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.

b. Verification

To verify a theorem means to check it works for specific examples. (It is not a proof.) On previous pages, we have already verified Stokes' Theorem for several functions and surfaces, with \(0\), \(1\) or more pieces to the boundary. So we here just give a few more exercises for practice.

1. LHS - Surface Integral:

   Always compute a divergence or curl in rectangular coordinates! It can be done in curvilinear coordinates, but it is much more complicated and we have not covered it in this class. When you are ready to do the integral, then you convert the divergence or curl into the coordinates needed for the integral.

   \(\displaystyle \iint_P \vec{\nabla}\times\vec{F}\cdot d\vec{S}=32\pi\)

   LHS:   Given the parametrization \[ (x,y,z)=\vec{R}(r,\theta) =\langle r\cos\theta,r\sin\theta,r^2\rangle \] we compute the tangent vectors \(\vec{e}_r\) and \(\vec{e}_{\theta}\) and the normal \(\vec{N}=\vec{e}_r\times\vec{e}_{\theta}\): \[ \begin{array}{c} \\ \vec{e}_r= \\ \vec{e}_{\theta}= \end{array} \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ (\quad \cos\theta & \sin\theta & \;2r\;) \\ (-r\sin\theta & r\cos\theta & \; 0\;) \end{vmatrix} \] \[\begin{aligned} \vec{N}&=\vec{e}_r\times\vec{e}_{\theta} \\ &=\hat{\imath}(-2r^2\cos\theta) -\hat{\jmath}(2r^2\sin\theta) +\hat{k}(r\cos^2\theta--r\sin^2\theta) \\ &=\langle-2r^2\cos\theta,-2r^2\sin\theta,r\rangle \end{aligned}\] In the first octant, all the trig functions are positive; so \(N_1\) and \(N_2\) are negative while \(N_3\) is positive; so \(\vec{N}\) points up and in. Since we need down and out, we reverse \(\vec{N}\): \[ \vec{N}=\langle 2r^2\cos\theta,2r^2\sin\theta,-r\rangle \] Next, we compute the curl of \(\vec{F}=\langle yz,-xz,z^2\rangle\): \[\begin{aligned} \vec{\nabla}\times\vec{F} &=\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ yz & -xz & z^2 \end{vmatrix} \\ &=\hat{\imath}(0--x)-\hat{\jmath}(0-y)+\hat{k}(-z-z) \\ &=\langle x,y,-2z\rangle \end{aligned}\] Now we evaluate the curl on the surface and compute its dot product with the normal: \[ \left.\vec{\nabla}\times\vec{F}\right|_{\vec{R}(r,\theta)} =\langle r\cos\theta,r\sin\theta,-2r^2\rangle \] \[ \left.\vec{\nabla}\times\vec{F}\right|_{\vec{R}(r,\theta)}\cdot\vec{N} =2r^3\cos^2\theta+2r^3\sin^2\theta+2r^3=4r^3 \] The limits for \(\theta\) are \(0 \le\theta \le 2\pi\). To find the limits on \(r\), we compute \[ 0 \le z=r^2 \le 4 \qquad \Longrightarrow \qquad 0 \le r \le 2 \] So the integral is: (Notice there is no extra \(r\) in front of the \(dr\,d\theta\).) \[\begin{aligned} \iint_P &\vec{\nabla}\times\vec{F}\cdot d\vec{S} =\int_0^{2\pi}\int_0^2 \left.\vec{\nabla}\times\vec{F}\right|_{\vec{R}(\phi,\theta)}\cdot\vec{N}\,dr\,d\theta \\ &=\int_0^{2\pi}\int_0^2 4r^3\,dr\,d\theta =2\pi\left[\rule{0pt}{10pt}r^4\right]_0^2=32\pi \end{aligned}\]

2. RHS - Line Integral(s):

   \(\displaystyle \oint_{\partial P} \vec{F}\cdot d\vec{s}=32\pi\)

   RHS:   The boundary, \(\partial P\), is the circle \(x^2+y^2=4\) at \(z=4\). It may be parametrized by setting \(r=2\) in the parametrization of the surface: \[ (x,y,z)=\vec{r}(\theta) =\langle 2\cos\theta,2\sin\theta,4\rangle \] The tangent vector is \[ \vec{v}=(-2\sin\theta,2\cos\theta,0) \] This is counterclockwise, but we need clockwise. So we reverse it: \[ \vec{v}=(2\sin\theta,-2\cos\theta,0) \]

   

   Next we evaluate \(\vec{F}=\langle yz,-xz,z^2\rangle\) on the curve and compute its dot product with the tangent vector: \[ \left.\vec{F}\right|_{\vec{r}(\theta)} =\langle 8\sin\theta,-8\cos\theta,16\rangle \] \[ \left.\vec{F}\right|_{\vec{r}(\theta)}\cdot\vec{v} =16\sin^2\theta+16\cos^2\theta=16 \] Finally we compute the line integral around the boundary circle: \[ \oint_{\partial P} \vec{F}\cdot d\vec{s} =\int_0^{2\pi} \vec{F}\cdot\vec{v}\,d\theta =\int_0^{2\pi} 16\,dt=32\pi \]

   Comparison:   The left and the right sides of the equation gave the same answer. Thus Stokes' Theorem is verified for this example. The line integral was easier.

1. LHS - Surface Integral:

   Always compute a divergence or curl in rectangular coordinates! It can be done in curvilinear coordinates, but it is much more complicated and we have not covered it in this class. When you are ready to do the integral, then you convert the divergence or curl into the coordinates needed for the integral.

   \(\displaystyle \iint_H \vec{\nabla}\times\vec{F}\cdot d\vec{S}=32\pi\)

   LHS:   We parametrize the hemisphere using spherical coordinates with \(\rho=4\): \[ (x,y,z)=\vec{R}(\phi,\theta) =(4\sin\phi\cos\theta,4\sin\phi\sin\theta,4\cos\phi) \] Then we compute the tangent vectors \(\vec{e}_\phi\) and \(\vec{e}_\theta\) and the normal \(\vec{N}=\vec{e}_\phi\times\vec{e}_\theta\): \[ \begin{array}{c} \\ \vec{e}_\phi= \\ \vec{e}_\theta= \end{array} \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ (4\cos\phi\cos\theta & 4\cos\phi\sin\theta & -4\sin\phi) \\ (-4\sin\phi\sin\theta & 4\sin\phi\cos\theta & \quad 0\quad) \end{vmatrix} \] \[\begin{aligned} \vec{N} &=\vec{e}_\phi\times\vec{e}_\theta \\ &=\hat{\imath}(0--16\sin^2\phi\cos\theta) -\hat{\jmath}(0-16\sin^2\phi\sin\theta) \\ &\quad+\hat{k}(16\sin\phi\cos\phi\cos^2\theta--16\sin\phi\cos\phi\sin^2\theta) \\ &=16\sin\phi\langle\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi\rangle \end{aligned}\] In the first octant, all the trig functions are positive; so all the components of \(\vec{N}\) are positive; so \(\vec{N}\) points out and up as desired. Also notice that \(\vec N\) is parallel to \(\vec R\) as it should be for a sphere.

   Next, we compute the curl of \(\vec{F}=\langle-4y,4x,z(x^2+y^2)\rangle\): \[\begin{aligned} \vec{\nabla}\times\vec{F} &=\begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ -4y & 4x & z(x^2+y^2) \end{vmatrix} \\ &=\hat{\imath}(2yz-0)-\hat{\jmath}(2xz-0)+\hat{k}(4--4) \\ &=\langle 2yz,-2xz,8\rangle \end{aligned}\] Now we evaluate the curl on the surface and compute its dot product with the normal: \[ \left.\vec{\nabla}\times\vec{F}\right|_{\vec{R}(\phi,\theta) } =\langle 32\sin\phi\cos\phi\sin\theta, -32\sin\phi\cos\phi\cos\theta,8\rangle \] \[\begin{aligned} \left.\vec{\nabla}\times\vec{F}\right|_{\vec{R}(\phi,\theta)}\cdot\vec{N} &=16\sin\phi(32\sin^2\phi\cos\phi\sin\theta\cos\theta \\ &\quad-32\sin^2\phi\cos\phi\sin\theta\cos\theta+8\cos\phi) \\ &=128\sin\phi\cos\phi \end{aligned}\] The limits for \(\theta\) are \(0 \le \theta \le 2\pi\). To find the limits on \(\phi\), we compute \[\begin{aligned} 0 \le z \le 2 \qquad 0 \le 4\cos\phi \le 2 \\ 0 \le \cos\phi \le \dfrac{1}{2} \qquad \phi=\dfrac{\pi}{2},\dfrac{\pi}{3} \end{aligned}\] So the integral is: \[\begin{aligned} \iint_H &\vec{\nabla}\times\vec{F}\cdot d\vec{S} =\int_0^{2\pi}\int_{\pi/3}^{\pi/2} \left.\vec{\nabla}\times\vec{F}\right|_{\vec{R}(\phi,\theta)} \cdot\vec{N}\,d\phi\,d\theta \\ &=\int_0^{2\pi}\int_{\pi/3}^{\pi/2} 128\sin\phi\cos\phi\,d\phi\,d\theta \\ &=256\pi\left[\dfrac{\sin^2\phi}{2}\right]_{\pi/3}^{\pi/2} =256\pi\left(\dfrac{1}{2}-\dfrac{3}{8}\right)=32\pi \end{aligned}\]

2. RHS - Line Integral(s):

   \(\displaystyle \oint_{\partial H} \vec{F}\cdot d\vec{s}=32\pi\)

   RHS:   The boundary of the surface consists of two pieces: a top circle at \(z=2\) or \(\phi=\dfrac{\pi}{3}\) and a bottom circle at \(z=0\) or \(\phi=\dfrac{\pi}{2}\). We must compute the line integral \(\displaystyle \oint_{\partial S} \vec{F}\cdot d\vec{s}\) for the two pieces and add them together.

   Since the normal to the surface points out, we must traverse the top circle clockwise and the bottom circle counterclockwise as seen from the positive \(z\)-axis. The traversal of the boundary curves are shown in the animation.

   

     Top:   We parametrize the top circle by substituting \(\phi=\dfrac{\pi}{3}\) into the parametrization of the surface: \[\begin{aligned} (x,y,z)=\vec{r}(\theta) &=\left(4\sin\dfrac{\pi}{3}\cos\theta,4\sin\dfrac{\pi}{3}\sin\theta, 4\cos\dfrac{\pi}{3}\right) \\ &=(2\sqrt{3}\cos\theta,2\sqrt{3}\sin\theta,2) \end{aligned}\] The tangent vector is \[ \vec{v}=(-2\sqrt{3}\sin\theta,2\sqrt{3}\cos\theta,0) \] This is counterclockwise, so we reverse it: \[ \vec{v}=(2\sqrt{3}\sin\theta,-2\sqrt{3}\cos\theta,0) \] Next we evaluate \(\vec{F}=\langle-4y,4x,z(x^2+y^2)\rangle\) on the curve and compute its dot product with the tangent vector: \[ \left.\vec{F}\right|_{\vec{r}(\theta)} =\langle-8\sqrt{3}\sin\theta,8\sqrt{3}\cos\theta,24\rangle \] \[ \left.\vec{F}\right|_{\vec{r}(\theta)}\cdot\vec{v} =-48\sin^2\theta-48\cos^2\theta=-48 \] Finally we compute the line integral around the top circle \[ \oint_{\text{top}} \vec{F}\cdot d\vec{s} =\int_0^{2\pi} \vec{F}\cdot\vec{v}\,d\theta =-\int_0^{2\pi} 48\,dt=-96\pi \]

     Bottom:   We parametrize the bottom circle by substituting \(\phi=\dfrac{\pi}{2}\) into the parametrization of the surface: \[\begin{aligned} (x,y,z)=\vec{r}(\theta) &=\left(4\sin\dfrac{\pi}{2}\cos\theta,4\sin\dfrac{\pi}{2}\sin\theta, 4\cos\dfrac{\pi}{2}\right) \\ &=(4\cos\theta,4\sin\theta,0) \end{aligned}\] The tangent vector is \[ \vec{v}=(-4\sin\theta,4\cos\theta,0) \] This is counterclockwise, so the orientation is correct. Next we evaluate \(\vec{F}=\langle-4y,4x,z(x^2+y^2)\rangle\) on the curve and compute its dot product with the tangent vector: \[ \left.\vec{F}\right|_{\vec{r}(\theta)} =\langle-16\sin\theta,16\cos\theta,0\rangle \] \[ \left.\vec{F}\right|_{\vec{r}(\theta)}\cdot\vec{v} =64\sin^2\theta+64\cos^2\theta=64 \] Finally we compute the line integral around the bottom circle \[ \oint_{\text{bot}} \vec{F}\cdot d\vec{s} =\int_0^{2\pi} \vec{F}\cdot\vec{v}\,d\theta =\int_0^{2\pi} 64\,dt=128\pi \]   Total:   We complete the computation of the right hand side by adding together the line integrals from the two pieces of the boundary: \[ \oint_{\partial S} \vec{F}\cdot d\vec{s} =\oint_{\text{top}} \vec{F}\cdot d\vec{s} +\oint_{\text{bot}} \vec{F}\cdot d\vec{s} =-96\pi+128\pi=32\pi \]

   Comparison:   The left and the right sides of the equation gave the same answer. Thus Stokes' Theorem is verified for this example.