Difficulty: Intermediate to Advanced | Prerequisites: Single-variable integration (MATH 125), partial derivatives
Double integrals extend the idea of "area under a curve" to "volume under a surface." Where a single integral sums up thin slices, a double integral sums up thin columns over a 2D region. This section covers how to set up and evaluate double integrals, how to reverse the order of integration, and the difference between Type I and Type II regions. It also consolidates the single-variable integration techniques (u-substitution, integration by parts, and the standard derivative/integral tables) that you will need when evaluating the inner and outer integrals.
Double integrals compute the volume under a surface z = f(x, y) over a region in the xy-plane. You choose to integrate dx first or dy first depending on the shape of the region (Type I or Type II). This section also reviews the single-variable integration tools (u-sub, by parts, standard formulas, trig identities) you will use inside those integrals.
Double integral
The integral ∬ f(x, y) dA, which sums the values of f over a 2D region. Think of it as stacking infinitely thin columns of height f(x, y) over the region and adding up their volumes.
Type I region
A region described by a ≤ x ≤ b, with y bounded between two functions of x: g₁(x) ≤ y ≤ g₂(x). The inner integral is dy and the outer is dx. In simple terms, you slice the region with vertical lines.
Type II region
A region described by c ≤ y ≤ d, with x bounded between two functions of y: h₁(y) ≤ x ≤ h₂(y). The inner integral is dx and the outer is dy. You slice the region with horizontal lines.
u-substitution
A technique for evaluating integrals by replacing a complicated expression with a single variable u. The "u" is the messy inner part of a composition. Do not forget to change the bounds when doing a definite integral.
Integration by parts
The integral version of the product rule: ∫ u dv = uv - ∫ v du. Use it when the integrand is a product of two unrelated function types (e.g., x·eˣ, x·ln(x)). Pick u as the part that simplifies when differentiated (LIATE order: logarithmic, inverse trig, algebraic, trig, exponential).
Sketch the region of integration in the xy-plane
Decide whether to use Type I (vertical slices, dy inner) or Type II (horizontal slices, dx inner)
Type I: outer bounds are constants on x, inner bounds are functions of x
Type II: outer bounds are constants on y, inner bounds are functions of y
For volume under z = f(x, y) above z = 0 over a rectangular region, the integral is straightforward: ∬ f(x, y) dA with constant bounds on both variables
Step 1: Sketch the region from the original limits
Step 2: If the original order has dx as the inner integral, the cross-sections are vertical (x-axis). The top function is the upper bound, the bottom function is the lower bound
Step 3: If the original order has dy as the inner integral, the cross-sections are horizontal (y-axis). The right function is the upper bound, the left function is the lower bound
Step 4: Re-express the bounds for the new order and rewrite the integral
Graph the region of integration in the xy-plane
Split the x-axis into m equal parts and the y-axis into n equal parts to create a grid of small rectangles
Choose sample points in each rectangle (given or midpoints)
Plug each sample point into f(x, y) to get heights
Volume ≈ (area of one small rectangle) × (sum of all heights)
Choose u as the complicated inner part of the integrand
Compute du and rewrite the integral entirely in terms of u
For definite integrals, change the bounds to u-values (find what u equals when x is at each bound)
Integrate, then substitute back if you kept the original variable
Formula: ∫ u dv = uv - ∫ v du
Use when the integrand is a product of two unrelated types
LIATE priority for choosing u: Logarithmic, Inverse trig, Algebraic, Trig, Exponential
Pick u as the type that appears earliest in LIATE (it will simplify when differentiated)
dv is whatever remains
u should be something that will eventually disappear after repeated differentiation
d/dx (xⁿ) = nxⁿ⁻¹
d/dx (eˣ) = eˣ
d/dx (aˣ) = aˣ ln(a)
d/dx (ln x) = 1/x
d/dx (log_a x) = 1/(x ln a)
d/dx (sin x) = cos x, d/dx (cos x) = -sin x
d/dx (tan x) = sec²x, d/dx (cot x) = -csc²x
d/dx (sec x) = sec x tan x, d/dx (csc x) = -csc x cot x
d/dx (tan⁻¹ x) = 1/(1 + x²), d/dx (sin⁻¹ x) = 1/√(1 - x²), d/dx (sec⁻¹ x) = 1/(x√(x² - 1))
Product rule: (FS)' = FS' + F'S
Quotient rule: (N/D)' = (DN' - ND')/D²
Chain rule: d/dx [f(g(x))] = f'(g(x)) · g'(x)
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C
∫ eᵃˣ dx = (1/a)eᵃˣ + C
∫ bˣ dx = bˣ/ln(b) + C
∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C
∫ cos(ax) dx = (1/a) sin(ax) + C
∫ sin(ax) dx = -(1/a) cos(ax) + C
∫ sec²x dx = tan x + C
∫ csc²x dx = -cot x + C
∫ sec x tan x dx = sec x + C
∫ csc x cot x dx = -csc x + C
∫ 1/(x² + a²) dx = (1/a) tan⁻¹(x/a) + C
∫ 1/√(a² - x²) dx = sin⁻¹(x/a) + C
∫ tan x dx = ln|sec x| + C
∫ cot x dx = ln|sin x| + C
∫ sec x dx = ln|sec x + tan x| + C
∫ csc x dx = ln|csc x - cot x| + C
sin²x + cos²x = 1
tan²x + 1 = sec²x
1 + cot²x = csc²x
sin²x = ½(1 - cos 2x)
cos²x = ½(1 + cos 2x)
sin x cos x = ½ sin 2x
ln 1 = 0, ln e = 1
ln(aᵇ) = b ln a
ln(ab) = ln a + ln b
xᵃ · xᵇ = xᵃ⁺ᵇ
(xᵃ)ᵇ = xᵃᵇ
ⁿ√x = x^(1/n)
1/eˣ = e⁻ˣ
Double integrals compute physical quantities like mass (when density varies across a plate), centre of mass, and moments of inertia. In probability, the joint probability over a 2D region is a double integral. Volume calculations under surfaces appear in terrain modelling, fluid flow, and heat distribution problems.
Students frequently mix up which bound is "top" vs "bottom" when setting up Type I integrals, and "left" vs "right" for Type II. Always sketch the region first.
When reversing integration order, students sometimes keep the original bounds unchanged. The bounds must be completely re-derived from the region sketch.
In u-substitution, forgetting to change the bounds of a definite integral is one of the most common errors. If the original bounds are in x, convert them to u-values.
Students sometimes pick dv as the part that disappears in integration by parts. It is the other way round: u is the part that simplifies when differentiated.
⚠️ Reversing integration order is a staple exam problem. Practise sketching the region and rewriting bounds.
⚠️ Know your basic integral and derivative tables cold. These are the building blocks for evaluating double integrals, and fumbling a trig integral wastes time under exam conditions.
⚠️ Integration by parts problems often appear where the integrand is a product of unrelated types (x eˣ, x² sin x, ln x). Recognise these patterns quickly.
⚠️ The Riemann sum approximation may appear as a conceptual question: what does the double integral represent geometrically?
True or false: In a Type I region, the inner integral is with respect to x. (False. In Type I, the inner integral is dy.)
Fill in the blank: When using u-substitution on a definite integral, you must change the ____ to match u. (bounds)
True or false: In integration by parts, u should be the function that gets more complicated when differentiated. (False. u should simplify when differentiated.)
Fill in the blank: ∫ sec²x dx = ____. (tan x + C)
Q: You are given ∫₀¹ ∫₀ˣ f(x, y) dy dx. Reverse the order of integration.
A: The region is 0 ≤ x ≤ 1, 0 ≤ y ≤ x. Sketch it: a triangle with vertices (0,0), (1,0), (1,1). In Type II form: for a given y, x runs from y to 1, and y runs from 0 to 1. Reversed: ∫₀¹ ∫_y¹ f(x, y) dx dy.
Q: Evaluate ∫ x eˣ dx using integration by parts.
A: Let u = x, dv = eˣ dx. Then du = dx, v = eˣ. By parts: x eˣ - ∫ eˣ dx = x eˣ - eˣ + C = eˣ(x - 1) + C.
Q: What is the LIATE rule and when do you use it?
A: LIATE (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) is a priority guide for choosing u in integration by parts. Pick u as the function type appearing earliest in the list, because those types simplify fastest when differentiated.
Q: When approximating a double integral with a Riemann sum, what does each term represent?
A: Each term is the height f(x_i, y_j) times the area of one small rectangle ΔA. The sum approximates the total volume under the surface.
Double integrals connect to triple integrals and other coordinate systems (polar, cylindrical, spherical) later in MATH 126. The integration techniques reviewed here (u-sub, by parts) are prerequisites for evaluating any iterated integral. The concept of reversing integration order reappears when switching to polar coordinates, where the region description changes form entirely.
Double integral, iterated integral, Type I region, Type II region, reversing integration order, changing order of integration, volume under surface, Riemann sum, u-substitution, integration by parts, LIATE, derivative table, integral table, trig identities, log rules, MATH 126, Calculus III, UW