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Stokes’ Theorem · Tutorial 937 of 999

Stokes' Theorem on Surfaces with Multiple Boundary Curves

Determine every induced boundary direction from the surface orientation, then use the correctly oriented curves in Stokes’ Theorem.

Advanced 10 min read

What You'll Learn

  • State Stokes’ Theorem for compact oriented surfaces with multiple boundary curves
  • Use the outward conormal to determine each induced boundary direction
  • Explain why a hole’s boundary direction is reversed relative to the outer boundary
  • Orient the two boundary curves of a cylindrical surface
  • Check boundary orientations by comparing circulation with curl flux

Why Multiple Boundary Curves Need Care

A surface can have more than one boundary curve. An annulus, for example, has an outer rim and an inner rim around its hole. Stokes’ Theorem uses the circulation along all boundary curves, but each curve must be traversed in the direction induced by the surface’s orientation.

The crucial point is that the induced directions are not all the same when viewed from one side. For an upward-oriented planar annulus, the outer boundary is counterclockwise when viewed from above, while the inner boundary is clockwise. The inner curve is not assigned its direction by the rule for an outer rim: its direction follows from the orientation of the surface and the fact that the surface lies outside the hole.

Key idea: Stokes’ Theorem sums the line integrals over every boundary component, each traversed with its induced orientation. On an annulus, the outer and inner curves therefore have opposite directions when viewed from the same side.

Induced Direction on a Boundary Curve

Suppose \(S\) is an oriented smooth surface with boundary. At a point of a boundary curve, the outward unit conormal is the unit vector that lies in the tangent plane of \(S\), is perpendicular to the boundary curve, and points out of the surface region. It is not generally the surface normal: it lies in the surface itself.

The induced tangent direction is defined by the orientation of \(S\). A boundary tangent vector \( \mathbf{T} \) points in the positive direction when the ordered pair consisting of the outward conormal and \( \mathbf{T} \) has the orientation of the surface. This is the outward-first convention for orienting the boundary.

Definition: At a boundary point, let \( \boldsymbol{\nu} \) be the outward unit conormal to the surface and let \( \mathbf{T} \) be a unit tangent to the boundary. The direction \( \mathbf{T} \) is the induced positive boundary direction when the ordered pair \( (\boldsymbol{\nu},\mathbf{T}) \) is positively oriented in the tangent plane of \(S\).

For a surface oriented by a chosen normal, the positive orientation of its tangent plane is the one consistent with that normal by the right-hand rule. On a horizontal plane oriented upward, a pair of tangent vectors is positive when its cross product points upward. At the rightmost point of a circular disk, the outward conormal points right; the positive boundary tangent points up. Thus the disk’s boundary runs counterclockwise when viewed from above.

Why an Annulus Has Opposite Boundary Directions

Consider the planar annulus \(1\le r\le 3\), oriented upward. At its outer boundary, the outward conormal points toward increasing \(r\). At its inner boundary, the outward conormal points toward decreasing \(r\), because leaving the annulus there means moving into the hole. That reversal of the conormal reverses the induced tangent direction as well.

The sign can be checked directly in polar coordinates. Use the parametrization \( \mathbf{X}(r,\theta)=(r\cos\theta,r\sin\theta,0) \). Its coordinate tangent vectors satisfy \( \mathbf{X}_r\times\mathbf{X}_\theta=(0,0,r) \), which points upward for \(r>0\). So the ordered coordinate directions \((\partial_r,\partial_\theta)\) agree with the surface orientation. Along the outer boundary the outward direction is \(+\partial_r\), so the positive tangent is \(+\partial_\theta\). Along the inner boundary it is \(-\partial_r\), so the positive tangent must be \(-\partial_\theta\).

$$ \begin{array}{c|c|c} \text{Boundary component} & \text{Outward conormal direction} & \text{Induced direction}\\ \hline r=3 & +\partial_r & +\partial_\theta\quad\text{(counterclockwise)}\\ r=1 & -\partial_r & -\partial_\theta\quad\text{(clockwise)} \end{array} $$

This is not a special extra rule for annuli. It is a consequence of the definition: the conormal pointing out of the surface changes direction at the inner edge, and the boundary tangent must change accordingly to preserve the surface orientation.

Stokes’ Theorem with Several Boundary Components

For a boundary with \(m\) disjoint closed curves, write \( \partial S=C_1\cup\cdots\cup C_m \). In the line integral over a closed curve, the chosen orientation specifies the direction of traversal. The integral over the whole boundary is the sum over its components.

Theorem: Let \(S\) be a compact, oriented, smooth surface whose boundary consists of finitely many disjoint smooth closed curves \(C_1,\ldots,C_m\). Let \( \mathbf{F} \) be a continuously differentiable vector field on an open set containing \(S\). Orient each \(C_j\) by the direction induced from \(S\). Then $$ \sum_{j=1}^{m}\oint_{C_j}\mathbf{F}\cdot d\mathbf{r} = \iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS. $$

Here \( \mathbf{n} \) is the unit normal consistent with the chosen surface orientation. Compactness is part of the stated hypotheses: it rules out unaccounted boundary contributions at infinity. Smoothness and the condition on \( \mathbf{F} \) ensure that the curl and the integrals in the theorem are defined in the required region.

The notation \( \oint_{\partial S} \) is sometimes used for the entire boundary integral, but it must be understood as a sum over components with their own induced directions. It does not mean that every component is automatically traversed counterclockwise in a common projection.

A Short Justification of the Boundary Signs

The sign convention can be read from an oriented parameter domain. For the annulus parametrization above, the positive coordinate order is \((\partial_r,\partial_\theta)\). At an outer edge, the outward coordinate direction is \(+\partial_r\). To make the ordered pair (outward direction, boundary direction) positive, the boundary direction is \(+\partial_\theta\). At an inner edge, the outward coordinate direction is \(-\partial_r\). Keeping the ordered pair positive now requires the boundary direction \(-\partial_\theta\).

More explicitly, the coefficient matrix of the pair \((+\partial_r,+\partial_\theta)\) in the ordered coordinate basis has determinant \(1\), so it preserves orientation. The coefficient matrix of \((-\partial_r,-\partial_\theta)\) also has determinant \(1\). By contrast, pairing \(-\partial_r\) with \(+\partial_\theta\) gives determinant \(-1\), which reverses orientation and is therefore the wrong induced direction.

$$ \det\begin{pmatrix}1&0\\0&1\end{pmatrix}=1, \qquad \det\begin{pmatrix}-1&0\\0&-1\end{pmatrix}=1, \qquad \det\begin{pmatrix}-1&0\\0&1\end{pmatrix}=-1. $$

This orientation check is also why the inner boundary contributes with a minus sign if one first writes both circles counterclockwise: the theorem requires the inner circle to be traversed in the reverse direction.

Worked Example: Stokes’ Theorem on a Planar Annulus

Worked Example: Stokes’ Theorem on a Planar Annulus

Let \(S\) be the annulus \(1\le x^2+y^2\le 9\) in the plane \(z=0\), oriented upward. Let \( \mathbf{F}=(-y,x,0) \). Find the induced boundary directions and check Stokes’ Theorem.

The outer circle has radius \(3\), and the inner circle has radius \(1\). By the outward-conormal rule, the outer circle is counterclockwise and the inner circle is clockwise, both viewed from above.

For a circle of radius \(a\), use \(x=a\cos\theta\), \(y=a\sin\theta\). Then \(dx=-a\sin\theta\,d\theta\) and \(dy=a\cos\theta\,d\theta\), so \( \mathbf{F}\cdot d\mathbf{r}=-y\,dx+x\,dy=a^2\,d\theta \).

For the outer circle, \(a=3\) and the positive traversal has \(0\le\theta\le2\pi\):

$$ \oint_{C_{\mathrm{out}}}\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}9\,d\theta = 18\pi. $$

For the inner circle, \(a=1\), but the induced traversal is clockwise. We can represent that by letting \(\theta\) decrease from \(0\) to \(-2\pi\):

$$ \oint_{C_{\mathrm{in}}}\mathbf{F}\cdot d\mathbf{r} = \int_0^{-2\pi}1\,d\theta = -2\pi. $$

The total boundary circulation is \(18\pi-2\pi=16\pi\). The curl is \( \nabla\times\mathbf{F}=(0,0,2) \). The upward curl flux is twice the annulus’s area, which is \( \pi(3^2-1^2)=8\pi \):

$$ \iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = 2\bigl(8\pi\bigr) = 16\pi. $$

Both sides equal \(16\pi\), as required. If the inner circle had incorrectly been traversed counterclockwise, its contribution would have been \(+2\pi\), giving \(20\pi\), not the curl flux.

Worked Example: The Two Ends of a Cylindrical Surface

Worked Example: The Two Ends of a Cylindrical Surface

Take the lateral surface of the cylinder \(x^2+y^2=4\), \(0\le z\le 2\), oriented outward. Let \( \mathbf{F}=(-y,x,0) \). Determine the induced directions on its two boundary circles and verify Stokes’ Theorem.

Parametrize the surface by \( \mathbf{X}(\theta,z)=(2\cos\theta,2\sin\theta,z) \). We have \( \mathbf{X}_\theta\times\mathbf{X}_z=(2\cos\theta,2\sin\theta,0) \), which points outward. Thus the coordinate order \((\partial_\theta,\partial_z)\) gives the surface orientation.

At the top edge \(z=2\), the outward conormal points \(+\partial_z\). For \((+\partial_z,\mathbf{T})\) to have the orientation \((\partial_\theta,\partial_z)\), the boundary tangent must point \(-\partial_\theta\). The top circle is clockwise when viewed from above. At the bottom edge \(z=0\), the outward conormal points \(-\partial_z\), so the induced boundary tangent is \(+\partial_\theta\). The bottom circle is counterclockwise.

Each circle has radius \(2\), so the circulation along a counterclockwise circle is \(2\pi(2^2)=8\pi\). The top circle is clockwise and contributes \(-8\pi\); the bottom circle contributes \(+8\pi\). Their sum is zero.

The curl is \( (0,0,2) \). The cylinder’s outward normal is horizontal, so its dot product with the vertical curl is zero at every point. Hence the curl flux is zero, matching the boundary sum.

Worked Example: A Downward-Oriented Annular Graph

Worked Example: A Downward-Oriented Annular Graph

Let \(S\) be the graph \(z=x+y\) above the planar annulus centered at \((1,-1)\) with inner radius \(1\) and outer radius \(2\). Orient \(S\) downward, and again take \( \mathbf{F}=(-y,x,0) \). Find the boundary directions and verify Stokes’ Theorem.

Because the orientation is downward, the projected outer boundary is clockwise when viewed from above, while the projected inner boundary is counterclockwise. To compute the circulation on a projected circle of radius \(a\), write \(x=1+a\cos\theta\) and \(y=-1+a\sin\theta\). Direct substitution gives

$$ -y\,dx+x\,dy = a(\cos\theta-\sin\theta)\,d\theta+a^2\,d\theta. $$

Over a full counterclockwise revolution, the sine and cosine terms integrate to zero, leaving \(2\pi a^2\). Reversing the direction changes the sign. The outer circle therefore contributes \(-2\pi(2^2)=-8\pi\), and the inner circle contributes \(+2\pi(1^2)=2\pi\). The total is \(-6\pi\).

For the graph, the upward-oriented area vector is \((-1,-1,1)\,dx\,dy\), so the downward-oriented area vector is \((1,1,-1)\,dx\,dy\). Since \( \nabla\times\mathbf{F}=(0,0,2) \), its dot product with that vector is \(-2\,dx\,dy\). The projected annulus has area \( \pi(2^2-1^2)=3\pi \), and therefore

$$ \iint_S(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = -2(3\pi) = -6\pi. $$

The boundary circulation and curl flux both equal \(-6\pi\). This check shows that reversing the surface orientation reverses every induced boundary direction and the sign of the curl flux.

Common Mistake: Giving Every Component the Same Direction

A frequent error is to traverse all circular boundaries counterclockwise merely because they look like circles in the plane. That direction is correct for the outer boundary of an upward-oriented annulus, but incorrect for the inner boundary. In the first example, this mistake would calculate

$$ 18\pi+2\pi=20\pi, $$

instead of the correct boundary sum \(18\pi-2\pi=16\pi\). The curl flux is \(16\pi\), exposing the sign error. The correction is to identify the outward conormal separately on each component: it points away from the annular surface into the hole at the inner curve, so that curve’s induced direction is clockwise.

A related error is to apply Stokes’ Theorem to an unbounded surface while assuming that its visible boundary curves are the only contributions. The compact-surface hypothesis stated here matters. For noncompact surfaces, one needs additional hypotheses that control behavior at infinity; the usual compact version does not by itself justify omitting any contribution there.

How to Orient Several Boundary Curves

1
Identify the surface orientation.
Record the chosen normal, or the positively oriented tangent basis that it determines.
2
Inspect one boundary component at a time.
At each component, find the conormal that lies in the surface and points out of the surface region.
3
Apply the outward-first convention.
Choose the boundary tangent so that (outward conormal, boundary tangent) agrees with the surface orientation.
4
Sum the line integrals with those directions.
Use each induced direction in Stokes’ Theorem; do not impose a shared clockwise or counterclockwise direction on all components.
5
Check the result when possible.
Compute the curl flux independently and compare it with the total boundary circulation.

Why the Orientation Accounting Matters

Stokes’ Theorem connects local rotation, represented by curl, to circulation along the entire boundary. With multiple boundary curves, the theorem remains a single equality, but its boundary side is a signed sum. An inner boundary has the opposite induced direction because the surface lies on the other side of that curve than it does at an outer edge.

Keeping that sign straight is essential both for applying the theorem and for interpreting its result. A reliable approach is to use the conormal definition rather than memorizing a picture: find which way is out of the surface at each boundary component, then select the tangent that preserves the surface orientation.

Check Your Understanding

Use the induced-orientation definition to answer these questions.

  1. For an upward-oriented planar annulus, why does the outward conormal point in opposite radial directions at the outer and inner boundaries?
  2. What are the induced directions on the two boundary circles of an upward-oriented annulus, when both are viewed from above?
  3. A compact cylindrical side is oriented outward. Why do its top and bottom boundary circles have opposite directions when viewed from above?
  4. If the orientation of an entire surface is reversed, what happens to the induced direction on each boundary component?
  5. What compactness and smoothness conditions are included in the stated version of Stokes’ Theorem?

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