Radial Integrands on Disks and Annuli

Why Radial Integrands Simplify

A function is radial when its value depends on a point only through its distance from the origin. Rotating a point around the origin does not change that distance, so the function has the same value all the way around each circle centered at the origin.

On a disk or annulus centered at the origin, this symmetry makes the angular part of the double integral especially simple. The radial function is constant as the angle varies, and the entire angular range contributes a factor of \(2\pi\). The remaining calculation is a one-dimensional integral in the radius.

The polar area element still matters: a thin ring at radius \(r\) has area proportional to \(r\). That is why the reduced integral contains \(r\,dr\), rather than just \(dr\).

Key formula: If the integrand is \(g(r)\) and the region is the centered annulus with inner radius \(a\) and outer radius \(b\), then

\iint_{D} g(r)\,dA = 2\pi\int_a^b g(r)\,r\,dr.

Here \(0\leq a\leq b\), and a disk of radius \(b\) is the special case \(a=0\).

Radial Functions and Centered Regions

In Cartesian coordinates, the distance from \((x,y)\) to the origin is \(\sqrt{x^2+y^2}\). Thus an integrand that depends only on this distance can be written as \(g(\sqrt{x^2+y^2})\). In polar coordinates, the same distance is \(r\), so the integrand becomes \(g(r)\).

Definition: A radial integrand centered at the origin is a function of the form \(g(r)\) in polar coordinates, or equivalently \(g(\sqrt{x^2+y^2})\) in Cartesian coordinates. A centered annulus with radii \(a\) and \(b\) is the region \(a\leq r\leq b\), with \(0\leq a\leq b\) and the full angular range \(0\leq\theta\leq2\pi\). When \(a=0\), the region is a disk.

For the formula below, assume \(g\) is continuous on the radial interval \([a,b]\). This condition guarantees that the integral is defined in the usual sense and that the reduction applies directly. Improper integrals with singularities require a separate convergence check, especially if the region includes the origin.

The polar area element is \(dA=r\,dr\,d\theta\). A centered annulus has the simple polar bounds \(a\leq r\leq b\) and \(0\leq\theta\leq2\pi\). These facts turn its double integral into an iterated integral over a rectangle in the \((r,\theta)\)-plane.

Deriving the Radial Formula

Write \(D\) for the centered annulus. Substituting the polar expression for the integrand and the polar area element gives

\iint_D g(r)\,dA = \int_0^{2\pi}\int_a^b g(r)\,r\,dr\,d\theta.

Because \(g(r)r\) does not depend on \(\theta\), the inner radial integral has the same value at every angle. Evaluating the angular integral therefore multiplies that value by the length of the angular interval, \(2\pi\):

\int_0^{2\pi}\left(\int_a^b g(r)\,r\,dr\right)d\theta = \left(\int_a^b g(r)\,r\,dr\right) \left(\int_0^{2\pi}d\theta\right) = 2\pi\int_a^b g(r)\,r\,dr.

This proves the formula for continuous \(g\) on \([a,b]\). In particular, the disk version follows by setting \(a=0\). The origin has no effect on the integral as an individual point, so the fact that its polar angle is not uniquely specified causes no problem.

Theorem: If \(g\) is continuous on \([a,b]\), where \(0\leq a\leq b\), then its integral over the full centered annulus \(a\leq r\leq b\) is \(2\pi\int_a^b g(r)r\,dr\). For the centered disk of radius \(b\), this becomes \(2\pi\int_0^b g(r)r\,dr\).

Worked Examples

Worked Example: A Quadratic Radial Integrand on a Disk

Evaluate \(\iint_D (x^2+y^2)\,dA\), where \(D\) is the disk of radius \(3\) centered at the origin.

First rewrite the integrand in polar coordinates. Since \(x^2+y^2=r^2\), the radial function is \(g(r)=r^2\). The disk has \(0\leq r\leq3\), so

\iint_D (x^2+y^2)\,dA = 2\pi\int_0^3 r^2r\,dr = 2\pi\int_0^3 r^3\,dr.

An antiderivative of \(r^3\) is \(r^4/4\), because \(\frac{d}{dr}(r^4/4)=r^3\). Therefore

2\pi\left[\frac{r^4}{4}\right]_0^3 = 2\pi\left(\frac{3^4}{4}-\frac{0^4}{4}\right) = 2\pi\left(\frac{81}{4}\right) = \frac{81\pi}{2}.

Check the antiderivative by differentiating \(r^4/4\); its derivative is \(r^3\), the radial integrand after including the area factor. Thus the integral is \(\frac{81\pi}{2}\).

Worked Example: A Polynomial on an Annulus

Evaluate \(\iint_A (1+x^2+y^2)\,dA\), where \(A\) is the annulus \(1\leq r\leq2\).

In polar coordinates, \(x^2+y^2=r^2\), so the integrand is \(1+r^2\). Including the polar area factor gives

\iint_A (1+x^2+y^2)\,dA = 2\pi\int_1^2 (1+r^2)r\,dr = 2\pi\int_1^2 (r+r^3)\,dr.

An antiderivative is \(r^2/2+r^4/4\), since its derivative is \(r+r^3\). Evaluate at both bounds:

2\pi\left[\frac{r^2}{2}+\frac{r^4}{4}\right]_1^2 = 2\pi\left[ \left(\frac{4}{2}+\frac{16}{4}\right) - \left(\frac{1}{2}+\frac{1}{4}\right) \right].

The upper-bound value is \(2+4=6\). The lower-bound value is \(1/2+1/4=3/4\). Their difference is \(6-3/4=21/4\), so the integral is \(2\pi(21/4)=21\pi/2\).

Check by differentiating the antiderivative: \(\frac{d}{dr}(r^2/2+r^4/4)=r+r^3\), as required. The result is \(\frac{21\pi}{2}\).

Worked Example: A Logarithmic Integrand on an Annulus

Evaluate \(\iint_A \ln(\sqrt{x^2+y^2})\,dA\) over the annulus \(1\leq r\leq e\). The inner radius is positive, so the logarithm is defined throughout the region.

Since \(\sqrt{x^2+y^2}=r\), the integrand is \(\ln r\). Applying the radial formula gives

\iint_A \ln(\sqrt{x^2+y^2})\,dA = 2\pi\int_1^e r\ln r\,dr.

An antiderivative of \(r\ln r\) is \(\frac{r^2}{2}\ln r-\frac{r^2}{4}\). Differentiating it gives \(r\ln r+r/2-r/2=r\ln r\), confirming the antiderivative. Hence

2\pi\left[\frac{r^2}{2}\ln r-\frac{r^2}{4}\right]_1^e = 2\pi\left[ \left(\frac{e^2}{2}-\frac{e^2}{4}\right) - \left(0-\frac{1}{4}\right) \right].

The upper-bound value is \(e^2/4\), and the lower-bound value is \(-1/4\). The difference is \((e^2+1)/4\). Multiplying by \(2\pi\) gives \(\pi(e^2+1)/2\). The antiderivative check verifies the radial calculation, so the integral is \(\frac{\pi(e^2+1)}{2}\).

Radial Powers

A useful special case is \(g(r)=r^p\). For a centered annulus with \(a>0\), every real exponent \(p\) gives a continuous integrand on \([a,b]\). For a disk, the formula below applies directly when \(p>-2\); this ensures the radial integral is finite at the origin. When \(p=-2\), the integral diverges at the origin, and when \(p<-2\), it also fails to converge there.

For \(p\ne-2\), the antiderivative of \(r^{p+1}\) is \(r^{p+2}/(p+2)\), so

\iint_D r^p\,dA = \frac{2\pi}{p+2}\left(b^{p+2}-a^{p+2}\right).

For a disk, take \(a=0\), with \(p>-2\), to obtain \(2\pi b^{p+2}/(p+2)\). The excluded exponent \(p=-2\) has radial integrand \(1/r\), whose integral from \(0\) to \(b\) is improper and divergent.

A Common Mistake: Omitting the Area Factor

The factor \(r\) in \(dA=r\,dr\,d\theta\) is essential. It accounts for how the area of a thin polar strip changes with its distance from the origin. Omitting it changes the value of the integral.

For example, a mistaken calculation for the first worked example might write

\int_0^{2\pi}\int_0^3 r^2\,dr\,d\theta = 2\pi\left[\frac{r^3}{3}\right]_0^3 = 18\pi.

This calculation treats the area element as \(dr\,d\theta\), rather than \(r\,dr\,d\theta\). The correct setup includes the missing factor:

\int_0^{2\pi}\int_0^3 r^2\,r\,dr\,d\theta = 2\pi\left[\frac{r^4}{4}\right]_0^3 = \frac{81\pi}{2}.

A reliable check is to write the area element explicitly before integrating. The integrand \(g(r)\) and the area factor \(r\) play different roles: the former describes the quantity being integrated, while the latter converts polar coordinate changes into area.

When the Reduction Applies

The formula is tailored to a full disk or annulus centered at the origin and an integrand depending only on \(r\). If the region has restricted angular bounds, or if the integrand varies with \(\theta\), the angular integral need not simply contribute \(2\pi\). The radial formula should not be applied unchanged to those situations.

Within its stated setting, the reduction is useful because it replaces a two-variable integral with a single integral while preserving the geometry through the factor \(r\). Always identify the inner and outer radii, express the integrand as a function of \(r\), include the area factor, and then evaluate the resulting one-dimensional integral.

Check Your Understanding

Use the radial formula and show the radial integral you would evaluate.

Write the double integral of \(r^4\) over a disk of radius \(2\) as a one-dimensional integral.
For an annulus \(2\leq r\leq5\), what are the radial bounds and the angular interval?
Evaluate the integral of \(1\) over a centered annulus with inner radius \(1\) and outer radius \(3\).
Why does an integral of \(r^{-2}\) over a disk containing the origin require a convergence check?
In the polar area element, what does the factor \(r\) account for?