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Section 2.7 Stokes's theorem

Definition 2.7.1. Induced orientation.

Let \(\mathcal{S}\) be a smooth surface with orientation \(\boldn\colon \mathcal{S}\rightarrow \R^3\text{.}\) Given a simple curve \(\mathcal{C}\subseteq \mathcal{S}\) lying partly on the boundary of \(\mathcal{S}\) the induced orientation on \(\mathcal{C}\) is the one given by the following right-hand rule: if your thumb points along \(\boldn\text{,}\) then your fingers curl in the direction of the orientation of \(\mathcal{C}\text{.}\) Alternatively, the orientation of \(\mathcal{C}\) is induced by \(\boldn\) if looking down along \(\boldn\text{,}\) the orientation of \(\mathcal{C}\) is such that the portion of \(\mathcal{S}\) the curve bounds lies on the left.

Definition 2.7.2. Orientation of piecewise smooth surface.

Let \(\mathcal{S}=\cup_{i=1}^k\mathcal{S}_k\) be a piecewise smooth surface, and assume each surface \(\mathcal{S}_k\) is orientable. An orientation of \(\mathcal{S}\) is a choice of orientations \(\boldn_k\) on each surface \(\mathcal{S}_k\) such that the following condition is satisfied: given any two surfaces \(\mathcal{S}_{k_1}, \mathcal{S}_{k_2}\) intersecting in the curve \(\mathcal{C}\text{,}\) the orientation induced on \(\mathcal{C}\) by the orientation \(\boldn_{k_1}\) is the opposite of the orientation induced by the orientation \(\boldn_{k_2}\text{.}\)

Example 2.7.4. Line integral over triangle.

Let \(\mathcal{C}\) be the triangle with vertices \((1,0,0), (0,1,0), (0,0,1)\) oriented counterclockwise from above. Compute \(\oint_\mathcal{C}\boldF\cdot \boldr\text{,}\) where \(\boldF(x,y,z)=\angvec{z^2, y^2, x}\text{.}\)

Solution.
  1. Identify as boundary of surface.

    We have \(\mathcal{C}=\partial \mathcal{S}\text{,}\) where \(\mathcal{S}\) is the triangular region enclosed by \(\mathcal{C}\text{.}\)

  2. Parametrize surface.

    The region \(\mathcal{S}\) is the region of the plane \(x+y+z=1\) lying above the triangular region \(\mathcal{R}\colon 0\leq x\leq 1, 0\leq y\leq 1-x\text{.}\) Writing \(z=1-x-y\text{,}\) we have parametrization

    \begin{align*} \boldr(x,y) \amp=(x,y,1-x-y) \\ \mathcal{R}\colon 0\amp\leq x\leq 1, 0\leq y\leq 1-x \text{.} \end{align*}
  3. Identify appropriate orientation of surface.

    For the induced orientation on \(\mathcal{C}\) to be counterclockwise from above, we want the upward orientation on \(\mathcal{S}\text{.}\) We know that for a parametrization of the form \(r(x,y)=(x,y,1-x-y)\) the orientation

    \begin{equation*} \boldn(x,y,z)=\frac{\boldr_x\times \boldr_y}{\abs{\boldr_x\times \boldr_y}} \end{equation*}

    is the upward one. For the step below we need just the denominator of this expression: \(\boldr_x\times \boldr_y=\angvec{1,1,1}\text{.}\)

  4. Integrate \(\curl\boldF\).

    A straightforward computation gives us \(\curl\boldF=\angvec{0,2z-1,0}\text{.}\) Using TheoremĀ 2.6.15 we have

    \begin{align*} \iint_\mathcal{S}\curl\boldF\cdot \boldn\, d\sigma \amp= \iint_{\mathcal{R}}\curl\boldF(\boldr(x,y))\cdot (\boldr_x\times \boldr_y)\, dA \\ \amp=\int_0^1\int_0^{1-x}\angvec{0,2(1-x-y)-1,0}\cdot\angvec{1,1,1}\, dy\, dx \\ \amp = \int_0^1\int_0^{1-x}1-2(x+y)\, dy\, dx\\ \amp \int_0^1 \left(y-(x+y)^2\Bigr_0^{1-x}\, dx\\ \amp =\int_0^1 x^2-x\, dx\\ \amp = \frac{1}{3}-\frac{1}{2}=-\frac{1}{6} \end{align*}

Example 2.7.5. Stokes for a pringle.

Let \(\mathcal{S}\) be a portion of the surface \(z=y^2-x^2\) lying within the cylinder \(x^2+y^2=1\text{,}\) and let \(\mathcal{C}\) be its boundary. Let \(\boldn\colon \mathcal{S}\rightarrow \R^3\) be the upward orientation on \(\mathcal{S}\text{.}\) Verify Stokes's theorem for \(\mathcal{S}, \mathcal{C}\text{,}\) and the vector field \(\boldF=y\boldi-x\boldj+x^2\boldk\text{.}\)

Solution.

I include this mainly as a nice parametrization example. The surface is parametrized using cylindrical coordinates as

\begin{align*} \boldr(r,\theta) \amp=(r\cos\theta, r\sin\theta, r^2(\sin^2\theta-\cos^2\theta))=(r\cos\theta,r\sin\theta,-r^2\cos 2\theta) \\ \mathcal{R}\colon 0\amp\leq r\leq 1, 0\leq \theta\leq 2\pi \text{,} \end{align*}

and the boundary curve \(\mathcal{C}\) with counterclockwise orientation has parametrization

\begin{equation*} \bolds(t)=\boldr(1,\theta)=(\cos\theta,\sin\theta,-r^2\cos 2\theta)\text{.} \end{equation*}

(Evaluate Sage cell below for a visualization.) Next compute

\begin{align*} \boldr_r \amp =\angvec{\cos\theta,\sin\theta,-2r\cos 2\theta}\\ \boldr_\theta\amp=\angvec{-r\sin\theta,r\cos\theta,2r^2\sin 2\theta} \\ \boldr_r\times\boldr_\theta \amp = \angvec{2r^2(\sin\theta\sin 2\theta+\cos\theta\cos 2\theta), -2r^2(\cos\theta\sin^2\theta+\sin\theta\cos 2\theta), r }\\ \curl\boldF \amp=\angvec{0,-2x,-2} \\ \curl\boldF(\boldr(r,\theta)) \amp =\angvec{0,-2r\cos\theta,-2}\text{.} \end{align*}

Since the \(z\)-component of \(\boldr_r\times\boldr_\theta\) is \(r\geq 0\text{,}\) we see that it gives rise to the given choice of upward orientation. With all the ingredients assembled, I leave it to you to verify that

\begin{align*} \iint_\mathcal{S}\curl\boldF\cdot \boldn\, d\sigma \amp=\iint_\mathcal{R}4r^3\cos\theta(\cos\theta\sin 2\theta+\sin\theta\cos 2\theta)-2r\, dA \\ \amp=-2\pi \\ \oint_\mathcal{C}\boldF\cdot d\boldr \amp = \int_0^{2\pi}\boldF(\bolds(t))\cdot \bolds'(t)\, dt\\ \amp =-2\pi\text{.} \end{align*}

Example 2.7.7. Stokes for sliced cylinder.

Let \(\boldF=\angvec{-y^2, x, z^2}\text{,}\) and let \(\mathcal{C}\) be the intersection of the cylinder \(x^2+y^2=1\) and the plane \(y+z=2\text{.}\) Compute \(\oint_\mathcal{C}\boldF\cdot \boldr\text{,}\) where \(\mathcal{C}\) is oriented counterclockwise, viewed from above.

Solution.

Solution 1.

The curve \(\mathcal{C}\) is the boundary of the elliptical region \(\mathcal{S}\) it encloses in the plane \(y+z=2\text{.}\) Since \(z=2-y\) for points on this plane, we can parametrize as

\begin{align*} \boldr(r,\theta) \amp =(r\cos\theta, r\sin\theta,2-r\sin\theta)\\ \mathcal{R}\colon 0\amp\leq r\leq 1, 0\leq \theta\leq 2\pi \text{.} \end{align*}

Assembling ingredients we have

\begin{align*} \boldr_r \amp =\angvec{\cos\theta, \sin\theta,-\sin\theta}\\ \boldr_\theta\amp =\angvec{-r\sin\theta,r\cos\theta,-r\cos\theta}\\ \boldr_r\times\boldr_\theta\amp=\angvec{0,r,r} \\ \curl\boldF \amp=\angvec{0,0,1+2y} \text{.} \end{align*}

Since the \(z\)-component of \(\boldr_r\times\boldr_\theta\) is nonnegative, we see that this gives rise to the upward orientation on \(\mathcal{S}\text{,}\) which induces the counterclockwise orientation on \(\mathcal{C}\text{.}\) We can now use Stokes's theorem to concude

\begin{align*} \oint_\mathcal{C}\boldF\cdot d\boldr \amp = \iint_\mathcal{S}\curl\boldF\cdot \boldn\, d\sigma\\ \amp=\iint_\mathcal{R}\angvec{0,0,1+2r\sin\theta}\cdot \angvec{0,r,r}\, dA \\ \amp =\int_0^{2\pi}\int_0^{1}r+2r^2\sin\theta\, dr\, d\theta\\ \amp =\int_0^{2\pi}\frac{1}{2}+\frac{2}{3}\sin\theta\, d\theta\\ \amp =\pi \end{align*}

Solution 2.

Alternatively, we can treat \(\mathcal{C}\) as the boundary of the piecewise smooth surface \(\mathcal{S}=\mathcal{S}_1\cup \mathcal{S}_2\text{,}\) where \(\mathcal{S}_1\) is the portion of the cylinder \(x^2+y^2=1\) above the \(xy\)-plane and below the plane \(y+z=2\text{,}\) and \(\mathcal{S}_2\) is the disc \(x^2+y^2\leq 1\) in the \(xy\)-plane. To have the given orientation of \(\mathcal{C}\) be induced by that of \(\mathcal{S}\text{,}\) we must choose the inward orientation \(\boldn\) on \(\mathcal{S}\) : this orientation is inward on the cylinder \(\mathcal{S}_1\) and upward on the disc \(\mathcal{S}_2\text{.}\) With this setup we have

\begin{align*} \oint_\mathcal{C}\boldF\cdot d\boldr \amp = \iint_\mathcal{S}\curl\boldF\cdot \boldn\, d\sigma\\ \amp= \cancelto{0}{\iint_\mathcal{S_1}\curl\boldF\cdot \boldn\, d\sigma}+\iint_\mathcal{S_2}\curl\boldF\cdot \boldn\, d\sigma\\ \amp =\iint\limits_{x^2+y^2\leq 1}\angvec{0,0,1+2y}\cdot \angvec{0,0,1}\, dA\\ \amp =\int_0^{2\pi}\int_0^1(1+2r\sin\theta)r\, dr\, d\theta\\\ \amp =\pi\text{.} \end{align*}

A few comments are in order. We have \(\iint_{\mathcal{S}_1}\curl\boldF\cdot \boldn\, d\sigma=0\) because the normal vectors to the cylinder \(\mathcal{S}_1\) point horizontally, and thus are orthogonal to the curl vectors \(\curl\boldF\text{,}\) which point vertically. Secondly, on the disc in the \(xy\)-plane we clearly have the upward normal vector equal to \(\angvec{0,0,1}\text{.}\)

Example 2.7.8. Stokes's implies Green's.

Show that Stokes's theorem implies Green's theorem.

Solution.

Let \(\mathcal{R}\) be the region enclosed by a simple, piecewise smooth planar curve \(\mathcal{C}\) oriented counterclockwise, and suppose \(\boldF(x,y)=\angvec{F_1(x,y),F_2(x,y)}\) satisfies the conditions of TheoremĀ 2.4.1. Define \(G(x,y,z)=\angvec{F_1(x,y),F_2(x,y),0}\text{,}\) and think of \(\mathcal{R}\subseteq\R^3\) as a planar surface living in the \(xy\)-plane, with \(\mathcal{C}\) is boundary. We choose the orientation \(\boldn(x,y,z)=\angvec{0,0,1}=\boldk\) on \(\mathcal{R}\text{,}\) which induces the counterclockwise orientation on \(\mathcal{C}\text{.}\) By Stokes's theorem we have

\begin{align*} \int_\mathcal{C}\boldF\cdot d\boldr\amp=\iint_\mathcal{R}\curl G\cdot \boldn \, d\sigma \amp \\ \amp =\iint_\mathcal{R}\angvec{0,0,\partial F_2/\partial x-\partial F_1/\partial y}\cdot \angvec{0,0,1} \, dA\\ \amp =\iint_\mathcal{R} \partial F_2/\partial x-\partial F_1/\partial y \, dA\text{.} \end{align*}