Radial Integrals Reduce to One Dimension
A function on the plane is radial when its value depends only on distance from the origin. For a radial function, integrating over a disk or a concentric annulus removes the angular dependence: the two-dimensional integral becomes a one-dimensional integral with a factor of \(r\). That factor is essential. It records how the area of a thin ring grows with its radius.
Throughout, \(m\) denotes Lebesgue measure on \(\mathbb{R}^2\). The space \((\mathbb{R}^2,\mathcal{L},m)\) is complete and sigma-finite, but not finite. Write \(D_R=\{x\in\mathbb{R}^2:|x|\le R\}\) for \(R\ge0\), and \(A_{a,b}=\{x\in\mathbb{R}^2:a\le |x|\le b\}\) for \(0\le a\le b<\infty\). The same integral formulas hold if either or both bounding circles are omitted.
We use the polar-coordinate integral formula established in the earlier material on polar coordinates. In the radial case it says that, for a Borel measurable \(g:[0,\infty)\to[0,\infty]\), integration over a disk or annulus gives the following formulas. The functions in these formulas are measurable on the Lebesgue measure space above; the values \(+\infty\) are allowed for nonnegative integrands.
Proof. In polar coordinates, the area element is \(r\,dr\,d\theta\), and the radial function \(g(|x|)\) becomes \(g(r)\), independent of \(\theta\). On a disk, the angular variable ranges from \(0\) to \(2\pi\), and the radial variable ranges from \(0\) to \(R\). The polar-coordinate formula therefore gives $$ \int_{D_R}g(|x|)\,dm(x) =\int_0^{2\pi}\int_0^R g(r)r\,dr\,d\theta =2\pi\int_0^R g(r)r\,dr. $$ For an annulus, the radial interval is \([a,b]\), so the same formula gives \(2\pi\int_a^b g(r)r\,dr\). These applications are valid for nonnegative measurable functions by Tonelli's theorem; the iterated integral may be infinite. For real-valued \(g\), apply the nonnegative formulas to its positive and negative parts when \(\int |g(|x|)|\,dm(x)<\infty\). Their integrals are then finite, so subtraction is defined and gives the stated formula. Finally, circles have area zero: the formula applied to their radial indicators gives an integral over a one-dimensional singleton, which is zero. Thus including or omitting bounding circles does not change these integrals. If \(a=b\), both the annulus and the one-dimensional interval have measure zero, as required. \(\square\)
In particular, for a real-valued Borel function \(g\), radial integrability on a disk is characterized exactly by $$ \int_{D_R}|g(|x|)|\,dm(x)=2\pi\int_0^R|g(r)|r\,dr<\infty. $$ On an annulus, replace the interval \([0,R]\) by \([a,b]\). This test is useful near the origin, where \(g\) may be unbounded: the relevant quantity is not just \(|g(r)|\), but \(r|g(r)|\).
Area and Polynomial Radial Weights
Worked Example: The Area of a Disk
Compute the area of the disk \(D_3\). Its indicator is the radial function \(g(r)=1\) for \(0\le r\le3\). The radial formula gives $$ m(D_3)=2\pi\int_0^3 r\,dr =2\pi\left[\frac{r^2}{2}\right]_0^3 =2\pi\left(\frac{9}{2}-0\right) =9\pi. $$ Check: the geometric area formula gives \(\pi\cdot3^2=9\pi\) as well. The agreement checks the radial factor and the disk radius.
Worked Example: A Polynomial Weight on an Annulus
Compute \(\displaystyle\int_{A_{1,3}}(|x|^2+2)\,dm(x)\). Here \(g(r)=r^2+2\), a nonnegative continuous radial weight, and the annulus has inner radius \(1\) and outer radius \(3\). Thus $$ \begin{aligned} \int_{A_{1,3}}(|x|^2+2)\,dm(x) &=2\pi\int_1^3(r^2+2)r\,dr\\ &=2\pi\left[\frac{r^4}{4}+r^2\right]_1^3\\ &=2\pi\left[\left(\frac{81}{4}+9\right)-\left(\frac14+1\right)\right]\\ &=2\pi\left(\frac{117}{4}-\frac54\right) =2\pi\cdot28 =56\pi. \end{aligned} $$ The antiderivative is verified by differentiating: \(\frac{d}{dr}(r^4/4+r^2)=r^3+2r=(r^2+2)r\). Check: the endpoint difference is \(112/4=28\), so multiplying by the angular factor \(2\pi\) gives \(56\pi\), as above.
More generally, for a real exponent \(p>-2\) and \(0\le a<b<\infty\), the power weight \(g(r)=r^p\) has annular integral $$ \int_{A_{a,b}}|x|^p\,dm(x) =2\pi\int_a^b r^{p+1}\,dr =\frac{2\pi}{p+2}\left(b^{p+2}-a^{p+2}\right). $$ When \(a=0\), the condition \(p>-2\) ensures the integral at the origin is finite. For \(p\le-2\), the integral over a disk of positive radius diverges, as the next example verifies in a particular case.
Exponential Weights and a Singularity at the Origin
Worked Example: An Exponential Weight on an Annulus
Let \(g(r)=e^{-2r^2}\), and compute its integral over \(A_{1,2}\). The radial formula and the substitution \(u=-2r^2\), so \(du=-4r\,dr\), give $$ \begin{aligned} \int_{A_{1,2}}e^{-2|x|^2}\,dm(x) &=2\pi\int_1^2e^{-2r^2}r\,dr\\ &=-\frac{\pi}{2}\left[e^{-2r^2}\right]_1^2\\ &=-\frac{\pi}{2}(e^{-8}-e^{-2})\\ &=\frac{\pi}{2}(e^{-2}-e^{-8}). \end{aligned} $$ Check: differentiating \(-\frac14e^{-2r^2}\) gives \(r e^{-2r^2}\), the integrand after taking out \(2\pi\). Also \(e^{-2}>e^{-8}\), so the final value is positive, as an integral of a positive function must be.
Worked Example: A Singular Power on a Disk
Consider \(f(x)=|x|^{-1}\) for \(x\ne0\) on \(D_1\), assigning any value, such as \(0\), at the origin. This is a nonnegative Borel function. The value at the single point \(0\) does not affect its Lebesgue integral. The radial formula gives $$ \int_{D_1}f\,dm =2\pi\int_0^1r^{-1}r\,dr =2\pi\int_0^1 1\,dr =2\pi. $$ Thus the singularity is integrable. Check: for \(q=1\), the power formula gives \(2\pi(1^{2-1}-0^{2-1})/(2-1)=2\pi\), the same value.
By contrast, for \(f(x)=|x|^{-2}\) away from the origin, the radial integral on \(D_1\) is \(2\pi\int_0^1r^{-1}\,dr=+\infty\). Indeed, for \(0<\varepsilon<1\), the integral from \(\varepsilon\) to \(1\) is \(-2\pi\ln\varepsilon\), which tends to \(+\infty\) as \(\varepsilon\) decreases to \(0\). This contrasts the finite \(q=1\) case with the logarithmic divergence at \(q=2\).
Proof. The value at \(0\) has no effect because a singleton has Lebesgue measure zero. On \(0<r\le R\), the radial integrand is \(r^{1-q}\). If \(q<2\), then \(2-q>0\), and for \(0<\varepsilon<R\), $$ 2\pi\int_\varepsilon^Rr^{1-q}\,dr =\frac{2\pi}{2-q}\left(R^{2-q}-\varepsilon^{2-q}\right). $$ As \(\varepsilon\) decreases to \(0\), \(\varepsilon^{2-q}\) tends to \(0\). Monotone Convergence applied to the increasing regions \(\{x:\varepsilon<|x|\le R\}\), or directly to the nonnegative radial integral, gives the claimed finite value. If \(q=2\), the truncated integral is \(2\pi\ln(R/\varepsilon)\), which tends to \(+\infty\) as \(\varepsilon\) decreases to \(0\). If \(q>2\), the same truncated integral equals \(\frac{2\pi}{q-2}(\varepsilon^{2-q}-R^{2-q})\), which tends to \(+\infty\) because \(2-q<0\). Thus the integral is infinite for every \(q\ge2\), proving both directions. \(\square\)
A Common Jacobian Error
A frequent mistake is to integrate \(g(r)\,dr\,d\theta\) instead of \(g(r)r\,dr\,d\theta\). For the unit weight on \(D_3\), that incorrect calculation would give \(2\pi\int_0^3 1\,dr=6\pi\), rather than the correct \(2\pi\int_0^3r\,dr=9\pi\). It fails even for the area of a disk. The factor \(r\) is the Jacobian in polar coordinates; it cannot be dropped when the integrand is radial. A second useful check is dimensional: the area of a disk scales as the square of its radius, whereas the incorrect answer \(2\pi R\) scales only linearly.
For disk and annulus problems, the practical procedure is therefore to identify the radial weight \(g\), set the correct inner and outer radii, include \(r\), and then evaluate the one-dimensional integral. The integrability test \( \int |g(r)|r\,dr<\infty \) should be checked whenever the weight is unbounded or the domain reaches the origin.
Check Your Understanding
For each problem, keep the radial Jacobian factor and check whether the domain includes the origin.
- Use the radial formula to compute the area of \(D_4\).
- Compute \(\displaystyle\int_{A_{2,5}}|x|^2\,dm(x)\).
- Evaluate \(\displaystyle\int_{A_{0,1}}e^{-|x|^2}\,dm(x)\).
- For which real values of \(p\) is \(|x|^p\) integrable on \(D_1\)? Explain the role of the radial factor.
- Identify the missing factor in the incorrect calculation \(m(D_R)=2\pi\int_0^R1\,dr\), and give the correct formula.