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      <text>29.1.2.7 further examples on scalar line integrals</text>
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    <questiontext format="moodle_auto_format">
      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<p>Find \(\displaystyle\int_C \left\{{\underline \nabla} (x^2 y )\right\} \cdot \underline{dr}\) where \(C\) is the contour \(y = 2x-x^2\) from \((0,0)\) to \((2,0)\). Here, \(\underline{\nabla}\) refers to the gradient operator, i.e. \(\underline{\nabla} \phi\equiv\mbox{grad}\;\phi\)</p>
<h4 class="HELM_solution">Solution</h4>
<p>Note that \({\underline \nabla} (x^2 y) = 2 x y \underline{i} + x^2 \underline{j}\) 
so the integral is \(\displaystyle\int_C \left\{ 2 x y \;dx + x^2 \;dy \right\}\).<br>
On \(y = 2x-x^2\), \(dy = (2 - 2 x) \;dx\) so the integral becomes 
\begin{eqnarray*} 
\displaystyle\int_C \left\{ 2 x y \;dx + x^2 \;dy \right\} &=& \displaystyle\int_{x=0}^2 
\left\{ 2 x (2x-x^2) \;dx + x^2 (2 - 2x)\;dx \right\} \\ &=& 
\displaystyle\int_0^2 (6x^2 - 4x^3) \;dx = \left[\phantom{\dfrac{1}{1}}\!\!\!2x^3 - x^4 \right]_0^2 = 0\qquad\qquad\qquad\qquad 
\end{eqnarray*}<p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Two paths from \((0,0)\) to \((4,2)\) are defined by \(\qquad\displaystyle \begin{array}{lll} \mathrm{(a)}&amp;C_1 :&amp;y=\displaystyle\frac{1}{2}x\qquad 0\le x\le 4\\\\ \mathrm{(b)}&amp;C_2 :&amp;\mbox{The straight line }y=0\mbox{ from }(0,0)\mbox{ to }(4,0)\mbox{ followed by}\\ \mbox{}&amp;C_3 :&amp;\mbox{The straight line }x=4\mbox{ from } (4,0)\mbox{ to }(4,2) \end{array}\)
<br>For each path find \(\displaystyle\displaystyle\int_C \underline{F} \cdot \underline{dr}\), where \(\underline{F}=2x \underline{i}+2y\underline{j}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>(a) For the straight line \(y=\displaystyle\frac{1}{2}x\) we have \(dy=\displaystyle\frac{1}{2}dx\)
Then, \(\displaystyle \int_{C_1}{\underline {F}\cdot \underline{dr}}=\int_{C_1}{2x\;dx+2y\;dy}=\int_0^4{\left(2x+\frac{x}{2}\right)\;dx}=\int_0^4\frac{5x}{2}\;dx=20\)
(b) For the straight line from \((0,0)\) to \((4,0)\) we have
 \( \displaystyle \int_{C_2}{\underline {F}\cdot \underline{dr}}=\int_0^4{2x\;dx}=16\).<br>
For the straight line from \((4,0)\) to \((4,2)\) we have
\(\displaystyle \int_{C_3}{\underline {F}\cdot \underline{dr}}=\int_0^2{2y\;dy}=4\).<br>
Adding these two results gives \(\displaystyle\int_C{\underline{F}\cdot\underline{dr}}=16+4=20\)</p>]]></text>
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