Unit 5, chapter 22

Let the circle choose the coordinates

How polar coordinates turn a double integral over a disk into an iterated integral in radius and angle, why the polar area element carries an extra factor r from a thin annular sector, and how radial symmetry helps.

  • Chapter 22
  • Polar coordinates, the Jacobian and spherical coordinates

What is the integral of x2+y2x^2 + y^2 over the unit disk DD, the set of points with x2+y2≤1x^2 + y^2 \le 1? In xx and yy the disk needs the curved limits y=±1−x2y = \pm\sqrt{1 - x^2}, yet membership in the disk and the value of the integrand both depend only on the distance from the origin. This lesson describes points by that distance and a direction angle, derives the polar area element from a thin annular sector, and shows that

∬D(x2+y2) dA=π2.\iint_D (x^2 + y^2)\,dA = \frac{\pi}{2}.

Key terms

Polar coordinates

Definition For r≥0r \ge 0 and an angle θ\theta in radians, the point with polar coordinates (r,θ)(r, \theta) is

x=rcos⁡θ,y=rsin⁡θ.x = r\cos\theta, \qquad y = r\sin\theta.

The radius rr is the distance from the origin, so r2=x2+y2r^2 = x^2 + y^2, and the angle θ\theta is measured counterclockwise from the positive xx axis. Every point other than the origin has exactly one pair with r>0r > 0 and θ\theta in a chosen half open interval of length 2π2\pi, such as 0≤θ<2π0 \le \theta < 2\pi. The origin has r=0r = 0 and any angle.

In plain words Instead of saying how far right and how far up a point is, say how far out it is and in which direction. Circles centered at the origin become r=constantr = \text{constant}, and rays from the origin become θ=constant\theta = \text{constant}.

Example In Cartesian coordinates the unit disk is

−1≤x≤1,−1−x2≤y≤1−x2,\begin{gathered} -1 \le x \le 1, \\[0.4em] -\sqrt{1 - x^2} \le y \le \sqrt{1 - x^2}, \end{gathered}

and the square roots come only from the circle. In polar coordinates the disk is 0≤r≤10 \le r \le 1 and 0≤θ≤2π0 \le \theta \le 2\pi, and its boundary is r=1r = 1. The integrand simplifies too, since cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 gives

x2+y2=r2cos⁡2θ+r2sin⁡2θ=r2.\begin{aligned} &x^2 + y^2 \\[0.4em] &\quad = r^2\cos^2\theta + r^2\sin^2\theta \\[0.4em] &\quad = r^2. \end{aligned}

Annular sector

Definition For 0≤r1<r20 \le r_1 < r_2 and θ1<θ2≤θ1+2π\theta_1 < \theta_2 \le \theta_1 + 2\pi, the annular sector is the set of points with polar coordinates r1≤r≤r2r_1 \le r \le r_2 and θ1≤θ≤θ2\theta_1 \le \theta \le \theta_2. It is the part of the ring between the circles of radii r1r_1 and r2r_2 that lies between two rays from the origin. A circular sector of radius ρ\rho and angle Δθ\Delta\theta in radians has area 12ρ2Δθ\tfrac12 \rho^2 \Delta\theta, so with Δθ=θ2−θ1\Delta\theta = \theta_2 - \theta_1 the annular sector has area

ΔA=12(r22−r12)Δθ.\Delta A = \tfrac12 \left( r_2^2 - r_1^2 \right) \Delta\theta.

Writing r1=rr_1 = r and r2=r+Δrr_2 = r + \Delta r gives the exact split

ΔA=r Δr Δθ+12(Δr)2 Δθ.\begin{aligned} \Delta A &= r\,\Delta r\,\Delta\theta \\[0.4em] &\quad + \tfrac12 (\Delta r)^2\,\Delta\theta. \end{aligned}

Equivalently, ΔA=rˉ Δr Δθ\Delta A = \bar{r}\,\Delta r\,\Delta\theta exactly, where rˉ=r+12Δr\bar{r} = r + \tfrac12 \Delta r is the middle radius.

In plain words A cell of a polar grid is a slightly curved box. Its straight sides have length Δr\Delta r and its arcs have lengths r Δθr\,\Delta\theta and (r+Δr) Δθ(r + \Delta r)\,\Delta\theta, so cells with the same Δr\Delta r and Δθ\Delta\theta get larger the farther they are from the center.

Example Take Δr=0.1\Delta r = 0.1 and Δθ=0.1\Delta\theta = 0.1. The cell from r=1r = 1 to r=1.1r = 1.1 has area

12(1.21−1)(0.1)=0.0105,\tfrac12 (1.21 - 1)(0.1) = 0.0105,

made of the leading term 1⋅0.1⋅0.1=0.011 \cdot 0.1 \cdot 0.1 = 0.01 and the correction 12(0.1)2(0.1)=0.0005\tfrac12 (0.1)^2 (0.1) = 0.0005. The cell from r=0.1r = 0.1 to r=0.2r = 0.2 with the same angle has area 12(0.04−0.01)(0.1)=0.0015\tfrac12 (0.04 - 0.01)(0.1) = 0.0015. The outer cell is 77 times larger, which is the ratio 1.05/0.151.05 / 0.15 of the middle radii.

Polar area element

Definition The polar area element is

dA=r dr dθ.dA = r\,dr\,d\theta.

It records that a polar grid cell with sides Δr\Delta r and Δθ\Delta\theta at radius rr has area r Δr Δθr\,\Delta r\,\Delta\theta up to the correction 12(Δr)2Δθ\tfrac12 (\Delta r)^2 \Delta\theta. In a Riemann sum of a bounded integrand over a polar grid, the corrections add up to at most a fixed multiple of the largest Δr\Delta r, so only the leading term survives in the limit. The factor rr is also the absolute value of the Jacobian determinant of the map (r,θ)↦(rcos⁡θ,rsin⁡θ)(r, \theta) \mapsto (r\cos\theta, r\sin\theta),

∣det⁡[cos⁡θ−rsin⁡θsin⁡θrcos⁡θ]∣=rcos⁡2θ+rsin⁡2θ=r,\begin{aligned} &\left\lvert \det \begin{bmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{bmatrix} \right\rvert \\[0.4em] &\quad = r\cos^2\theta + r\sin^2\theta \\[0.4em] &\quad = r, \end{aligned}

the area scale of a change of variables, which is the subject of Build coordinates from the boundaries.

In plain words After rewriting the integrand in rr and θ\theta, multiply by rr before integrating. The factor is not part of the height. It measures how large the floor tiles are, and tiles far from the center are larger.

Example The area of the unit disk is

∫02π∫01r dr dθ=2π⋅12=π.\begin{aligned} \int_0^{2\pi} \int_0^1 r\,dr\,d\theta &= 2\pi \cdot \frac12 \\[0.4em] &= \pi. \end{aligned}

Without the factor, ∫02π∫01dr dθ=2π\int_0^{2\pi} \int_0^1 dr\,d\theta = 2\pi, which is the area of the rectangle [0,1]×[0,2π][0, 1] \times [0, 2\pi] in the (r,θ)(r, \theta) plane and not the area of the disk.

Double integral in polar coordinates

Definition Let DD be the region given in polar coordinates by α≤θ≤β\alpha \le \theta \le \beta and h1(θ)≤r≤h2(θ)h_1(\theta) \le r \le h_2(\theta), where 0<β−α≤2π0 < \beta - \alpha \le 2\pi and the functions h1h_1 and h2h_2 are continuous on [α,β][\alpha, \beta] with 0≤h1≤h20 \le h_1 \le h_2. If ff is continuous on DD and g(r,θ)=f(rcos⁡θ,rsin⁡θ)g(r, \theta) = f(r\cos\theta, r\sin\theta), then

∬Df dA=∫αβ∫h1(θ)h2(θ)g r dr dθ,\begin{aligned} &\iint_D f\,dA \\[0.4em] &\quad = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} g\,r\,dr\,d\theta, \end{aligned}

Here gg is evaluated at (r,θ)(r, \theta). The map from (r,θ)(r, \theta) to (x,y)(x, y) is one-to-one on these parameters except where r=0r = 0 and, when β−α=2π\beta - \alpha = 2\pi, on the edges θ=α\theta = \alpha and θ=β\theta = \beta, which land on the same ray. These exceptions lie on sets of zero area, so they do not affect the integral. For an annular sector the limits on rr are constants.

In plain words Rewrite the integrand with x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta, rewrite the region as limits on rr and θ\theta, and multiply by rr. All three changes go together.

Example For the unit disk DD the limits are 0≤r≤10 \le r \le 1 and 0≤θ≤2π0 \le \theta \le 2\pi, the integrand x2+y2x^2 + y^2 is continuous and nonnegative and equals r2r^2, and the area factor is rr. So

∬D(x2+y2) dA=∫02π∫01r2⋅r dr dθ=∫02π14 dθ=π2.\begin{aligned} &\iint_D (x^2 + y^2)\,dA \\[0.4em] &\quad = \int_0^{2\pi} \int_0^1 r^2 \cdot r\,dr\,d\theta \\[0.4em] &\quad = \int_0^{2\pi} \frac14\,d\theta = \frac{\pi}{2}. \end{aligned}

The power r3r^3 combines two factors of rr from the height and one from the area element. The radial integral ∫01r3 dr=14\int_0^1 r^3\,dr = \tfrac14 does not depend on θ\theta, because the integrand looks the same in every direction. The Cartesian setup with the limits ±1−x2\pm\sqrt{1 - x^2} gives the same value π2\tfrac{\pi}{2}, after considerably more work.

Radial symmetry

Definition A function on a region of the plane is radially symmetric about the origin when its value depends only on the distance to the origin, so that f(x,y)=φ(r)f(x, y) = \varphi(r) with r=x2+y2r = \sqrt{x^2 + y^2} for some function φ\varphi. A disk r≤br \le b and an annulus a≤r≤ba \le r \le b centered at the origin are radially symmetric regions. If DD is the region a≤r≤ba \le r \le b with 0≤a<b0 \le a < b, and f=φ(r)f = \varphi(r) with φ\varphi continuous on [a,b][a, b], then

∬Df dA=2π∫abφ(r) r dr,\begin{aligned} &\iint_D f\,dA \\[0.4em] &\quad = 2\pi \int_a^b \varphi(r)\,r\,dr, \end{aligned}

where a=0a = 0 gives the disk.

In plain words When nothing changes with direction, the angular integral only contributes a factor 2π2\pi, and the problem becomes one radial integral. The weight rr remains, because the thin ring at radius rr has circumference 2πr2\pi r and therefore area about 2πr dr2\pi r\,dr. Polar coordinates also help when only the region is round, but then the angular integral has to be computed.

Example On the annulus 1≤x2+y2≤41 \le x^2 + y^2 \le 4, the function 1x2+y2\frac{1}{x^2 + y^2} equals 1r2\frac{1}{r^2} and is continuous, since the origin is not in the region. So

∬DdAx2+y2=2π∫121r2 r dr=2πln⁡2≈4.355.\begin{aligned} &\iint_D \frac{dA}{x^2 + y^2} \\[0.4em] &\quad = 2\pi \int_1^2 \frac{1}{r^2}\,r\,dr \\[0.4em] &\quad = 2\pi \ln 2 \approx 4.355. \end{aligned}

For an integrand that depends on the angle, such as x2x^2 on the unit disk, the angular integral no longer reduces to 2π2\pi, but symmetry still helps. Exchanging xx and yy maps the disk onto itself, so the integrals of x2x^2 and of y2y^2 over it are equal. They add up to the integral of x2+y2x^2 + y^2, so each one is half of π2\tfrac{\pi}{2}, which is π4\tfrac{\pi}{4}.

Check yourself

  1. Evaluate the integral of x2+y2\sqrt{x^2 + y^2} over the disk x2+y2≤4x^2 + y^2 \le 4.
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The integrand is radially symmetric with φ(r)=r\varphi(r) = r, which is continuous on [0,2][0, 2]. So the integral is

2π∫02r⋅r dr=2π⋅83=16π3.\begin{aligned} 2\pi \int_0^2 r \cdot r\,dr &= 2\pi \cdot \frac83 \\[0.4em] &= \frac{16\pi}{3}. \end{aligned}

This is the volume under the cone z=x2+y2z = \sqrt{x^2 + y^2} over the disk. As a check, the cylinder of radius 22 and height 22 has volume 8π8\pi, and the part of it above the cone is a solid cone of radius 22 and height 22, with volume 13π⋅22⋅2=8π3\tfrac13 \pi \cdot 2^2 \cdot 2 = \tfrac{8\pi}{3}. The difference is 8π−8π3=16π38\pi - \tfrac{8\pi}{3} = \tfrac{16\pi}{3}.

  1. Evaluate ∬Dxy dA\iint_D xy\,dA, where DD is the part of the ring 1≤x2+y2≤41 \le x^2 + y^2 \le 4 in the first quadrant.
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The region is the annular sector 1≤r≤21 \le r \le 2 and 0≤θ≤π20 \le \theta \le \tfrac{\pi}{2}. The integrand is xy=r2cos⁡θsin⁡θxy = r^2 \cos\theta \sin\theta, and with the area factor it becomes r3cos⁡θsin⁡θr^3 \cos\theta \sin\theta. The limits are constants and this is a function of θ\theta times a function of rr, so the integral is the product of ∫0π/2cos⁡θsin⁡θ dθ=12\int_0^{\pi/2} \cos\theta \sin\theta\,d\theta = \tfrac12 and ∫12r3 dr=154\int_1^2 r^3\,dr = \tfrac{15}{4},

∬Dxy dA=12⋅154=158.\iint_D xy\,dA = \frac12 \cdot \frac{15}{4} = \frac{15}{8}.

Leaving out the factor rr would give ∫12r2 dr=73\int_1^2 r^2\,dr = \tfrac73 in place of 154\tfrac{15}{4} and the wrong value 76\tfrac76.

  1. A classmate says the average distance from the center of the unit disk to its points is 12\tfrac12, because the distances run from 00 to 11 and ∫01r dr=12\int_0^1 r\,dr = \tfrac12. Find the correct average, the integral of x2+y2\sqrt{x^2 + y^2} over the disk divided by the area of the disk.
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The disk has area π\pi, and the integrand is radially symmetric, so

1π∬Dx2+y2 dA=2∫01r⋅r dr=23.\begin{aligned} &\frac{1}{\pi} \iint_D \sqrt{x^2 + y^2}\,dA \\[0.4em] &\quad = 2 \int_0^1 r \cdot r\,dr = \frac23. \end{aligned}

The classmate’s integral gives every radius the same weight. The thin ring at radius rr has area about 2πr dr2\pi r\,dr, so larger distances cover more of the disk and pull the average up. In fact half of the disk’s area lies outside the circle of radius 12≈0.707\tfrac{1}{\sqrt{2}} \approx 0.707, since the disk of radius ρ\rho has area πρ2\pi \rho^2.

How this lesson connects