ∈ [0, 2x].

$∫₀¹ ∫₀^(2x) e^(x²) dy dx

Inner: ∫₀^(2x) e^(x²) dy = e^(x²)·2x

Outer: ∫₀¹ 2x e^(x²) dx
Let u = x², du = 2x dx.
= ∫₀¹ e^u du = [e^u]₀¹ = e − 1
$

Worked Example 4: Volume Between Two Surfaces

Problem: Find the volume of the solid bounded by the paraboloid z = 4 − x² − y² and the plane z = 0.

Solution: The intersection of z = 4 − x² − y² with z = 0 gives x² + y² = 4, a circle of radius 2. R is the disk x² + y² ≤ 4.

Volume = ∬_R (4 − x² − y²) dA.

This is easiest in polar coordinates (next subject), but we can do it in rectangular:

As Type I: x ∈ [−2, 2], y ∈ [−√(4−x²), √(4−x²)].

$V = ∫_(−2)² ∫_(−√(4−x²))^(√(4−x²)) (4 − x² − y²) dy dx

Inner w.r.t. y:
∫ (4 − x² − y²) dy = (4−x²)y − y³/3 evaluated at ±√(4−x²).
= 2(4−x²)√(4−x²) − 2(4−x²)^(3/2)/3
= 2(4−x²)^(3/2) − (2/3)(4−x²)^(3/2) = (4/3)(4−x²)^(3/2)

Outer: (4/3) ∫_(−2)² (4−x²)^(3/2) dx
Let x = 2 sin θ, dx = 2 cos θ dθ.
x = −2 → θ = −π/2. x = 2 → θ = π/2.
= (4/3) ∫_(−π/2)^(π/2) (4 − 4sin²θ)^(3/2) · 2 cos θ dθ
= (4/3) ∫_(−π/2)^(π/2) (4cos²θ)^(3/2) · 2 cos θ dθ
= (4/3) ∫_(−π/2)^(π/2) 8 cos³θ · 2 cos θ dθ
= (4/3)·16 ∫_(−π/2)^(π/2) cos⁴θ dθ
$

Using ∫_(−π/2)^(π/2) cos⁴θ dθ = (3π)/8: V = (64/3) · (3π/8) = 8π.

(We'll verify this much more easily using polar coordinates in 06-10: V = ∫₀^(2π) ∫₀² (4 − r²) r dr dθ = 2π∫₀² (4r − r³)dr = 2π[2r² − r⁴/4]₀² = 2π(8 − 4) = 8π. ✓)

Quiz

Q1: The double integral ∬_R f(x, y) dA, when f(x, y) ≥ 0, represents:

A) The area of region R B) The volume under the surface z = f(x, y) above region R C) The average value of f over R D) The arc length of the boundary of R

Correct: B)

• If you chose B: The double integral of a nonnegative function gives the volume of the solid between the xy-plane and the surface. Correct!
• If you chose A: That's ∬_R 1 dA, not ∬_R f(x, y) dA in general.
• If you chose C: The average value is (1/Area) ∬_R f dA, not the integral itself.
• If you chose D: Arc length is a line integral, not a double integral.

Q2: Fubini's theorem states that for a continuous f on a rectangle [a, b] × [c, d]:

A) ∬_R f dA = f(b, d) − f(a, c) B) ∬_R f dA = ∫_a^b ∫_c^d f(x, y) dy dx = ∫_c^d ∫_a^b f(x, y) dx dy C) ∬_R f dA = f_x(b, d) · f_y(a, c) D) The order of integration always gives different results

Correct: B)

• If you chose B: Fubini's theorem says the double integral equals both iterated integrals, and they give the same result. Correct!
• If you chose A: That resembles the Fundamental Theorem for single integrals, not Fubini.
• If you chose C: This is not a valid integration formula.
• If you chose D: Fubini guarantees equality — the two orders give the same value.

Q3: For a Type I region, the double integral is set up as:

A) ∫c^d ∫{h₁(y)}^{h₂(y)} f(x, y) dx dy B) ∫a^b ∫{g₁(x)}^{g₂(x)} f(x, y) dy dx C) ∫a^b ∫_c^d f(x, y) dx dy D) ∫{g₁(x)}^{g₂(x)} ∫_a^b f(x, y) dy dx

Correct: B)

• If you chose B: Type I (vertically simple): x-limits are constants, y-limits are functions of x. Correct!
• If you chose A: That's Type II (horizontally simple), with x-limits as functions of y.
• If you chose C: That's a rectangle, not a general Type I region.
• If you chose D: The order is wrong — outer integral must have constant limits.

Q4: Why might you reverse the order of integration?

A) To make the region smaller B) To avoid an inner integral that has no elementary antiderivative in the given order C) Because Type II is always easier than Type I D) To change the value of the integral

Correct: B)

• If you chose B: Reversing the order can transform an impossible inner integral (like ∫ sin(y²) dy) into a tractable one (∫ y sin(y²) dy). Correct!
• If you chose A: The region stays the same — only the description changes.
• If you chose C: Neither type is universally easier; it depends on the integrand and region.
• If you chose D: Reversing order doesn't change the value of the integral (Fubini's theorem).

Q5: To find the area of a region R using double integrals, you compute:

A) ∬_R f(x, y) dA for f(x, y) = x + y B) ∬_R 1 dA C) ∫_a^b f(x) dx D) ∬_R ∇f · n dA

Correct: B)

• If you chose B: Setting f(x, y) = 1 makes the double integral equal to the area of R. Correct!
• If you chose A: That computes the integral of (x + y), not the area.
• If you chose C: Single integrals give area under a curve, not area of a 2D region generally.
• If you chose D: That's a flux integral (divergence theorem), not an area computation.

Q6: For a region bounded by y = x² and y = 2x − x², the correct Type I limits are:

A) x ∈ [0, 2], y ∈ [x², 2x − x²] B) x ∈ [0, 1], y ∈ [x², 2x − x²] C) x ∈ [0, 1], y ∈ [2x − x², x²] D) x ∈ [−1, 1], y ∈ [x², 2x − x²]

Correct: B)

• If you chose B: The curves intersect at x = 0 and x = 1; on (0,1), 2x − x² > x². Correct!
• If you chose A: The intersection is at x = 1, not x = 2.
• If you chose C: For x ∈ (0,1), 2x − x² > x², so the upper bound should be 2x − x².
• If you chose D: The intersection is only at x = 0 and x = 1; for negative x the curves don't enclose a region.

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Evaluate ∬_R (2x + 3y) dA over R = [0, 1] × [0, 2].

Problem 2: Evaluate ∫₀¹ ∫₀^x (x² + y²) dy dx.

Problem 3: Evaluate ∬_R x dA where R is the region bounded by y = x² and y = 4.

Problem 4: Reverse the order of integration for ∫₀¹ ∫_(y²)^y f(x, y) dx dy.

Problem 5: Evaluate ∫₀¹ ∫_(√x)¹ sin(π y³) dy dx by reversing the order.

Problem 6: Find the volume of the solid under z = x² + y² above the square [0, 1] × [0, 1].

Problem 7: Evaluate ∬_R e^(y/x) dA where R is the triangle bounded by y = 0, y = x, and x = 2.

Summary

1. The double integral ∬_R f(x, y) dA is the limit of Riemann sums and gives the volume under z = f(x, y) over region R when f ≥ 0; for f = 1, it gives the area of R
2. Fubini's theorem lets us compute double integrals as iterated integrals: integrate w.r.t. one variable while treating the other as constant, then integrate the result — the order can be swapped freely over rectangles
3. For general regions, classify as Type I (vertical, dy dx with y-limits as functions of x) or Type II (horizontal, dx dy with x-limits as functions of y) — sketch the region first
4. When an inner integral is impossible to evaluate in closed form, try reversing the order of integration — this can transform an intractable integral into an easy one
5. The average value of f over R is f_avg = (1/Area)∬_R f dA; setting up the correct bounds is the most common source of errors

Pitfalls

• Getting the bounds order wrong. For dy dx (Type I), the outer limits must be constants (x-limits) and the inner limits (y-limits) can be functions of x. Swapping this — putting functions on the outer integral — produces a nonsensical iterated integral that depends on a variable that was supposed to be integrated out.
• Skipping the sketch. Setting up double integrals without sketching the region is the #1 source of bound errors. Without a picture, it's easy to use the wrong curve as the upper bound, miss intersection points, or fail to notice that the region needs to be split into subregions.
• Mixing up the variable being held constant. In ∫∫ f(x,y) dy dx, the inner integration treats x as constant and integrates with respect to y. Students sometimes incorrectly treat y as constant in the inner integral or substitute x-dependent limits into the wrong variable. Always explicitly note which variable is "live" and which is "frozen" at each stage.
• Failing to reverse the order of integration when needed. When the inner integral cannot be evaluated in closed form (e.g., ∫ sin(y²) dy, ∫ e^(x²) dx), reversing the order often transforms it into an easy integral. Recognizing this situation — and having the skill to re-describe the region — is critical.
• Treating variable limits as constants in the outer evaluation. After evaluating the inner integral, the result is a function of the outer variable only. A common mistake is to leave y in the expression after completing the inner dy integral and then trying to integrate with respect to x.

Next Steps

Move on to 06-10 — Double Integrals in Polar Coordinates to learn how polar coordinates simplify integration over circular and radially symmetric regions, the role of the Jacobian factor r, and applications to computing area, mass, and center of mass.