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    <name>
      <text>29.3.1.1 Stokes' Theorem Information</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3> Stokes' Theorem </h3>
<p>This is a theorem that equates a line integral to a surface integral. For any vector field \(\underline{F}\) and a contour \(C\) which bounds an area \(S\),</p>
<p>\(\qquad\displaystyle \displaystyle\int\!\! \displaystyle\int_S ({\underline \nabla} \times \underline{F}) \cdot \underline{dS} = \displaystyle\oint_C \underline{F} \cdot \underline{dr}\) </p>
<br>
<figure class="HELM_figure"> 
<img src="@@PLUGINFILE@@/29-3-figure16-Strokes.svg" alt="" width="312" height="153" class="img-responsive atto_image_button_text-bottom"> 
<figcaption><strong>A surface for Stokes' theorem.</strong></figcaption> 
</figure>
<p>Notes</p>
<ol>
<li>\(\underline{dS}\) is a vector perpendicular to the surface \(S\) and \(\underline{dr}\) is a line element along the contour \(C\). The sense of \(\underline{dS}\) is linked to the direction of travel along \(C\) by a right hand screw rule.</li>
<li>Both sides of the equation are scalars.</li>
<li>The theorem is often a useful way of calculating a line integral along a contour composed of several distinct parts (e.g. a square or other figure).</li>
<li>\({\underline \nabla} \times \underline{F}\) is a vector field representing the curl of the vector field \(\underline{F}\) and may, alternatively, be written as curl \(\underline{F}\).</li>
</ol>


<h5> Justification of Stokes' Theorem</h5>
<p>Imagine that the surface \(S\) is divided into a set of infinitesimally small rectangles \(ABCD\) where the axes are adjusted so that \(AB\) and \(CD\) lie parallel to the new \(x\)-axis i.e. \(AB = \delta x\) and \(BC\) and \(AD\) lie parallel to the new \(y\)-axis i.e. \(BC = \delta y\).</p>
<p>Now, \(\displaystyle\oint_C \underline{F} \cdot \underline{dr}\) is calculated, where \(C\) is the boundary of a typical such rectangle. </p>
<p>The contributions along \(AB\), \(BC\), \(CD\) and \(DA\) are </p>

<p>\(\qquad \displaystyle\underline{F}(x,y,0) \cdot \underline{\delta x} = F_x(x,y,z) \delta x\), </p>
<p> \(\qquad \displaystyle \underline{F}(x+\delta x,y,0) \cdot \underline{\delta y} = F_y(x+\delta x,y,z) \delta y\), </p>
<p> \(\qquad \displaystyle\underline{F}(x,y+\delta y,0) \cdot (- \underline{\delta x}) = - F_x(x,y+\delta y,z) \delta x\) </p>
<p> \(\qquad \displaystyle \underline{F}(x,y,0) \cdot ( - \underline{ \delta x}) = - F_y(x,y,z) \delta y\). </p>

<p> Thus, 
\[\begin{aligned} \displaystyle\oint_C \underline{F} \cdot \underline{dr} &amp;\approx (F_x(x,y,z) - F_x(x,y+\delta y,z))\delta x + (F_y(x+\delta x,y,z) - F_y(x,y,z))\delta y\qquad \\ 
&amp;\approx \dfrac{\partial F_y}{\partial x} \delta x \delta y - \dfrac{\partial F_x}{\partial y} \delta x \delta y \\ 
&amp;\approx({\underline \nabla} \times \underline{F})_z \delta S\\ 
&amp;=({\underline \nabla} \times \underline{F}) \cdot \underline{dS} \end{aligned}\] 
as \(\underline{dS}\) is perpendicular to the \(x\)- and \(y\)- axes.</p>

<p>Thus, for each small rectangle, \(\quad\displaystyle \displaystyle\oint_C \underline{F} \cdot \underline{dr} \approx ({\underline \nabla} \times \underline{F}) \cdot \underline{dS}\)</p>

<p>When the contributions over all the small rectangles are summed, the line integrals along the inner parts of the rectangles cancel and all that remains is the line integral around the outside of the surface \(S\). The surface integrals sum. Hence, the theorem applies for the area \(S\) bounded by the contour \(C\).While the above does not constitute a formal proof of Stokes' theorem, it does give an appreciation of the origin of the theorem. </p>
<figure class="HELM_figure"> 
<img src="@@PLUGINFILE@@/29-3-figure17-contributions.svg" alt="&quot;&quot;" width="300" height="147" class="img-fluid atto_image_button_text-bottom">
<figcaption><strong>Line integral cancellation and non-cancellation. </strong></figcaption>
</figure>

<br>
<div class="HELM_keypoint">
<h4>Key Point</h4>
<p> <strong> Stokes' Theorem </strong> </p>
<p> \(\qquad\displaystyle  \displaystyle\oint_C \underline{F} \cdot \underline{dr} = \displaystyle\int\!\! \displaystyle\int_S ({\underline \nabla} \times \underline{F}) \cdot \underline{dS}  \)</p>
<p>The closed contour integral of the scalar product of a vector function with the vector along the contour is equal to the integral of the scalar product of the curl of that vector function and the unit normal, over the corresponding surface. </p>
</div>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Verify Stokes' theorem for the vector function \(\underline{F} = y^2 \underline{i} - (x + z) \underline{j} + yz \underline{k}\) and the unit square \(0 \le x \le 1\), \(0 \le y \le 1\), \(z=0\).</p>

<h4 class="HELM_solution">Solution</h4>
<p> If \(\underline{F} = y^2 \underline{i} - (x + z) \underline{j} + yz \underline{k}\) then \({\underline \nabla} \times 
\underline{F} = (z+1)\underline{i} + (-1-2y)\underline{k} = 
\underline{i} + (-1-2y)\underline{k}\) (as \(z=0\)). Note that 
\(\underline{dS} =  \, \mathrm{d}x  \, \mathrm{d}y \underline{k}\) so that \(({\underline \nabla} \times 
\underline{F}) \cdot \underline{dS} = (-1 - 2y)  \, \mathrm{d}y  \, \mathrm{d}x\)  </p>

<p>\[\begin{align} 
\text{Thus}\qquad \displaystyle\int\!\! \displaystyle\int_S ({\underline \nabla} \times \underline{F}) \cdot \underline{dS} &amp;=  \displaystyle\int_{x=0}^1  \displaystyle\int_{y=0}^1 (-1 -2y)  \, \mathrm{d}y \, \mathrm{d}x \\ 
&amp;=  \displaystyle\int_{x=0}^1 \left[(-y -y^2) \right]_{y=0}^1  \, \mathrm{d}x = \displaystyle\int_{x=0}^1 (-2)  \, \mathrm{d}x\qquad\qquad\qquad\qquad\qquad\qquad \\ 
&amp;= \left[-2 x \right]_0^1 = -2 + 0 = -2 
\end{align}\]</p>

<p>To evaluate \( \displaystyle\oint_C \underline{F} \cdot \underline{dr}\), we must consider the four sides separately. </p>
<p> When \(y=0\), \(\underline{F} = -x \underline{j}\) and \(\underline{dr} = \, \mathrm{d}x\underline{i}\) so \(\underline{F} \cdot \underline{dr} = 0\) i.e. the contribution of this side to the integral is zero. </p>
<p> When \(x=1\), \(\underline{F} = y^2 \underline{i} - \underline{j}\) and \(\underline{dr} = \, \mathrm{d}y\underline{j}\) so \(\underline{F} \cdot \underline{dr} = - \, \mathrm{d}y\) so the contribution to the integral is</p>
<p> \(\qquad \displaystyle\int_{y=0}^1 (- \, \mathrm{d}y) = \left[- y \right]_0^1 = -1\). </p>
<p> When \(y=1\), \(\underline{F} =  \underline{i} - x\underline{j}\) and \(\underline{dr} = -\, \mathrm{d}x\underline{i}\) so \(\underline{F} \cdot \underline{dr} = - \, \mathrm{d}x\) so the contribution to the integral is </p>
<p> \(\qquad \displaystyle\int_{x=0}^1 (- \, \mathrm{d}x) = \left[ - x \right]_0^1 = -1\).</p>
<p> When \(x=0\), \(\underline{F} = y^2 \underline{i}\) and \(\underline{dr} = -\, \mathrm{d}y\underline{j}\) so \(\underline{F} \cdot  \underline{dr} = 0\) so the contribution to the integral is zero. </p>
<p> The integral \(\displaystyle \oint_C \underline{F} \cdot \underline{dr}\) is the sum of the contributions i.e. \(0 - 1 - 1 + 0 = -2\). </p>
<p> Thus \(\quad\displaystyle   \displaystyle\int\!\! \displaystyle\int_S ({\underline \nabla} \times \underline{F}) \cdot \underline{dS} =  \displaystyle\oint_C \underline{F} \cdot \underline{dr} = -2, \quad\) i.e. Stokes' theorem has been verified. </p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Using cylindrical polar coordinates verify Stokes' theorem for the function \(\underline{F} = \rho^2 \underline{\hat{\phi}}\) the circle \(\rho=a,z=0\) and the surface \(\rho\le a,z=0\).</p>

<h4 class="HELM_solution">Solution</h4>
<p> Firstly, find \( \displaystyle\oint_C \underline{F} \cdot \underline{dr}\). This can be done by integrating along the contour \(\rho = a\) from \(\phi = 0\) to \(\phi = 2 \pi\). Here \(\underline{F} = a^2 \underline{\hat{\phi}}\) (as \(\rho = a\)) and \(\underline{dr} = a \; d\phi \; \underline{\hat{\phi}}\)(remembering the scale factor) so \(\underline{F} \cdot 
\underline{dr} = a^3  d\phi\) and hence </p>
<p> \(\qquad\displaystyle\oint_C \underline{F} \cdot \underline{dr} =  \displaystyle\int_0^{2 \pi} a^3 d\phi = 2 \pi a^3\) </p>
<p> As \(\underline{F} = \rho^2 \underline{\hat{\phi}}\), \({\underline \nabla} \times \underline{F} = 3 \rho \underline{\hat{z}}\) and \(({\underline \nabla} \times \underline{F}) \cdot \underline{dS} = 3 \rho\) as \(\underline{dS} = \underline{\hat{z}}\). </p>
<p>Thus 
\(\begin{align}  
\qquad \displaystyle\int\!\! \displaystyle\int_S ({\underline \nabla} \times \underline{F}) \cdot \underline{dS} &amp;=  \displaystyle\int_{\phi = 0}^{2 \pi}  \displaystyle\int_{\rho = 0}^1 3 \rho \times \rho  d\rho  d\phi =  \displaystyle\int_{\phi = 0}^{2 \pi}  \displaystyle\int_{\rho = 0}^a 3 \rho^2  d\rho  d\phi\qquad\qquad\qquad\qquad \\
&amp;=  \displaystyle\int_{\phi = 0}^{2 \pi} \left[rho^3 \right]_{\rho = 0}^a d\phi =  \displaystyle\int_0^{2\pi} a^3 d\phi = 2 \pi a^3 
\end{align}\) </p>
<p> Hence </p>
<p>\(\qquad\displaystyle \displaystyle\oint_C \underline{F} \cdot \underline{dr} =  \displaystyle\int\!\! \displaystyle\int_S ({\underline \nabla} \times \underline{F}) 
\cdot \underline{dS} =  2 \pi a^3 \)</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Find the closed line integral \(\displaystyle\oint_C \underline{F} \cdot \underline{dr}\) for the vector field \(\underline{F} = y^2 \underline{i} + (x^2 - z)\underline{j} + 2 x y \underline{k}\) and for the contour \(ABCDEFGHA\) in the below figure. </p>
<figure class="HELM_figure"> 
<img src="@@PLUGINFILE@@/29-3-figure18-contour.svg" alt="&quot;&quot;" width="320" height="161" class="img-fluid atto_image_button_text-bottom">
<figcaption><strong>Closed contour \(ABCDEFGHA\)</strong></figcaption>
</figure>
<br>

<h4 class="HELM_solution">Solution</h4>
<p> To find the line integral directly would require eight line integrals i.e. along \(AB\), \(BC\), \(CD\), \(DE\), \(EF\), \(FG\), \(GH\)and \(HA\). It is easier to carry out a surface integral to find \(\displaystyle\int \displaystyle\int_S ({\underline \nabla} \times \underline{F}) \cdot \underline{dS}, \, \) which is equal to the required line integral \(\displaystyle\oint_C \underline{F} \cdot \underline{dr}\) by Stokes' theorem. </p>
<p> As \(\underline{F} = y^2 \underline{i} + (x^2 - z)\underline{j} + 2 x y \underline{k} \quad \), \({\underline \nabla} \times \underline{F} = \left| \begin{array}{ccc} 
  \underline{i} &amp; \underline{j} &amp; \underline{k} \\ 
  \frac{\partial}{\partial x} &amp; \frac{\partial}{\partial y} &amp; \frac{\partial}{\partial z} 
  \\   y^2 &amp; x^2 - z &amp; 2 x y 
\end{array} \right| = (2x+1)\underline{i} - 2y \underline{j} + (2x - 2y)\underline{k}\) </p> 

<p> As the contour lies in the \(x\)-\(y\) plane, the unit normal is \(\underline{k}\) and \(\underline{dS} = \, \mathrm{d}x   \, \mathrm{d}y   \underline{k}\) </p>
<p> Hence \(({\underline \nabla} \times \underline{F}) \cdot \underline{dS} = (2x - 2y)  \, \mathrm{d}x  \, \mathrm{d}y\). </p>
<p>To work out \(\displaystyle\int\!\! \displaystyle\int_S ({\underline \nabla} \times \underline{F}) \cdot \underline{dS}\), it is necessary to divide the area inside the contour into two smaller areas i.e. the rectangle \(ABCDGH\) and the trapezium \(DEFG\). </p>
<p> On \(ABCDGH\), the integral is 
\(\begin{align} 
 \displaystyle\int_{y=0}^4  \displaystyle\int_{x=0}^6 (2x - 2y)  \, \mathrm{d}x  \, \mathrm{d}y &amp;=  \displaystyle\int_{y=0}^4 \left[\phantom{\dfrac{}{}} x^2 - 2 x y \right]_{x=0}^6  \, \mathrm{d}y =  \displaystyle\int_{y=0}^4 ( 36 - 12 y 
)  \, \mathrm{d}y \\ 
&amp;= \left[ 36 y - 6 y^2 \right]_0^4 = 36 \times 4 - 6 \times 16 - 0 = 48 \end{align}\) </p>

<p> On \(DEFG\), the integral is 
\(\begin{align} 
 \displaystyle\int_{y=4}^7  \displaystyle\int_{x=1}^{y-2} (2x - 2y)  \, \mathrm{d}x  \, \mathrm{d}y &amp;=  \displaystyle\int_{y=4}^7 \left[x^2 - 2 x y \right]_{x=1}^{y-2}  \, \mathrm{d}y = \displaystyle\int_{y=4}^7 ( -y^2 + 2y + 3 )  \, \mathrm{d}y \\ 
&amp;= \left[ -\frac{1}{3}y^3 + y^2 + 3 y \right]_4^7 = -\frac{343}{3} + 49 + 21 + \frac{64}{3} - 16 - 12 = -51 
\end{align}\) </p>

<p> So the full integral is \(\displaystyle\int \displaystyle\int_S ({\underline \nabla} \times \underline{F}) \cdot \underline{dS} = 48 - 51 = -3\). </p>
<p> \(\therefore\) By Stokes' theorem, \(\quad\displaystyle \int\int_S ({\underline \nabla} \times \underline{F}) \cdot \underline{dS} =  \displaystyle\oint_C \underline{F} \cdot \underline{dr} = -3 \) </p>
<hr>
 
<p>From Stokes' theorem, it can be seen that surface integrals of the form \(\displaystyle\int\! \displaystyle\int_S({\underline \nabla}\times{\underline F})\cdot{\underline {d S}}\) depend only on the contour bounding the surface and not on the internal part of the surface.</p>]]></text>
<file name="29-3-figure16-Strokes.svg" path="/" encoding="base64">[encoded data, 5772 characters]</file>
<file name="29-3-figure17-contributions.svg" path="/" encoding="base64">[encoded data, 27432 characters]</file>
<file name="29-3-figure18-contour.svg" path="/" encoding="base64">[encoded data, 21680 characters]</file>
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