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      <text>29.1.4.17 Vector line integrals</text>
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      <text><![CDATA[<h3>Vector line integrals</h3>
<p>It is also possible to form less commonly used integrals of the types: \(\displaystyle\int_C f(x,y,z)\; \underline{dr}\) and \(\displaystyle\int_C \underline{F}(x,y,z) \times \underline{dr}\). Each of these integrals evaluates to a vector. <br>Remembering that \(\underline{dr} = \mathrm{d}x\;\underline{i}+\mathrm{d}y\;\underline{j}+\mathrm{d}z\;\underline{k}\), an integral of the form \(\displaystyle\int_C f(x,y,z)\; \underline{dr}\) becomes <br>\(\displaystyle\int_C f(x,y,z) \mathrm{d}x \; \underline{i} + \displaystyle\int_C f(x,y,z) \;\mathrm{d}y \; \underline{j} + \displaystyle\int_C f(x,y,z) \mathrm{d}z\; \underline{k}\). The first term can be evaluated by expressing \(y\) and \(z\) in terms of \(x\). Similarly the second and third terms can be evaluated by expressing all terms as functions of \(y\) and \(z\) respectively. Alternatively, all variables can be expressed in terms of a parameter \(t\). If an integral is two-dimensional, the term in \(z\) will be absent.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate the integral \(\displaystyle\int_C x y^2 \underline{dr}\) where \(C\) represents the contour \(y = x^2\) from \((0,0)\) to \((1,1)\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>This is a two-dimensional integral so the term in \(z\) will be absent. \begin{eqnarray*} I &amp;=&amp; \displaystyle\int_C x y^2 \underline{dr} \\ &amp;=&amp; \displaystyle\int_C x y^2 (\mathrm{d}x\underline{i} + \mathrm{d}y\underline{j}) \\ &amp;=&amp; \displaystyle\int_C x y^2 \mathrm{d}x \;\underline{i} + \displaystyle\int_C x y^2 \;\mathrm{d}y \;\underline{j} \\ &amp;=&amp; \displaystyle\int_{x=0}^1 x (x^2)^2 \mathrm{d}x \;\underline{i} + \displaystyle\int_{y=0}^1 y^{1/2} y^2 \;\mathrm{d}y\;\underline{j} \\ &amp;=&amp; \displaystyle\int_0^1 x^5 \mathrm{d}x\;\underline{i} + \displaystyle\int_0^1 y^{5/2}\;\mathrm{d}y\;\underline{j}\qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad \qquad\qquad \\ &amp;=&amp; \left[\frac{1}{6}x^6\right]_0^1 \;\underline{i} + \left[\frac{2}{7}x^{7/2}\right]_0^1 \;\underline{j}\\ &amp;=&amp; \frac{1}{6}\underline{i} + \frac{2}{7}\underline{j} \end{eqnarray*}</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Find \(I = \displaystyle\int_C x \underline{dr}\) for the contour \(C\) given parametrically by \(x = \cos t\), \(y = \sin t\), \(z = t - \pi\) starting at \(t = 0\) and going to \(t = 2 \pi\), i.e. the contour starts at \((1,0,-\pi)\) and finishes at \((1,0,\pi)\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>The integral becomes \(\displaystyle\int_C x (\mathrm{d}x\,\underline{i} + \mathrm{d}y\,\underline{j} + \mathrm{d}z\,\underline{k}). \) Now, \(x = \cos t\), \(y = \sin t\), \(z = t - \pi\) so \(\mathrm{d}x = -\sin t \;dt\), \(\mathrm{d}y = \cos t \;dt\) and \(\mathrm{d}z = dt\). So \begin{eqnarray*} I &amp;=&amp; \displaystyle\int_0^{2 \pi} \cos t (-\sin t \;dt\; \underline{i} + \cos t \;dt\; \underline{j} + dt \;\underline{k}) \\ &amp;=&amp; - \displaystyle\int_0^{2 \pi} \cos t \sin t \;dt\;\underline{i} + \displaystyle\int_0^{2 \pi} \cos^2 t \;dt\;\underline{j} + \displaystyle\int_0^{2 \pi} \cos t \;dt\;\underline{k} \\ &amp;=&amp; -\frac{1}{2} \displaystyle\int_0^{2 \pi} \sin 2t \;dt\;\underline{i} + \frac{1}{2} \displaystyle\int_0^{2\pi} (1 + \cos 2t)\;dt\;\underline{j} + \left[\phantom{\dfrac{ }{ }}\sin t \right]_0^{2\pi}\;\underline{k}\qquad\qquad\qquad\qquad \\ &amp;=&amp; \frac{1}{4}\left[\phantom{\dfrac{ }{ }}\cos 2t\right]_0^{2\pi}\;\underline{i}\ + \ \frac{1}{2}\left[ t + \frac{1}{2}\sin 2t\right]_0^{2\pi}\;\underline{j}\ + \ 0\underline{k} \\ &amp;=&amp; 0\underline{i} + \pi\;\underline{j} = \pi \underline{j} \end{eqnarray*}</p>
<p>Integrals of the form \(\displaystyle\int_C \underline{F} \times \underline{dr}\) can be evaluated as follows. If the vector field \(\underline{F} = F_1\underline{i} + F_2\underline{j} + F_3\underline{k}\) and \(\underline{dr} = \mathrm{d}x\;\underline{i}+\mathrm{d}y\;\underline{j}+\mathrm{d}z\;\underline{k}\) then: \[\begin{aligned} \underline{F} \times \underline{dr} &amp;= \left| \begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ &amp; &amp; \\ F_1 &amp; F_2 &amp; F_3 \\ &amp; &amp; \\ \mathrm{d}x &amp; \mathrm{d}y &amp; \mathrm{d}z \end{array}\right| \\ &amp;=(F_2\;\mathrm{d}z - F_3\;\mathrm{d}y)\underline{i} + (F_3\;\mathrm{d}x - F_1\;\mathrm{d}z)\underline{j} + (F_1\;\mathrm{d}y - F_2\;\mathrm{d}x)\underline{k} \\ &amp;= (F_3\underline{j}-F_2\underline{k})\mathrm{d}x + (F_1\underline{k} - F_3\underline{i})\mathrm{d}y + (F_2\underline{i} - F_1\underline{j})\mathrm{d}z \end{aligned}\]</p>
<p>There are thus a maximum of six terms involved in one such integral; the exact details may dictate which method to use.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate the integral \(\displaystyle\int_C (x^2\underline{i} + 3 x y \underline{j}) \times \underline{dr}\) where \(C\) represents the curve \(y = 2x^2\) from \((0,0)\) to \((1,2)\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>Note that the \(z\) components of both \(\underline{F}\) and \(\underline{dr}\) are zero. \(\underline{F}\times \underline{dr} = \left| \begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ &amp; &amp; \\ x^2 &amp; 3xy &amp; 0 \\ &amp; &amp; \\ \mathrm{d}x &amp; \mathrm{d}y &amp; 0 \end{array} \right| = (x^2 \mathrm{d}y - 3 x y\ \mathrm{d}x)\underline{k}\) and \(\displaystyle\int_C (x^2\underline{i} + 3 x y \underline{j}) \times \underline{dr} = \displaystyle\int_C (x^2 \mathrm{d}y - 3 x y \mathrm{d}x)\underline{k} \)</p>
<p>Now, on \(C\), \(y = 2x^2\) \(\mathrm{d}y = 4x\mathrm{d}x\) and \begin{eqnarray*} \displaystyle\int_C (x^2\underline{i} + 3 x y \underline{j}) \times \underline{dr} &amp;\!\!\!=\!\!\!&amp; \displaystyle\int_C \{x^2 \mathrm{d}y - 3 x y \mathrm{d}x\}\underline{k} \\ &amp;\!=\!&amp; \displaystyle\int_{x=0}^1 \left\{ x^2 \times 4x \mathrm{d}x - 3 x \times 2 x^2 \mathrm{d}x \right\}\underline{k}\qquad\qquad\qquad\qquad\qquad\qquad \\ &amp;=&amp; \displaystyle\int_0^1\!\!\! -2 x^3 \mathrm{d}x \underline{k} \\ &amp;=&amp;- \left[ \frac{1}{2}x^4 \right]_0^1 \underline{k} \\ &amp;=&amp;- \frac{1}{2}\underline{k} \end{eqnarray*}</p>
<hr>
<h4 class="HELM_example">Engineering Example</h4>
<p><strong>Force on a loop due to a magnetic field</strong></p>
<p><strong>Introduction</strong></p>
<p>A current \(\underline{I}\) in a magnetic field \(\underline{B}\) is subject to a force \(\underline{F}\) given by \(\displaystyle \underline{F}= I \; \underline{dr} \times \underline{B}\) where the current can be regarded as having magnitude \(I\) and flowing (positive charge) in the direction given by the vector \(\underline{dr}\). The force is known as the <strong>Lorentz force</strong> and is responsible for the workings of an electric motor. If current flows around a loop, the total force on the loop is given by the integral of \(\underline{F}\) around the loop, i.e. \(\qquad\displaystyle \underline{F} = \oint (I \; \underline{dr} \times \underline{B}) = -I \oint (\underline{B} \times \underline{dr})\) where the closed path of the integral represents one circuit of the loop.</p>
<p></p>
<figure class="HELM_figure"><img class="img-fluid" src="@@PLUGINFILE@@/loop-1.svg" alt="The magnetic field through a loop of current" width="300" height="214">
<figcaption>The magnetic field through a loop of current.</figcaption>
</figure>
<p></p>
<p><strong>Problem in words</strong></p>
<p>A current of 1 amp flows around a circuit in the shape of the unit circle in the \(Oxy\) plane. A magnetic field of 1 tesla (T) in the positive \(z\)-direction is present. Find the total force on the circuit loop.</p>
<p><strong>Mathematical statement of problem</strong></p>
<p>Choose an origin at the centre of the circuit and use polar coordinates to describe the position of any point on the circuit and the length of a small element.</p>
<p>Calculate the line integral around the circuit to give the force required using the given values of current and magnetic field.</p>
<p><strong>Mathematical analysis</strong></p>
<p>The circuit is described parametrically by \(\displaystyle x = \cos \theta \;\;\; y = \sin \theta \;\;\; z=0\) with \[\begin{aligned} \underline{dr} &amp;= -\sin \theta \; d\theta \; \underline{i} + \cos \theta \; d\theta \; \underline{j} \\ \underline{B} &amp; = B \; \underline{k} \end{aligned}\]</p>
<p>since \(\underline{B}\) is constant. Therefore, the force on the circuit is given by \[\begin{aligned} \underline{F} &amp;= -I B \oint \underline{k} \times \underline{dr}\\ &amp;= -\oint \underline{k} \times \underline{dr} \mbox{ (since \(I=1\) A and \(B=1\) T)} \end{aligned}\] where \[\begin{aligned} \underline{k} \times \underline{dr} &amp;; = \left|\begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k}\\ 0 &amp; 0 &amp; 1\\ -\sin \theta \; d\theta &amp; \cos \theta \; d\theta &amp; 0\\ \end{array}\right|\\ &amp;=\left(- \cos \theta \; \underline{i} - \sin \theta \; \underline{j} \; \right) d\theta \end{aligned}\]</p>
<p>So \[\begin{aligned} \underline{F} &amp;= -\int_{\theta=0}^{2\pi} \left( - \cos \theta \; \underline{i} - \sin \theta \; \underline{j} \; \right) d\theta \\ &amp;= \Bigl[ \sin \theta \; \underline{i} - \cos \theta \; \underline{j} \; \Bigr]_{\theta=0}^{2\pi}\\ &amp; = (0 - 0) \; \underline{i} - (1 - 1) \; \underline{j} = \underline{0}\end{aligned}\]</p>
<p>Hence there is no net force on the loop.</p>
<p><strong>Interpretation</strong></p>
<p>At any given point of the circle, the force on the point opposite is of the same magnitude but opposite direction, and so cancels, leaving a zero net force. Tip: Use symmetry arguments to avoid detailed calculations whenever possible! <br>A scalar or vector involved in a vector line integral may itself be a vector derivative as this next Example illustrates.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Find the vector line integral \(\displaystyle\int_C ({\underline \nabla} \cdot \underline{F}) \;\underline{dr}\) where \(\underline{F}\) is the vector \(x^2\underline{i}+2xy\underline{j}+2xz\underline{k}\) and \(C\) is the curve \(y = x^2\), \(z = x^3\) from \(x=0\) to \(x=1\) i.e. from \((0,0,0)\) to \((1,1,1)\). Here \(\underline{\nabla}\cdot\underline{F}\) is the (scalar) divergence of the vector \(\underline{F}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>As \(\underline{F} = x^2\underline{i}+2xy\underline{j}+2xz\underline{k}\), \({\underline \nabla} \cdot \underline{F} = 2x + 2x + 2x = 6x\). The integral \begin{eqnarray*}\int_C ({\underline \nabla} \cdot \underline{F}) \;\underline{dr} &amp;=&amp; \displaystyle\int_C 6x (\mathrm{d}x\;\underline{i} + \mathrm{d}y\;\underline{j} + \mathrm{d}z\;\underline{k})\\ &amp;=&amp; \displaystyle\int_C 6x\;\mathrm{d}x\;\underline{i} + \displaystyle\int_{C} 6x\;\mathrm{d}y\;\underline{j} + \displaystyle\int_C 6x\;\mathrm{d}z\;\underline{k}\qquad\qquad\qquad\qquad\qquad\qquad\quad\end{eqnarray*} The first term is \(\qquad\displaystyle \displaystyle\int_C 6x\;\mathrm{d}x\;\underline{i} = \displaystyle\int_{x=0}^1 6x\;\mathrm{d}x\;\underline{i} = \left[\phantom{\dfrac{1}{1}}\!\!\!3x^2 \right]_0^1\underline{i} = 3\underline{i} \) In the second term, as \(y = x^2\) on \(C\), \(\mathrm{d}y\) may be replaced by \(2x\;\mathrm{d}x\) so \(\qquad\displaystyle \displaystyle\int_C 6x\;\mathrm{d}y\;\underline{j} = \displaystyle\int_{x=0}^1 6x\times 2x\;\mathrm{d}x\;\underline{j} = \displaystyle\int_0^1 12x^2\;\mathrm{d}x\;\underline{j} = \left[\phantom{\dfrac{1}{1}}\!\!\!4x^3 \right]_0^1\underline{j} = 4\underline{j} \) In the third term, as \(z = x^3\) on \(C\), \(\mathrm{d}z\) may be replaced by \(3x^2\;\mathrm{d}x\) so \(\qquad\displaystyle \displaystyle\int_C 6x\;\mathrm{d}z\;\underline{k} = \displaystyle\int_{x=0}^1 6x\times 3x^2\;\mathrm{d}x\;\underline{k} = \displaystyle\int_0^1 18x^3\;\mathrm{d}x\;\underline{k} = \left[\frac{9}{2}x^4 \right]_0^1\underline{k} = \frac{9}{2}\underline{k} \) On summing, \(\displaystyle\int_C ({\underline \nabla} \cdot \underline{F}) \;\underline{dr} = 3\underline{i} + 4\underline{j} + \frac{9}{2}\underline{k}\).</p>]]></text>
<file name="loop-1.svg" path="/" encoding="base64">[encoded data, 11796 characters]</file>
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