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Stokes’ Theorem · Tutorial 933 of 999

Using Hemispheres and Caps as Spanning Surfaces

Compare curl flux through hemispheres and planar caps by matching each surface normal to its induced direction around their common circle.

Advanced 10 min read

What You'll Learn

  • Identify the common circular boundary of an upper hemisphere and its planar cap.
  • Determine the boundary direction induced by an outward or inward hemisphere normal.
  • Match the hemisphere orientation with the upward or downward orientation of its cap.
  • Apply Stokes’ theorem to compute and compare curl fluxes through the two surfaces.
  • Detect and correct sign errors caused by incompatible orientations.

One Circle, Two Spanning Surfaces

A circle in space can bound many different surfaces. For a circle in the plane \(z=0\), two natural choices are the disk it encloses and the upper hemisphere whose rim is that circle. Stokes’ theorem relates the curl flux through each surface to a line integral around its boundary—but only when the direction of that boundary agrees with the surface’s orientation.

This orientation condition is the central point. An outward-oriented upper hemisphere and an upward-oriented disk induce the same direction around their common circle. If the disk is oriented downward instead, it induces the opposite direction. The surfaces can share a geometric boundary while having different oriented boundaries.

Definition: The induced orientation of a boundary curve is the direction determined by the chosen normal to its surface using the right-hand rule. For a planar disk with upward normal, the induced direction is counterclockwise when viewed from above.

Set-Up and Orientation of the Common Circle

Let \(R>0\). Write the upper hemisphere, its circular boundary, and the planar cap as

$$ H_R=\{(x,y,z):x^2+y^2+z^2=R^2,\ z\geq 0\}, \qquad C_R=\{(x,y,0):x^2+y^2=R^2\}, $$
$$ D_R=\{(x,y,0):x^2+y^2\leq R^2\}. $$

Choose the outward normal on the hemisphere and the upward normal on the disk. A convenient parametrization of the circle is

$$ \mathbf{c}(t)=(R\cos t,R\sin t,0), \qquad 0\leq t\leq 2\pi. $$

As \(t\) increases, this parametrization travels counterclockwise when viewed from the positive \(z\)-axis. The upward-oriented disk induces exactly this direction. The outward-oriented hemisphere does too. At the point \((R,0,0)\), the hemisphere’s outward normal points in the positive \(x\)-direction. A tangent in the positive \(y\)-direction makes the inward direction along the surface point toward negative \(z\), into the hemisphere. Thus the increasing-\(t\) tangent is consistent with the hemisphere’s outward orientation.

The same fact can be checked using a parametrization of the hemisphere:

$$ \mathbf{r}(\theta,\phi) = (R\sin\theta\cos\phi,\ R\sin\theta\sin\phi,\ R\cos\theta), \quad 0\leq\theta\leq\frac{\pi}{2}, \quad 0\leq\phi\leq2\pi. $$

The cross product \(\mathbf{r}_{\theta}\times\mathbf{r}_{\phi}\) points outward. At the rim \(\theta=\pi/2\), increasing \(\phi\) travels counterclockwise around \(C_R\). The parametrization degenerates at the pole, but the hemisphere itself is smooth there; the pole does not change the boundary orientation.

Orientation rule for this pair: The outward-oriented upper hemisphere and the upward-oriented planar cap both induce counterclockwise travel around their common circle. Reversing either surface normal reverses its induced boundary direction.

Why the Two Curl-Flux Integrals Agree

Stokes’ theorem states that if an oriented surface \(S\) has a suitably smooth boundary and \(\mathbf{F}\) has continuous first partial derivatives on an open region containing \(S\), then

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

where the boundary is traversed in the direction induced by the chosen surface normal. Apply this theorem to both \(H_R\), with its outward normal, and \(D_R\), with its upward normal. Both surfaces induce the same counterclockwise traversal of \(C_R\), so both right-hand sides are the same line integral:

$$ \iint_{H_R}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS = \oint_{C_R,\ \mathrm{CCW}}\mathbf{F}\cdot d\mathbf{r} = \iint_{D_R}(\nabla\times\mathbf{F})\cdot\mathbf{k}\,dA. $$

Here \(\mathbf{k}=(0,0,1)\) is the upward unit normal. This equality is a direct consequence of applying Stokes’ theorem twice with compatible boundary directions; it does not say that the curl has the same values on the two surfaces.

1
Choose the hemisphere orientation.
For the upper hemisphere, outward means the normal points away from the sphere’s center.
2
Find its induced circle direction.
An outward-oriented upper hemisphere induces counterclockwise travel around its rim when viewed from above.
3
Orient the cap to match.
The upward disk induces the same counterclockwise direction, so Stokes’ theorem gives equal curl fluxes.

Worked Example: A Constant Curl on a Unit Hemisphere

Worked Example: A Constant Curl on a Unit Hemisphere

Let \(H_1\) be the outward-oriented upper unit hemisphere and \(D_1\) the upward unit disk. For \(\mathbf{F}(x,y,z)=(-y,x,0)\), compute the curl flux through each surface and verify that they agree.

First compute the curl from its components:

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

On the disk, the upward unit normal is \(\mathbf{k}\), so the curl flux is

$$ \iint_{D_1}(\nabla\times\mathbf{F})\cdot\mathbf{k}\,dA = \iint_{D_1}2\,dA = 2\cdot\pi(1)^2 = 2\pi. $$

On the unit sphere, the outward unit normal at \((x,y,z)\) is \((x,y,z)\). Thus the integrand is \(2z\). Using the spherical parametrization above, with \(R=1\), the area element is \(dS=\sin\theta\,d\theta\,d\phi\) and \(z=\cos\theta\). Therefore,

$$ \begin{aligned} \iint_{H_1}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS &=\int_0^{2\pi}\int_0^{\pi/2} 2\cos\theta\sin\theta\,d\theta\,d\phi\\ &=2\pi\left[\sin^2\theta\right]_0^{\pi/2}\\ &=2\pi(1-0)=2\pi. \end{aligned} $$

As a check, the common boundary is \(\mathbf{c}(t)=(\cos t,\sin t,0)\), so \(d\mathbf{c}=(-\sin t,\cos t,0)\,dt\). Along it, \(\mathbf{F}=(-\sin t,\cos t,0)\), giving

$$ \oint_{C_1,\ \mathrm{CCW}}\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} \bigl(\sin^2t+\cos^2t\bigr)\,dt = 2\pi. $$

Both surface fluxes match the counterclockwise boundary integral.

Worked Example: A Nonconstant Curl on the Upper Hemisphere

Worked Example: A Nonconstant Curl on the Upper Hemisphere

Use \(\mathbf{F}(x,y,z)=(-y(1+z),x(1+z),0)\), the outward upper unit hemisphere, and the upward unit disk. Compute each curl flux and check it using the common boundary.

The curl is

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

On the upward disk, \(z=0\), so the vertical component of the curl is \(2\). Hence

$$ \iint_{D_1}(\nabla\times\mathbf{F})\cdot\mathbf{k}\,dA = \iint_{D_1}2\,dA = 2\pi. $$

On the unit hemisphere, the outward unit normal is \((x,y,z)\). The dot product with the curl is

$$ (-x,-y,2+2z)\cdot(x,y,z) = -x^2-y^2+2z+2z^2. $$

Substitute \(x=\sin\theta\cos\phi\), \(y=\sin\theta\sin\phi\), and \(z=\cos\theta\). Since \(x^2+y^2=\sin^2\theta\), the flux is

$$ \begin{aligned} \iint_{H_1}(\nabla\times\mathbf{F})\cdot\mathbf{n}\,dS &=\int_0^{2\pi}\int_0^{\pi/2} \left(-\sin^2\theta+2\cos\theta+2\cos^2\theta\right) \sin\theta\,d\theta\,d\phi\\ &=2\pi\left( -\int_0^{\pi/2}\sin^3\theta\,d\theta +2\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta +2\int_0^{\pi/2}\cos^2\theta\sin\theta\,d\theta \right)\\ &=2\pi\left(-\frac{2}{3}+1+\frac{2}{3}\right)\\ &=2\pi. \end{aligned} $$

For a direct check, on the rim \(z=0\), the field becomes \((-y,x,0)\). The boundary line integral is therefore the one computed in the preceding example, namely \(2\pi\). The hemisphere and disk fluxes both equal this value, even though the curl is not constant on the hemisphere.

Worked Example: Reversing the Cap Orientation

Worked Example: Reversing the Cap Orientation

Take the sphere of radius \(2\), the field \(\mathbf{F}(x,y,z)=(-y,x,0)\), and orient the upper hemisphere outward. Compare its boundary integral with the flux through the downward-oriented disk of radius \(2\).

The curl remains \((0,0,2)\). The outward hemisphere induces counterclockwise travel. On its boundary, use \(\mathbf{c}(t)=(2\cos t,2\sin t,0)\) for \(0\leq t\leq2\pi\). Then \(d\mathbf{c}=(-2\sin t,2\cos t,0)\,dt\), and \(\mathbf{F}=(-2\sin t,2\cos t,0)\). Thus

$$ \oint_{C_2,\ \mathrm{CCW}}\mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi}4\,dt = 8\pi. $$

By Stokes’ theorem, the outward hemisphere’s curl flux is \(8\pi\). The downward disk has unit normal \(-\mathbf{k}\), so its curl flux is

$$ \iint_{D_2}(\nabla\times\mathbf{F})\cdot(-\mathbf{k})\,dA = \iint_{D_2}(-2)\,dA = -2\cdot\pi(2)^2 = -8\pi. $$

This negative flux agrees with the disk’s induced clockwise direction: reversing the normal reverses the boundary direction and changes the line integral from \(8\pi\) to \(-8\pi\). The downward disk matches an inward-oriented upper hemisphere, not the outward-oriented one.

Common Pitfall: Matching the Circle but Not Its Direction

A common error is to notice that two surfaces have the same geometric boundary and conclude that their Stokes integrals must have the same sign. The theorem equates a surface integral with the line integral in its induced direction; it does not disregard orientation.

For example, in the radius-\(2\) calculation above, a tempting but incorrect statement is: “The outward hemisphere has flux \(8\pi\), and the downward cap has flux \(-8\pi\), so the two fluxes agree because the surfaces share the circle.” They do not agree. The correct comparison is either outward hemisphere with upward disk, giving \(8\pi\) for each, or inward hemisphere with downward disk, giving \(-8\pi\) for each. In each valid comparison, the induced boundary directions match.

Check before applying Stokes: Record the chosen normal on each surface, determine the resulting direction around the circle, and compare those directions before equating the line integrals.

Check Your Understanding

For each question, keep track of both the geometric circle and its induced direction.

  1. Viewed from above, which direction is induced on the rim of an outward-oriented upper hemisphere?
  2. Which orientation of a planar disk induces the same boundary direction as an outward-oriented upper hemisphere?
  3. If the normal of the upward disk is reversed, what happens to its induced boundary direction and its curl flux?
  4. For \(\mathbf{F}=(-y,x,0)\) on the unit circle, what is the counterclockwise line integral, and what is the curl flux through the upward unit disk?
  5. Why is it not enough to know that a hemisphere and a cap have the same boundary curve before comparing their Stokes integrals?

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