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      <text>29.1.1.1 Line integrals intro</text>
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      <text><![CDATA[<h3>Line integrals</h3>
<p>Workbook 28 was concerned with evaluating an integral over <strong>all</strong> points within a rectangle or other shape (or over a cuboid or other volume). In a related manner, an integral can take place over a line or curve running through a two-dimensional (or three-dimensional) region. Line integrals may involve scalar or vector fields. Those involving scalar fields are dealt with first. A line integral in two dimensions may be written as</p>
<p>\(\qquad\displaystyle \displaystyle\int_{C}F(x,y)dw\)</p>
<p>There are three main features determining this integral:</p>
<p>\(\qquad F(x,y)\):This is the scalar function to be integrated e.g. \(F(x,y)=x^{2}+4y^{2}.\)</p>
<p>\(\qquad C\): This is the curve along which integration takes place. e.g. \(y=x^{2}\) or \(x=\sin y\) or \(x=t-1\); \(y=t^{2}\) (where \(x\) and \(y\) are expressed in terms of a parameter \(t\)).</p>
<p>\(\qquad d w\):This gives the variable of the integration. Three main cases are \(dx\), \(dy\) and \(ds\). Here ‘\(s\)’ is arc length and so indicates position along the curve \(C\).</p>
<p>\(ds\) may be written as \(ds=\sqrt{\left( dx\right) ^{2}+\left( dy\right) ^{2}}\) or \(ds=\sqrt{1+\left( \dfrac{dy}{dx}\right) ^{2}}dx\). A fourth case is when \(F(x,y)\ d w\) has the form: \(F_{1}dx+F_{2}dy\). This is a combination of the cases \(d x\) and \(d y\).</p>
<p></p>
<p>The integral \(\displaystyle\int_C F(x,y) \;ds\) represents the area beneath the surface \(z=F(x,y)\) but above the curve \(C\).</p>
<p>The integrals \(\displaystyle\int_C F(x,y) \;dx\) and \(\displaystyle\int_C F(x,y) \;dy\) represent the projections of this area onto the \(xz\) and \(yz\) planes respectively.</p>
<p>A particular case of the integral \(\displaystyle\int_C F(x,y) \;ds\) is the integral \(\displaystyle\int_C 1 \;ds\). This is a means of calculating the length along a curve i.e. an arc length.</p>
<p></p>
<figure class="HELM_figure"><img class="img-fluid" src="@@PLUGINFILE@@/Fig29page3-1.svg" alt="Representation of a line integral and its projections onto the xz and yz planes" width="589" height="285">
<figcaption>Representation of a line integral and its projections onto the \(xz\) and \(yz\) planes</figcaption>
</figure>
<p></p>
<p>The technique for evaluating a line integral is to express all quantities in the integral in terms of a single variable. If the integral is with respect to ’\(x\)’ or ’\(y\)’, then the curve ’\(C\)’ and the function ’\(F\)’ may be expressed in terms of the relevant variable. If the integral is with respect to \(ds\), normally all quantities are expressed in terms of \(x\). If \(x\) and \(y\) are given in terms of a parameter \(t\), then \(t\) is used as the variable.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Find \(\displaystyle\int_{c}x\left( 1+4y\right) dx\) where \(C\) is the curve \(% y=x^{2}\), starting from \(x=0,y=0\) and ending at \(x=1,y=1\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>As this integral concerns only points along \(C\) and the integration is carried out with respect to \(x\), \(y\) may be replaced by \(x^{2}\). The limits on \(x\) will be 0 to 1. So the integral becomes \[\begin{eqnarray*} \displaystyle\int_C x (1 + 4 y) \; dx &amp;=&amp; \displaystyle\int_{x=0}^{1}x\left( 1+4x^{2}\right) dx =\displaystyle\int_{x=0}^{1}\left( x+4x^{3}\right) dx\qquad\qquad\qquad\qquad \\ &amp;=&amp; \left[ \frac{x^{2}}{2}+x^{4}\right] _{0}^{1}=\left( \frac{1}{2}+1\right) -\left( 0\right) =\frac{3}{2} \end{eqnarray*}\]</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Find \(\displaystyle\int_{c}x\left( 1+4y\right) \;dy\) where \(C\) is the curve \(% y=x^{2}\), starting from \(x=0,y=0\) and ending at \(x=1,y=1\). This is the same as Example 1 other than \(dx\) being replaced by \(dy\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>As this integral concerns only points along \(C\) and the integration is carried out with respect to \(y\), everything may be expressed in terms of \(y\), i.e. \(x\) may be replaced by \(y^{1/2}\). The limits on \(y\) will be 0 to 1. So the integral becomes \[\begin{eqnarray*} \displaystyle\int_C x (1 + 4 y) \; dy &amp;=&amp; \displaystyle\int_{y=0}^{1}y^{1/2}\left( 1+4y\right) dy =\displaystyle\int_{y=0}^{1}\left( y^{1/2}+4y^{3/2}\right) dy\qquad\qquad\qquad\qquad \\ &amp;=&amp; \left[ \frac{2}{3}y^{3/2}+\frac{8}{5}y^{5/2}\right] _{0}^{1}=\left( \frac{2}{3}+\frac{8}{5} \right) -\left( 0\right) =\frac{34}{15} \end{eqnarray*}\]</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Find \(\displaystyle\int_{c}x\left( 1+4y\right) ds\) where \(C\) is the curve \(y=x^{2}\), starting from \(x=0,y=0\) and ending at \(x=1,y=1\). This is the same integral and curve as the previous two examples but the integration is now carried out with respect to \(s\), the arc length parameter.</p>
<h4 class="HELM_solution">Solution</h4>
<p>As this integral is with respect to \(x\), all parts of the integral can be expressed in terms of \(x\), Along \(y=x^{2}\), \(ds=\sqrt{1+\left( \dfrac{dy}{dx}\right) ^{2}}dx=\sqrt{1+\left( 2x\right)^{2}% }dx=\sqrt{1+4x^{2}}dx\) <br>So, the integral is <br>\(\qquad\displaystyle \displaystyle\int_{c}x\left( 1+4y\right) ds = \displaystyle\int_{x=0}^{1}x\left( 1+4x^{2}\right) \sqrt{1+4x^{2}}\ dx= \displaystyle\int_{x=0}^{1}x\left( 1+4x^{2}\right) ^{3/2} dx \)This can be evaluated using the transformation \(u=1+4x^{2}\) so \(du=8xdx\) i.e. \(x\; dx=\dfrac{du}{8}\). When \(x=0\), \(u=1\) and when \(x=1\), \(% u=5\). <br>Hence, \[\begin{eqnarray*} \displaystyle\int_{x=0}^{1}x\left( 1+4x^{2}\right) ^{3/2}dx &amp;=&amp; \frac{1}{8}\displaystyle\int_{u=1}^{5}u^{3/2}du \\ &amp;=&amp; \frac{1}{8}\times\frac{2}{ 5}\left[ \phantom{\dfrac{1}{1}}u^{5/2}\right] _{1}^{5}=\frac{1}{20}\left[ 5^{5/2}-1\right] \approx 2.745\qquad\qquad \qquad\qquad\end{eqnarray*}\]</p>
<p>Note that the results for Examples 1,2 and 3 are all different: Example 3 is the area between a curve and a surface above; Examples 1 and 2 give projections of this area onto other planes.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Find \(\displaystyle\int_C xy\;dx\) where, on \(C\), \(x\) and \(y\) are given in terms of a parameter \(t\) by <br>\(x=3t^2\), \(y = t^3 - 1\) for \(t\) varying from \(0\) to \(1\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>Everything can be expressed in terms of \(t\), the parameter. Here \(x=3t^2\) so \(dx = 6t\;dt\). The limits on \(t\) are \(t=0\) and \(t=1\). The integral becomes \[\begin{eqnarray*} \displaystyle\int_C x y \;dx &amp;=&amp; \displaystyle\int_{t=0}^1 3t^2 \;(t^3-1)\;6t\;dt = \displaystyle\int_{t=0}^1 (18t^6 - 18t^3) \; dt\qquad\qquad\qquad\qquad\qquad\qquad \\ &amp;=&amp; \left[\frac{18}{7}t^7 - \frac{18}{4}t^4 \right]_0^1 = \frac{18}{7}-\frac{9}{2}-0 = -\frac{27}{14} \end{eqnarray*}\]</p>
<div class="HELM_keypoint">
<p>A line integral is normally evaluated by expressing all variables in terms of one variable. In general \[\displaystyle\int_Cf(x,y)\ d s\ne \displaystyle\int_Cf(x,y)\ d y\ne \displaystyle\int_Cf(x,y)\ d x\]</p>
</div>]]></text>
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