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Stokes’ Theorem · Tutorial 939 of 999

Stokes' Theorem for Nonplanar Boundary Curves

Use an appropriately chosen spanning surface to convert circulation around a spatial boundary curve into a curl-flux integral.

Advanced 10 min read

What You'll Learn

  • State the hypotheses needed to apply Stokes’ Theorem to a closed space curve.
  • Choose a spanning graph surface for a genuinely nonplanar curve.
  • Use an oblique plane patch for a plane-cylinder intersection curve.
  • Determine the surface orientation from the direction of the boundary curve.
  • Check curl-flux answers by directly computing the circulation when practical.

Introduction

A closed curve in three-dimensional space need not lie in a horizontal plane, or even in any plane. Directly parameterizing such a curve and calculating its circulation can be awkward. Stokes’ Theorem offers another route: find an oriented surface whose boundary is the curve, then integrate the curl of the vector field over that surface.

The useful choice is often not the surface on which the curve was first described. A curve might be specified as the intersection of a cylinder and another surface, while a graph over a disk provides a simpler spanning surface. The curve’s position in space determines its boundary; it does not force us to use a particular surface to fill it.

The central idea: For a suitable oriented surface \(S\) with oriented boundary \(C\), Stokes’ Theorem equates circulation around \(C\) with the flux of the curl through \(S\). Choose a surface that makes that flux integral manageable.

Stokes’ Theorem and Surface Choice

Let \(\mathbf{F}\) be a vector field with continuous first partial derivatives on an open set containing an oriented, piecewise smooth surface \(S\). If the boundary \(C=\partial S\) is a piecewise smooth closed curve, and its direction agrees with the orientation of \(S\), then Stokes’ Theorem states:

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

Here \(\mathbf{n}\) is the chosen unit normal to the surface, and the boundary direction is the one induced by that normal using the right-hand rule. The equality permits us to calculate either side using whichever description is simpler. In this tutorial, the emphasis is on using a convenient surface to evaluate the curl flux associated with a spatial boundary.

A particularly useful surface is a graph \(z=g(x,y)\) over a region \(D\) in the \(xy\)-plane. Parameterize it by \(\mathbf{r}(x,y)=(x,y,g(x,y))\). With the upward orientation, its oriented area element is:

$$ \mathbf{r}_x\times\mathbf{r}_y = (1,0,g_x)\times(0,1,g_y) = (-g_x,-g_y,1), \qquad \mathbf{n}\,dS=(-g_x,-g_y,1)\,dA. $$

Thus the curl flux is obtained by taking the dot product with the entire oriented area vector and integrating over the base region \(D\). The upward orientation induces a boundary direction whose projection onto the \(xy\)-plane runs counterclockwise around \(D\).

A Genuinely Nonplanar Boundary

Consider a curve lying on the cylinder \(x^2+y^2=1\) and the graph \(z=xy\). Its parameterization is \(\mathbf{r}(t)=(\cos t,\sin t,\cos t\sin t)\), for \(0\le t\le2\pi\). This is genuinely a space curve: it does not lie in one plane. Its spanning surface can be the graph \(z=xy\) over the unit disk. That surface has a simple parameterization even though its boundary’s height varies as the curve goes around.

Worked Example: A Space Curve on a Graph

Evaluate \(\oint_C \mathbf{F}\cdot d\mathbf{r}\) for \(\mathbf{F}=(0,0,x^2)\), where \(C\) is \(\mathbf{r}(t)=(\cos t,\sin t,\cos t\sin t)\), \(0\le t\le2\pi\), with increasing \(t\).

First check orientation. The curve projects to \((\cos t,\sin t)\), which runs counterclockwise around the unit disk. So it is the positive boundary direction for the upward-oriented graph \(z=xy\). The vector field’s curl is:

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

For \(g(x,y)=xy\), we have \(g_x=y\) and \(g_y=x\). The upward-oriented area vector is \((-y,-x,1)\). Therefore:

$$ (\nabla\times\mathbf{F})\cdot(-y,-x,1) = (0,-2x,0)\cdot(-y,-x,1) = 2x^2. $$

Integrate over the unit disk \(D\). By polar coordinates, \(x=r\cos\theta\), so:

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D 2x^2\,dA = \int_0^{2\pi}\int_0^1 2r^2\cos^2\theta\,r\,dr\,d\theta = 2\left(\frac14\right)\pi = \frac{\pi}{2}. $$

Check by direct circulation. Along the curve, \(x=\cos t\), \(z=\cos t\sin t\), and \(dz/dt=\cos^2t-\sin^2t\). Since only the third component of \(\mathbf{F}\) is nonzero, \(\mathbf{F}\cdot d\mathbf{r}=x^2\,dz\). Hence:

$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}\cos^2t(\cos^2t-\sin^2t)\,dt = \int_0^{2\pi}\cos^4t\,dt - \int_0^{2\pi}\cos^2t\sin^2t\,dt = \frac{3\pi}{4}-\frac{\pi}{4} = \frac{\pi}{2}. $$

The direct calculation agrees with the curl-flux result.

Plane-Cylinder Intersection Curves

A plane intersecting a circular cylinder can produce a boundary curve that looks spatial when viewed in the usual coordinate directions, though it lies in the plane of intersection. A disk in that oblique plane is a convenient spanning surface. It can be parameterized using the cylinder’s \(x\)- and \(y\)-coordinates, with the plane supplying the third coordinate.

Worked Example: An Oblique Plane Cutting a Cylinder

Let \(C\) be the intersection of the cylinder \(x^2+y^2=4\) and the plane \(z=x+y\). Give \(C\) the direction traced by \(\mathbf{r}(t)=(2\cos t,2\sin t,2\cos t+2\sin t)\), \(0\le t\le2\pi\). For \(\mathbf{F}=(0,0,x)\), evaluate the circulation by using the planar disk in \(z=x+y\) that spans \(C\).

The disk is parameterized by \(\mathbf{r}(u,v)=(u,v,u+v)\), where \(u^2+v^2\le4\). Its oriented area vector for increasing \(u,v\) is:

$$ \mathbf{r}_u\times\mathbf{r}_v = (1,0,1)\times(0,1,1) = (-1,-1,1). $$

The boundary direction given in the question projects to a counterclockwise circle in the \(uv\)-plane, so it agrees with this orientation. Compute the curl and the flux integrand:

$$ \nabla\times\mathbf{F}=(0,-1,0), \qquad (\nabla\times\mathbf{F})\cdot(-1,-1,1) = (0,-1,0)\cdot(-1,-1,1) = 1. $$

The projection domain is a disk of radius \(2\), with area \(4\pi\). Therefore:

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \iint_{u^2+v^2\le4}1\,dA = 4\pi. $$

As a check, direct parameterization gives \(x=2\cos t\) and \(dz/dt=2(-\sin t+\cos t)\). Thus:

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}x\frac{dz}{dt}\,dt = 4\int_0^{2\pi}(\cos^2t-\cos t\sin t)\,dt = 4(\pi-0) = 4\pi. $$

The oblique disk turns the surface integral into a constant integrand over a circular region. The cylinder describes the boundary, but the plane disk supplies the spanning surface.

Another Graph Filling a Cylinder Curve

The same strategy applies when the boundary height varies in a different way. If a curve is given by its intersection with a vertical cylinder and a graph, the graph over the cylinder’s disk may fill it directly. Compute the curl flux using the graph’s oriented area vector, and make sure the boundary direction agrees with the chosen normal.

Worked Example: A Different Nonplanar Cylinder Boundary

Let \(C\) be given by \(x^2+y^2=4\) and \(z=(x^2-y^2)/4\), oriented by increasing \(t\) in \(\mathbf{r}(t)=(2\cos t,2\sin t,\cos^2t-\sin^2t)\). For \(\mathbf{F}=(0,0,xy)\), find the circulation using the graph surface \(z=(x^2-y^2)/4\) over the disk \(D:x^2+y^2\le4\).

The boundary projects counterclockwise, so use the upward graph orientation. Here \(g_x=x/2\) and \(g_y=-y/2\), giving the oriented area vector \((-x/2,y/2,1)\). The curl is:

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

Taking the dot product with the graph area vector gives:

$$ (x,-y,0)\cdot\left(-\frac{x}{2},\frac{y}{2},1\right) = -\frac{x^2}{2}-\frac{y^2}{2} = -\frac{x^2+y^2}{2}. $$

In polar coordinates, the disk has \(0\le r\le2\), and \(x^2+y^2=r^2\). Thus the integral is:

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = -\int_0^{2\pi}\int_0^2\frac{r^2}{2}r\,dr\,d\theta = -\frac12(2\pi)\left(\frac{2^4}{4}\right) = -4\pi. $$

For a direct check, along \(C\) we have \(xy=4\cos t\sin t\) and \(dz/dt=-4\cos t\sin t\). Therefore:

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}xy\frac{dz}{dt}\,dt = -16\int_0^{2\pi}\cos^2t\sin^2t\,dt = -16\left(\frac{\pi}{4}\right) = -4\pi. $$

The result matches the flux calculation.

Why the Graph Formula Works

The graph computation follows from its parameterization, rather than from a special rule for a particular example. With \(\mathbf{r}(x,y)=(x,y,g(x,y))\), moving a small amount in the \(x\)-direction gives the tangent vector \(\mathbf{r}_x=(1,0,g_x)\); moving in the \(y\)-direction gives \(\mathbf{r}_y=(0,1,g_y)\). Their cross product is perpendicular to both tangent directions and has the orientation determined by their order:

$$ \mathbf{r}_x\times\mathbf{r}_y=(-g_x,-g_y,1). $$

The magnitude of this cross product accounts for the surface area of a small patch, while its direction accounts for the orientation. Consequently, for any continuously differentiable vector field \(\mathbf{G}\), the flux through the upward graph is:

$$ \iint_S\mathbf{G}\cdot\mathbf{n}\,dS = \iint_D\mathbf{G}(x,y,g(x,y))\cdot(-g_x,-g_y,1)\,dA. $$

Applying this formula with \(\mathbf{G}=\nabla\times\mathbf{F}\) gives the curl flux in Stokes’ Theorem. The boundary direction must still be compatible with this orientation; the cross product alone does not settle the direction of a curve that was specified independently.

A Common Mistake: Dropping Components of the Normal

In the first example, the upward-oriented graph area vector is \((-y,-x,1)\), not merely \((0,0,1)\). A tempting but incorrect calculation is to use only the vertical component of the normal:

$$ (0,-2x,0)\cdot(0,0,1)=0. $$

This would incorrectly predict zero curl flux. The surface is tilted, so its oriented area vector has horizontal components as well. The correct dot product uses every component:

$$ (0,-2x,0)\cdot(-y,-x,1) = 2x^2, $$

which gives the nonzero flux \(\pi/2\). For a graph surface, use \((-g_x,-g_y,1)\) with the upward orientation, rather than replacing it with a vertical vector.

A Practical Surface-Selection Routine

1
Read the boundary description.
Look for a plane, graph, cylinder, or other simple surface on which the curve lies.
2
Choose a filling surface.
For a graph over a disk, use that graph over the entire disk. For a plane-cylinder intersection, consider the disk in the intersection plane.
3
Match the orientation.
Check that the curve’s stated direction is induced by the chosen normal. For an upward graph, its projection should be counterclockwise.
4
Calculate the curl flux.
Compute the curl, dot it with the oriented area vector, and integrate over the projection region.
5
Check the result.
When practical, calculate the circulation directly along the boundary and compare it with the flux.

The surface should have precisely the given curve as its boundary and should meet the smoothness and orientation requirements of Stokes’ Theorem. A convenient parameterization is not enough if it describes a different boundary or induces the opposite direction.

Check Your Understanding

Use the definitions and examples above to answer these questions.

  1. For the upward graph \(z=g(x,y)\), what oriented area vector is used in the curl-flux integral?
  2. What direction does an upward graph induce on the boundary of its projection region?
  3. Why can a plane disk be a convenient spanning surface for the intersection of a plane and a cylinder?
  4. In the first worked example, why does using only the vertical component of the graph’s normal give an incorrect flux?
  5. What must be checked between a chosen surface orientation and the direction of the specified boundary curve?

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