Tutorials › Calculus III › Replacing a Difficult Surface with an Easier One

Home  /  Tutorials  /  Stokes’ Theorem

Stokes’ Theorem · Tutorial 935 of 999

Replacing a Difficult Surface with an Easier One

Use Stokes’ Theorem to justify replacing a difficult spanning surface with an easier one, provided the oriented boundary is unchanged and the field is smooth throughout the region between them.

Advanced 12 min read

What You'll Learn

  • State the conditions under which two spanning surfaces have equal curl flux
  • Track the boundary orientations required for a valid surface replacement
  • Justify the replacement using Stokes’ Theorem and cancellation on a closed surface
  • Compute and check curl flux on a hemisphere, a graph, and a warped surface
  • Recognize why different geometric boundaries invalidate the replacement argument

Why Replace a Surface?

A circulation integral around a curve can be related by Stokes’ Theorem to the flux of a curl through any suitable oriented surface spanning that curve. Some spanning surfaces make the flux integral much easier to calculate than others.

The important point is that the choice is not arbitrary. The surfaces must have the same geometric boundary with the same orientation, and the vector field must be smooth throughout the region between them. Under those conditions, the curl flux is independent of which spanning surface is used.

Central idea: If two suitably oriented surfaces have the same oriented boundary and the vector field is continuously differentiable throughout the region between them, their fluxes of the curl are equal.

The Surface-Replacement Theorem

Theorem: Let \(S_1\) and \(S_2\) be oriented, piecewise-smooth surfaces with the same oriented boundary curve \(C\). Suppose a vector field \(\mathbf{F}\) has continuous first partial derivatives on an open region containing both surfaces and the region between them. Then \[ \iint_{S_1}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = \iint_{S_2}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS, \] where each surface orientation induces the specified direction on \(C\).

Here, “the same oriented boundary” requires more than having boundary curves of the same shape. The curves must occupy the same points in space, and the orientations induced by the two surfaces must agree. The smoothness assumption ensures that Stokes’ Theorem applies to the surfaces and that the field has no failure of differentiability in the intervening region.

Why the Fluxes Are Equal

Stokes’ Theorem states that for an oriented, piecewise-smooth surface \(S\) with positively oriented boundary \(\partial S\), and a vector field with continuous first partial derivatives on a neighborhood of \(S\), \[ \iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = \oint_{\partial S}\mathbf{F}\cdot d\mathbf{r}. \] Apply this result separately to \(S_1\) and \(S_2\). Since each has the same oriented boundary \(C\), both curl-flux integrals equal the same circulation.

$$ \iint_{S_1}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = \oint_C\mathbf{F}\cdot d\mathbf{r} = \iint_{S_2}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS. $$

This is the complete justification for replacing one surface by the other. The surfaces need not have the same shape or area. What matters is the common oriented boundary and the hypotheses that allow Stokes’ Theorem to be used.

There is also a useful closed-surface interpretation. Give \(S_1\) its original orientation and reverse the orientation of \(S_2\). The combined surface has no boundary: the two copies of \(C\) run in opposite directions and cancel. Its curl flux is the flux through \(S_1\) minus the flux through \(S_2\). Stokes’ Theorem on a closed surface says that this total is zero, giving the same equality.

$$ \partial(S_1\cup(-S_2))=C-C=0, \qquad \iint_{S_1\cup(-S_2)}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS=0. $$

Worked Example: Replacing a Hemisphere with a Disk

Worked Example: Replacing a Hemisphere with a Disk

Let \(S_1\) be the upper hemisphere \(x^2+y^2+z^2=4\), oriented outward, and let \(S_2\) be the disk \(x^2+y^2\leq 4\) in the plane \(z=0\), oriented upward. Their common boundary is the circle \(x^2+y^2=4,\ z=0\), oriented counterclockwise as viewed from above. For \(\mathbf{F}=(-y/2,x/2,0)\), compute the curl flux through both surfaces.

First calculate the curl:

$$ \nabla\times\mathbf{F} = \left( \frac{\partial 0}{\partial y}-\frac{\partial (x/2)}{\partial z}, \frac{\partial (-y/2)}{\partial z}-\frac{\partial 0}{\partial x}, \frac{\partial (x/2)}{\partial x}-\frac{\partial (-y/2)}{\partial y} \right) =(0,0,1). $$

For the disk, the upward unit normal is \((0,0,1)\), so the integrand is \(1\). The disk has radius \(2\) and area \(4\pi\). Therefore,

$$ \iint_{S_2}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = \iint_{S_2}1\,dS = 4\pi. $$

For the hemisphere, use the spherical parametrization \(\mathbf{r}(\phi,\theta)=(2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi)\), where \(0\leq\phi\leq\pi/2\) and \(0\leq\theta\leq2\pi\). The outward vector area element is \[ \mathbf{r}_{\phi}\times\mathbf{r}_{\theta} = (4\sin^2\phi\cos\theta,4\sin^2\phi\sin\theta,4\sin\phi\cos\phi). \] Its dot product with the curl \((0,0,1)\) is \(4\sin\phi\cos\phi\). Hence

$$ \begin{aligned} \iint_{S_1}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS &=\int_0^{2\pi}\int_0^{\pi/2}4\sin\phi\cos\phi\,d\phi\,d\theta\\ &=\int_0^{2\pi}2\,d\theta\\ &=4\pi. \end{aligned} $$

The two fluxes agree. As a check using Stokes’ Theorem, on the common circle write \(x=2\cos t,\ y=2\sin t\), for \(0\leq t\leq2\pi\). Then \(dx=-2\sin t\,dt\), \(dy=2\cos t\,dt\), and \[ \mathbf{F}\cdot d\mathbf{r} =-\frac{y}{2}\,dx+\frac{x}{2}\,dy =2\sin^2t\,dt+2\cos^2t\,dt =2\,dt. \] Thus \(\oint_C\mathbf{F}\cdot d\mathbf{r}=\int_0^{2\pi}2\,dt=4\pi\), matching both flux calculations.

Worked Example: A Paraboloid and Its Boundary Disk

Worked Example: A Paraboloid and Its Boundary Disk

Let \(S_1\) be the graph \(z=1-x^2-y^2\) over \(x^2+y^2\leq1\), oriented upward. Let \(S_2\) be the unit disk \(z=0\), also oriented upward. The graph meets the plane \(z=0\) when \(x^2+y^2=1\), so the two surfaces have exactly the same boundary circle, with the same counterclockwise orientation. Take \(\mathbf{F}=(-y/2,x/2,xy)\).

Its curl is

$$ \nabla\times\mathbf{F} = \left( \frac{\partial(xy)}{\partial y}-\frac{\partial(x/2)}{\partial z}, \frac{\partial(-y/2)}{\partial z}-\frac{\partial(xy)}{\partial x}, \frac{\partial(x/2)}{\partial x}-\frac{\partial(-y/2)}{\partial y} \right) =(x,-y,1). $$

For the graph \(z=f(x,y)=1-x^2-y^2\), an upward vector area element is \((-f_x,-f_y,1)\,dx\,dy=(2x,2y,1)\,dx\,dy\). Therefore the curl-flux integrand is \[ (x,-y,1)\cdot(2x,2y,1)=2x^2-2y^2+1. \]

$$ \begin{aligned} \iint_{S_1}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS &=\iint_{x^2+y^2\leq1}(2x^2-2y^2+1)\,dx\,dy\\ &=2\left(\frac{\pi}{4}\right)-2\left(\frac{\pi}{4}\right)+\pi\\ &=\pi. \end{aligned} $$

In the second line, symmetry gives \(\iint x^2\,dA=\iint y^2\,dA\), and each integral is \(\pi/4\). On the flat disk, the upward normal is \((0,0,1)\), so the integrand is the third curl component, \(1\). Its area is \(\pi\), giving \[ \iint_{S_2}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS =\iint_{x^2+y^2\leq1}1\,dA=\pi. \] Both surfaces therefore give flux \(\pi\). As a boundary check, on \(z=0\) the field’s third component contributes nothing to the line integral because \(dz=0\). The first two components give the circulation \(\pi\) around the unit circle, consistent with Stokes’ Theorem.

Worked Example: A Warped Surface Above a Square

Worked Example: A Warped Surface Above a Square

Consider the square \(0\leq x\leq1,\ 0\leq z\leq1\) in the plane \(y=0\), and the warped surface \[ \mathbf{r}(x,z)=(x,g(x,z),z), \qquad g(x,z)=x(1-x)z(1-z). \] Orient both surfaces so that the flat square has normal \((0,-1,0)\). Since \(g=0\) on every edge of the square, the warped surface and the flat square have the same geometric boundary. Use \(\mathbf{F}=(0,0,x)\).

The curl is \(\nabla\times\mathbf{F}=(0,-1,0)\). For the flat square, the dot product of the curl with the chosen normal is \(1\). Its area is \(1\), so its curl flux is \(1\).

For the warped surface, the tangent vectors and their oriented cross product are \[ \mathbf{r}_x=(1,g_x,0),\qquad \mathbf{r}_z=(0,g_z,1),\qquad \mathbf{r}_x\times\mathbf{r}_z=(g_x,-1,g_z). \] The curl dotted with this vector area element is \((0,-1,0)\cdot(g_x,-1,g_z)=1\). Thus the flux is

$$ \iint_{S}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = \int_0^1\int_0^1 1\,dz\,dx = \left(\int_0^1dx\right)\left(\int_0^1dz\right) =1. $$

The warped surface and the flat square both have curl flux \(1\). The field is polynomial, so its first partial derivatives are continuous throughout the region between the surfaces. The smoothness requirement is satisfied.

Common Pitfall: Similar-Looking Boundaries Are Not Enough

A surface-replacement argument fails if the two surfaces do not have the same boundary curve in space. For instance, consider the graph \(z=x(2-x)\) over \(0\leq x\leq2,\ 0\leq y\leq1\), and compare it with the rectangle in the plane \(z=0\) over that same range. The graph meets \(z=0\) at \(x=0\) and \(x=2\), but its edges over \(y=0\) and \(y=1\) are elevated. At the point \((1,0)\), the graph has height \(z=1\), while the rectangle’s corresponding point is \((1,0,0)\). Their boundaries are different curves, so the surface-replacement theorem does not apply.

The correction is to check the actual boundary points, not merely the shape of the parameter domain or the surfaces’ projections. If the boundaries differ, one cannot infer equality of curl flux from this theorem. Separate calculations might happen to give equal values, but that equality would need its own justification.

Before replacing a surface, check: the geometric boundary is identical, the induced boundary directions agree, the surfaces are admissible for Stokes’ Theorem, and the field is continuously differentiable on a region containing the surfaces and the space between them.

Why Surface Replacement Matters

A complicated surface may require an awkward parametrization, while another surface with the same oriented boundary may have a simple normal or a simpler parameter domain. The theorem lets you select the convenient surface without changing the curl flux. This is useful only after the hypotheses have been checked; convenience does not compensate for a mismatched boundary or a field that lacks the required smoothness.

The theorem concerns the flux of \(\nabla\times\mathbf{F}\), not the flux of an arbitrary vector field. Its reason is Stokes’ Theorem: the curl flux through each spanning surface equals the circulation of \(\mathbf{F}\) around that surface’s oriented boundary. The shared boundary makes those circulations identical.

Check Your Understanding

Use the boundary, orientation, and smoothness conditions to decide whether a surface replacement is justified.

  1. Two oriented spanning surfaces meet along the same circle, but induce opposite directions around it. Can their curl fluxes be set equal as stated?
  2. Why does applying Stokes’ Theorem to both surfaces prove that their curl fluxes agree?
  3. A graph meets a plane along its boundary circle. What must be checked before replacing the graph by the planar disk?
  4. In the warped-square example, why is the vector area element \((g_x,-1,g_z)\,dx\,dz\) consistent with the chosen orientation?
  5. Why does the surface-replacement theorem not apply when two surfaces have boundary curves that only look alike in projection?

Previous

Next

Back to Tutorials