When a Surface Integral Is Improper
An ordinary scalar surface integral is taken over a patch where the surface and integrand are sufficiently well behaved. Two common situations fall outside that setting: the surface patch may extend without bound, or the scalar function may become unbounded at one or more points of the patch. In either case, the integral is defined by first restricting to bounded pieces where the usual surface integral applies, then taking a limit.
The limit may be finite, in which case the improper integral converges, or infinite or nonexistent, in which case it diverges. The truncation and the limiting process matter: they must match the way the surface or integrand becomes improper.
The factor \(\left\lVert \mathbf{r}_u\times\mathbf{r}_v\right\rVert\) is the surface-area factor from the parametrized surface integral formula. Thus, convergence depends both on the behavior of \(f\) and on how much surface area lies in each truncation. A function that decreases toward zero does not automatically have a finite integral.
Truncating an Unbounded Surface
Suppose an unbounded surface is parametrized over an unbounded domain \(D\). Choose bounded domains \(D_R\) that increase to fill \(D\) as \(R\) increases. For instance, when \(D=\mathbb{R}^2\), one may take the disks \(D_R=\{(u,v):u^2+v^2\le R^2\}\). The improper integral is defined by the limit of the integrals over the corresponding bounded surface pieces, if that limit is finite.
For a nonnegative integrand, the truncated integrals increase as the regions increase. Their limit can be finite or \(+\infty\); a finite value means convergence, while \(+\infty\) means divergence. If the integrand changes sign, a finite limit for one particular family of truncations may depend on the chosen limiting procedure. Absolute convergence avoids this ambiguity: if the integral of \(|f|\) over the surface is finite, the signed integral is well-defined independently of the exhaustion.
Worked Example: A Decaying Function on the Entire Plane
Consider the plane \(z=0\), parametrized by \(\mathbf{r}(x,y)=(x,y,0)\), and the function \(f(x,y,0)=(1+x^2+y^2)^{-2}\). Determine the improper integral over the whole plane.
First compute the surface-area factor. We have \(\mathbf{r}_x=(1,0,0)\) and \(\mathbf{r}_y=(0,1,0)\), so \(\mathbf{r}_x\times\mathbf{r}_y=(0,0,1)\) and its magnitude is \(1\). Truncate the plane to the disk of radius \(R\). In polar coordinates, \(x^2+y^2=r^2\) and \(dS=r\,dr\,d\theta\), so \[ \iint_{S_R} f\,dS = \int_0^{2\pi}\int_0^R \frac{r}{(1+r^2)^2}\,dr\,d\theta. \]
For the radial integral, set \(w=1+r^2\), so \(dw=2r\,dr\). The lower and upper limits for \(w\) are \(1\) and \(1+R^2\). Therefore, \[ \int_0^R \frac{r}{(1+r^2)^2}\,dr = \frac12\int_1^{1+R^2}w^{-2}\,dw = \frac12\left[-w^{-1}\right]_1^{1+R^2} = \frac12\left(1-\frac{1}{1+R^2}\right). \] Integrating over \(\theta\) multiplies this by \(2\pi\), giving \(\pi\left(1-\frac{1}{1+R^2}\right)\). Taking \(R\to\infty\), the term \(1/(1+R^2)\) tends to zero. Thus the integral converges to \(\pi\).
As a check, the truncated expression tends to \(\pi\) from below and increases as \(R\) increases, as expected for a nonnegative integrand on expanding disks.
Power Laws on a Plane
A useful convergence test comes from comparing the integrand to a power of the distance from the origin. On a plane, polar coordinates convert the area element into \(r\,dr\,d\theta\). The factor \(r\) is essential: it records how the area of a circular band grows with its radius.
Here “behaves like” means that, sufficiently far out (or sufficiently near the origin), the integrand is bounded above and below by positive constant multiples of \(r^{-p}\). The criterion follows by comparing its polar-coordinate integral with \[ \int r^{-p}r\,dr=\int r^{1-p}\,dr. \]
For the tail from \(R_0>0\) to infinity, if \(p\ne 2\), integration gives \[ \int_{R_0}^T r^{1-p}\,dr = \frac{T^{2-p}-R_0^{2-p}}{2-p}. \] When \(p>2\), \(T^{2-p}\to 0\), so the limit is finite. When \(p<2\), \(T^{2-p}\to\infty\), so the tail diverges. At \(p=2\), the integral is \(\log T-\log R_0\), which also diverges as \(T\to\infty\). Near the origin, the same antiderivative shows that the limit as the lower endpoint tends to zero is finite exactly when \(2-p>0\), or \(p<2\). At \(p=2\), the logarithm diverges there as well.
Worked Example: A Power Singularity on a Disk
On the unit disk in the plane \(z=0\), let \(f(x,y,0)=(x^2+y^2)^{-1/2}\) away from the origin. The function is unbounded at the origin. Determine whether its surface integral converges and, if so, find its value.
Exclude a disk of radius \(\varepsilon>0\) around the origin. The plane has surface-area factor \(1\). In polar coordinates the truncated integral is \[ \int_0^{2\pi}\int_\varepsilon^1 r^{-1}r\,dr\,d\theta = \int_0^{2\pi}\int_\varepsilon^1 1\,dr\,d\theta. \] The inner integral is \(1-\varepsilon\), so the full truncated integral is \(2\pi(1-\varepsilon)\). Taking \(\varepsilon\to0^+\) gives \(2\pi\), a finite value.
This agrees with the near-origin power test: here \(p=1<2\). The singularity is unbounded, but its contribution over disks shrinking to the origin tends to zero. Indeed, the integral over the excluded disk would be \(2\pi\varepsilon\), which vanishes as \(\varepsilon\to0^+\).
Convergence Depends on Surface Area Growth
The planar power-law threshold is not universal for every surface. A different surface-area factor changes the radial integral and can change the convergence condition. An infinite cylinder, for example, has one unbounded direction rather than two expanding planar directions.
Worked Example: An Improper Integral on an Infinite Cylinder
Take the unit cylinder parametrized by \(\mathbf{r}(\theta,z)=(\cos\theta,\sin\theta,z)\), with \(0\le\theta\le2\pi\) and \(z\in\mathbb{R}\). Let \(f(\theta,z)=(1+z^2)^{-1}\). Compute the integral over the whole cylinder.
The tangent vectors are \(\mathbf{r}_\theta=(-\sin\theta,\cos\theta,0)\) and \(\mathbf{r}_z=(0,0,1)\). Their cross product is \((\cos\theta,\sin\theta,0)\), with magnitude \(1\). Truncate to \(-R\le z\le R\). Then \[ \iint_{S_R}f\,dS = \int_0^{2\pi}\int_{-R}^R\frac{1}{1+z^2}\,dz\,d\theta. \] An antiderivative with respect to \(z\) is \(\arctan z\), so the inner integral is \(\arctan R-\arctan(-R)=2\arctan R\). The full truncated integral is \(4\pi\arctan R\). Since \(\arctan R\to\pi/2\), the improper integral converges to \(2\pi^2\).
As a check, this is \(2\pi\) times the integral along the cylinder’s axis, \(\int_{-\infty}^{\infty}(1+z^2)^{-1}\,dz=\pi\). The surface-area factor is \(1\), so each unit of axial length contributes the same circumference factor \(2\pi\).
More generally, on this cylinder an integrand behaving like \(|z|^{-p}\) for large \(|z|\) has a tail comparable to \(\int_R^\infty z^{-p}\,dz\). That integral converges exactly when \(p>1\). The threshold differs from the planar threshold \(p>2\), because the cylinder’s unbounded area grows in one direction, not across two radial directions.
A Reliable Procedure and a Common Mistake
When setting up an improper scalar surface integral, make the source of impropriety explicit. For an unbounded patch, choose a bounded exhaustion and send its truncation parameter to infinity. For a singular integrand, remove a neighborhood of the singular set and shrink that neighborhood. If both occur, identify both limits and state the order or joint exhaustion being used. Then calculate the bounded surface integrals and test whether the resulting limit is finite.
Determine whether the patch is unbounded, the integrand is unbounded, or both.
Truncate the unbounded directions or exclude neighborhoods of singular points.
Compute the ordinary scalar surface integral on each bounded piece, including the parametrization’s area factor.
Conclude convergence only if the limit exists and is finite.
A common mistake is to check only whether the integrand tends to zero. For example, on the plane \(z=0\), the function \((1+x^2+y^2)^{-1/2}\) tends to zero as \(r\to\infty\). But on a disk of radius \(R\), its integral is \[ \int_0^{2\pi}\int_0^R (1+r^2)^{-1/2}r\,dr\,d\theta = 2\pi\left(\sqrt{1+R^2}-1\right). \] This tends to infinity, not to a finite value. The function decays, but not fast enough to compensate for the growing area of distant annuli.
The correction is to analyze the complete radial integrand, including the area factor. For a planar power \(r^{-p}\), the radial integrand is \(r^{1-p}\), not just \(r^{-p}\). This is why the threshold is \(p>2\), rather than merely \(p>0\).
Check Your Understanding
For each question, identify the truncation or limiting process before deciding whether the integral converges.
- On the plane, an integrand behaves like \(r^{-3/2}\) at infinity. Does its surface integral over the plane have a finite tail? Explain using the radial power test.
- On the unit disk, an integrand behaves like \(r^{-5/2}\) near the origin. Does its integral converge near the singular point?
- Why does an integrand behaving like \(|z|^{-3/2}\) at infinity converge on an infinite cylinder, even though the planar threshold is different?
- What additional condition ensures that an improper integral of a signed function is independent of the chosen exhaustion?
- On the plane, why is it not enough to know that the integrand tends to zero at infinity?