26. Stokes' Theorem

a. The Theorem

What happens to Stokes' Theorem when the boundary of the surface consists of more than one curve? The theorem still holds, but there are some special directional considerations.

2. Surfaces whose Boundaries have Several Curves

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.

The boundary, \(\partial S,\) of the surface, \(S,\) is usually a single closed curve surrounding the surface. However, if the boundary of the surface has more than one closed curve, (as in the figure at the right) then each piece must be traversed "counterclockwise" as seen from the tip of the normal to \(S\). This notion of counterclockwise needs more explanation which is given in the "More Details" below.

The vector field \(\vec{F}\) is nice if it has continuous first partial derivatives. Similarly, the surface, \(S,\) and its boundary, \(\partial S,\) are nice if they are defined by parametric functions with continuous first partial derivatives.

Show Hand

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, if there are two or more pieces to the boundary, there is no way to say which piece is the outer curve. For example, the cylinder shown at the right has two pieces to its boundary and neither can be regarded as the outer curve.

Assuming the cylinder is oriented by an outward normal, to determine which direction to traverse each piece of the boundary, draw a small circle on the surface with counterclockwise arrows as seen from the tip of the normal. Equivalently, put the fist of your right hand on the small circle with your thumb pointing in the direction of the normal. Then your fingers point counterclockwise on the circle. (Click on Show Hand and try it.) Move the small circle close to each boundary and see which direction the arrows say you should travel along the boundary. In this sense, both pieces of the boundary are traversed counterclockwise.

However, if we look at this cylinder from above, then the top circle is traversed clockwise, while the bottom circle is traversed counterclockwise.



Show Hand

Compute \(\displaystyle \iint_C \vec{\nabla}\times\vec{F}\cdot d\vec{S}\) for the vector field \(\vec{F} =\left\langle \dfrac{y}{z},-\,\dfrac{x}{z},\dfrac{x+y}{z}\right\rangle\) over the cylinder, \(C\), given by \(x^2+y^2=4\) for \(1 \le z \le 4\) oriented outward.

eg_cyl_arrows_anim

We use Stokes' Theorem to convert the surface integral over the cylinder into a line integral around the boundary of the cylinder which is \(2\) circles. \[\begin{aligned} \iint_C &\vec{\nabla}\times\vec{F}\cdot d\vec{S} =\oint_{\partial C} \vec{F}\cdot d\vec{s} \\ &=\oint_\text{lower} \vec{F}\cdot d\vec{s} +\oint_\text{upper} \vec{F}\cdot d\vec{s} \end {aligned}\]   Lower Circle:   The lower circle with radius \(2\) at \(z=1\), may be parametrized by \(\vec r_1(\theta)=(2\cos\theta,2\sin\theta,1)\) for \(0 \le \theta \le 2\pi\). So on the curve, the vector field \(\vec{F} =\left\langle \dfrac{y}{z},-\,\dfrac{x}{z},\dfrac{x+y}{z}\right\rangle\) becomes \[ \vec{F}(\vec r_1) =\left\langle 2\sin\theta,-2\cos\theta,2\cos\theta+2\sin\theta\right\rangle \] The velocity is: \[ \vec v_1=\langle -2\sin\theta,2\cos\theta,0\rangle \] This points counterclockwise as seen from above (since the \(x\) component is negative and the \(y\) component is positive in the \(1^\text{st}\) quadrant). This is the correct orientation. (See the figure above.) Then the dot product is: \[ \vec{F}(\vec r_1)\cdot\vec v_1 =-2\sin\theta\,2\sin\theta-2\cos\theta\,2\cos\theta=-4 \] And the line integral is: \[\begin{aligned} \int_{\vec r_1} \vec{F}\cdot d\vec{s} &=\int_0^{2\pi} \vec{F}(\vec r_1)\cdot\vec v_1\,d\theta \\ &=\int_0^{2\pi} -4\,d\theta=-8\pi \end{aligned}\]   Upper Circle:   The upper circle with radius \(2\) at \(z=4\), may be parametrized by \(\vec r_2(\theta)=(2\cos\theta,2\sin\theta,4)\) for \(0 \le \theta \le 2\pi\). So on the curve, the vector field \(\vec{F} =\left\langle \dfrac{y}{z},-\,\dfrac{x}{z},\dfrac{x+y}{z}\right\rangle\) becomes \[\begin{aligned} \vec{F}(\vec r_2) &=\left\langle \dfrac{2\sin\theta}{4},-\,\dfrac{2\cos\theta}{4}, \dfrac{2\cos\theta+2\sin\theta}{4}\right\rangle \\ &=\left\langle \dfrac{\sin\theta}{2},-\,\dfrac{\cos\theta}{2}, \dfrac{\cos\theta+\sin\theta}{2}\right\rangle \end{aligned}\] The velocity is: \[ \vec v_2=\langle -2\sin\theta,2\cos\theta,0\rangle \] This points counterclockwise as seen from above (since the \(x\) component is negative and the \(y\) component is positive in the \(1^\text{st}\) quadrant). Since the normal points outward, the upper circle must be traversed clockwise as seen from above. (See the figure above.) So we reverse the velocity: \[ \text{Reverse:}\qquad\vec v_2=\langle 2\sin\theta,-2\cos\theta,0\rangle \] Then the dot product is: \[ \vec{F}(\vec r_2)\cdot\vec v_2 =\dfrac{2\sin\theta}{4}2\sin\theta+\dfrac{2\cos\theta}{4}2\cos\theta=1 \] And the line integral is: \[\begin{aligned} \int_{\vec r_2} \vec{F}\cdot d\vec{s} &=\int_0^{2\pi} \vec{F}(\vec r_2)\cdot\vec v_2\,d\theta \\ &=\int_0^{2\pi} 1\,d\theta=2\pi \end{aligned}\]

  Total:   We add the line integrals: \[\begin{aligned} \oint_{\partial C} \vec{F}\cdot d\vec{s} &=\oint_\text{lower} \vec{F}\cdot d\vec{s} +\oint_\text{upper} \vec{F}\cdot d\vec{s} \\ &=-8\pi+2\pi=-6\pi \end{aligned}\]

Check: Surface Integral Directly

To check, we compute the surface integral directly. We start by computing the curl of \(\vec F\): \[\begin{aligned} \vec{\nabla}\times\vec{F} &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ \dfrac{y}{z} & -\,\dfrac{x}{z} & \dfrac{x+y}{z} \end{vmatrix} \\ &=\hat\imath\left(\dfrac{1}{z}-\dfrac{x}{z^2}\right) -\hat\jmath\left(\dfrac{1}{z}--\dfrac{y}{z^2}\right) +\hat k\left(-\dfrac{1}{z}-\dfrac{1}{z}\right) \\ &=\left\langle \dfrac{1}{z}-\dfrac{x}{z^2},-\dfrac{1}{z}-\dfrac{y}{z^2}, -\dfrac{2}{z}\right\rangle \end {aligned}\] In cylindrical coordinates, the surface is \(r=2\). So a parametrization is: \[ \vec R(\theta,z)=\langle 2\cos\theta,2\sin\theta,z\rangle \] On the surface, the curl of \(\vec F\) is: \[\begin{aligned} \left.\vec{\nabla}\times\vec{F}\right|_{R(\theta,z)} &=\left\langle \dfrac{1}{z}-\dfrac{2\cos\theta}{z^2}, -\dfrac{1}{z}-\dfrac{2\sin\theta}{z^2}, -\dfrac{2}{z}\right\rangle \end {aligned}\] The normal is: \[\begin{aligned} \vec N&=\vec e_\theta\times\vec e_z =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ -2\sin\theta & 2\cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &=\hat\imath(2\cos\theta-0)-\hat\jmath(-2\sin\theta-0) +\hat k(0) \\ &=\langle 2\cos\theta,2\sin\theta,0 \rangle \end {aligned}\] This is horizontal (since the \(z\)-component is \(0\)) and points outward (since the \(x\) and \(y\) components are positive in the first quadrant). So the normal is oriented correctly. Next we compute the dot product: \[\begin{aligned} \left.\vec{\nabla}\times\vec{F}\right|_{R(\theta,z)}\cdot\vec N &=\dfrac{2\cos\theta}{z}-\dfrac{4\cos^2\theta}{z^2} -\dfrac{2\sin\theta}{z}-\dfrac{4\sin^2\theta}{z^2} \\ &=\dfrac{2\cos\theta-2\sin\theta}{z}-\dfrac{4}{z^2} \end {aligned}\] Finally, we can compute the integral: \[\begin{aligned} \iint_C \vec{\nabla}\times\vec{F}\cdot d\vec{S} &=\int_0^{2\pi}\int_1^4 \left.\vec{\nabla}\times\vec{F}\right|_{R(\theta,z)} \cdot\vec N\,d\theta\,dz \\ &=\int_0^{2\pi}\int_1^4 \left(\dfrac{2\cos\theta-2\sin\theta}{z}-\dfrac{4}{z^2}\right)\,d\theta\,dz \end {aligned}\] We do the \(\theta\) integral first, noting that \[ \int_0^{2\pi} \cos\theta\,d\theta=\int_0^{2\pi} \sin\theta\,d\theta=0 \] So \[\begin{aligned} \iint_C &\vec{\nabla}\times\vec{F}\cdot d\vec{S} =-2\pi\int_1^4 \dfrac{4}{z^2}\,dz =8\pi\left[\dfrac{1}{z}\right]_1^4 \\ &=8\pi\left(\dfrac{1}{4}-1\right)=2\pi-8\pi=-6\pi \end {aligned}\]

Both methods gave the same answer. This checks both computations and also verifies Stokes' Theorem for this function and surface. This time the two methods were about equal difficulty..

Use Stokes' Theorem to compute \(\displaystyle \iint_P \vec{\nabla}\times\vec{F}\cdot d\vec{S}\) for the vector field \(\vec{F} =\langle yz,-xz,x^2+y^2\rangle\) over the piece of the paraboloid, \(P\), given by \(z=14-x^2-y^2\) for \(5 \le z \le 10\) oriented down and in.

ex_parab_arrows_anim

\(\displaystyle \iint_P \vec{\nabla}\times\vec{F}\cdot d\vec{S} =\oint_{\partial P} \vec{F}\cdot d\vec{s}=10\pi\)

By Stokes' Theorem: \[ \iint_P \vec{\nabla}\times\vec{F}\cdot d\vec{S} =\oint_{\partial P} \vec{F}\cdot d\vec{s} \] So we now compute the line integral around the boundary which is \(2\) circles. The lower one, at height \(z=5\), has radius \(r=3\) and must be traversed clockwise. The upper one, at height \(z=10\), has radius \(r=2\) and must be traversed counterclockwise.

  Lower Circle:   The lower circle may be parametrized as: \[ \vec r_1(\theta)=(3\cos\theta,3\sin\theta,5) \] The velocity is: \[ \vec v_1=\langle -3\sin\theta,3\cos\theta,0\rangle \] This is counterclockwise. So we reverse it: \[ \text{Reverse:}\qquad\vec v_1=\langle 3\sin\theta,-3\cos\theta,0\rangle \] On the circle, the vector field \(\vec{F}=\langle yz,-xz,x^2+y^2\rangle\) becomes: \[ \left.\vec{F}\right|_{\vec r_1(\theta)} =\langle 15\sin\theta,-15\cos\theta,9\rangle \] So the line integral is: \[\begin{aligned} \oint_{\vec r_1} \vec{F}\cdot d\vec{s} &=\int_0^{2\pi} \vec{F}\cdot\vec v_1\,d\theta \\ &=\int_0^{2\pi} (45\sin^2\theta+45\cos^2\theta)\,d\theta \\ &=45(2\pi)=90\pi \end {aligned}\]   Upper Circle:   The upper circle may be parametrized as: \[ \vec r_1(\theta)=(2\cos\theta,2\sin\theta,10) \] The velocity is: \[ \vec v_1=\langle -2\sin\theta,2\cos\theta,0\rangle \] This is counterclockwise as required. On the circle, the vector field \(\vec{F}=\langle yz,-xz,x^2+y^2\rangle\) becomes: \[ \left.\vec{F}\right|_{\vec r_2(\theta)} =\langle 20\sin\theta,-20\cos\theta,4\rangle \] So the line integral is: \[\begin{aligned} \oint_{\vec r_2} \vec{F}\cdot d\vec{s} &=\int_0^{2\pi} \vec{F}\cdot\vec v_2\,d\theta \\ &=\int_0^{2\pi} (-40\sin^2\theta-40\cos^2\theta)\,d\theta \\ &=-40(2\pi)=-80\pi \end {aligned}\]   Total:   We add the line integrals: \[\begin{aligned} \oint_{\partial P} \vec{F}\cdot d\vec{s} &=\oint_\text{lower} \vec{F}\cdot d\vec{s} +\oint_\text{upper} \vec{F}\cdot d\vec{s} \\ &=90\pi-80\pi=10\pi \end{aligned}\]

We check by computing the surface integral directly. We start by computing the curl of \(\vec F\): \[\begin{aligned} \vec{\nabla}\times\vec{F} &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \partial_x & \partial_y & \partial_z \\ yz & -xz & x^2+y^2 \end{vmatrix} \\ &=\hat\imath(2y--x)-\hat\jmath(2x-y)+\hat k(-z-z) \\ &=\langle 2y+x,-2x+y,-2z\rangle \end {aligned}\] Since the surface is the parabola, \(z=14-x^2-y^2=14-r^2\), when \(z=5\) we have \(r=3\) and when \(z=10\) we have \(r=2\). We parametrize starting with cylindrical coordinates: \[ \vec R(r,\theta) =\langle r\cos\theta,r\sin\theta,14-r^2\rangle \] On the surface, the curl of \(\vec F\) is: \[ \left.\vec{\nabla}\times\vec{F}\right|_{R(r,\theta)} =\langle 2r\sin\theta+r\cos\theta,-2r\cos\theta+r\sin\theta,-28+2r^2\rangle \] The normal is: \[\begin{aligned} \vec N&=\vec e_r\times\vec e_\theta =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ \cos\theta & \sin\theta & -2r \\ -r\sin\theta & r\cos\theta & 0 \end{vmatrix} \\ &=\hat\imath(0--2r^2\cos\theta)-\hat\jmath(0-2r^2\sin\theta) \\ &\quad+\hat k(r\cos^2\theta--r\sin^2\theta) \\ &=\langle 2r^2\cos\theta,2r^2\sin\theta,r\rangle \end {aligned}\] This points up and out but we need down and in. So we reverse it: \[ \text{Reverse:}\qquad\vec N=\langle -2r^2\cos\theta,-2r^2\sin\theta,-r\rangle \] Finally, we can compute the integral: \[\begin{aligned} \iint_P &\vec{\nabla}\times\vec{F}\cdot d\vec{S} =\int_0^{2\pi}\int_2^3 \left.\vec{\nabla}\times\vec{F}\right|_{R(r,\theta)} \cdot\vec N\,dr\,d\theta \\ &=\int_0^{2\pi}\int_2^3 (-4r^3\sin\theta\cos\theta-2r^3\cos^2\theta \\ &\quad+4r^3\cos\theta\sin\theta-2r^3\sin^2\theta +28r-2r^3)\,dr\,d\theta \\ &=\int_0^{2\pi}\int_2^3 (-2r^3+28r-2r^3)\,dr\,d\theta \\ &=2\pi\int_2^3 28r-4r^3\,dr =2\pi\left[14r^2-r^4\right]_2^3 \\ &=2\pi(126-81)-2\pi(56-16) =10\pi \end {aligned}\]

Both methods gave the same answer. This checks both computations and also verifies Stokes' Theorem for this function and surface. This time the surface integral was easier.

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