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Deriving the Polar-Coordinate Integral Formula

See how sector geometry yields the polar area factor and why general Lebesgue-measurable integrands require a careful interpretation of the formula.

Advanced 12 min read

What You'll Learn

  • Derive the factor \(r\) from the determinant of the polar-coordinate map and from annular-sector area.
  • Identify the origin and angular seam excluded by a one-to-one polar parametrization.
  • Prove the integral formula for nonnegative Borel functions using measure uniqueness and simple-function approximation.
  • Extend the formula to nonnegative Lebesgue-measurable functions using a Borel representative.
  • Compute polar integrals over a square and verify them in Cartesian coordinates.
  • Recognize why a literal polar pullback of a Lebesgue-measurable function can have a nonmeasurable radial section.

From Polar Coordinates to an Area Measure

Polar coordinates replace planar position by radius and angle. The area element is not \(dr\,d\theta\): a small angular change sweeps out a wider arc at larger radius. The resulting factor \(r\) can be seen in the Jacobian, but to justify an integral formula for measurable functions we must also account for the origin, the angular seam, and measurability.

Write \(m\) for Lebesgue measure on \(\mathbb R^2\), and let \(dr\,d\theta\) denote product Lebesgue measure on \((0,\infty)\times(0,2\pi)\). All integrands in the main formula will first be nonnegative and Borel measurable, with \(+\infty\) allowed. The measure spaces involved are sigma-finite. We use Tonelli's theorem for nonnegative measurable functions on these product spaces; it permits the iterated integrals below to equal the product-space integral, including when that value is infinite.

Consider the map \(T(r,\theta)=(r\cos\theta,r\sin\theta)\) on \(D=(0,\infty)\times(0,2\pi)\). It is a continuous bijection from \(D\) onto \(U=\mathbb R^2\setminus\{(x,0):x\ge0\}\), with continuous inverse. The excluded positive horizontal ray is the angular seam, and the origin is excluded because \(r>0\). On this open domain, the derivative matrix and its determinant are

$$ DT(r,\theta)= \begin{pmatrix} \cos\theta&-r\sin\theta\\ \sin\theta&r\cos\theta \end{pmatrix}, \qquad \det DT(r,\theta)=r(\cos^2\theta+\sin^2\theta)=r. $$

This determinant explains the local area factor. We will establish the global measure identity directly from sector areas, so the justification does not depend on applying a general change-of-variables theorem to the polar map.

Deriving the Measure Identity

Lemma: For every Borel set \(A\subseteq D\), the image \(T(A)\) is Borel in \(U\), and $$ m(T(A))=\int_A r\,dr\,d\theta. $$

Proof. Because \(T\) is a homeomorphism from \(D\) to \(U\), it maps Borel sets in \(D\) to Borel sets in \(U\); since \(U\) is open, these are also Borel sets in \(\mathbb R^2\). Thus \(\nu(A)=m(T(A))\) defines a measure on the Borel subsets of \(D\): disjoint sets have disjoint images, and \(T\) maps their union to the union of their images.

First take a rectangle \(Q=(a,b)\times(\alpha,\beta)\) whose closure lies in \(D\), where \(0<a<b<\infty\) and \(0<\alpha<\beta<2\pi\). The image is an annular sector, up to its boundary. A circle of radius \(R\) has area \(\pi R^2\), by scaling of Lebesgue measure from the unit disk. A sector of angle \(\gamma\) in that disk has area \(\gamma/(2\pi)\) times the disk's area: equal angular sectors have equal area by rotation invariance, so this holds first for rational fractions of a full turn; approximation by increasing and decreasing rational angles and continuity of finite measure gives it for every \(\gamma\in[0,2\pi]\). Scaling the sector to radius \(R\) gives area \(\gamma R^2/2\). Subtract the inner sector of radius \(a\) from the outer one of radius \(b\). The annular-sector area is therefore $$ \nu(Q)=\frac{\beta-\alpha}{2}(b^2-a^2) =\int_\alpha^\beta\int_a^b r\,dr\,d\theta. $$ The rectangle boundaries do not affect this calculation: their images are finitely many line segments or circular arcs, which have planar Lebesgue measure zero.

For completeness, the rectangles with rational endpoints and closure in \(D\), together with the empty set, form a pi-system generating the Borel sigma-algebra of \(D\). The two measures \(\nu\) and \(\lambda(A)=\int_A r\,dr\,d\theta\) agree on these rectangles. Both are sigma-finite: \(D\) is covered by countably many bounded rectangles with closure in \(D\), and both measures are finite on each such rectangle. The uniqueness theorem for sigma-finite measures agreeing on a generating pi-system (applied on a countable finite-measure cover) gives \(\nu=\lambda\) on all Borel subsets of \(D\). This proves the lemma. \(\square\)

Theorem (Polar-coordinate integral formula for Borel functions): If \(f:\mathbb R^2\to[0,\infty]\) is Borel measurable, then $$ \int_{\mathbb R^2} f(x,y)\,dm(x,y) = \int_0^{2\pi}\int_0^\infty f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. $$ Both sides may equal \(+\infty\). The displayed iterated integral is defined: the pullback is Borel measurable on \(D\), and Tonelli's theorem applies.

Proof. The seam \(\{(x,0):x\ge0\}\) has planar measure zero, as does the single origin. Thus removing them does not change the integral of a nonnegative Borel function. On \(U\), the function \(f\) is Borel measurable. For a Borel set \(A\subseteq D\), the lemma gives \(m(T(A))=\int_A r\,dr\,d\theta\). Consequently the measure on \(D\) obtained by pulling back planar Lebesgue measure under \(T\) agrees with the measure having density \(r\) with respect to \(dr\,d\theta\).

For a nonnegative Borel simple function on \(U\), the equality of its integrals with respect to these two measures follows by writing the function as a finite sum of nonnegative constants times indicators of Borel sets and applying the lemma to each set. For general nonnegative Borel \(f\), choose an increasing sequence of nonnegative Borel simple functions converging pointwise to \(f|_U\). The Monotone Convergence Theorem, applied to each of the two measures, passes the equality from the simple functions to \(f|_U\). Finally, Tonelli's theorem identifies the integral on \(D\) with the stated iterated integral. The omitted seam and origin are null, so the left side is unchanged. \(\square\)

Worked Applications

Worked Example: The Indicator of the Unit Square

Let \(f(x,y)=\chi_{[0,1]\times[0,1]}(x,y)\). This is a nonnegative Borel function. Apart from boundary points, which do not affect either integral, its polar preimage has \(0<\theta<\pi/2\) and \(0<r<\min(\sec\theta,\csc\theta)\). For \(0<\theta<\pi/4\), the limiting side is \(x=1\), so \(r<\sec\theta\). For \(\pi/4<\theta<\pi/2\), it is \(y=1\), so \(r<\csc\theta\). Therefore

$$ \begin{aligned} \int_{\mathbb R^2} f\,dm &=\int_0^{\pi/4}\int_0^{\sec\theta}r\,dr\,d\theta +\int_{\pi/4}^{\pi/2}\int_0^{\csc\theta}r\,dr\,d\theta\\ &=\frac12\int_0^{\pi/4}\sec^2\theta\,d\theta +\frac12\int_{\pi/4}^{\pi/2}\csc^2\theta\,d\theta\\ &=\frac12(1+1)=1. \end{aligned} $$

Indeed, \(\int_0^{\pi/4}\sec^2\theta\,d\theta=[\tan\theta]_0^{\pi/4}=1\), and \(\int_{\pi/4}^{\pi/2}\csc^2\theta\,d\theta=[-\cot\theta]_{\pi/4}^{\pi/2}=1\). Check: the Cartesian area is \(1\cdot1=1\), agreeing with the polar calculation.

Worked Example: Integrating \(x^2+y^2\) Over the Unit Square

Use the same polar bounds, now with \(f(x,y)=(x^2+y^2)\chi_{[0,1]\times[0,1]}(x,y)\). On the polar preimage \(x^2+y^2=r^2\), so multiplication by the area factor makes the integrand \(r^3\). Thus

$$ \begin{aligned} \int_{\mathbb R^2} f\,dm &=\frac14\int_0^{\pi/4}\sec^4\theta\,d\theta +\frac14\int_{\pi/4}^{\pi/2}\csc^4\theta\,d\theta\\ &=\frac12\int_0^{\pi/4}\sec^4\theta\,d\theta. \end{aligned} $$

The last equality follows by substituting \(u=\pi/2-\theta\) in the second integral: its limits become \(\pi/4\) and \(0\), and \(\csc(\pi/2-u)=\sec u\). For the remaining integral, set \(v=\tan\theta\); then \(dv=\sec^2\theta\,d\theta\) and \(\sec^2\theta=1+v^2\). Hence $$ \int_0^{\pi/4}\sec^4\theta\,d\theta =\int_0^1(1+v^2)\,dv =\left[v+\frac{v^3}{3}\right]_0^1 =\frac43. $$ The polar answer is therefore \(\frac12\cdot\frac43=\frac23\). Check: direct integration gives \(\int_0^1\int_0^1(x^2+y^2)\,dy\,dx=\frac13+\frac13=\frac23\).

Worked Example: Integrating \(x\) Over the Unit Square

Take \(f(x,y)=x\chi_{[0,1]\times[0,1]}(x,y)\). Under the polar map, \(x=r\cos\theta\); including the area factor gives \(r^2\cos\theta\). With the same split at \(\pi/4\),

$$ \begin{aligned} \int_{\mathbb R^2} f\,dm &=\frac13\int_0^{\pi/4}\cos\theta\,\sec^3\theta\,d\theta +\frac13\int_{\pi/4}^{\pi/2}\cos\theta\,\csc^3\theta\,d\theta\\ &=\frac13\int_0^{\pi/4}\sec^2\theta\,d\theta +\frac13\int_{\pi/4}^{\pi/2}\frac{\cos\theta}{\sin^3\theta}\,d\theta\\ &=\frac13+\frac16=\frac12. \end{aligned} $$

For the second angular integral, an antiderivative of \(\cos\theta/\sin^3\theta\) is \(-1/(2\sin^2\theta)\). Its values at \(\pi/2\) and \(\pi/4\) are \(-1/2\) and \(-1\), so that integral is \(1/2\), contributing \(1/6\). Check: direct integration gives \(\int_0^1\int_0^1x\,dy\,dx=\int_0^1x\,dx=1/2\).

Lebesgue-Measurable Integrands and the Seam

The theorem above deliberately assumes Borel measurability. The Lebesgue sigma-algebra \(\mathcal L\) on \(\mathbb R^2\) is the completion of the Borel sigma-algebra: each Lebesgue-measurable set differs from a Borel set by a subset of a Borel null set. The corresponding completion property for functions says that every nonnegative \(\mathcal L\)-measurable function \(f\) has a nonnegative Borel-measurable representative \(g\) with \(f=g\) \(m\)-a.e. This follows, for example, by applying the set completion property to the level sets used in simple-function approximations of \(f\).

Extension to Lebesgue-measurable functions: For nonnegative \(\mathcal L\)-measurable \(f:\mathbb R^2\to[0,\infty]\), choose a Borel representative \(g\) equal to \(f\) \(m\)-a.e. Then define the right side of the polar formula using \(g\). Its value is independent of the chosen Borel representative, and $$ \int_{\mathbb R^2}f\,dm = \int_0^{2\pi}\int_0^\infty g(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta. $$

To check independence, let \(g_1,g_2\) be Borel representatives of \(f\). They agree outside an \(m\)-null Borel set \(N\). The polar-coordinate theorem applied to \(\chi_N\) gives \(\int_D\chi_N(T(r,\theta))r\,dr\,d\theta=m(N)=0\). Thus their pullbacks agree for \(r\,dr\,d\theta\)-almost every \((r,\theta)\), and their weighted integrals are equal. The Borel representative also ensures that Tonelli's theorem applies and that every fixed-angle radial section is measurable.

Common pitfall: composing the original function pointwise. A Lebesgue-measurable function need not have a measurable polar pullback at every angle. To see this, choose a non-Lebesgue-measurable set \(A\subset(1,2)\) and put \(E=\{(-r,0):r\in A\}\). The set \(E\) lies on a planar null line, so it is Lebesgue measurable and \(m(E)=0\). Hence \(f=\chi_E\) is Lebesgue measurable and \(\int f\,dm=0\). But at the angle \(\theta=\pi\), the radial section \(r\mapsto f(r\cos\pi,r\sin\pi)\) is \(\chi_A(r)\), which is not Lebesgue measurable. The literal iterated integral using this pullback is therefore not defined as written. The correction is to use a Borel representative in the formula, or to formulate an almost-everywhere-section version and disregard exceptional angles; one must not claim that every section of the original pullback is measurable.

The origin and seam cause a different, harmless issue. They are planar null sets, so omitting them does not change a Lebesgue integral. In the parameter domain, \(\theta=0\) and \(\theta=2\pi\) are endpoint sets of product measure zero, and \(r=0\) is also a null boundary. A one-to-one parametrization avoids double-counting by using \(0<\theta<2\pi\); writing the integral with endpoints \(0\) and \(2\pi\) gives the same value.

Check Your Understanding

In each question, distinguish the measure identity from the pointwise behavior of a particular representative.

  1. Compute the determinant of the derivative of \(T(r,\theta)=(r\cos\theta,r\sin\theta)\), and explain why its absolute value is the area factor.
  2. Why is the polar map restricted to \(0<\theta<2\pi\) not a one-to-one parametrization of the entire plane?
  3. For \(f(x,y)=\chi_{[0,1]\times[0,1]}(x,y)\), identify the radial upper bound on each side of \(\theta=\pi/4\).
  4. State precisely how the polar formula is interpreted for a nonnegative Lebesgue-measurable function that is not Borel measurable.
  5. Explain why the nonmeasurable-set example on the negative horizontal axis does not contradict the formula using a Borel representative.

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