Double Integrals in Polar Coordinates

Cut a region with circles about the origin and rays from it, and each cell is a piece of a ring whose area is its middle radius times Δr times Δθ. That middle radius is the r in r dr dθ. Mark a cell in either picture and compare its area in the two planes.

Double integral in polar coordinates	$\iint_R f\,dA=\int_{\alpha}^{\beta}\!\int_{r_1(\theta)}^{r_2(\theta)}f\,r\,dr\,d\theta$
The Gaussian integral	$\int_{-\infty}^{\infty}e^{-x^{2}}\,dx=\sqrt{\pi}$

Worked example — Double Integrals in Polar Coordinates

Marked cell in the rθ-plane	0.1745
Marked cell in the xy-plane	0.16
The double integral	3.1416
The same without the r (wrong)	6.2832

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Lesson

The theory — Double Integrals in Polar Coordinates

A double integral in polar coordinates integrates over a region described by an angle θ and a distance r from the origin, with the area element dA = r dr dθ.

What each symbol means

r, θ: the distance from the origin and the angle from the positive x-axis: x = r cos θ, y = r sin θ
r̄, Δr, Δθ: a cell's middle radius, its width in r and its width in θ

Where the formula comes from

Cut the region into polar cells. A cell between the radii r₁ and r₂ and the angles θ and θ + Δθ is the difference of two sectors, and its area factors into the middle radius times Δr times Δθ.
$\begin{aligned}\Delta A&=\tfrac{1}{2}\left(r_2^{2}-r_1^{2}\right)\Delta\theta\\&=\tfrac{r_1+r_2}{2}\,(r_2-r_1)\,\Delta\theta=\bar r\,\Delta r\,\Delta\theta\end{aligned}$
Sum and take the limit. f at each cell times the cell's area, summed, tends to an iterated integral in which r̄ has become r. The limits of r may depend on θ.
$\begin{aligned}\iint_R f\,dA&=\int_{\alpha}^{\beta}\!\int_{r_1(\theta)}^{r_2(\theta)}f\,r\,dr\,d\theta,\\ x&=r\cos\theta,\quad y=r\sin\theta\end{aligned}$
Square the Gaussian. A product of two one-variable integrals is a double integral over the plane, and in polar form the r gives e^(−r²) an antiderivative with a formula.
$\begin{aligned}I^{2}&=\iint_{\mathbb{R}^{2}}e^{-(x^{2}+y^{2})}\,dA\\&=\int_{0}^{2\pi}\!\int_{0}^{\infty}e^{-r^{2}}\,r\,dr\,d\theta\\&=2\pi\left[-\tfrac{1}{2}e^{-r^{2}}\right]_{0}^{\infty}=\pi\\ I&=\int_{-\infty}^{\infty}e^{-x^{2}}\,dx=\sqrt{\pi}\end{aligned}$

Assumes: The region is θ between two angles and, along each ray, r between two curves r₁(θ) ≤ r₂(θ). The integral over the whole plane is improper; growing discs and growing squares give the same π, because each square lies between two discs.

Problems solved in full

The area of a disc 2 steps
Find the area of the disc of radius R as a double integral in polar coordinates, and what you get if you forget the r.

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1. The disc is 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, and dA = r dr dθ:
  
  $\begin{aligned}A&=\int_{0}^{2\pi}\!\int_{0}^{R}r\,dr\,d\theta\\&=\int_{0}^{2\pi}\tfrac{R^{2}}{2}\,d\theta=\pi R^{2}\end{aligned}$
2. Without the r:
  
  $\int_{0}^{2\pi}\!\int_{0}^{R}dr\,d\theta=2\pi R$
Answer

πR², which is 3.1416 at R = 1. Without the r, 2πR: the circumference, a length.

$A=\pi R^{2},\qquad R=1:\ A\approx3{.}1416$
Inside a cardioid 2 steps
Find the area inside the cardioid r = 1 + cos θ.

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1. For each θ in [0, 2π], r runs from 0 to 1 + cos θ, and the inner integral is half the square of that limit:
  
  $\begin{aligned}A&=\int_{0}^{2\pi}\!\int_{0}^{1+\cos\theta}r\,dr\,d\theta\\&=\tfrac{1}{2}\int_{0}^{2\pi}\left(1+\cos\theta\right)^{2}d\theta\end{aligned}$
2. Over a full turn cos θ integrates to 0 and cos² θ to π:
  
  $\begin{aligned}A&=\tfrac{1}{2}\int_{0}^{2\pi}\left(1+2\cos\theta+\cos^{2}\theta\right)d\theta\\&=\tfrac{1}{2}\left(2\pi+0+\pi\right)=\tfrac{3\pi}{2}\end{aligned}$
Answer

3π/2 ≈ 4.7124.

$A=\tfrac{3\pi}{2}\approx4{.}7124$
From x and y to r and θ 2 steps
Evaluate ∫₀¹ ∫₀^√(1−x²) (x² + y²) dy dx.

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1. y runs from 0 up to the circle x² + y² = 1 and x from 0 to 1: the quarter of the unit disc in the first quadrant, which is 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2. And x² + y² = r²:
  
  $\begin{aligned}&\int_{0}^{1}\!\int_{0}^{\sqrt{1-x^{2}}}\left(x^{2}+y^{2}\right)dy\,dx\\&\quad=\int_{0}^{\pi/2}\!\int_{0}^{1}r^{2}\cdot r\,dr\,d\theta\end{aligned}$
2. Integrate:
  
  $\int_{0}^{\pi/2}\left[\tfrac{r^{4}}{4}\right]_{0}^{1}d\theta=\int_{0}^{\pi/2}\tfrac{1}{4}\,d\theta=\tfrac{\pi}{8}$
Answer

π/8 ≈ 0.3927.

$\tfrac{\pi}{8}\approx0{.}3927$
A disc that does not sit on the origin 3 steps
Find ∬ x dA over the disc x² + y² ≤ 2x, and from it the x-coordinate of the disc's centroid.

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1. In polar coordinates the edge is r² = 2r cos θ, so r = 2 cos θ, and r ≥ 0 needs cos θ ≥ 0. θ runs from −π/2 to π/2, once round the circle; a full turn would trace it twice and double the answer.
  
  $x^{2}+y^{2}\le2x\iff r\le2\cos\theta,\quad -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}$
2. With x = r cos θ:
  
  $\begin{aligned}\iint x\,dA&=\int_{-\pi/2}^{\pi/2}\!\int_{0}^{2\cos\theta}r\cos\theta\cdot r\,dr\,d\theta\\&=\int_{-\pi/2}^{\pi/2}\tfrac{8}{3}\cos^{4}\theta\,d\theta=\tfrac{8}{3}\cdot\tfrac{3\pi}{8}=\pi\end{aligned}$
3. The disc has radius 1 and area π:
  
  $\bar x=\frac{\iint x\,dA}{A}=\frac{\pi}{\pi}=1$
Answer

∬ x dA = π ≈ 3.1416, so the centroid is at x = 1, the centre of the disc.

$\iint x\,dA=\pi\approx3{.}1416,\qquad \bar x=1$
The volume of a ball 2 steps
Find the volume of a ball of radius R as a double integral in polar coordinates.
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1. The upper half is the volume under z = √(R² − x² − y²) over the disc r ≤ R:
  
  $V=2\int_{0}^{2\pi}\!\int_{0}^{R}\sqrt{R^{2}-r^{2}}\,r\,dr\,d\theta$
2. Substitute u = R² − r², so du = −2r dr; the r from the area element is what makes the substitution work:
  
  $\begin{aligned}V&=2\cdot2\pi\int_{0}^{R^{2}}\tfrac{1}{2}\sqrt{u}\,du\\&=2\pi\cdot\tfrac{2}{3}R^{3}=\tfrac{4\pi R^{3}}{3}\end{aligned}$
Answer

4πR³/3, the volume found by slices earlier in the chapter.

$V=\tfrac{4\pi R^{3}}{3}$
The Gaussian over a disc 2 steps
Evaluate ∬ e^(−(x²+y²)) dA over the disc of radius R, and find its limit as R → ∞.

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1. In polar form the integrand is e^(−r²), and the r beside it makes the antiderivative one with a formula:
  
  $\int_{0}^{R}e^{-r^{2}}\,r\,dr=\left[-\tfrac{1}{2}e^{-r^{2}}\right]_{0}^{R}=\tfrac{1}{2}\left(1-e^{-R^{2}}\right)$
2. The θ-integral multiplies by 2π:
  
  $\int_{0}^{2\pi}\tfrac{1}{2}\left(1-e^{-R^{2}}\right)d\theta=\pi\left(1-e^{-R^{2}}\right)$
Answer

π(1 − e^(−R²)), which is 3.0841 at R = 2 and tends to π.

$\pi\left(1-e^{-R^{2}}\right)\xrightarrow[R\to\infty]{}\pi,\qquad R=2:\ \approx3{.}0841$
∫e^(−x²) dx = √π 3 steps
Show that the integral of e^(−x²) over the whole line is √π.

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1. Let I(b) be the integral from −b to b. Its square is a double integral over the square |x|, |y| ≤ b:
  
  $\begin{aligned}I(b)^{2}&=\int_{-b}^{b}e^{-x^{2}}dx\int_{-b}^{b}e^{-y^{2}}dy\\&=\iint_{|x|,|y|\le b}e^{-(x^{2}+y^{2})}\,dA\end{aligned}$
2. The square contains the disc of radius b and lies inside the disc of radius b√2. The integrand is positive, so by problem 6:
  
  $\pi\left(1-e^{-b^{2}}\right)\le I(b)^{2}\le\pi\left(1-e^{-2b^{2}}\right)$
3. Both bounds tend to π as b → ∞, so I(b)² does too, and I(b) is positive:
  
  $\int_{-\infty}^{\infty}e^{-x^{2}}\,dx=\lim_{b\to\infty}I(b)=\sqrt{\pi}$
Answer

√π ≈ 1.7725. At b = 2 the bounds are 3.0841 and 3.1405, and I(2)² ≈ 3.1123 lies between them.

$\int_{-\infty}^{\infty}e^{-x^{2}}\,dx=\sqrt{\pi}\approx1{.}7725$
The constant in the normal distribution 2 steps
Find the integral of e^(−x²/2) over the whole line, and explain the factor 1/√(2π) in the standard normal density.
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1. Substitute x = √2 u, so dx = √2 du:
  
  $\begin{aligned}\int_{-\infty}^{\infty}e^{-x^{2}/2}\,dx&=\sqrt{2}\int_{-\infty}^{\infty}e^{-u^{2}}\,du\\&=\sqrt{2}\cdot\sqrt{\pi}=\sqrt{2\pi}\end{aligned}$
2. Divide by √(2π) and the total is 1:
  
  $\int_{-\infty}^{\infty}\tfrac{1}{\sqrt{2\pi}}e^{-x^{2}/2}\,dx=\tfrac{\sqrt{2\pi}}{\sqrt{2\pi}}=1$
Answer

√(2π). So (1/√(2π)) e^(−x²/2) has total area 1, as a probability density must.

$\int_{-\infty}^{\infty}e^{-x^{2}/2}\,dx=\sqrt{2\pi}$

Supplementary problems, answers only

Find the area of one petal of r = sin 3θ.

Show answer

π/12 ≈ 0.2618. One petal runs from θ = 0 to θ = π/3.

$A=\int_{0}^{\pi/3}\tfrac12\sin^{2}3\theta\,d\theta=\frac{\pi}{12}\approx0{.}2618$
Evaluate ∬ (x² + y²) dA over the unit disc.

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π/2 ≈ 1.5708, from ∫₀^2π ∫₀¹ r² · r dr dθ.

$\int_{0}^{2\pi}\!\int_{0}^{1}r^{3}\,dr\,d\theta=\frac{\pi}{2}\approx1{.}5708$
Evaluate ∬ dA/(x² + y²) over the ring 1 ≤ x² + y² ≤ 4.

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2π ln 2 ≈ 4.3552. The r in dA leaves 1/r to integrate.

$\int_{0}^{2\pi}\!\int_{1}^{2}\frac{dr}{r}\,d\theta=2\pi\ln2\approx4{.}3552$
Find the volume under z = 4 − x² − y² and above the plane z = 0.

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8π ≈ 25.1327. The surface meets z = 0 on the circle r = 2.

$\begin{aligned}V&=\int_{0}^{2\pi}\!\int_{0}^{2}\left(4-r^{2}\right)r\,dr\,d\theta\\&=8\pi\approx25{.}1327\end{aligned}$
Find the area inside the cardioid r = 1 + cos θ and outside the circle r = 1.

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2 + π/4 ≈ 2.7854. The curves cross at θ = ±π/2, and between those angles the cardioid is the outer edge.

$\begin{aligned}A&=\int_{-\pi/2}^{\pi/2}\tfrac12\left((1+\cos\theta)^{2}-1\right)d\theta\\&=2+\frac{\pi}{4}\approx2{.}7854\end{aligned}$

Leave out the r and the disc's area comes out as its circumference

Press Unit disc. With the r the integral is π ≈ 3.1416; without it the same limits give 2π ≈ 6.2832, the circumference, which is a length. ∫∫ dr dθ measures the rectangle the disc becomes in the rθ-plane, 1 by 2π, and in that plane every cell has the same area Δr Δθ, while in the disc the cells near the centre are slivers and those at the rim are broad. A cell between the radii r₁ and r₂ is a difference of two sectors, ½(r₂² − r₁²)Δθ, which factors as ½(r₁ + r₂) · (r₂ − r₁) · Δθ: the middle radius times Δr times Δθ, exactly. Mark a cell at the rim and then one at the centre and compare the two cell rows in the results. Because nothing was approximated, the sum over the grid prints π for this disc even with n = 1, a single cell.

The r you might forget is what makes ∫e^(−x²) dx come out

e^(−x²) has no elementary antiderivative, none built from powers, roots, exponentials, logarithms and trigonometric functions: Liouville proved it in the 1830s, and the page on non-elementary integrals runs his test. Over the whole line its integral is still exactly √π. Call that integral I. Then I² is ∫e^(−x²) dx times ∫e^(−y²) dy, which is the integral of e^(−(x²+y²)) over the whole plane, and in polar coordinates the integrand e^(−r²) arrives with an r beside it. r e^(−r²) has one: −½e^(−r²), so the inner integral is ½, the θ-integral multiplies it by 2π, and I² = π. Press Gaussian on a disc: the inner-integral row shows that bracket. Silvanus Thompson's life of Kelvin (1910) tells how Kelvin once wrote this result on the blackboard and told his class: “A mathematician is one to whom that is as obvious as that twice two makes four is to you. Liouville was a mathematician.”

The whole plane is a limit of squares, and the discs still give the right answer

I² was defined over squares, |x|, |y| ≤ b, and computed over discs. Press Gaussian on a square. The square holds the disc of radius b and sits inside the disc of radius b√2, and e^(−(x²+y²)) is positive, so its integral over the square lies between π(1 − e^(−b²)) and π(1 − e^(−2b²)). Both close on π, so the square's value must too. At b = 2 the three numbers are 3.0841, 3.1123 and 3.1405. Press Gaussian, b = 1: 1.9859, 2.231 and 2.7164, and the corners outside the inner disc carry 11% of the square's value. That is why the argument gives the integral over the whole line and only bounds over a finite interval: ∫e^(−x²) dx from −1 to 1 is 1.4936, and the discs place it between √1.9859 ≈ 1.4092 and √2.7164 ≈ 1.6482.

Learning path

Applications of integration

References (2)

The double integral over a polar region as the iterated integral of f(r, θ) r dr dθ (equation 5.8 and Theorem 5.8), the polar subrectangle's area r*ΔrΔθ with r* the middle radius, and the improper integral of e^(−10(x²+y²)) over the plane in polar coordinates (Example 5.35): G. Strang and E. Herman, Calculus Volume 3, OpenStax (2016), section 5.3, “Double Integrals in Polar Coordinates”.
Insight block 2: Kelvin wrote ∫e^(−x²) dx = √π on the blackboard and told his class “A mathematician is one to whom that is as obvious as that twice two makes four is to you. Liouville was a mathematician.” (vol. 2, chapter XXV, p. 1139, footnote 1): S. P. Thompson, The Life of William Thomson, Baron Kelvin of Largs, vol. 2, Macmillan, London (1910).

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.