26. Stokes' Theorem

Let \(S\) be a nice surface in \(\mathbb{R}^3\) with a nice properly oriented boundary, \(\partial S\), and let \(\vec{F}\) be a nice vector field on \(S\). Then \[ \iint_S \vec{\nabla}\times\vec{F}\cdot d\vec{S} =\oint_{\partial S} \vec{F}\cdot d\vec{s} \] Each piece of the boundary of the surface must be traversed counterclockwise as seen from the tip of the normal vector to the surface.

a. The Theorem

What happens to Stokes' Theorem when there is no boundary? The theorem still holds, but both sides are now \(0\).

3. Surfaces with No Boundary

In discussing Green's Theorem, the region always has an outer boundary curve which is traversed counterclockwise. In addition, it can have one or more inner boundary curves surounding each hole traversed clockwise.

In discussing Stokes' Theorem, the surface usually has one or more boundary curves, and there is no way to say which is an outer or inner boundary. However, it is also possible for there to be no pieces to the boundary (e.g. a sphere or donut surface).

StokesThm2

Compute \(\displaystyle \iint_S \vec{\nabla}\times\vec{F}\cdot d\vec{S}\) for the vector field \(\vec{F}=\langle yz^2,-xz^2,z^3\rangle\) over the sphere, \(S\), given by \(x^2+y^2+z^2=4\) oriented outward as in the figure above.

Try doing this problem yourself before you read the two solutions.

In the first solution, using a surface integral, parametrize the surface starting from spherical coordinates and setting \(\rho=2\). Compute the curl of \(\vec F\) and evaluate it on the surface. Find the normal vector and check it points outward. Find their dot product and compute the integral for \(0 \lt \theta \lt 2\pi\) and \(0 \lt \phi \lt \pi\).

In the second solution, using a line integral, parametrize any boundary curve, evaluate \(\vec F\) on the curve, dot it into the velocity of the curve and compute the integral.

Surface Integral:

We first compute the integral directly as a surface integral. We parametrize the surface in spherical coordinates with \(\rho=2\): \[ \vec R(\phi,\theta) =\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\rangle \] Next the curl of \(\vec F=\langle yz^2,-xz^2,z^3\rangle\) is: \[\begin{aligned} \vec{\nabla}\times\vec{F} &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ yz^2 & -xz^2 & z^3 \end{vmatrix} \\ &=\hat\imath(0--2xz) -\hat\jmath(0-2yz) +\hat k(-z^2-z^2) \\ &=\langle 2xz,2yz,-2z^2\rangle \end {aligned}\] On the surface, the curl of \(\vec F\) is: \[\begin{aligned} \left.\vec{\nabla}\times\vec{F}\right|_{R(\phi,\theta)} &=\langle 8\sin\phi\cos\phi\cos\theta, 8\sin\phi\cos\phi\sin\theta,-8\cos^2\phi\rangle \end {aligned}\] The normal is: \[\begin{aligned} \vec N&=\vec e_\phi\times\vec e_\theta =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 2\cos\phi\cos\theta & 2\cos\phi\sin\theta & -2\sin\phi \\ -2\sin\phi\sin\theta & 2\sin\phi\cos\theta & 0 \\ \end{vmatrix} \\ &=\hat\imath(0--4\sin^2\phi\cos\theta) -\hat\jmath(0-4\sin^2\phi\sin\theta) \\ &\quad+\hat k(4\sin\phi\cos\phi\cos^2\theta--4\sin\phi\cos\phi\sin^2\theta) \\ &=\langle 4\sin^2\phi\cos\theta,4\sin^2\phi\sin\theta,4\sin\phi\cos\phi\rangle \end {aligned}\] This is radially outward since it is just \(2\sin\phi\) times the position vector \(\vec R(\phi,\theta)\). So the normal is oriented correctly. Next we compute the dot product: \[\begin{aligned} \left.\vec{\nabla}\times\vec{F}\right|_{R(\phi,\theta)}\cdot\vec N &=32(\sin^3\phi\cos\phi\cos^2\theta+\sin^3\phi\cos\phi\sin^2\theta -\cos^3\phi\sin\phi) \\ &=32(\sin^3\phi\cos\phi-\cos^3\phi\sin\phi) \\ \end {aligned}\] Finally, we can compute the integral: \[\begin{aligned} \iint_S \vec{\nabla}\times\vec{F}\cdot d\vec{S} &=\int_0^{2\pi}\int_0^\pi \left.\vec{\nabla}\times\vec{F}\right|_{R(\phi,\theta)} \cdot\vec N\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_0^\pi 32(\sin^3\phi\cos\phi-\cos^3\phi\sin\phi)\,d\phi\,d\theta \\ &=2\pi\left[\dfrac{}{}8\sin^4\phi+8\cos^4\phi\right]_0^\pi \\ &=16\pi(\cos^4\pi-\cos^4 0)=0 \end {aligned}\] All that work to get zero!

Line Integral:

There is no boundary! So there are no integrals to compute. So the line integral is trivially \(0\): \[ \oint_{\partial S} \vec{F}\cdot d\vec{s}=0 \]

What happened in this problem, always happens. Since there is no boundary, there are no line integrals to compute and the line integral side of Stokes' Theorem always gives \(0\). By Stokes' Theorem, the surface integral side must also be \(0\). This proves:

If we integrate the curl of any vector field over a closed surface (i.e. with no boundary), then the integral will always be \(0\).

The torus, \(T\), (donut) shown at the right may be parametrized as: \[ R(\theta,\phi) =((5+3\cos\phi)\cos\theta,(5+3\cos\phi)\sin\theta,3\sin\phi) \] for \(0 \lt \theta \lt 2\pi\) and \(0 \lt \phi \lt 2\pi\). Here \(\phi\) goes around the small circle of radius \(3\), while \(\theta\) goes around the big circle of radius \(5\) measured from the center of the hole to the center of the ring. In cylindrical coordinates, the equation of the torus is \((r-5)^2+z^2=9\) where \(r=5+3\cos\phi\) and \(z=3\sin\phi\).

NoBoundary-exercise

Compute \(\displaystyle \iint_T \vec{\nabla}\times\vec{F}\cdot d\vec{S}\) over this torus, oriented outward, for the vector field \(\vec{F}=\langle yz,-xz,x^2+y^2\rangle\).

  1. Surface Integral:  First solve it directly as a surface integral.

    \(\displaystyle \iint_T \vec{\nabla}\times\vec{F}\cdot d\vec{S}=0\)

    See the remark from the previous example. Since there is no boundary, the integral of the curl of \(\vec F\) is automatically \(0\): \[ \iint_T \vec{\nabla}\times\vec{F}\cdot d\vec{S}=0 \]

  2. Line Integral:  Next use Stokes' Theorem to convert the surface integral into a line integral.

    \(\displaystyle \oint_{\partial T} \vec{F}\cdot d\vec{s}=0\)

    There is no boundary! So there are no integrals to compute. So the line integral is trivially \(0\): \[ \oint_{\partial T} \vec{F}\cdot d\vec{s}=0 \]

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