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Stokes’ Theorem · Tutorial 932 of 999

Stokes' Theorem on Disks and Planar Patches

Learn to set up and evaluate curl-flux integrals on disks and planar patches, with orientations and projected bounds handled explicitly.

Advanced 13 min read

What You'll Learn

  • Determine the oriented normal to a planar patch from its parametrization.
  • Set up curl-flux integrals on horizontal and tilted disks.
  • Use polar bounds for circular regions and rectangular bounds for graph patches.
  • Convert a graph’s unit normal and surface area element into a single oriented area factor.
  • Check flux calculations against the boundary circulation in Stokes’ Theorem.
  • Recognize sign and area-factor errors in planar surface integrals.

Introduction

Stokes’ Theorem relates circulation around a boundary curve to the flux of curl through a surface spanning that curve. For a disk or another planar patch, the surface is simple geometrically, but the calculation still depends on two choices: the orientation of the surface and the coordinates used to describe the patch.

A horizontal disk can often be handled with polar coordinates. A tilted disk is naturally described using coordinates in its own plane. A planar patch given as a graph can instead be integrated over its projection onto a coordinate plane. In every case, the normal and the bounds must describe the same oriented surface.

The key calculation: If an oriented surface is parametrized by \(\mathbf r(u,v)\), its curl flux is the integral of \((\nabla\times\mathbf F)\cdot(\mathbf r_u\times\mathbf r_v)\) over the parameter region. The order of the cross product determines the orientation.

Oriented Planar Patches

Let a planar patch be parametrized by \(\mathbf r(u,v)=\mathbf p+u\mathbf a+v\mathbf b\), where \(\mathbf a\) and \(\mathbf b\) are constant, nonparallel vectors. The cross product \(\mathbf a\times\mathbf b\) is perpendicular to the plane and specifies an orientation. Reversing the order of the vectors reverses that orientation.

Definition: For a regular parametrized surface \(\mathbf r(u,v)\) over a region \(D\), the oriented curl flux is \(\iint_D(\nabla\times\mathbf F)(\mathbf r(u,v))\cdot(\mathbf r_u\times\mathbf r_v)\,du\,dv\). Reversing the chosen orientation changes the sign of the integral. The parametrization must be regular on the patch, and the vector field must have continuous first partial derivatives on an open region containing the surface.

The factor \(\mathbf r_u\times\mathbf r_v\) combines the direction of the normal with the local area scaling. Thus, one does not need to use a unit normal and a separate surface-area factor if this cross product is used directly.

For a graph \(z=g(x,y)\), take \(\mathbf r(x,y)=(x,y,g(x,y))\). Direct differentiation gives

$$ \mathbf r_x=(1,0,g_x),\qquad \mathbf r_y=(0,1,g_y),\qquad \mathbf r_x\times\mathbf r_y=(-g_x,-g_y,1). $$

This normal has positive \(z\)-component, so it gives the upward orientation. The downward orientation uses its negative. The graph’s curl flux is therefore an integral over the projected region \(D\) in the \(xy\)-plane, using the appropriate signed normal factor.

$$ \iint_S(\nabla\times\mathbf F)\cdot d\mathbf S = \iint_D(\nabla\times\mathbf F)(x,y,g(x,y)) \cdot(-g_x,-g_y,1)\,dx\,dy $$

Why the Cross-Product Formula Works

Over a small parameter rectangle with side lengths \(\Delta u\) and \(\Delta v\), the surface is approximated by a parallelogram with side vectors \(\mathbf r_u\Delta u\) and \(\mathbf r_v\Delta v\). Its oriented area vector is approximately \((\mathbf r_u\times\mathbf r_v)\Delta u\Delta v\). The curl flux over that small patch is approximately the curl at the patch times this area vector. Adding these contributions and taking the limit gives the surface integral in the definition.

Stokes’ Theorem then states that, for a smooth oriented surface \(S\) with positively oriented boundary \(\partial S\),

$$ \oint_{\partial S}\mathbf F\cdot d\mathbf r = \iint_S(\nabla\times\mathbf F)\cdot d\mathbf S. $$

The boundary direction must be the one compatible with the chosen normal by the right-hand rule. This theorem gives a useful check on a surface-flux calculation: the surface integral and the circulation around its induced boundary must agree.

Worked Example: Curl Flux Through a Horizontal Disk

Worked Example: Curl Flux Through a Horizontal Disk

Let \(\mathbf F(x,y,z)=(-y/2,x/2,0)\), and let \(S\) be the disk \(x^2+y^2\leq 4\) in the plane \(z=0\), oriented upward. Find the curl flux.

First compute the curl:

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

The upward unit normal is \((0,0,1)\). Use polar coordinates on the disk, with \(0\leq r\leq 2\) and \(0\leq\theta\leq2\pi\). The area factor is \(r\), so

$$ \iint_S(\nabla\times\mathbf F)\cdot d\mathbf S = \int_0^{2\pi}\int_0^2 r\,dr\,d\theta = \int_0^{2\pi}\left[\frac{r^2}{2}\right]_0^2d\theta = \int_0^{2\pi}2\,d\theta = 4\pi. $$

Check using the boundary circle with its induced counterclockwise orientation: \(\mathbf r(\theta)=(2\cos\theta,2\sin\theta,0)\), \(0\leq\theta\leq2\pi\). Then \(d\mathbf r=(-2\sin\theta,2\cos\theta,0)\,d\theta\), and \(\mathbf F(\mathbf r(\theta))=(-\sin\theta,\cos\theta,0)\). Hence \(\mathbf F\cdot d\mathbf r=2\,d\theta\), giving circulation \(\int_0^{2\pi}2\,d\theta=4\pi\), as required.

Worked Example: A Tilted Circular Disk

Worked Example: A Tilted Circular Disk

Let \(\mathbf F(x,y,z)=(0,x+x^2/2,0)\). Consider the radius-one disk centered at \(\mathbf p=(1,0,0)\), lying in the plane with oriented unit normal \(\mathbf n=(1,1,1)/\sqrt{3}\). Find the curl flux.

Choose orthonormal vectors in the plane:

$$ \mathbf e_1=\frac{(1,-1,0)}{\sqrt{2}}, \qquad \mathbf e_2=\frac{(1,1,-2)}{\sqrt{6}}. $$

They satisfy \(\mathbf e_1\times\mathbf e_2=\mathbf n\), so the parametrization \(\mathbf r(u,v)=\mathbf p+u\mathbf e_1+v\mathbf e_2\) has the required orientation. The parameter region is the unit disk \(u^2+v^2\leq1\), and \(\mathbf r_u\times\mathbf r_v=\mathbf n\).

The curl and the \(x\)-coordinate on the disk are

$$ \nabla\times\mathbf F=(0,0,1+x), \qquad x=1+\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{6}}. $$

Therefore the flux is

$$ \begin{aligned} \iint_{u^2+v^2\leq1} (\nabla\times\mathbf F)(\mathbf r(u,v))\cdot\mathbf n\,du\,dv &= \frac{1}{\sqrt{3}} \iint_{u^2+v^2\leq1} \left(2+\frac{u}{\sqrt{2}}+\frac{v}{\sqrt{6}}\right)\,du\,dv. \end{aligned} $$

In polar coordinates \(u=\rho\cos\theta,\ v=\rho\sin\theta\), with \(0\leq\rho\leq1\) and \(0\leq\theta\leq2\pi\), the terms containing \(\cos\theta\) and \(\sin\theta\) integrate to zero. The remaining term gives

$$ \frac{1}{\sqrt{3}}\int_0^{2\pi}\int_0^1 2\rho\,d\rho\,d\theta = \frac{1}{\sqrt{3}}\int_0^{2\pi}1\,d\theta = \frac{2\pi}{\sqrt{3}}. $$

For a check, the induced boundary is \(\mathbf r(\cos\theta,\sin\theta)\). Along it, the \(y\)-coordinate is \(-\cos\theta/\sqrt{2}+\sin\theta/\sqrt{6}\), so its derivative is \(\sin\theta/\sqrt{2}+\cos\theta/\sqrt{6}\). Since \(\mathbf F\cdot d\mathbf r=(x+x^2/2)\,dy\), integrating around \(0\leq\theta\leq2\pi\) gives \(2\pi/\sqrt{3}\): the constant and cubic trigonometric terms integrate to zero, and the remaining quadratic terms give \(4\pi(1/\sqrt{2})(1/\sqrt{6})=2\pi/\sqrt{3}\). This agrees with the flux.

Worked Example: A Rectangular Patch Given as a Graph

Worked Example: A Rectangular Patch Given as a Graph

Let \(\mathbf F(x,y,z)=(0,0,xy)\), and let \(S\) be the graph \(z=2x-y\) over the rectangle \(0\leq x\leq1,\ 0\leq y\leq2\). Use the upward orientation to compute the curl flux.

The curl is

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

For \(g(x,y)=2x-y\), the upward oriented area factor is \((-2,1,1)\). Thus the integrand is \((x,-y,0)\cdot(-2,1,1)=-2x-y\), and

$$ \begin{aligned} \iint_S(\nabla\times\mathbf F)\cdot d\mathbf S &=\int_0^1\int_0^2(-2x-y)\,dy\,dx\\ &=\int_0^1\left[-2xy-\frac{y^2}{2}\right]_0^2dx\\ &=\int_0^1(-4x-2)\,dx\\ &=\left[-2x^2-2x\right]_0^1\\ &=-4. \end{aligned} $$

Check the boundary circulation. The boundary runs counterclockwise in the projected \(xy\)-rectangle. Since \(\mathbf F\cdot d\mathbf r=xy\,dz\), the bottom and left edges contribute zero. On the right edge \(x=1,\ y:0\to2\), \(dz=-dy\), so the contribution is \(\int_0^2-y\,dy=-2\). On the top edge \(y=2,\ x:1\to0\), \(dz=2\,dx\), so the contribution is \(\int_1^0 4x\,dx=-2\). The total is \(-2-2=-4\), agreeing with the surface integral.

A Common Area-Factor Mistake

For the graph \(z=2x-y\), the upward unit normal is \((-2,1,1)/\sqrt{6}\). If this unit normal is used, it must be paired with the surface-area element \(dS=\sqrt{6}\,dx\,dy\). Using the unit normal but integrating against \(dx\,dy\) alone would incorrectly give

$$ \int_0^1\int_0^2 (x,-y,0)\cdot\frac{(-2,1,1)}{\sqrt{6}}\,dy\,dx = -\frac{4}{\sqrt{6}}. $$

This is wrong because \(dx\,dy\) measures area in the projection, not area on the tilted graph. The correct calculation includes the surface-area factor:

$$ \frac{(-2,1,1)}{\sqrt{6}}\,dS = \frac{(-2,1,1)}{\sqrt{6}}\sqrt{6}\,dx\,dy = (-2,1,1)\,dx\,dy. $$

That restores the correct flux, \(-4\). Equivalently, use the oriented cross product \(\mathbf r_x\times\mathbf r_y=(-2,1,1)\) from the start.

A Reliable Setup

Before integrating over a planar patch, identify both its orientation and its parameter region. For a circular disk, coordinates in the disk’s own plane provide a circular parameter region even when the disk is tilted. For a graph over a rectangle or another region in the \(xy\)-plane, the graph parametrization supplies the oriented area factor and the projected region supplies the bounds.

1
Compute the curl.
Find \(\nabla\times\mathbf F\) before substituting the surface parametrization.
2
Choose coordinates for the patch.
Use polar coordinates for a circular parameter region or projected coordinates for a graph patch.
3
Determine the oriented area factor.
Compute \(\mathbf r_u\times\mathbf r_v\), or use the graph formula with the required orientation.
4
Substitute and integrate.
Evaluate the curl on the surface, take its dot product with the oriented area factor, and use bounds covering the patch once.
5
Check the result.
When practical, compute the circulation along the boundary with the direction induced by the chosen normal.

These steps matter because a correct curl with the wrong normal, a correct normal with incompatible bounds, or a missing area factor can each change the result. Stokes’ Theorem provides a particularly effective final check: the computed flux must equal the circulation on the compatibly oriented boundary.

Check Your Understanding

Use the orientation and the bounds explicitly in each setup.

  1. For a disk parametrized by \(\mathbf r(u,v)=\mathbf p+u\mathbf e_1+v\mathbf e_2\), what condition on \(\mathbf e_1\times\mathbf e_2\) ensures the chosen normal orientation?
  2. What are the upward oriented area factor and the projected region for the graph \(z=3x+2y\) over a rectangle in the \(xy\)-plane?
  3. Why does a tilted graph require a surface-area factor when using its unit normal with projected coordinates?
  4. For a tilted circular disk, why can coordinates within the disk’s own plane make the integration bounds simpler than projection onto the \(xy\)-plane?
  5. What boundary direction must be used to check a curl-flux calculation with Stokes’ Theorem?

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