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Stokes’ Theorem · Tutorial 945 of 999

A Worked Stokes Problem with a Circular Boundary

Learn to compute circulation around an oriented circle using the curl over its disk, and check that the circle’s direction matches the disk’s normal.

Advanced 10 min read

What You'll Learn

  • Set up Stokes’ Theorem for a circular boundary and its spanning disk
  • Compute a curl and integrate its normal component over a disk
  • Parameterize a circle with the direction induced by a chosen disk normal
  • Check circulation results with direct boundary line integrals
  • Identify and correct sign errors caused by inconsistent orientations

From a Circular Boundary to Its Disk

A circulation integral around a circle can involve a lengthy parameterized line integral. If the vector field is smooth on the disk bounded by that circle, Stokes’ Theorem offers another route: compute the curl, then integrate its component normal to the disk. The choice of normal matters, because it determines the direction in which the boundary must be traversed.

This tutorial develops that calculation through circular disks. Each example checks the boundary direction against the disk normal and verifies the resulting circulation by direct integration around the circle.

Stokes’ Theorem for a disk: Let \(\mathbf{F}\) be a vector field with continuous first partial derivatives on an open set containing an oriented disk \(D\) and its boundary \(C\). If \(C\) has the orientation induced by the chosen unit normal \(\mathbf{n}\) on \(D\), then
$$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_D (\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS. $$

The left side is the circulation of \(\mathbf{F}\) along the oriented circle. The right side is the flux of the curl through the disk. The regularity condition ensures that the curl is defined and that Stokes’ Theorem applies throughout the disk, not merely along its boundary.

Orienting a Circular Disk and Its Boundary

For a disk of radius \(a\), center \(\mathbf{c}\), and unit normal \(\mathbf{n}\), choose perpendicular unit vectors \(\mathbf{e}_1\) and \(\mathbf{e}_2\) in the disk so that \(\mathbf{e}_1\times\mathbf{e}_2=\mathbf{n}\). A boundary parameterization consistent with this normal is

$$ \mathbf{r}(t) = \mathbf{c} + a\bigl(\cos t\,\mathbf{e}_1+\sin t\,\mathbf{e}_2\bigr), \qquad 0\le t\le 2\pi. $$

As \(t\) increases, the parameterization moves around the boundary in the direction induced by \(\mathbf{n}\). Replacing \(\mathbf{n}\) with \(-\mathbf{n}\) reverses that induced direction. Consequently, the circulation changes sign when the boundary orientation is reversed.

For a planar disk with constant unit normal \(\mathbf{n}\), the surface integral can also be written using ordinary area in the disk’s plane:

$$ \iint_D (\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = \iint_D \bigl((\nabla\times\mathbf{F})\cdot\mathbf{n}\bigr)\,dA. $$

Here \(dA\) denotes area measured within the plane of the disk. If the curl is constant, its dot product with \(\mathbf{n}\) is constant too, and the surface integral is that value multiplied by the disk’s area, \(\pi a^2\).

A Reliable Calculation Sequence

1
Specify the disk and its normal.
Identify the radius, plane, center, and chosen unit normal.
2
Check the boundary direction.
Choose a circle parameterization whose increasing parameter agrees with the normal.
3
Compute the curl.
Find \(\nabla\times\mathbf{F}\) from the component partial derivatives.
4
Integrate its normal component.
Evaluate \((\nabla\times\mathbf{F})\cdot\mathbf{n}\) over the disk.
5
Verify the result.
When practical, substitute the boundary parameterization into the line integral and compare.

Worked Examples

Worked Example: A Horizontal Disk

Let \(\mathbf{F}(x,y,z)=\langle -y,x,0\rangle\). Find the circulation around the circle \(x^2+y^2=4,\ z=0\), oriented counterclockwise as viewed from the positive \(z\)-axis.

The disk has radius \(2\), lies in the \(xy\)-plane, and has upward unit normal \(\mathbf{n}=\langle 0,0,1\rangle\). The counterclockwise direction viewed from above is the direction induced by this normal.

Write \(\mathbf{F}=\langle P,Q,R\rangle\), where \(P=-y\), \(Q=x\), and \(R=0\). Then

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

The normal component is \(\langle 0,0,2\rangle\cdot\langle 0,0,1\rangle=2\). The disk area is \(\pi(2)^2=4\pi\), so Stokes’ Theorem gives

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \iint_D 2\,dA = 2(4\pi) = 8\pi. $$

To check, parameterize the oriented boundary by \(\mathbf{r}(t)=\langle 2\cos t,2\sin t,0\rangle\), for \(0\le t\le 2\pi\). Then \(\mathbf{r}'(t)=\langle-2\sin t,2\cos t,0\rangle\), and \(\mathbf{F}(\mathbf{r}(t))=\langle-2\sin t,2\cos t,0\rangle\). Therefore

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_0^{2\pi} \bigl(4\sin^2t+4\cos^2t\bigr)\,dt = \int_0^{2\pi}4\,dt = 8\pi. $$

The direct calculation agrees with the disk integral, and the positive sign matches the stated orientation.

Worked Example: A Circle in a Vertical Plane

Let \(\mathbf{F}(x,y,z)=\langle 0,0,x\rangle\). Find the circulation around the unit circle \(x^2+z^2=1,\ y=0\), with the disk oriented by the normal \(\mathbf{n}=\langle 0,1,0\rangle\).

Use \(\mathbf{r}(u,v)=\langle u,0,-v\rangle\), where \(u^2+v^2\le 1\), to parameterize the disk. Its tangent vectors are \(\mathbf{r}_u=\langle1,0,0\rangle\) and \(\mathbf{r}_v=\langle0,0,-1\rangle\), so \(\mathbf{r}_u\times\mathbf{r}_v=\langle0,1,0\rangle\). This confirms the chosen normal. The corresponding positively oriented boundary is \(\mathbf{r}(t)=\langle\cos t,0,-\sin t\rangle\), \(0\le t\le2\pi\).

For \(\mathbf{F}=\langle P,Q,R\rangle=\langle0,0,x\rangle\), the curl is

$$ \nabla\times\mathbf{F} = \left\langle \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right\rangle = \langle 0,0-1,0\rangle = \langle0,-1,0\rangle. $$

Its component along the chosen normal is \(\langle0,-1,0\rangle\cdot\langle0,1,0\rangle=-1\). The disk has area \(\pi\), giving

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \iint_D (-1)\,dA = -\pi. $$

For a direct check, \(\mathbf{r}'(t)=\langle-\sin t,0,-\cos t\rangle\) and \(\mathbf{F}(\mathbf{r}(t))=\langle0,0,\cos t\rangle\). Thus

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \langle0,0,\cos t\rangle\cdot \langle-\sin t,0,-\cos t\rangle\,dt = -\int_0^{2\pi}\cos^2t\,dt = -\pi. $$

The negative circulation is consistent with the upward-pointing normal in the \(y\)-direction and the resulting boundary orientation.

Worked Example: A Tilted Circular Disk

Let \(\mathbf{F}(x,y,z)=\langle0,3x,2y\rangle\). Find the circulation around the radius-\(2\) disk centered at the origin in the plane spanned by \(\mathbf{e}_1=\langle1,0,0\rangle\) and \(\mathbf{e}_2=\langle0,4/5,-3/5\rangle\), oriented by the unit normal \(\mathbf{n}=\langle0,3/5,4/5\rangle\).

The vectors \(\mathbf{e}_1,\mathbf{e}_2\) are perpendicular unit vectors, and

$$ \mathbf{e}_1\times\mathbf{e}_2 = \left\langle0,\frac35,\frac45\right\rangle = \mathbf{n}. $$

Thus the circle parameterization \(\mathbf{r}(t)=2(\cos t\,\mathbf{e}_1+\sin t\,\mathbf{e}_2)\) has the required direction. The curl is

$$ \nabla\times\mathbf{F} = \left\langle \frac{\partial (2y)}{\partial y}-\frac{\partial (3x)}{\partial z}, \frac{\partial 0}{\partial z}-\frac{\partial (2y)}{\partial x}, \frac{\partial (3x)}{\partial x}-\frac{\partial 0}{\partial y} \right\rangle = \langle2,0,3\rangle. $$

The curl’s normal component is \(\langle2,0,3\rangle\cdot\langle0,3/5,4/5\rangle=12/5\). The disk area is \(\pi(2)^2=4\pi\), so

$$ \oint_C\mathbf{F}\cdot d\mathbf{r} = \iint_D \frac{12}{5}\,dA = \frac{12}{5}(4\pi) = \frac{48\pi}{5}. $$

For an explicit check, the boundary parameterization and derivative are

$$ \mathbf{r}(t) = \left\langle2\cos t,\frac85\sin t,-\frac65\sin t\right\rangle, \qquad \mathbf{r}'(t) = \left\langle-2\sin t,\frac85\cos t,-\frac65\cos t\right\rangle. $$

Along the boundary, \(\mathbf{F}(\mathbf{r}(t))=\langle0,6\cos t,\frac{16}{5}\sin t\rangle\). Hence

$$ \begin{aligned} \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) &= \frac{48}{5}\cos^2t-\frac{96}{25}\sin t\cos t,\\ \oint_C\mathbf{F}\cdot d\mathbf{r} &= \frac{48}{5}\int_0^{2\pi}\cos^2t\,dt -\frac{96}{25}\int_0^{2\pi}\sin t\cos t\,dt\\ &= \frac{48}{5}\pi-\frac{96}{25}(0) = \frac{48\pi}{5}. \end{aligned} $$

Here \(\int_0^{2\pi}\cos^2t\,dt=\pi\), while \(\int_0^{2\pi}\sin t\cos t\,dt=0\). The direct line integral confirms the surface calculation.

Why the Boundary Direction Must Match the Normal

The normal does more than specify which side of the disk is positive: it fixes the orientation of the boundary. The parameterizations above demonstrate the rule concretely. For a horizontal disk with upward normal, increasing the circle parameter gives counterclockwise motion when viewed from above. For a different disk, the direction must be checked against its own normal rather than borrowed from a picture of the \(xy\)-plane.

For example, the vertical-disk calculation used \(\mathbf{n}=\langle0,1,0\rangle\) and the boundary \(\langle\cos t,0,-\sin t\rangle\). Using \(\langle\cos t,0,\sin t\rangle\) instead would reverse the direction. The direct integral would then be

$$ \int_0^{2\pi}\cos^2t\,dt=\pi, $$

which has the wrong sign for the stated normal. The correction is to use the boundary direction induced by the normal, or equivalently to reverse the normal as well. Reversing just one of them makes the calculation inconsistent.

When the Curl Is Not Constant

The method does not require a constant curl. For a variable curl, the integrand \((\nabla\times\mathbf{F})\cdot\mathbf{n}\) must be integrated over the disk rather than replaced by one value times the area. A polar parameterization of a horizontal disk of radius \(a\) is \(\langle r\cos\theta,r\sin\theta,0\rangle\), with \(0\le r\le a\) and \(0\le\theta\le2\pi\); its area element is \(r\,dr\,d\theta\). For other planes, use coordinates in an orthonormal basis of that plane, keeping the orientation consistent with the chosen normal.

A useful check on any calculation is to ask whether the sign is plausible. A positive normal component of the curl throughout a disk gives positive circulation for its induced boundary orientation; reversing the orientation reverses the sign. When the normal component varies, positive and negative parts may offset one another, so the final sign depends on the full integral.

Key takeaway: For a circular boundary, Stokes’ Theorem turns circulation into the integral of the curl’s normal component over the disk. Specify the disk normal, use the boundary direction it induces, and keep that orientation consistent throughout the calculation.

Check Your Understanding

Use Stokes’ Theorem and the orientation rules to answer these questions.

  1. What is the unit normal to a disk in the \(xy\)-plane oriented counterclockwise when viewed from above?
  2. If a disk’s normal is reversed but its boundary parameterization is left unchanged, what happens to the agreement between the two sides of Stokes’ Theorem?
  3. For a constant curl on a disk of radius \(a\), what two quantities determine the curl flux through the disk?
  4. For \(\mathbf{F}=\langle0,0,x\rangle\), what circulation results around the unit circle in \(y=0\) when the disk normal is \(\langle0,-1,0\rangle\)?
  5. Why is the circle direction in a tilted plane not determined merely by whether it looks clockwise from a fixed viewpoint?

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