<?xml version="1.0" encoding="UTF-8"?>
<quiz>
  <question type="description">
    <name>
      <text>29.1.2.5 Line integrals of scalar products</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>Line integrals of scalar products</h3>
<p>Integrals of the form \(\displaystyle\int_C \underline{F} \cdot \underline{dr}\) occur in applications such as the following. </p>
<p></p>
<figure class="HELM_figure"><img class="img-fluid" src="@@PLUGINFILE@@/37_3_5-1.svg" alt="Schematic for cyclist travelling from A to B into a head wind" width="350" height="206">
<figcaption>Schematic for cyclist travelling from \(A\) to \(B\) into a head wind.</figcaption>
</figure>
<p></p>
<p>Consider a cyclist riding along the road from \(A\) to \(B\). Suppose it is necessary to find the total work the cyclist has to do in overcoming a wind of velocity \(\underline{v}\).</p>
<p>On moving from \(S\) to \(T\), along an element \(\underline{\delta r}\) of road, the work done is given by ‘Force \(\times\) distance’ \(=|\underline{F}|\times |\underline{\delta r}|\cos \theta\) where \(\underline{F}\), the force, is directly proportional to \(\underline{v}\), but in the opposite direction, and \(|\underline{\delta r}|\cos\theta\) is the component of the distance travelled in the direction of the wind.</p>
<p>So, the work done travelling \(\underline{\delta r}\) is \(-k\underline{v}\cdot\underline{\delta r}\). Letting \(\underline{\delta r}\) become infinitesimally small, the work done becomes \(-k\underline{v} \cdot \underline{dr}\) and the total work is \(-k\displaystyle\int_A^B \underline{v}\cdot \underline{dr}\).</p>
<p>This is an example of the integral along a line, of the scalar product of a vector field, with a vector element of the line. The term <strong>scalar line integral</strong> is often used for integrals of this form. The vector \(\underline{dr}\) may be considered to be \(dx\underline{i} +dy\underline{j}+dz\underline{k}\).</p>
<p>Multiplying out the scalar product, the ’scalar line integral’ of the vector \(\underline{F}\) along contour \(C\), is given by \(\displaystyle\int_C \underline{F} \cdot \underline{dr}\) and equals \(\displaystyle\int_C \{F_x \; dx + F_y \;dy + F_z\;dz\}\) in three dimensions, and \(\displaystyle\int_C \{F_x \; dx + F_y \;dy\}\) in two dimensions, where \(F_x\), \(F_y\), \(F_z\) are the components of \(\underline{F}\). <br>If the contour \(C\) has its start and end points in the same positions i.e. it represents a closed contour, the symbol \(\displaystyle\oint_C\) rather than \(\displaystyle\int_C\) is used, i.e. \(\displaystyle\oint_C \underline{F} \cdot \underline{dr}\) .</p>
<p>As before, to evaluate the line integral, express the path and the function \(\underline{F}\) in terms of either \(x\), \(y\) and \(z\), or in terms of a parameter \(t\). Note that \(t\) often represents time.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>5 Find \(\displaystyle\int_C \{2xy \;dx - 5x\;dy\}\) where \(C\) is the curve \(y=x^3\)\(0\le x \le 1\). [This is the integral \(\displaystyle\int_C\underline{F}\cdot \underline{dr}\) where \(\underline{F}=2xy\underline{i}-5x\underline{j}\) and \(\underline{dr}=dx\;\underline{i}+dy\;\underline{j}\).]</p>
<h4 class="HELM_solution">Solution</h4>
<p>It is possible to split this integral into two different integrals and express the first term as a function of \(x\) and the second term as a function of \(y\). However, it is also possible to express everything in terms of \(x\). Note that on \(C\), \(y = x^3\) so \(dy = 3x^2\;dx\) and the integral becomes \[ \begin{eqnarray*} \displaystyle\int_C \{2xy \;dx - 5x\;dy\} &amp;=&amp; \displaystyle\int_{x=0}^1 \left(2 x \; x^3 \;dx - 5 x \; 3x^2\;dx \right) = \displaystyle\int_0^1 (2x^4 - 15x^3)\;dx \\ &amp;=&amp; \left[\frac{2}{5}x^5 - \frac{15}{4}x^4 \right]_0^1 = \frac{2}{5} - \frac{15}{4} - 0 = -\frac{67}{20} \end{eqnarray*}\]</p>
<div class="HELM_keypoint">
<h4>Key Point</h4>
<p>An integral of the form \(\displaystyle\int_C \underline{F} \cdot \underline{dr}\) may be expressed as \(\displaystyle\int_C \{F_x\;dx + F_y\;dy + F_z\;dz\}\). Knowing the expression for the path \(C\), every term in the integral can be further expressed in terms of one of the variables \(x\), \(y\) or \(z\) or in terms of a parameter \(t\) and hence integrated.</p>
<p>If an integral is two-dimensional there are no terms involving \(z\).</p>
<p>The integral \(\displaystyle\int_C \underline{F} \cdot \underline{dr}\) evaluates to a scalar.</p>
</div>
<hr>
<h4 class="HELM_example">Example</h4>
<p>6Three paths from \((0,0)\) to \((1,2)\) are defined by \(\qquad\displaystyle \begin{array}{lll} \mathrm{(a)}&amp;C_1 :&amp;y=2x\\ \mathrm{(b)}&amp;C_2 :&amp;y=2x^2\\ \mathrm{(c)}&amp;C_3 :&amp;y=0\mbox{ from }(0,0)\mbox{ to }(1,0)\mbox{ and}\ x=1\mbox{ from }(1,0)\mbox{ to }(1,2) \end{array}\) Sketch each path, and along each path find \(\displaystyle\displaystyle\int \underline{F} \cdot \underline{dr}\), where \(\underline{F}=y^2 \underline{i}+xy\underline{j}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>(a) \(\displaystyle\displaystyle\int \underline{F} \cdot \underline{dr}=\displaystyle\int\left\{y^2\mathrm{d} x+xy\mathrm{d} y\right\}\).</p>
<p>Along \(y=2x\), \(\displaystyle\frac{d y}{d x}=2\) so \(d y=2\mathrm{d} x\).</p>
<p>Then \[\begin{aligned} \displaystyle\int_{C_1} \underline{F} \cdot \underline{dr}&amp;=\displaystyle\int_{x=0}^{1}\left\{\left(2x\right)^2\mathrm{d} x+x\left(2x\right)\left(2\mathrm{d} x\right)\right\}\\ &amp;= \displaystyle\int_0^1\left(4x^2+4x^2\right)\mathrm{d} x=\displaystyle\int_0^1 8x^2\mathrm{d} x\\ &amp;= \left[\frac{8}{3}x^2\right]_0^1 = \frac{8}{3} \qquad\qquad\qquad\qquad \end{aligned} \]</p>
<p>\[\incl\degraphics{37_3_4(a)}\] \centerline{{\bf Figure 3(a):} Integration along path \(C_1\)}</p>
<p></p>
<p>(b) \(\ \displaystyle\displaystyle\int \underline{F} \cdot \underline{dr}=\displaystyle\int\left\{y^2\mathrm{d} x+xy\mathrm{d} y\right\}\).</p>
<p>Along \(y=2x^2\), \(\displaystyle\frac{d y}{d x}=4x\) so \(d y=4x\mathrm{d} x\).</p>
<p>Then \(\displaystyle\int_{C_2} \underline{F} \cdot \underline{dr}=\displaystyle\int_{x=0}^{1}\left\{\left(2x^2\right)^2\mathrm{d} x+x\left(2x^2\right)\left(4x\mathrm{d} x\right)\right\} = \displaystyle\int_0^1 12x^4\mathrm{d} x = \left[\dfrac{12}{5}x^5\right]_0^1 = \dfrac{12}{5}\) <br>\([\incl\mathrm{d}egraphics{37_3_4(b)}\] \centerline{{\bf Figure 3(b):} Integration along path \(C_2\)}</p>
<p>Note that the answer is different to part (a), i.e., the line integral depends upon the path taken.</p>
<p>(c) As the contour \(C_3\), has two distinct parts with different equations, it is necessary to break the full contour \(OA\) into the two parts, namely \(OB\) and \(BA\) where \(B\) is the point \((1,0)\). Hence \(\qquad\displaystyle\displaystyle\int_{C_3} \underline{F} \cdot \underline{dr}=\displaystyle\int_O^B\underline{F} \cdot \underline{dr}+\displaystyle\int_B^A\underline{F} \cdot \underline{dr}\)</p>
<div class="HELM_keypoint">
<h4>Key Point</h4>
In general, the value of a line integral depends on the path of integration as well as upon the end points.</div>
<hr>
<h4 class="HELM_example">Example</h4>
<p>7 Find \(\displaystyle\displaystyle\int_A^O \underline{F} \cdot \underline{dr}\), where \(\underline{F} = y^2 \underline{i} + x y \underline{j}\) and the path \(C_4\) from \(A\) to \(O\) is the straight line from \((1,2)\) to \((0,0)\), that is the reverse of \(C_1\) in (a) of the last Example.</p>
<p>Deduce \(\displaystyle\oint_C \underline{F} \cdot \underline{dr}\), the integral around the closed path \(C\) formed by the parabola \(y=2x^2\) from \((0,0)\) to \((1,2)\) and the line \(y=2x\) from \((1,2)\) to \((0,0)\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>Reversing the path interchanges the limits of integration, which results in a change of sign for the value of the integral. \(\qquad\displaystyle \displaystyle\int_A^O \underline{F}\cdot \underline{d r}=-\displaystyle\int_O^A \underline{ F}\cdot \underline{d r}=-\frac{8}{3} \) The integral along the parabola (calculated in (b) of the previous Example) evaluates to \(\displaystyle\frac{12}{5}\), then \(\qquad\displaystyle \displaystyle\oint_C \underline{ F}\cdot \underline{d r}=\displaystyle\int_{C_2} \underline{ F}\cdot \underline{d r}+\displaystyle\int_{C_4} \underline{F}\cdot \underline{d r}=\frac{12}{5}-\frac{8}{3}=-\frac{4}{15}\approx -0.267\)</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>8Consider the vector field \(\displaystyle \underline{F}=y^2z^3\underline{i}+2xyz^3\underline{j}+3xy^2z^2\underline{k}\). Let \(C_1\) and \(C_2\) be the curves from \(O=(0,0,0)\) to \(A=(1,1,1)\), given by \[\begin{aligned} C_1&amp;:&amp;x=t,\qquad y=t,\qquad z=t\qquad (0\leq t\leq 1)\\ C_2&amp;:&amp;x=t^2,\qquad y=t,\qquad z=t^2\qquad (0\leq t\leq 1) \end{aligned}\]</p>
<ol>
<li>Evaluate the scalar integral of the vector field along each path.</li>
<li>Find the value of \(\displaystyle\displaystyle\oint_C \underline{F} \cdot \underline{dr}\) where \(C\) is the closed path along \(C_1\) from \(O\) to \(A\) and back along \(C_2\) from \(A\) to \(O\).</li>
</ol>
<h4 class="HELM_solution">Solution</h4>
<p>The path \(C_1\) is given in terms of the parameter \(t\) by \(x=t\), \(y=t\) and \(z=t\). Hence \(\qquad\displaystyle \frac{d x}{d t}=\frac{d y}{d t}=\frac{d z}{d t}=1\mbox{ and }\frac{d\underline{r}}{d t}=\frac{d x}{d t}\underline{i}+\frac{d y}{d t}\underline{j}+\frac{d z}{d t}\underline{k}=\underline{i}+\underline{j}+\underline{k}\).</p>
<p>Now by substituting for \(x=y=z=t\) in \(\underline{F}\) we have\(\displaystyle \underline{F}=t^5\underline{i}+2t^5\underline{j}+3t^5\underline{k}\).</p>
<p>Hence \(\displaystyle\underline{F}\cdot\frac{d\underline{r}}{d t}=t^5+2t^5+3t^5=6t^5\).</p>
<p>The values of \(t=0\) and \(t=1\) correspond to the start and end point of \(C_1\) and so these are the required limits of integration.</p>
<p>Now \(\qquad\displaystyle \displaystyle\int_{C_1}\underline{F} \cdot\underline{dr}=\displaystyle\int_0^1\underline{F}\cdot\frac{d\underline{r}}{d t}\mathrm{d} t=\displaystyle\int_0^1 6t^5\mathrm{d} t=\left[\phantom{\dfrac{1}{1}}\!\!\!t^6\right]_0^1=1\)</p>
<p>For the path \(C_2\) the parameterisation is \(x=t^2\), \(y=t\) and \(z=t^2\) so \(\displaystyle\frac{d\underline{r}}{d t}=2t\underline{i}+\underline{j}+2t\underline{k}\).<br>Substituting \(x=t^2\), \(y=t\) and \(z=t^2\) in \(\underline{F}\) we have<br>\(\qquad\displaystyle \underline{F} =t^8\underline{i}+2t^9\underline{j}+3t^8\underline{k}\mbox{ and } \underline{F}\cdot\frac{d\underline{r}}{d t}=2t^9+2t^9+6t^9=10t^9 \qquad\displaystyle \displaystyle\int_{C_2}\underline{F}\cdot\underline{d r}=\displaystyle\int_0^1 10t^9\mathrm{d} t=\left[\phantom{\dfrac{1}{1}}\!\!\!t^{10}\right]_0^1=1\) For the closed path \(C\) \(\qquad \displaystyle\oint_C\underline{F}\cdot \underline{dr}=\displaystyle\int_{C_1}\underline{F}\cdot \underline{dr} -\displaystyle\int_{C_2} \underline{F}\cdot\underline{dr}=1-1=0\) (Note: A line integral round a closed path is not necessarily zero - see Example 7.)</p>
<p><strong>Further points on Example 8</strong> \[\begin{array}{|c|c|c|} \hline \mbox{Vector Field} &amp; \mbox{Path} &amp; \mbox{Line Integral}\\ \hline \underline{F} &amp; C_1 &amp;1\\ \underline{F} &amp; C_2 &amp; 1\\ \underline{F} &amp; \mbox{closed} &amp; 0\\ \hline \end{array}\] Note that the value of the line integral of \(\underline{ F}\) is \(1\) for both paths \(C_1\) and \(C_2\). In fact, this result would hold for any path from \((0,0,0)\) to \((1,1,1)\). <br>The field \(\underline{ F}\) is an example of a <strong>conservative vector field</strong>; these are discussed in detail in the next subsection. <br>In \(\displaystyle\int_C \underline{F} \cdot \underline{dr}\), the vector field \(\underline{F}\) may be the gradient of a scalar field or the curl of a vector field.</p>]]></text>
<file name="37_3_5-1.svg" path="/" encoding="base64">[encoded data, 19028 characters]</file>
    </questiontext>
    <generalfeedback format="html">
      <text/>
    </generalfeedback>
    <defaultgrade>0</defaultgrade>
    <penalty>0</penalty>
    <hidden>0</hidden>
    <idnumber/>
  </question>
</quiz>
