Difficulty: Intermediate | Prerequisites: Single-variable integration (MATH 124/125), familiarity with polar coordinates from precalculus or Calculus II.
Double integrals extend single-variable integration to functions of two variables, letting you compute volumes, masses, and accumulated quantities over two-dimensional regions. Two key skills tested in MATH 126 are converting a Cartesian double integral into polar coordinates (essential when the integrand or the region involves x² + y²) and switching the order of integration (dx dy vs dy dx) to make an otherwise impossible integral computable. Both require sketching the region of integration to read off the correct bounds.
When the integrand or region involves circles or x² + y², convert to polar coordinates (x = r cos θ, y = r sin θ, dA = r dr dθ). When an integral is written dy dx and the inner integral has no closed-form antiderivative, sketch the region and rewrite as dx dy (or vice versa). Always sketch the region first to get the limits right.
Double integral
The integral ∬_R f(x, y) dA, which sums f over a two-dimensional region R. When f ≥ 0, this gives the volume under the surface z = f(x, y) above R.
Iterated integral
A double integral evaluated as two nested single integrals. The inner integral is computed first (treating the outer variable as constant), then the result is integrated over the outer variable.
Order of integration
Whether you integrate with respect to x first (dx dy) or y first (dy dx). Fubini's theorem guarantees both give the same answer, but one order may be far easier to compute.
Polar coordinates
x = r cos θ, y = r sin θ, where r ≥ 0 is the distance from the origin and θ is the angle from the positive x-axis. The area element transforms as dA = r dr dθ. The extra factor of r is essential and easily forgotten.
Type I region (dy dx)
A region described by a ≤ x ≤ b, g₁(x) ≤ y ≤ g₂(x). The outer integral runs over x, and for each x the inner integral sweeps y between two curves.
Type II region (dx dy)
A region described by c ≤ y ≤ d, h₁(y) ≤ x ≤ h₂(y). The outer integral runs over y, and for each y the inner integral sweeps x between two curves.
When to convert
The integrand contains x² + y², √(x² + y²), or similar radial expressions.
The region of integration is (part of) a circle, annulus, or sector.
The Cartesian limits involve √(a² - x²) (upper half of a circle) or similar.
How to convert
Replace x with r cos θ, y with r sin θ.
Replace dA (or dy dx) with r dr dθ. Do not forget the extra r.
Rewrite the integrand in terms of r and θ.
Determine the r and θ limits from the region.
Worked example from the exam
Compute ∫₀¹ ∫_{√3·x}^{√(4-x²)} x·√(x² + y²) dy dx.
Step 1: Sketch the region. The outer integral has x from 0 to 1. The inner limits are y = √3·x (a line through the origin with slope √3, i.e. θ = π/3) and y = √(4 - x²) (the upper half of the circle x² + y² = 4, i.e. r = 2). The region is a wedge of the disk of radius 2, between angles θ = π/3 and θ = π/2 (since x goes from 0 to 1 and x = r cos θ with r up to 2, the right boundary is x = 1, but looking more carefully: when x = 1, y = √3 gives the point (1, √3), and θ = arctan(√3) = π/3; the upper boundary y = √(4 - x²) is the circle r = 2; when x = 0, we are on the y-axis, θ = π/2).
Step 2: Convert. The integrand x·√(x² + y²) = r cos θ · r = r² cos θ. With dA = r dr dθ, the full integrand becomes r³ cos θ.
Step 3: Set up polar limits. θ goes from π/3 to π/2. r goes from 0 to 2.
∫_{π/3}^{π/2} ∫₀² r³ cos θ dr dθ
Step 4: Evaluate.
Inner: ∫₀² r³ dr = r⁴/4 |₀² = 4.
Outer: 4 ∫{π/3}^{π/2} cos θ dθ = 4[sin θ]{π/3}^{π/2} = 4(1 - √3/2) = 4 - 2√3.
Why change the order
Some iterated integrals cannot be evaluated because the inner integral has no elementary antiderivative (e.g. ∫ e^{y²} dx is fine, but ∫ e^{y²} dy is not). Switching the order may resolve this.
How to change the order
Sketch the region described by the original limits.
Re-describe the same region with the other variable as the outer integral.
Write the new iterated integral with the swapped limits.
Worked example from the exam
The triangle T has vertices (0, 0), (0, 1), (1, 1). The solid S is bounded above by z = e^{y²} and below by T.
Setting up dx dy:
For a fixed y (from 0 to 1), x runs from 0 to y (left boundary x = 0, right boundary x = y, since the hypotenuse of the triangle is y = x, i.e. x = y).
∫₀¹ ∫₀ʸ e^{y²} dx dy
Setting up dy dx:
For a fixed x (from 0 to 1), y runs from x to 1 (lower boundary y = x, upper boundary y = 1).
∫₀¹ ∫ₓ¹ e^{y²} dy dx
Evaluating: use dx dy
The dy dx version requires integrating e^{y²} with respect to y, which has no elementary antiderivative. The dx dy version integrates e^{y²} with respect to x first, treating y as constant.
Inner: ∫₀ʸ e^{y²} dx = y · e^{y²}.
Outer: ∫₀¹ y · e^{y²} dy. Let u = y², du = 2y dy. This gives (1/2) ∫₀¹ eᵘ du = (1/2)(e - 1).
Volume of S = (e - 1)/2.
Polar coordinate substitution
x = r cos θ, y = r sin θ, x² + y² = r², dA = r dr dθ
Double integral as iterated integral (Fubini's theorem)
∬_R f(x, y) dA = ∫a^b ∫{g₁(x)}^{g₂(x)} f(x, y) dy dx = ∫c^d ∫{h₁(y)}^{h₂(y)} f(x, y) dx dy
Both orders give the same value, provided f is continuous on R.
Volume under a surface
Volume = ∬_R f(x, y) dA, where f(x, y) ≥ 0 is the height of the surface above the region R.
Double integrals compute physical quantities spread over flat regions: the mass of a plate with variable density, the charge on a surface, the average temperature across a sensor array. Polar coordinates are the natural choice for anything with circular symmetry, from antenna radiation patterns to the probability distribution of darts on a target.
The most common error in polar conversion is forgetting the extra factor of r in dA = r dr dθ. Every polar integral must include this r.
Students often assume the limits of r always go from 0 to some constant. In many problems, the lower limit of r depends on θ (or the region is not centred at the origin).
When switching the order of integration, students sometimes swap the limits mechanically without sketching. This leads to wrong bounds. Always draw the region.
Confusing "dy dx" with "dx dy": the inner integral variable must match the inner differential. If you write ∫∫ f dx dy, the inner integral is with respect to x.
Polar conversion problems are near-guaranteed on a MATH 126 midterm. The telltale sign is x² + y² in the integrand or limits shaped like √(a² - x²).
"Set up the integral in two different orders" is a classic exam question. It tests whether you understand the region, not just the computation. You must be able to describe the same region both ways.
When one order leads to an integral you cannot evaluate (e.g. ∫ e^{y²} dy), the exam is testing whether you recognise the need to switch.
Always double-check your limits by picking a sample point in the region and verifying it satisfies both the original and the new bounds.
Fill in the blank: When converting to polar coordinates, dA = dy dx becomes ______. (r dr dθ.)
True or false: ∫₀¹ ∫₀^{√(1-x²)} f(x, y) dy dx integrates over a full circle of radius 1. (False. It is the quarter-circle in the first quadrant.)
True or false: Switching the order of integration changes the value of the integral. (False, provided the function is continuous on the region.)
Fill in the blank: The line y = √3·x makes an angle of ______ with the positive x-axis. (π/3, or 60 degrees.)
True or false: If you cannot find an antiderivative for the inner integral, you should try switching the order. (True.)
Q: Convert the integral ∫₀² ∫₀^{√(4-y²)} (x² + y²) dx dy to polar coordinates and evaluate.
A: The region is the quarter-disk x ≥ 0, y ≥ 0, x² + y² ≤ 4, so r from 0 to 2, θ from 0 to π/2. The integrand x² + y² = r². With dA = r dr dθ: ∫₀^{π/2} ∫₀² r² · r dr dθ = ∫₀^{π/2} ∫₀² r³ dr dθ = ∫₀^{π/2} 4 dθ = 2π.
Q: Set up ∫₀¹ ∫ₓ¹ sin(y²) dy dx with the order of integration reversed.
A: Sketch the region: x from 0 to 1, y from x to 1. This is the triangle below y = 1, above y = x, with x ≥ 0. Reversing: y from 0 to 1, x from 0 to y. The integral becomes ∫₀¹ ∫₀ʸ sin(y²) dx dy = ∫₀¹ y sin(y²) dy = [-cos(y²)/2]₀¹ = (1 - cos 1)/2.
Q: Why does ∫₀¹ ∫ₓ¹ e^{y²} dy dx require a change of order?
A: The function e^{y²} has no elementary antiderivative with respect to y. Switching to dx dy makes e^{y²} a constant for the inner x-integral, and the resulting y-integral can be solved by substitution.
Q: In polar coordinates, what region does 0 ≤ r ≤ 1, π/4 ≤ θ ≤ 3π/4 describe?
A: A sector (wedge) of the unit disk from 45 degrees to 135 degrees, covering the upper part of the disk between the lines y = x and y = -x.
Double integrals in polar coordinates extend naturally to triple integrals in cylindrical and spherical coordinates later in MATH 126. The technique of switching the order of integration is closely related to Fubini's theorem and reappears when evaluating line integrals and surface integrals. The idea of choosing the right coordinate system to simplify a problem is a recurring theme throughout multivariable calculus and physics.
Double integral, iterated integral, polar coordinates, polar conversion, r dr dθ, area element, Fubini's theorem, order of integration, Type I region, Type II region, dy dx, dx dy, change of variables, Jacobian, volume under surface, MATH 126, Calculus III, multivariable integration, switching integration order