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The Divergence Theorem · Tutorial 968 of 999

Singularities Inside a Closed Surface

Excise an interior singularity, track the inner boundary’s reversed normal, and use its limiting flux to recover the flux through the original closed surface.

Advanced 10 min read

What You'll Learn

  • Identify why a field singular inside a solid does not meet the Divergence Theorem’s hypotheses there.
  • Set up a punctured solid by removing a small ball around the singularity.
  • Relate the original outer flux to the divergence integral and the small sphere’s outward flux.
  • Track the inner boundary’s normal direction and avoid the associated sign error.
  • Distinguish a singularity with nonzero limiting flux from one whose small-sphere flux tends to zero.

Why an Interior Singularity Changes the Flux Calculation

A closed surface may surround a point where a vector field is undefined or unbounded. In that case, the Divergence Theorem cannot be applied directly to the entire solid: one of its requirements is that the field be continuously differentiable throughout the solid and on its boundary. A singularity inside the solid violates that condition, even if the field is smooth at every point of the surface itself.

The remedy is to remove a small ball around the singularity. The resulting punctured solid no longer contains the singular point, so the theorem can be applied there. Its boundary has two pieces: the original outer surface and the surface of the removed ball. The crucial detail is that the outward normal for the punctured solid points into the removed ball along that inner boundary.

Core idea: Excising a singularity creates an inner boundary. Its normal points opposite to the usual outward normal of the small sphere, so the original outer flux equals the divergence integral over the punctured solid plus the small sphere’s flux with its usual outward normal.

The Excision Formula

Let \(E\) be a solid with a piecewise smooth, closed boundary \(S\), oriented with the outward unit normal \(\mathbf{n}\). Suppose a vector field \(\mathbf{F}\) has an isolated singularity at an interior point \(\mathbf{a}\). Choose \(\varepsilon>0\) small enough that the closed ball of radius \(\varepsilon\) centered at \(\mathbf{a}\) lies entirely inside \(E\). Write \(B_\varepsilon(\mathbf{a})\) for that ball and \(\Sigma_\varepsilon=\partial B_\varepsilon(\mathbf{a})\) for its spherical boundary.

Assume \(\mathbf{F}\) is continuously differentiable on the punctured solid and its boundary, where the punctured solid is \(E_\varepsilon=E\setminus B_\varepsilon(\mathbf{a})\). Let \(\mathbf{u}\) be the usual outward unit normal on \(\Sigma_\varepsilon\), pointing away from the center \(\mathbf{a}\). The outward normal of \(E_\varepsilon\), however, is \(-\mathbf{u}\) on this inner boundary.

Excision formula: Under the conditions above, the Divergence Theorem on the punctured solid gives
$$ \int_S \mathbf{F}\cdot\mathbf{n}\,dS = \iiint_{E_\varepsilon}\nabla\cdot\mathbf{F}\,dV + \int_{\Sigma_\varepsilon}\mathbf{F}\cdot\mathbf{u}\,dS. $$
The last integral uses the small sphere’s usual outward normal, not the outward normal of the punctured solid.

To see the sign, the boundary of \(E_\varepsilon\) consists of \(S\) and \(\Sigma_\varepsilon\). Its total outward flux is therefore \(\int_S\mathbf{F}\cdot\mathbf{n}\,dS-\int_{\Sigma_\varepsilon}\mathbf{F}\cdot\mathbf{u}\,dS\). The Divergence Theorem equates this to the integral of the divergence over \(E_\varepsilon\). Rearranging gives the formula.

The equality holds for each admissible positive \(\varepsilon\). If the two terms on the right have limits as \(\varepsilon\) tends to zero, then the fixed flux through \(S\) is the sum of those limits. In particular, a nonzero limiting flux through the shrinking sphere contributes to the outer flux even when the divergence is zero everywhere away from the singularity.

Worked Example: The Inverse-Square Field on a Sphere

Worked Example: The Inverse-Square Field on a Sphere

Let \(E\) be the ball of radius \(R>0\) centered at the origin, and consider \(\mathbf{F}(\mathbf{x})=\mathbf{x}/\|\mathbf{x}\|^3\), where \(\mathbf{x}=(x,y,z)\). The field is singular at the origin, so the Divergence Theorem cannot be applied directly to \(E\). Find the outward flux through its boundary \(S\) using excision.

Remove the ball of radius \(\varepsilon\), where \(0<\varepsilon<R\). Away from the origin, compute the divergence component by component. Since \(\|\mathbf{x}\|^2=x^2+y^2+z^2\), the components of the field are \(x/\|\mathbf{x}\|^3\), \(y/\|\mathbf{x}\|^3\), and \(z/\|\mathbf{x}\|^3\). Thus

$$ \nabla\cdot\mathbf{F} = \frac{1}{\|\mathbf{x}\|^3} -\frac{3x^2}{\|\mathbf{x}\|^5} +\frac{1}{\|\mathbf{x}\|^3} -\frac{3y^2}{\|\mathbf{x}\|^5} +\frac{1}{\|\mathbf{x}\|^3} -\frac{3z^2}{\|\mathbf{x}\|^5} = \frac{3}{\|\mathbf{x}\|^3} -\frac{3(x^2+y^2+z^2)}{\|\mathbf{x}\|^5} =0. $$

The divergence integral over the punctured ball is therefore zero. On the small sphere, \(\mathbf{u}=\mathbf{x}/\varepsilon\) and \(\|\mathbf{x}\|=\varepsilon\), so \(\mathbf{F}=\mathbf{u}/\varepsilon^2\). Hence \(\mathbf{F}\cdot\mathbf{u}=1/\varepsilon^2\), and the sphere has area \(4\pi\varepsilon^2\). Its outward flux is

$$ \int_{\Sigma_\varepsilon}\mathbf{F}\cdot\mathbf{u}\,dS = \frac{1}{\varepsilon^2}(4\pi\varepsilon^2) = 4\pi. $$

The excision formula now gives \(\int_S\mathbf{F}\cdot\mathbf{n}\,dS=0+4\pi=4\pi\). Check directly on the outer sphere: its radius is \(R\), so \(\mathbf{F}=\mathbf{n}/R^2\); therefore the flux is \((1/R^2)(4\pi R^2)=4\pi\), as required.

Worked Example: A Shifted Singularity Inside a Cube

Worked Example: A Shifted Singularity Inside a Cube

Let \(E=[-2,2]^3\), and let \(\mathbf{a}=(1/2,0,0)\). For \(\mathbf{x}\ne\mathbf{a}\), define \(\mathbf{F}(\mathbf{x})=(\mathbf{x}-\mathbf{a})/\|\mathbf{x}-\mathbf{a}\|^3\). Find the outward flux through the cube’s boundary \(S\).

The singular point lies inside the cube. Excise the ball \(B_\varepsilon(\mathbf{a})\) with \(\varepsilon=1/10\). This ball lies inside the cube: the distances from \(\mathbf{a}\) to the nearest faces are \(3/2\) in the \(x\)-direction and \(2\) in each of the other coordinate directions, all greater than \(1/10\). The same component calculation as in the previous example, with \(\mathbf{x}\) replaced by \(\mathbf{x}-\mathbf{a}\), shows that \(\nabla\cdot\mathbf{F}=0\) away from \(\mathbf{a}\). Thus the divergence integral over the punctured cube is zero.

On the small sphere, let \(\mathbf{u}=(\mathbf{x}-\mathbf{a})/\varepsilon\). Then \(\mathbf{F}=\mathbf{u}/\varepsilon^2\), so its outward flux is \((1/\varepsilon^2)(4\pi\varepsilon^2)=4\pi\). The excision formula yields

$$ \int_S\mathbf{F}\cdot\mathbf{n}\,dS = 0+4\pi = 4\pi. $$

As a sign check, the punctured cube’s flux through its inner boundary is \(-4\pi\), because that boundary’s outward normal is \(-\mathbf{u}\). Its total boundary flux is \(4\pi-4\pi=0\), which agrees with the zero divergence integral. This verifies the result without requiring a separate surface integral over the cube’s six faces.

Worked Example: A Singularity with Vanishing Small-Sphere Flux

Worked Example: A Singularity with Vanishing Small-Sphere Flux

On the ball of radius \(R>0\) centered at the origin, consider \(\mathbf{G}(\mathbf{x})=\mathbf{x}/\|\mathbf{x}\|^2\) for \(\mathbf{x}\ne\mathbf{0}\). This field is still singular at the origin, but its small-sphere flux behaves differently. Find its flux through the outer sphere.

Writing \(r=\|\mathbf{x}\|\), its components are \(x/r^2,y/r^2,z/r^2\). Differentiating gives

$$ \nabla\cdot\mathbf{G} = \left(\frac{1}{r^2}-\frac{2x^2}{r^4}\right) + \left(\frac{1}{r^2}-\frac{2y^2}{r^4}\right) + \left(\frac{1}{r^2}-\frac{2z^2}{r^4}\right) = \frac{1}{r^2}. $$

Remove a ball of radius \(\varepsilon\), with \(0<\varepsilon<R\). In spherical coordinates the volume element is \(r^2\sin\theta\,dr\,d\theta\,d\phi\). The divergence integral over the punctured ball is

$$ \int_0^{2\pi}\int_0^\pi\int_\varepsilon^R \frac{1}{r^2}r^2\sin\theta\,dr\,d\theta\,d\phi = (R-\varepsilon) \left(\int_0^\pi\sin\theta\,d\theta\right) \left(\int_0^{2\pi}d\phi\right) = 4\pi(R-\varepsilon). $$

On the small sphere, \(\mathbf{G}=\mathbf{u}/\varepsilon\), so its outward flux is \((1/\varepsilon)(4\pi\varepsilon^2)=4\pi\varepsilon\). Consequently, the outer flux is \(4\pi(R-\varepsilon)+4\pi\varepsilon=4\pi R\), independent of \(\varepsilon\). Directly, on the sphere of radius \(R\), \(\mathbf{G}=\mathbf{n}/R\); its flux is \((1/R)(4\pi R^2)=4\pi R\), confirming the calculation. As \(\varepsilon\) tends to zero, the small-sphere flux tends to zero, unlike the inverse-square field’s flux.

The Inner Normal and a Common Sign Error

The two normals associated with the small sphere serve different purposes. The vector \(\mathbf{u}\) points outward from the removed ball. The outward normal of the remaining punctured solid points into the hole, so it equals \(-\mathbf{u}\). Keeping those roles distinct is essential when applying the theorem.

Common mistake: For the inverse-square field, one might incorrectly write “outer flux plus small-sphere outward flux equals zero” because the divergence vanishes. That gives outer flux \(=-4\pi\). But the theorem uses the punctured solid’s inner normal, which is \(-\mathbf{u}\). Its inner-boundary flux is \(-4\pi\), so \( \text{outer flux}-4\pi=0\), and the outer flux is \(+4\pi\).

Excision repairs the domain of application; it does not make the original field smooth at the singularity. For each chosen \(\varepsilon\), check that the removed ball lies entirely inside the solid and that the field is continuously differentiable on the punctured region and its boundary. The flux through the small sphere may then be calculated and compared as \(\varepsilon\) shrinks.

The limiting formula is useful when both the divergence integral over the punctured solid and the small-sphere flux have limits. A singular field can contribute a finite, nonzero amount through arbitrarily small spheres, as the inverse-square example shows. Another singular field can have small-sphere flux tending to zero, as in the third example. The singularity alone does not determine that limit; the field’s behavior near it must be calculated.

Check Your Understanding

Use the excision formula and the orientation of the inner boundary to answer these questions.

  1. Why can the Divergence Theorem not be applied directly to a solid containing a point where the field is singular?
  2. How does the outward normal of a punctured solid’s inner boundary relate to the usual outward normal of the removed ball?
  3. For \(\mathbf{F}(\mathbf{x})=\mathbf{x}/\|\mathbf{x}\|^3\), what is the outward flux through a sphere of radius \(\varepsilon\) centered at the origin?
  4. If the divergence is zero away from the singularity and the small-sphere outward flux is \(C\), what does the excision formula give for the outer flux?
  5. For \(\mathbf{G}(\mathbf{x})=\mathbf{x}/\|\mathbf{x}\|^2\), what happens to its small-sphere flux as the radius tends to zero?

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