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	$\iint_{R} f d A = \int_{α}^{β} \int_{r_{1} (θ)}^{r_{2} (θ)} f r d r d θ$
The Gaussian integral	$\int_{- \infty}^{\infty} e^{- x^{2}} d x = π$

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.
$Δ A = \frac{1}{2} (r_{2}^{2} - r_{1}^{2}) Δ θ = \frac{r _{1} + r _{2}}{2} (r_{2} - r_{1}) Δ θ = \overset{r}{ˉ} Δ r Δ θ$
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.
$\iint_{R} f d A x = \int_{α}^{β} \int_{r_{1} (θ)}^{r_{2} (θ)} f r d r d θ, = r cos θ, y = r sin θ$
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.
$I^{2} I = \iint_{R^{2}} e^{- (x^{2} + y^{2})} d A = \int_{0}^{2 π} \int_{0}^{\infty} e^{- r^{2}} r d r d θ = 2 π [- \frac{1}{2} e^{- r^{2}}]_{0}^{\infty} = π = \int_{- \infty}^{\infty} e^{- x^{2}} d x = π$

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:
  
  $A = \int_{0}^{2 π} \int_{0}^{R} r d r d θ = \int_{0}^{2 π} \frac{R ^{2}}{2} d θ = π R^{2}$
2. Without the r:
  
  $\int_{0}^{2 π} \int_{0}^{R} d r d θ = 2 π R$
Answer

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

$A = π R^{2}, R = 1 : A \approx 3.1416$
Inside a cardioid 2 steps
Find the area inside the cardioid r = 1 + cos θ.

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Show solution
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:
  
  $A = \int_{0}^{2 π} \int_{0}^{1 + c o s θ} r d r d θ = \frac{1}{2} \int_{0}^{2 π} (1 + cos θ)^{2} d θ$
2. Over a full turn cos θ integrates to 0 and cos² θ to π:
  
  $A = \frac{1}{2} \int_{0}^{2 π} (1 + 2 cos θ + cos^{2} θ) d θ = \frac{1}{2} (2 π + 0 + π) = \frac{3 π}{2}$
Answer

3π/2 ≈ 4.7124.

$A = \frac{3 π}{2} \approx 4.7124$
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:
  
  $\int_{0}^{1} \int_{0}^{1 - x^{2}} (x^{2} + y^{2}) d y d x = \int_{0}^{π /2} \int_{0}^{1} r^{2} \cdot r d r d θ$
2. Integrate:
  
  $\int_{0}^{π /2} [\frac{r ^{4}}{4}]_{0}^{1} d θ = \int_{0}^{π /2} \frac{1}{4} d θ = \frac{π}{8}$
Answer

π/8 ≈ 0.3927.

$\frac{π}{8} \approx 0.3927$
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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Show solution
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.
  
  $x^{2} + y^{2} \leq 2 x ⟺ r \leq 2 cos θ, - \frac{π}{2} \leq θ \leq \frac{π}{2}$
2. With x = r cos θ:
  
  $\iint x d A = \int_{- π /2}^{π /2} \int_{0}^{2 c o s θ} r cos θ \cdot r d r d θ = \int_{- π /2}^{π /2} \frac{8}{3} cos^{4} θ d θ = \frac{8}{3} \cdot \frac{3 π}{8} = π$
3. The disc has radius 1 and area π:
  
  $\overset{x}{ˉ} = \frac{\iint x d A}{A} = \frac{π}{π} = 1$
Answer

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

$\iint x d A = π \approx 3.1416, \overset{x}{ˉ} = 1$
The volume of a ball 2 steps
Find the volume of a ball of radius R as a double integral in polar coordinates.
Show solution
1. The upper half is the volume under z = √(R² − x² − y²) over the disc r ≤ R:
  
  $V = 2 \int_{0}^{2 π} \int_{0}^{R} R^{2} - r^{2} r d r d θ$
2. Substitute u = R² − r², so du = −2r dr; the r from the area element is what makes the substitution work:
  
  $V = 2 \cdot 2 π \int_{0}^{R^{2}} \frac{1}{2} u d u = 2 π \cdot \frac{2}{3} R^{3} = \frac{4 π R ^{3}}{3}$
Answer

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

$V = \frac{4 π R ^{3}}{3}$
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:
  
  $\int_{0}^{R} e^{- r^{2}} r d r = [- \frac{1}{2} e^{- r^{2}}]_{0}^{R} = \frac{1}{2} (1 - e^{- R^{2}})$
2. The θ-integral multiplies by 2π:
  
  $\int_{0}^{2 π} \frac{1}{2} (1 - e^{- R^{2}}) d θ = π (1 - e^{- R^{2}})$
Answer

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

$π (1 - e^{- R^{2}}) R \to \infty π, R = 2 : \approx 3.0841$
∫e^(−x²) dx = √π 3 steps
Show that the integral of e^(−x²) over the whole line is √π.

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Show solution
1. Let I(b) be the integral from −b to b. Its square is a double integral over the square |x|, |y| ≤ b:
  
  $I (b)^{2} = \int_{- b}^{b} e^{- x^{2}} d x \int_{- b}^{b} e^{- y^{2}} d y = \iint_{∣ x ∣, ∣ y ∣ \leq b} e^{- (x^{2} + y^{2})} d A$
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:
  
  $π (1 - e^{- b^{2}}) \leq I (b)^{2} \leq π (1 - e^{- 2 b^{2}})$
3. Both bounds tend to π as b → ∞, so I(b)² does too, and I(b) is positive:
  
  $\int_{- \infty}^{\infty} e^{- x^{2}} d x = b \to \infty lim I (b) = π$
Answer

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

$\int_{- \infty}^{\infty} e^{- x^{2}} d x = π \approx 1.7725$
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.
Show solution
1. Substitute x = √2 u, so dx = √2 du:
  
  $\int_{- \infty}^{\infty} e^{- x^{2} /2} d x = 2 \int_{- \infty}^{\infty} e^{- u^{2}} d u = 2 \cdot π = 2 π$
2. Divide by √(2π) and the total is 1:
  
  $\int_{- \infty}^{\infty} \frac{1}{2 π} e^{- x^{2} /2} d x = \frac{2 π}{2 π} = 1$
Answer

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

$\int_{- \infty}^{\infty} e^{- x^{2} /2} d x = 2 π$

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 42Math team · Reviewed against published sources. Corrections welcome.