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Flux and Circulation Applications · Tutorial 997 of 999

Spotting Invalid Theorem Applications

Practice identifying exactly why a flux or circulation argument fails before choosing a valid surface or region.

Advanced 10 min read

What You'll Learn

  • State the smoothness and domain conditions needed for Stokes’ theorem and the divergence theorem.
  • Distinguish an open surface from a closed surface and a closed curve from a closed surface.
  • Check whether a field is defined and continuously differentiable throughout the required domain.
  • Match the orientation of a surface with the direction of its boundary curve.
  • Diagnose invalid arguments involving a vortex singularity and a point-source field.

Why Theorem Checks Matter

Stokes’ theorem and the divergence theorem can replace one integral with another that is often easier to calculate. But the theorems are conditional: a calculation may be correct while the theorem invoked to justify it does not apply. The most serious errors often come from overlooking a singularity, using an open surface where a closed boundary is required, or pairing incompatible orientations.

A useful habit is to pause before calculating and identify the objects involved: the vector field, the full surface or solid region, its boundary, and the chosen orientation. Then check that the field is sufficiently smooth and defined everywhere the theorem requires—not merely where the final integral is taken.

The key principle: A theorem applies to the entire geometric domain specified in its hypotheses. Smoothness on the boundary alone does not make up for a singularity inside the surface or solid.

What the Theorems Require

For Stokes’ theorem, let \(S\) be an oriented, piecewise-smooth surface with boundary curve or curves \(\partial S\). The vector field \(\mathbf{F}\) must have continuously differentiable components on an open set containing \(S\). The orientation of \(\partial S\) must be induced by the orientation of \(S\): when viewed from the side toward which the chosen normal points, the positive direction follows the right-hand rule.

$$ \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS. $$

For the divergence theorem, let \(E\) be a bounded solid region with a piecewise-smooth, closed boundary \(\partial E\). The field \(\mathbf{F}\) must have continuously differentiable components on an open set containing \(E\), including its boundary. The normal on \(\partial E\) points outward from \(E\).

$$ \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_E \nabla\cdot\mathbf{F}\,dV. $$

The different boundary requirements matter. A surface used in Stokes’ theorem can be open: a disk, for instance, has a boundary circle. By contrast, the divergence theorem uses the entire closed boundary of a solid. A curve being closed does not make a surface closed, and an open surface cannot be substituted for the complete boundary of a volume.

A Quick Justification for the Divergence Theorem

The divergence theorem expresses a general cancellation: flux across internal faces cancels, leaving only flux through the outside boundary. This cancellation can be seen directly for a rectangular box \(E=[a,b]\times[c,d]\times[e,f]\) and a continuously differentiable field \(\mathbf{F}=(P,Q,R)\). Consider first the \(P\) component of the divergence integral:

$$ \iiint_E \frac{\partial P}{\partial x}\,dV = \int_c^d\int_e^f \bigl(P(b,y,z)-P(a,y,z)\bigr)\,dz\,dy. $$

This follows by integrating with respect to \(x\) and applying the fundamental theorem of calculus. The two terms on the right are exactly the outward flux of the \(P\) component through the faces \(x=b\) and \(x=a\). Applying the same calculation to \(Q\) in the \(y\) direction and \(R\) in the \(z\) direction accounts for the other four faces. Adding the three equalities gives total boundary flux equal to the integral of \(\nabla\cdot\mathbf{F}\) over the box. For more general regions the theorem extends this boundary accounting; the smoothness and domain conditions ensure that the derivatives and integrals are valid throughout the region.

Worked Example: A Vortex Singular at the Disk’s Centre

Worked Example: A Vortex Singular at the Disk’s Centre

Consider \(\mathbf{F}(x,y,z)=\left(-\dfrac{y}{x^2+y^2},\dfrac{x}{x^2+y^2},0\right)\) and the unit disk \(S\) in the plane \(z=0\). Its boundary is the unit circle with counterclockwise orientation as viewed from above. A tempting argument says that the curl is zero, so Stokes’ theorem gives zero circulation. Check whether that argument is valid.

First, the field is undefined wherever \(x^2+y^2=0\), in particular at the origin. The origin lies inside the disk, so the field is not continuously differentiable on an open set containing all of \(S\). Stokes’ theorem therefore does not apply to the disk.

The circulation can nevertheless be calculated directly. Parameterize the boundary by \(\mathbf{r}(t)=(\cos t,\sin t,0)\), for \(0\leq t\leq2\pi\). Then \(\mathbf{r}'(t)=(-\sin t,\cos t,0)\) and \(\mathbf{F}(\mathbf{r}(t))=(-\sin t,\cos t,0)\). Hence

$$ \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \bigl((-\sin t)(-\sin t)+(\cos t)(\cos t)\bigr)\,dt = \int_0^{2\pi}1\,dt = 2\pi. $$

For \(x^2+y^2>0\), the \(z\)-component of the curl is

$$ \frac{\partial}{\partial x}\left(\frac{x}{x^2+y^2}\right) - \frac{\partial}{\partial y}\left(\frac{-y}{x^2+y^2}\right) = \frac{y^2-x^2}{(x^2+y^2)^2} - \frac{y^2-x^2}{(x^2+y^2)^2} = 0. $$

This calculation verifies zero curl away from the origin, but not at the origin, where the field is undefined. As a check on the domain issue, remove a disk of radius \(\varepsilon\), where \(0<\varepsilon<1\), and use the resulting annulus. The field is smooth on a neighborhood of that annulus and its curl is zero there. The outer boundary circulation is \(2\pi\). The annulus’s induced inner boundary orientation is clockwise, so its circulation is \(-2\pi\). Their sum is \(2\pi-2\pi=0\), matching the zero curl integral on the annulus. The missing origin is precisely why this valid annulus calculation does not imply zero circulation around the original disk.

Worked Example: The Normal and Boundary Orientations Must Agree

Worked Example: The Normal and Boundary Orientations Must Agree

Let \(\mathbf{F}(x,y,z)=(-y,x,0)\), and let \(S\) be the disk \(x^2+y^2\leq4\) in the plane \(z=0\). Give \(S\) the upward normal. A common mistake is to use the clockwise boundary direction with the surface integral computed using that upward normal.

First calculate the curl. Its components are

$$ \nabla\times\mathbf{F} = \left( \frac{\partial 0}{\partial y}-\frac{\partial x}{\partial z}, \frac{\partial(-y)}{\partial z}-\frac{\partial 0}{\partial x}, \frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y} \right) = (0,0,2). $$

With upward unit normal \(\mathbf{n}=(0,0,1)\), the surface integral is

$$ \iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = \iint_S 2\,dS = 2\cdot\pi(2)^2 = 8\pi. $$

The upward normal induces a counterclockwise boundary direction. Parameterizing that direction by \(\mathbf{r}(t)=(2\cos t,2\sin t,0)\), \(0\leq t\leq2\pi\), gives \(\mathbf{r}'(t)=(-2\sin t,2\cos t,0)\) and \(\mathbf{F}(\mathbf{r}(t))=(-2\sin t,2\cos t,0)\). Thus

$$ \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \bigl(4\sin^2 t+4\cos^2 t\bigr)\,dt = \int_0^{2\pi}4\,dt = 8\pi. $$

The two sides agree, as Stokes’ theorem requires. Reversing the curve direction changes the line integral to \(-8\pi\), while leaving the upward normal and its surface integral unchanged at \(8\pi\). The mismatched calculation is not a failure of the theorem: the curve orientation no longer matches the chosen surface orientation. If the normal is reversed as well, both integrals change sign and agree again.

Worked Example: An Open Hemisphere Is Not a Closed Boundary

Worked Example: An Open Hemisphere Is Not a Closed Boundary

Consider the upper hemisphere of the unit sphere and the constant field \(\mathbf{F}=(0,0,1)\). A mistaken use of the divergence theorem might claim that the flux through the hemisphere equals the volume integral of the divergence. The hemisphere alone is not the closed boundary of a solid: its flat circular base is missing.

On the unit sphere, the outward normal is \(\mathbf{n}=(x,y,z)\), so \(\mathbf{F}\cdot\mathbf{n}=z=\cos\phi\), where \(\phi\) is the angle from the positive \(z\)-axis. The upper hemisphere has \(0\leq\phi\leq\pi/2\), \(0\leq\theta\leq2\pi\), and \(dS=\sin\phi\,d\phi\,d\theta\). Therefore its flux is

$$ \iint_{\text{hemisphere}}\mathbf{F}\cdot\mathbf{n}\,dS = \int_0^{2\pi}\int_0^{\pi/2} \cos\phi\sin\phi\,d\phi\,d\theta = 2\pi\left[\frac{\sin^2\phi}{2}\right]_0^{\pi/2} = \pi. $$

The divergence of this constant field is zero, so its integral over the upper half-ball is \(0\). That does not contradict the divergence theorem: the curved hemisphere is not the full boundary. The missing base disk has outward normal \((0,0,-1)\), and its flux is \(\iint_{\text{disk}}-1\,dA=-\pi\). Including it gives total boundary flux \(\pi-\pi=0\), which agrees with the volume integral.

Worked Example: A Point Source Inside a Sphere

Worked Example: A Point Source Inside a Sphere

Let \(\mathbf{F}(x,y,z)=\dfrac{(x,y,z)}{(x^2+y^2+z^2)^{3/2}}\), and let \(E\) be the ball of radius \(a>0\) centred at the origin. On the sphere, the outward unit normal is \(\mathbf{n}=(x,y,z)/a\), and the field is \(\mathbf{F}=\mathbf{n}/a^2\). Thus \(\mathbf{F}\cdot\mathbf{n}=1/a^2\), and

$$ \iint_{\partial E}\mathbf{F}\cdot\mathbf{n}\,dS = \frac{1}{a^2}\cdot4\pi a^2 = 4\pi. $$

For \(r=\sqrt{x^2+y^2+z^2}>0\), compute the divergence by differentiating each component:

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

It would be invalid to conclude that the sphere’s flux is zero using the divergence theorem on the whole ball. The field is undefined at the origin inside \(E\), so it is not continuously differentiable on a neighborhood of the ball. As a check, apply the theorem instead to the shell \(\varepsilon\leq r\leq a\), where \(0<\varepsilon<a\). The field is smooth there and its divergence is zero. The outer sphere contributes \(4\pi\). The shell’s inner boundary has normal pointing toward the origin, so its flux is \(-4\pi\). The total is \(4\pi-4\pi=0\), matching the divergence integral over the shell. Again, the singularity inside the original solid prevents the tempting application.

A Practical Audit Before Calculating

When an argument invokes Stokes’ theorem or the divergence theorem, check the hypotheses in a fixed order. In particular, do not infer that a field is regular throughout a domain just because it behaves well on the surface where an integral is evaluated.

1
Identify the theorem’s domain.
For Stokes’ theorem, locate the entire surface. For the divergence theorem, identify the solid and its complete boundary.
2
Search for singularities.
Check where the field is undefined or fails to have continuous first partial derivatives. Determine whether any such point lies on or inside the required domain.
3
Check the boundary geometry.
Stokes’ theorem uses the boundary of the chosen surface. The divergence theorem requires the whole closed boundary of the solid.
4
Match orientations.
For Stokes’ theorem, ensure the boundary direction is induced by the selected surface normal. For the divergence theorem, use outward normals.
5
Only then compare integrals.
If a condition fails, the theorem gives no equality for that setup. A coincidental numerical agreement does not repair the missing hypothesis.

Common Mistakes and Why the Checks Matter

A frequent error is to calculate a curl or divergence away from a singular point and then treat that calculation as valid on a region containing the point. Derivatives computed on the field’s domain say nothing about whether the field is defined at an excluded point. The vortex and point-source examples show why such holes can account for nonzero circulation or flux even when curl or divergence vanishes everywhere else in the region.

Another error is to think that every use of an integral theorem requires a closed surface. Stokes’ theorem relates circulation around the boundary of an oriented surface, so the surface may be open. The divergence theorem is different: its surface is the closed boundary of a solid. A hemisphere without its base is not such a boundary.

Orientation is not a cosmetic choice. Reversing the boundary direction reverses a line integral, and reversing the surface normal reverses the corresponding oriented surface integral. In Stokes’ theorem those choices must be linked by the right-hand rule. A reliable application checks geometry, domain, smoothness, and orientation before doing the calculation.

Check Your Understanding

For each question, identify the theorem condition that needs to be checked before trusting the proposed calculation.

  1. A vector field is smooth on a circle but undefined at the centre of the disk it bounds. Can Stokes’ theorem be applied to the disk? Why or why not?
  2. What boundary direction is induced by an upward normal on a horizontal disk?
  3. Why can the divergence theorem not be applied to the curved surface of a hemisphere alone?
  4. A field has zero divergence everywhere except at a point inside a proposed solid. What must be checked before using the divergence theorem on that solid?
  5. If the normal on a Stokes surface is reversed, what must happen to the induced boundary orientation?

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