Source: James Stewart, Calculus: Early Transcendentals, Ch. 15.3
Difficulty: Intermediate | Prerequisites: Polar coordinates (Ch. 10.3), double integrals over rectangular regions (Ch. 15.1–15.2).
When a region of integration is circular, annular, or wedge-shaped, expressing it in rectangular coordinates produces ugly, hard-to-evaluate limits. Polar coordinates let you describe these regions naturally: circles become constant-r lines, and rays become constant-θ lines. This section teaches you how to convert a double integral from Cartesian to polar form, pick the right limits, and avoid the single most common mistake (forgetting the extra r in the integrand). You should already be comfortable converting between (x, y) and (r, θ) and setting up iterated double integrals in rectangular coordinates.
To evaluate a double integral over a circular or radially symmetric region, switch from dA = dx dy to dA = r dr dθ. The extra factor of r comes from the geometry of polar area elements. Set your limits on r and θ to trace out the region, convert f(x, y) into f(r cos θ, r sin θ), and integrate.
Polar coordinates (r, θ)
A coordinate system where a point in the plane is located by its distance r from the origin and the angle θ its radius makes with the positive x-axis. Think of it as: instead of "how far right and how far up," you ask "how far out and at what angle."
Polar rectangle
The region R = {(r, θ) | a ≤ r ≤ b, α ≤ θ ≤ β}. In simple terms, this is the area between two concentric circles (radii a and b) and two rays (angles α and β). It looks like a curved strip or a slice of an annulus.
Polar subrectangle Rᵢⱼ
One small piece of a polar rectangle after you subdivide the r-interval into m parts and the θ-interval into n parts. In simple terms, these are the tiny curved "tiles" you chop the region into when building the Riemann sum.
Area element in polar coordinates (dA = r dr dθ)
The infinitesimal area of a polar subrectangle. The factor r appears because, at larger radii, a given angular width Δθ sweeps out a bigger arc. Think of it as: the further out you go, the more ground a small angle covers, so area scales with r.
Jacobian (the extra r)
When changing variables in a double integral from Cartesian to polar, the Jacobian of the transformation is r. This is why dA = dx dy becomes r dr dθ. In simple terms, it is the correction factor that accounts for the stretching of coordinates.
Tags: polar rectangle, polar coordinates conversion, r-theta region, annular region
Rectangular ↔ polar conversion
x = r cos θ, y = r sin θ
r² = x² + y²
These let you rewrite any f(x, y) in terms of r and θ
What a polar rectangle looks like
Defined by R = {(r, θ) | a ≤ r ≤ b, α ≤ θ ≤ β}
The curves r = a and r = b are concentric circles; the curves θ = α and θ = β are rays from the origin
The result is a curved strip (if the angular range is small) or a full annulus (if β – α = 2π)
Special cases: a = 0 gives a sector of a disk; β – α = 2π gives a full annulus or disk
Tags: polar Riemann sum, polar area element, dA in polar, r dr dtheta
Building the Riemann sum
Partition [a, b] into m subintervals of width Δr = (b – a)/m
Partition [α, β] into n subintervals of width Δθ = (β – α)/n
This chops the polar rectangle into m × n small polar subrectangles Rᵢⱼ
Finding the area of each subrectangle
The area of a circular sector with radius r and central angle θ is ½r²θ
Subtracting two sectors (outer radius rᵢ, inner radius rᵢ₋₁, same angle Δθ) gives ΔAᵢ = rᵢ* Δr Δθ, where rᵢ* = ½(rᵢ₋₁ + rᵢ) is the midpoint radius
This is where the factor r comes from: polar tiles are bigger at larger radii
From Riemann sum to integral
Evaluate f at the centre of each subrectangle: (rᵢ* cos θⱼ*, rᵢ* sin θⱼ*)
The double sum ΣΣ f(rᵢ* cos θⱼ*, rᵢ* sin θⱼ*) rᵢ* Δr Δθ is a Riemann sum for the iterated integral
Define g(r, θ) = r · f(r cos θ, r sin θ), and the limit of the Riemann sum gives ∫∫ g(r, θ) dr dθ
Tags: change to polar coordinates formula, general polar region, Type III region, variable r limits
The polar rectangle formula (Equation 2 in Stewart)
If f is continuous on a polar rectangle R = {(r, θ) | a ≤ r ≤ b, α ≤ θ ≤ β} where 0 ≤ β – α ≤ 2π, then:
∫∫_R f(x, y) dA = ∫_α^β ∫_a^b f(r cos θ, r sin θ) · r dr dθ
The key steps: replace x with r cos θ, replace y with r sin θ, replace dA with r dr dθ.
General polar regions (Equation 3 in Stewart)
If the region is not a polar rectangle but has the form D = {(r, θ) | α ≤ θ ≤ β, h₁(θ) ≤ r ≤ h₂(θ)}, then:
∫∫_D f(x, y) dA = ∫α^β ∫{h₁(θ)}^{h₂(θ)} f(r cos θ, r sin θ) · r dr dθ
Here the r-limits depend on θ. This is the polar analogue of a Type I or Type II region in Cartesian coordinates.
How to set up the limits
Draw the region and identify its boundary curves
Express each boundary curve in polar form (e.g. x² + y² = 2x becomes r = 2 cos θ)
For each angle θ, determine where a ray from the origin enters the region (that is h₁(θ)) and where it exits (that is h₂(θ))
Determine the range of θ that sweeps across the entire region
Coordinate conversion
x = r\cos\theta, \quad y = r\sin\theta, \quad r^2 = x^2 + y^2Area element
dA = r\,dr\,d\thetaDouble integral over a polar rectangle (Eq. 2)
\iint_R f(x,y)\,dA = \int_\alpha^\beta \int_a^b f(r\cos\theta,\, r\sin\theta)\; r\,dr\,d\thetaDouble integral over a general polar region (Eq. 3)
\iint_D f(x,y)\,dA = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,\, r\sin\theta)\; r\,dr\,d\thetaArea of a polar subrectangle
\Delta A_i = r_i^* \,\Delta r\,\Delta\thetawhere rᵢ* = ½(rᵢ₋₁ + rᵢ) is the midpoint radius.
Sector area (used in the derivation)
A = \tfrac{1}{2}r^2\thetaExample 1: Upper half-annulus
Problem: Evaluate ∫∫_R (3x + 4y²) dA where R is the region in the upper half-plane bounded by the circles x² + y² = 1 and x² + y² = 4.
Step 1, identify the region in polar coordinates:
The circles become r = 1 and r = 2
"Upper half-plane" means y ≥ 0, so θ runs from 0 to π
Polar rectangle: 1 ≤ r ≤ 2, 0 ≤ θ ≤ π
Step 2, convert the integrand:
3x + 4y² = 3r cos θ + 4r² sin² θ
Multiply by r (the Jacobian): integrand becomes 3r² cos θ + 4r³ sin² θ
Step 3, integrate with respect to r first (inner integral):
∫₁² (3r² cos θ + 4r³ sin² θ) dr = [r³ cos θ + r⁴ sin² θ] from r = 1 to r = 2
= (8 cos θ + 16 sin² θ) – (cos θ + sin² θ) = 7 cos θ + 15 sin² θ
Step 4, integrate with respect to θ (outer integral):
Use sin² θ = ½(1 – cos 2θ)
∫₀^π (7 cos θ – 15/2 (1 – cos 2θ)) dθ = 15π/2
Example 2: Volume under a paraboloid inside a cylinder
Problem: Find the volume of the solid under z = x² + y², above the xy-plane, and inside the cylinder x² + y² = 2x.
Step 1, rewrite the cylinder in polar form:
x² + y² = 2x becomes r² = 2r cos θ, so r = 2 cos θ
Completing the square: (x – 1)² + y² = 1, a circle of radius 1 centred at (1, 0)
Step 2, set limits:
For each θ, r runs from 0 to 2 cos θ (the boundary of the disk)
θ runs from –π/2 to π/2 (to sweep the full disk)
D = {(r, θ) | –π/2 ≤ θ ≤ π/2, 0 ≤ r ≤ 2 cos θ}
Step 3, set up the integral:
z = x² + y² = r² in polar, and dA = r dr dθ
V = ∫{-π/2}^{π/2} ∫₀^{2cosθ} r² · r dr dθ = ∫{-π/2}^{π/2} ∫₀^{2cosθ} r³ dr dθ
Step 4, evaluate:
Inner integral: [r⁴/4] from 0 to 2 cos θ = 4 cos⁴ θ
By symmetry: 2 · 4 ∫₀^{π/2} cos⁴ θ dθ = 8 ∫₀^{π/2} ((1 + cos 2θ)/2)² dθ
Result: V = 3π/2
Polar double integrals turn up whenever a physical quantity is distributed over a circular or radially symmetric region. Engineers use them to calculate the moment of inertia of a circular plate or disk, which determines how the object resists rotational acceleration. In probability and statistics, the famous Gaussian integral (the one that gives the normal distribution its normalising constant) is evaluated by converting to polar coordinates, turning an otherwise impossible integral into a straightforward one.
Students often forget to include the extra factor of r when converting dA. Writing dA = dr dθ instead of dA = r dr dθ is the single most common error on exams. Every polar double integral needs that r.
Students sometimes confuse the roles of r and θ when setting limits. The inner integral is almost always with respect to r (for a fixed angle, how far does the ray travel?), and the outer integral is with respect to θ.
When a region is described by a curve like r = 2 cos θ, students sometimes set the θ-limits to 0 to 2π by default. This is wrong: r = 2 cos θ is negative for θ in (π/2, 3π/2), so the correct limits are –π/2 to π/2 (or equivalently 0 to π if you use symmetry carefully).
Students sometimes try to convert every double integral to polar. Polar coordinates only help when the region or the integrand has circular symmetry. For rectangular or triangular regions, Cartesian coordinates are usually simpler.
⚠️ Forgetting the r factor in dA = r dr dθ is tested constantly. If your answer is off by a factor related to r, check this first.
⚠️ Converting boundary curves to polar form is a required skill. You will be expected to go from x² + y² = 2x to r = 2 cos θ quickly and confidently.
⚠️ Setting correct θ-limits for non-centred circles (like x² + y² = 2x) is a common exam question. Sketch the region first.
⚠️ Trig identities (sin² θ = ½(1 – cos 2θ), cos⁴ θ via double-angle) are needed to finish most polar integrals. Have these memorised.
True or false: When converting a double integral to polar coordinates, dA = dr dθ. (False: dA = r dr dθ.)
Fill in the blank: The polar rectangle R = {(r, θ) | 1 ≤ r ≤ 3, 0 ≤ θ ≤ π/2} describes a region shaped like a ______. (Quarter-annulus, or a quarter of a ring.)
True or false: The equation x² + y² = 4 becomes r = 4 in polar coordinates. (False: it becomes r = 2.)
Fill in the blank: The area element ΔAᵢ for a polar subrectangle equals ______ . (rᵢ Δr Δθ.)*
True or false: For the region inside r = 2 cos θ, the correct θ-limits are 0 to 2π. (False: –π/2 to π/2, or equivalently 0 to π with care.)
Q: Set up (but do not evaluate) the double integral of f(x, y) = x over the disk x² + y² ≤ 9 using polar coordinates.
A: The disk is 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π. Since x = r cos θ, the integral is ∫₀^{2π} ∫₀³ (r cos θ) · r dr dθ = ∫₀^{2π} ∫₀³ r² cos θ dr dθ.
Q: Explain why the factor r appears in dA = r dr dθ.
A: A small polar subrectangle at radius r has arc length r Δθ in the angular direction and width Δr in the radial direction. Its area is approximately r Δr Δθ. The further from the origin, the more ground a given angular increment covers.
Q: Convert the Cartesian region x² + y² ≤ 2y into polar form and state the limits of integration.
A: x² + y² = 2y becomes r² = 2r sin θ, so r = 2 sin θ. This is a circle of radius 1 centred at (0, 1). The limits are 0 ≤ θ ≤ π, 0 ≤ r ≤ 2 sin θ.
Q: Evaluate ∫∫_R (x² + y²) dA where R is the annulus 1 ≤ x² + y² ≤ 4.
A: In polar: ∫₀^{2π} ∫₁² r² · r dr dθ = ∫₀^{2π} ∫₁² r³ dr dθ = 2π · [r⁴/4]₁² = 2π · (4 – 1/4) = 15π/2.
Q: A student writes ∫₀^π ∫₀^{2cosθ} r² · r dr dθ for the volume under z = x² + y² inside x² + y² = 2x. What did they do wrong?
A: They used θ-limits of 0 to π, but r = 2 cos θ is negative for θ in (π/2, π). The correct θ-limits are –π/2 to π/2. With the 0 to π limits, the integral double-counts part of the region and misses the rest.
This section builds directly on Ch. 15.1–15.2 (double integrals over rectangular regions) by extending the same Riemann-sum logic to a different coordinate system. The technique of changing variables reappears in Ch. 15.10 (Change of Variables in Multiple Integrals), where the Jacobian generalises the r factor you see here. Polar double integrals are also the foundation for triple integrals in cylindrical coordinates (Ch. 15.7), which simply add a z-axis to the polar setup.
Double integrals in polar coordinates, polar double integral, change of variables polar, dA = r dr dtheta, polar rectangle, polar region, Jacobian polar coordinates, area element polar, converting Cartesian to polar integral, iterated integral polar, r dr d-theta, annular region integral, sector area, polar Riemann sum, Stewart 15.3, Calculus III polar integrals, MATH 215 polar coordinates, multivariable calculus polar