Extend single-variable integration to functions of two variables. Master iterated integrals, Fubini’s Theorem, Type I & II regions, switching order of integration, and double integrals in polar coordinates.
12 Worked Examples10-Question Quiz1 700+ WordsFubini’s TheoremPolar Coordinates
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§ 01What Is a Double Integral? The Big Idea
A single integral measures the signed area under a curve. A double integral measures the signed volume under a surface — and the leap from one to two dimensions is simpler than it looks.
Recall that a single definite integral is defined as the limit of a Riemann sum: we cut the interval [a, b] into thin strips, multiply the function value by the strip width, and add everything up as the strips become infinitely thin. A double integral does exactly the same thing, but now our region of integration lives in the xy-plane, and we sum up rectangular columns of height f(x, y) and base area ΔA = Δx Δy.
Definition — Double Integral as a Limit of a Double Riemann Sum
Here R is a rectangular region in the plane, (xij*, yij*) is any sample point in the (i, j)-th sub-rectangle, and ΔA = Δx Δy. If f(x, y) ≥ 0 throughout R, the double integral equals the volume of the solid that sits above R and below the surface z = f(x, y).
The Volume Interpretation
Imagine a lumpy hillside. The xy-plane is the ground; the surface z = f(x, y) is the height of the hill at each point. The double integral counts the total “material” stacked between the ground and the hill over the region R. When f takes negative values, the solid below the xy-plane is subtracted, making the answer a signed volume — exactly the 2D analogue of signed area.
Beyond pure volume, double integrals compute mass (when f is density), electric charge, probability, moments of inertia, and a great many other physical and engineering quantities. The technique is genuinely one of the most powerful in all of applied mathematics.
§ 02Iterated Integrals — How We Actually Compute Them
The limit-of-a-Riemann-sum definition tells us what a double integral is. An iterated integral tells us how to compute it — by doing two single integrals one at a time.
For a rectangular region R = [a, b] × [c, d], the double integral can always be broken into two nested single integrals:
Iterated Integral — Rectangular Region
∫∫R f(x, y) dA = ∫cd ∫ab f(x, y) dx dy = ∫ab ∫cd f(x, y) dy dx
How to Read an Iterated Integral
Work inside out. In the expression ∫cd ∫ab f(x, y) dx dy:
Step 1 — Inner integral: Integrate f(x, y) with respect to x from a to b. During this step, treat y as a constant. The result is a function of y alone — call it A(y).
Step 2 — Outer integral: Integrate A(y) with respect to y from c to d. This produces a number.
Example 01 — Simplest Iterated Integral
Evaluate ∫02 ∫13 (x + y) dx dy
Step 1
Inner integral (fix y, integrate over x from 1 to 3): ∫13 (x + y) dx = [x²/2 + xy]13 = (9/2 + 3y) − (1/2 + y) = 4 + 2y
Step 2
Outer integral (integrate over y from 0 to 2): ∫02 (4 + 2y) dy = [4y + y²]02 = 8 + 4 = 12
∫∫R (x + y) dA = 12
Example 02 — Switching the Order on a Rectangle
Verify Example 01 by integrating in the order dy dx
Step 1
Inner integral (fix x, integrate over y from 0 to 2): ∫02 (x + y) dy = [xy + y²/2]02 = 2x + 2
Step 2
Outer integral (integrate over x from 1 to 3): ∫13 (2x + 2) dx = [x² + 2x]13 = (9 + 6) − (1 + 2) = 12 ✓
Same answer — 12. This is no coincidence: see Fubini’s Theorem (§ 03).
§ 03Fubini’s Theorem — Why Order Doesn’t Matter on Rectangles
✦ Fubini’s Theorem (rectangular region)
If f(x, y) is continuous on the rectangle R = [a, b] × [c, d], then:
∫∫R f(x, y) dA = ∫cd ∫ab f(x, y) dx dy = ∫ab ∫cd f(x, y) dy dx
Fubini’s Theorem guarantees that on a rectangle, both orders of integration give the same answer. This is enormously practical: if one order produces a messy inner integral, simply switch to the other order. The theorem extends to non-rectangular regions (§ 04) and is the backbone of almost every double-integral calculation you will ever do.
Separable Functions — A Very Useful Special Case
If f(x, y) can be written as a product g(x) · h(y), then the double integral over a rectangle separates cleanly:
Separable Integrand
∫ab ∫cd g(x)h(y) dy dx = ∫ab g(x) dx · ∫cd h(y) dy
Example 03 — Separable Integrand
Evaluate ∫0π ∫01 x sin(y) dy dx
Separate
f(x,y) = x · sin(y), which is a product. Apply the separable formula.
Sometimes the inner integral as written is impossible to evaluate in closed form. Switching the order of integration often transforms an intractable problem into a straightforward one — without changing the answer.
The process has two parts: (1) determine the region D from the original limits, then (2) describe the same region D using the opposite variable as outer. Always sketch D first — the picture tells you everything.
⚠ Common Mistake — Swapping Without Sketching
Simply swapping the “dx dy” symbols and the limits is wrong. You must re-derive both sets of limits by describing the same region from a different perspective. Always draw the region before switching.
Example 07 — Switching Order to Unlock the Integral
Switch the order of integration and evaluate: ∫01 ∫x1 ey² dy dx
Problem
The inner integral ∫ ey² dy has no elementary antiderivative. We must switch order.
Identify D
Current limits: x from 0 to 1; y from x to 1. The region D is: 0 ≤ x ≤ 1, x ≤ y ≤ 1 — this is the triangle above the line y = x in the unit square.
Re-describe
Looking at D differently (Type II): y from 0 to 1; for each fixed y, x ranges from 0 to y. New integral: ∫01 ∫0y ey² dx dy
Inner
∫0y ey² dx = y ey² (ey² is constant w.r.t. x)
Outer
∫01 y ey² dy = [ey²/2]01 = (e − 1)/2
∫01 ∫x1 ey² dy dx = (e − 1)/2 ≈ 0.859
Example 08 — Switching Order on a Triangular Region
Switch order and evaluate: ∫02 ∫y/21 ex² dx dy
Identify D
Outer: y from 0 to 2. Inner: x from y/2 to 1. Region D: 0 ≤ y ≤ 2, y/2 ≤ x ≤ 1.
Re-describe
Solve for y-range in terms of x: x from 0 to 1; for each x, y ranges from 0 to 2x. New integral: ∫01 ∫02x ex² dy dx
Inner
∫02x ex² dy = 2x ex²
Outer
∫01 2x ex² dx = [ex²]01 = e − 1
Result: e − 1 ≈ 1.718
§ 06Double Integrals in Polar Coordinates
When the region D or the integrand f(x, y) has circular symmetry, converting to polar coordinates can turn a difficult integral into a simple one.
Recall: x = r cos θ, y = r sin θ, and r² = x² + y². When we substitute into a double integral, the area element dA changes:
Polar Area Element — The Critical Factor
dA = dx dy = r dr dθ
The extra factor of r is not optional. It comes from the Jacobian of the polar coordinate transformation and accounts for the fact that polar “rectangles” (sectors) have area r Δr Δθ, not Δr Δθ. Forgetting this factor is one of the most common errors at this level.
Double Integral in Polar Coordinates
∫∫D f(x, y) dA = ∫αβ ∫g1(θ)g2(θ) f(r cos θ, r sin θ) r dr dθ
Example 09 — Area of a Disk via Double Integral
Use a double integral in polar coordinates to find the area of the disk x² + y² ≤ a²
Setup
Area = ∫∫D 1 dA. In polar: r from 0 to a; θ from 0 to 2π.
Integral
∫02π ∫0a r dr dθ = ∫02π [r²/2]0a dθ = ∫02π a²/2 dθ
Outer
a²/2 · 2π = πa²
Area = πa² — exactly the formula we know! ✓
Example 10 — Gaussian-Type Integral
Evaluate ∫∫D e−(x²+y²) dA where D is the disk of radius R
Substitute
x²+y² = r². The integrand becomes e−r². Limits: r from 0 to R; θ from 0 to 2π.
Integral
∫02π ∫0R e−r² r dr dθ = 2π ∫0R r e−r² dr
Sub u = r²
du = 2r dr, so ∫ r e−r²dr = −e−r²/2. Evaluate from 0 to R: (1 − e−R²)/2
Result: π(1 − e−R²) → π as R → ∞
Example 11 — Volume Under a Paraboloid
Find the volume under z = 4 − x² − y² above the xy-plane (where z ≥ 0)
Boundary
z = 0 when x² + y² = 4: a circle of radius 2. Region D: disk r ≤ 2.
Polar form
z = 4 − r². Volume = ∫02π ∫02 (4−r²) r dr dθ
Inner
∫02 (4r−r³) dr = [2r² − r⁴/4]02 = 8 − 4 = 4
Outer
∫02π 4 dθ = 8π
Volume = 8π ≈ 25.13
§ 07Applications — Mass, Average Value & Probability
Mass of a Lamina
If a flat plate (lamina) occupies region D with density function ρ(x, y) (mass per unit area), its total mass is:
Mass of a Lamina
M = ∫∫D ρ(x, y) dA
Average Value
The average value of f over a region D with area A(D) is:
Average Value
favg = 1/A(D) ∫∫D f(x, y) dA
Example 12 — Mass of a Variable-Density Plate
A plate occupies 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 with density ρ(x,y) = 3x + 2y. Find its total mass.
Setup
M = ∫01 ∫02 (3x + 2y) dx dy
Inner
∫02 (3x+2y) dx = [3x²/2 + 2xy]02 = 6 + 4y
Outer
∫01 (6 + 4y) dy = [6y + 2y²]01 = 8
Total mass = 8 units
§ 08Common Mistakes — What to Watch Out For
⚠ Mistake 1 — Forgetting the r in Polar Coordinates
dA = r dr dθ, not dr dθ. This is the most frequent error in polar integrals. Always write the r immediately when switching to polar form.
⚠ Mistake 2 — Treating y as a Variable in the Inner Integral
When integrating with respect to x in the inner integral, y is a constant. It does not get integrated. Write “treat y as constant” in your working until it becomes automatic.
⚠ Mistake 3 — Reversing the Curve Inequalities
For a Type I region, the inner (y) limits go from the lower curve to the upper curve. If you swap them, you get the negative of the correct answer. Sketch the region and label which curve is on top.
⚠ Mistake 4 — Swapping Limits Without Re-deriving Them
You cannot switch order of integration by simply swapping the integral signs and limits. You must re-describe the region with the new outer variable fixed. Sketching is non-negotiable here.
✦ ✦ ✦
§ 09Test Your Knowledge — 10-Question Quiz
Answer all 10 questions. Select your option and click Check. Correct answers are shown in green; wrong answers reveal the correct choice.
Score: 0 / 0
Question 1 of 10
Evaluate ∫01 ∫02 (x + y) dx dy
Question 2 of 10
According to Fubini’s Theorem, on a rectangle R, which of the following is always true for a continuous f?
Question 3 of 10
For a Type I region, the limits on the inner integral are:
Question 4 of 10
What is the area element dA in polar coordinates?
Question 5 of 10
The integral ∫01 ∫x1 ey² dy dx cannot be evaluated as written. After switching the order of integration, the new integral is:
Question 6 of 10
Evaluate ∫∫D dA where D is the disk x² + y² ≤ 9
Question 7 of 10
For f(x,y) = g(x) · h(y), the double integral over rectangle [0,2]×[0,1] satisfies:
Question 8 of 10
A lamina occupies the unit square [0,1]×[0,1] with density ρ(x,y) = 2x. Its total mass is:
Question 9 of 10
The volume under z = x² + y² and above the rectangle [0,1]×[0,1] is:
Question 10 of 10
Which integral correctly sets up the volume under z = 1 above the quarter-disk x²+y²≤4, x≥0, y≥0?