From Radial Boundaries to Planar Integrals
In polar coordinates, a region can be described by the angles it occupies and the distance it reaches along each ray. This description is especially useful when the boundary is not circular: the outer radius may vary with the angle. The essential accounting is that planar area is \(r\,dr\,d\theta\), not \(dr\,d\theta\). That factor determines both the area formula and the radial integrals over the region.
Throughout, planar area means Lebesgue measure \(m\) on \((\mathbb{R}^2,\mathcal{L},m)\). Let \(a<b\), with \(b-a\le 2\pi\), and let \(R:[a,b]\to[0,\infty)\) be a bounded Borel-measurable function. The associated radial region is described in polar coordinates by \(a\le\theta\le b\) and \(0\le r\le R(\theta)\). The endpoints in angle and radius are included; changing these boundary conventions does not change the integrals below.
We use the polar-coordinate change-of-variables formula established earlier. In the form needed here, for a nonnegative Lebesgue-measurable function \(f:\mathbb{R}^2\to[0,\infty]\), or for a real-valued \(f\) integrable on the region in question, it gives
The factor \(r\) is the Jacobian determinant in polar coordinates. For nonnegative \(f\), the iterated integral is justified by the polar-coordinate formula and Tonelli's theorem; the value \(+\infty\) is allowed. For a real-valued \(f\), the formula applies when \(\int_D|f|\,dm<\infty\), so the integrals are defined and Fubini's theorem applies. The same formula applies to a sector by using constant radial limits, or by subtracting the integral over the inner radial region.
Area and Radial Moments
Proof. For area, apply the polar-coordinate formula to the indicator of \(D\), which is nonnegative and measurable. The polar-coordinate expression for the indicator is \(1\) when \(a\le\theta\le b\) and \(0\le r\le R(\theta)\), and is \(0\) otherwise. Therefore $$ m(D)=\int_a^b\int_0^{R(\theta)}r\,dr\,d\theta =\int_a^b\left[\frac{r^2}{2}\right]_{0}^{R(\theta)}d\theta =\frac12\int_a^bR(\theta)^2\,d\theta. $$ This also covers \(R(\theta)=0\), since the corresponding inner integral is zero. For the moment formula, use \(x^2+y^2=r^2\), so the polar integrand including its Jacobian is \(r^q r=r^{q+1}\). Because \(q>-2\), the integral at \(r=0\) is finite, and direct integration gives $$ \int_a^b\int_0^{R(\theta)}r^{q+1}\,dr\,d\theta =\int_a^b\frac{R(\theta)^{q+2}}{q+2}\,d\theta. $$ Both conclusions follow. \(\square\)
For a sector, these formulas reduce to familiar expressions. If \(\rho\le r\le R_0\) and \(a\le\theta\le b\), its area is \(\tfrac12(R_0^2-\rho^2)(b-a)\). For a varying radial boundary, the square \(R(\theta)^2\) must be integrated over the angle; using a single average radius generally gives the wrong area.
Worked Example: Area and Second Moment of a Circular Sector
Let \(D\) be the sector \(0\le\theta\le\pi/3\), \(0\le r\le3\). Compute its area and \(\int_D(x^2+y^2)\,dm\).
Here \(R(\theta)=3\). The area formula gives $$ m(D)=\frac12\int_0^{\pi/3}3^2\,d\theta =\frac92\cdot\frac{\pi}{3} =\frac{3\pi}{2}. $$ For the second moment, the polar integrand is \(r^2\cdot r=r^3\), so $$ \int_D(x^2+y^2)\,dm =\int_0^{\pi/3}\int_0^3r^3\,dr\,d\theta =\int_0^{\pi/3}\frac{3^4}{4}\,d\theta =\frac{81}{4}\cdot\frac{\pi}{3} =\frac{27\pi}{4}. $$ Check: the sector area also equals half the angle in radians times the squared radius: \(\tfrac12(\pi/3)(3^2)=3\pi/2\). The moment has the expected radial integral \(\int_0^3r^3\,dr=81/4\), multiplied by the angle \(\pi/3\), giving \(27\pi/4\).
Integrating Over a Nonconstant Boundary
When \(R\) varies with \(\theta\), the radial integration is still performed first. For example, if the integrand is \(x\), then \(x=r\cos\theta\), and the Jacobian makes the radial integrand \(r^2\cos\theta\). Thus $$ \int_Dx\,dm=\frac13\int_a^b\cos\theta\,R(\theta)^3\,d\theta. $$ This formula is not an area formula: the extra power of \(R\) comes from the factor \(r\) in \(x\) and the factor \(r\) in the area element.
Worked Example: Area and Horizontal First Moment
Take \(D\) with \(0\le\theta\le\pi\) and \(R(\theta)=1+\tfrac12\cos\theta\). This boundary is positive because \(-1\le\cos\theta\le1\), so \(\tfrac12\le R(\theta)\le\tfrac32\). Compute the area and \(\int_Dx\,dm\).
First expand the square: $$ R(\theta)^2=1+\cos\theta+\frac14\cos^2\theta. $$ Using \(\int_0^\pi1\,d\theta=\pi\), \(\int_0^\pi\cos\theta\,d\theta=0\), and \(\int_0^\pi\cos^2\theta\,d\theta=\pi/2\), the area is $$ m(D)=\frac12\left(\pi+0+\frac14\cdot\frac{\pi}{2}\right) =\frac12\cdot\frac{9\pi}{8} =\frac{9\pi}{16}. $$ For the first moment, expand the cube and multiply by \(\cos\theta\): $$ \cos\theta\,R(\theta)^3 =\cos\theta+\frac32\cos^2\theta+\frac34\cos^3\theta+\frac18\cos^4\theta. $$ On \([0,\pi]\), the integrals of \(\cos\theta\) and \(\cos^3\theta\) are both zero, while \(\int_0^\pi\cos^2\theta\,d\theta=\pi/2\) and \(\int_0^\pi\cos^4\theta\,d\theta=3\pi/8\). Therefore $$ \int_Dx\,dm =\frac13\left(\frac32\cdot\frac{\pi}{2}+\frac18\cdot\frac{3\pi}{8}\right) =\frac13\left(\frac{3\pi}{4}+\frac{3\pi}{64}\right) =\frac13\cdot\frac{51\pi}{64} =\frac{17\pi}{64}. $$ Check: the larger radial extent occurs near \(\theta=0\), where \(x\) is positive; the computed horizontal moment is positive. The area computation also follows directly from the theorem using the expanded \(R^2\), with the odd cosine term integrating to zero.
Worked Example: A Symmetric Star-Shaped Region
Let \(D\) have \(-\pi/2\le\theta\le\pi/2\) and \(R(\theta)=1+\cos\theta\). Since \(0\le\cos\theta\le1\) on this interval, the radius ranges from \(1\) at the endpoints to \(2\) at \(\theta=0\). Find the area.
Expand \(R^2=1+2\cos\theta+\cos^2\theta\). The three angular integrals are $$ \int_{-\pi/2}^{\pi/2}1\,d\theta=\pi,\qquad \int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta=2,\qquad \int_{-\pi/2}^{\pi/2}\cos^2\theta\,d\theta=\frac{\pi}{2}. $$ Consequently, $$ m(D)=\frac12\left(\pi+2\cdot2+\frac{\pi}{2}\right) =\frac12\left(\frac{3\pi}{2}+4\right) =\frac{3\pi}{4}+2. $$ Check: the middle term in the area integral contributes \(\tfrac12\int_{-\pi/2}^{\pi/2}2\cos\theta\,d\theta=2\); the constant and squared-cosine terms contribute \(\pi/2\) and \(\pi/4\), respectively. Their sum is \(2+3\pi/4\), as above. The formula also respects the symmetry about the \(x\)-axis: the two half-regions have equal area.
Common Pitfall: Omitting the Jacobian
A frequent error is to replace planar area by \(dr\,d\theta\). Polar coordinates do not preserve area uniformly: an angular interval of fixed width sweeps out a larger arc at larger radius. The Jacobian factor \(r\) accounts for this change.
For a concrete correction, consider the sector \(0\le\theta\le\pi/3\), \(0\le r\le3\). Omitting the Jacobian would give $$ \int_0^{\pi/3}\int_0^3 1\,dr\,d\theta =3\cdot\frac{\pi}{3} =\pi, $$ which is not its area. With the required factor \(r\), the calculation is $$ \int_0^{\pi/3}\int_0^3r\,dr\,d\theta =\frac{9}{2}\cdot\frac{\pi}{3} =\frac{3\pi}{2}. $$ The sector formula provides an independent check: half of the angle times the squared radius is \(\tfrac12(\pi/3)9=3\pi/2\). The discrepancy is not an endpoint issue; it comes from using the wrong area element.
When setting up a radial integral, check each ingredient separately: the angular interval describes the directions in the region, the radial limits describe the points along each ray, and the Jacobian is \(r\). Then check that the integrand is in the right class: nonnegative measurable functions permit Tonelli's theorem, while signed integrands require integrability before using Fubini's theorem. These checks prevent both geometric errors and unjustified integral manipulations.
Check Your Understanding
For each question, identify the angular interval, radial boundary, and Jacobian factor before evaluating an integral.
- Find the area of the sector \(0\le\theta\le\pi/4\), \(0\le r\le2\).
- For \(0\le\theta\le\pi/2\) and \(R(\theta)=2+\sin\theta\), write the area integral. Which powers of \(R\) appear after integrating in \(r\)?
- For a radial-boundary region with \(R(\theta)=1\) on \([0,\pi]\), compute \(\int_D(x^2+y^2)\,dm\).
- Explain why the polar-coordinate formula for a signed real-valued integrand requires an integrability hypothesis, while a nonnegative integrand can be treated by Tonelli's theorem.
- A calculation for a sector uses \(\int_a^b\int_0^{R_0}1\,dr\,d\theta\) as its area. Identify the missing factor and state the corrected integral.