Why Change the Spanning Surface?
A curve in space can bound many different surfaces. If a vector field is given, calculating the flux of its curl over one surface may be easier than calculating it over another. The key question is whether changing the surface changes the answer.
For surfaces with the same oriented boundary, Stokes’ theorem says that the curl flux is the same, provided the field is smooth enough on each surface. A second argument uses the Divergence Theorem on the region between the surfaces. That argument has an additional requirement: the field must be sufficiently smooth throughout that region. A singularity between the surfaces can block this volume argument even when Stokes’ theorem applies separately to both surfaces.
The Surface-Comparison Result
Suppose \(S_1\) and \(S_2\) are oriented, piecewise-smooth surfaces whose induced boundary orientations are both the same oriented curve \(C\). Assume \(\mathbf{F}\) has continuous first partial derivatives on open neighborhoods of both surfaces. Stokes’ theorem applies separately to each surface:
Thus the equality of curl fluxes does not require the field to be defined everywhere between the surfaces. It does require the field to be defined and sufficiently smooth on each surface, including their common boundary.
Why the Bridging-Volume Argument Works
There is another useful way to see the equality when the surfaces form the boundary of a suitable region. Suppose a bounded piecewise-smooth volume \(V\) lies between \(S_1\) and \(S_2\). Choose the boundary orientation so that \(S_1\) is oriented outward from \(V\) and \(S_2\) is oriented inward. The closed boundary is then \(S_1\cup(-S_2)\). If \(\mathbf{F}\) has continuous second partial derivatives on an open neighborhood of \(V\) and its boundary, the Divergence Theorem applies to \(\nabla\times\mathbf{F}\).
The integrand on the right is zero. In coordinates, if \(\mathbf{F}=(P,Q,R)\), then
The last equality follows by equality of mixed partial derivatives, which is guaranteed by the assumed continuity of the second partial derivatives. Therefore the volume integral is zero, and the two curl fluxes are equal. This proof needs smoothness throughout the bridging volume. The direct Stokes argument needs smoothness on the two surfaces, but not throughout the space between them.
Worked Example: A Disk and a Paraboloid Cap
Worked Example: A Disk and a Paraboloid Cap
Let \(\mathbf{F}=(-y/2,x/2,0)\). Compare the upward-oriented unit disk \(S_1\) in the plane \(z=0\) with the upward-oriented paraboloid cap \(S_2\) given by \(z=1-x^2-y^2\), where \(x^2+y^2\leq 1\). Both have the unit circle as their boundary. Compute and check the curl flux through each.
First calculate the curl:
On the unit disk, the upward unit normal is \((0,0,1)\), so the integrand is \(1\). The disk has area \(\pi\), giving \(\iint_{S_1}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS=\pi\).
Parametrize the cap by \(\mathbf{X}(u,v)=(u,v,1-u^2-v^2)\) for \(u^2+v^2\leq1\). Then \(\mathbf{X}_u=(1,0,-2u)\) and \(\mathbf{X}_v=(0,1,-2v)\), so \(\mathbf{X}_u\times\mathbf{X}_v=(2u,2v,1)\), which points upward. The flux is
For a check, parametrize the common boundary counterclockwise by \(\mathbf{r}(t)=(\cos t,\sin t,0)\), \(0\leq t\leq2\pi\). Then \(\mathbf{F}(\mathbf{r}(t))=(-\sin t/2,\cos t/2,0)\) and \(\mathbf{r}'(t)=(-\sin t,\cos t,0)\). Their dot product is \(\sin^2t/2+\cos^2t/2=1/2\), so the circulation is \(\int_0^{2\pi}(1/2)\,dt=\pi\). It agrees with both flux calculations, as Stokes’ theorem requires.
A Singularity Can Block the Volume Proof
A singularity in the region between two surfaces prevents applying the Divergence Theorem there if the field is not defined throughout that region. It does not, by itself, prevent comparing the fluxes: if the field is smooth on each surface, Stokes’ theorem still applies to each one separately.
Worked Example: A Singular Point Between Two Valid Surfaces
Consider \(\mathbf{F}=(0,0,1/[x^2+y^2+(z-1)^2])\), which is singular at \((0,0,1)\). Let \(S_1\) be the upward unit disk \(z=0\), and let \(S_2\) be the upward cap \(z=2(1-x^2-y^2)\) over the unit disk. Both have the unit circle at \(z=0\) as their boundary. Compute both fluxes and determine why the volume argument is unavailable.
Write \(D=x^2+y^2+(z-1)^2\). Away from the singular point, \(\nabla\times\mathbf{F}=(-2y/D^2,2x/D^2,0)\). On \(S_1\), the upward normal is \((0,0,1)\), so the dot product with the curl is zero everywhere on the disk. Hence the flux through \(S_1\) is \(0\).
For \(S_2\), use \(\mathbf{X}(u,v)=(u,v,2(1-u^2-v^2))\). Its upward-oriented area vector is \(\mathbf{X}_u\times\mathbf{X}_v=(4u,4v,1)\). At each point of the cap, the curl-area-vector dot product is \[ \left(-\frac{2v}{D^2},\frac{2u}{D^2},0\right)\cdot(4u,4v,1) = -\frac{8uv}{D^2}+\frac{8uv}{D^2} = 0. \] The cap does not contain \((0,0,1)\): at \(u=v=0\) its height is \(2\), and at every other point its horizontal coordinates are nonzero. Thus the field is smooth on both surfaces, and the flux through \(S_2\) is also \(0\).
The common boundary lies in \(z=0\), where \(dz=0\), so \(\mathbf{F}\cdot d\mathbf{r}=0\) along it. Its circulation is \(0\), checking both results by Stokes’ theorem. However, the point \((0,0,1)\) lies inside the region between the disk and the cap. The field is singular there, so the Divergence Theorem cannot be applied to the curl throughout that region. The surface comparison remains valid because each surface separately avoids the singularity.
Orientation Is Part of the Comparison
Having the same geometric boundary curve is not enough. The induced directions on that curve must agree. Reversing a surface orientation reverses its normal and its induced boundary direction; consequently, it reverses the sign of the curl flux.
Worked Example: Reversing the Cap Orientation
Use the field and surfaces from the disk-and-cap example, but orient the paraboloid cap downward while keeping the disk upward. The upward disk has curl flux \(\pi\). The downward cap has the negative of the upward flux, so its curl flux is \(-\pi\).
The upward disk induces a counterclockwise boundary direction when viewed from above. The downward cap induces a clockwise direction. Its boundary circulation is therefore the negative of the counterclockwise circulation: \(\int_0^{2\pi}(-1/2)\,dt=-\pi\). This agrees with its downward curl flux. The two fluxes, \(\pi\) and \(-\pi\), differ because the surfaces do not have the same oriented boundary. Comparing them as if they did would violate the orientation requirement.
Practical Checks and Common Mistakes
When replacing one spanning surface by another, first check the boundary orientation, then check that the field is smooth on both surfaces. If using the Divergence Theorem as a separate comparison, also check that the field is smooth throughout the volume between them and that the two surfaces form the appropriately oriented boundary of that volume.
Surface independence is useful because it allows a difficult surface integral to be replaced by an easier one without changing its value. That simplification is justified by a shared oriented boundary and Stokes’ theorem, not by the shape or relative position of the surfaces alone.
Check Your Understanding
Use the hypotheses and orientation checks in this tutorial to answer the following questions.
- What smoothness conditions allow Stokes’ theorem to compare the curl fluxes on two surfaces with the same oriented boundary?
- Why does the Divergence Theorem proof require a stronger smoothness condition on the field?
- If the orientation of one spanning surface is reversed, what happens to its induced boundary direction and curl flux?
- In the singular-point example, why is the volume argument unavailable even though the fluxes agree?
- Can a singularity between two surfaces alone establish that their curl fluxes differ? Explain.