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  <question type="description">
    <name>
      <text>29.2.2.6 Integrating a vector field</text>
    </name>
    <questiontext format="moodle_auto_format">
      <text><![CDATA[<h4>Integrating a vector field</h4>
<p>In a similar manner to the case of a scalar field, a vector field may be integrated over a surface. Two common types of integral are \(\displaystyle\int_S \underline{F}(\underline{r}) \cdot \underline{dS}\) and \(\displaystyle\int_S \underline{F}(\underline{r}) \times \underline{dS}\) which integrate to a scalar and a vector respectively. Again, when \(\underline{dS}\) is expressed appropriately, the expression will reduce to a double integral. The form \(\displaystyle\int_S \underline{F}(r)\cdot \underline{dS}\) has many important applications, e.g. the flux of a vector field such as an electric or magnetic field.</p>
 
<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate the integral \[\displaystyle\int_A ( x^2 y \underline{i} + z \underline{j} + (2x+y) \underline{k}) \cdot \underline{dS}\] over the area \(A\) where \(A\) is the square \(0 \le x \le 1\), \(0 \le y \le 1\), \(z=0\).</p>
 <h4 class="HELM_solution">Solution</h4>
<p>On \(A\), the unit normal is \(dx\,   dy\,   \underline{k}\)
 \begin{eqnarray*} &\therefore&\displaystyle\int_A ( x^2 y \underline{i} + z \underline{j} + (2x+y) \underline{k}) \cdot
(\underline{k} \;dx\, dy)  \\
&=& \displaystyle\int_{y=0}^1 \displaystyle\int_{x=0}^1 (2x+y) \;  dx\,   dy = \displaystyle\int_{y=0}^1
\left[\phantom{\dfrac{1}{1}}\!\!\! x^2 + xy \right]_{x=0}^1  dy\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\
&=& \displaystyle\int_{y=0}^1 (1+y)   dy = \left[ y + \frac{1}{2}y^2
\right]_0^1 = \frac{3}{2}
\end{eqnarray*}</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Evaluate \(\displaystyle\int_A \underline{r} \cdot \underline{dS}\) where \(A\) represents the surface of the unit cube \(0\le x \le 1\),\(0 \le y \le 1\),\(0 \le z \le 1\) and \(\underline{r}\) represents the vector \(x\underline{i}+y\underline{j}+z\underline{k}\) .</p>
<h4 class="HELM_solution">Solution</h4>
<p>The vector \(\underline{dS}\) (in the direction of the normal vector) will be a constant
vector on each face, but will be different for each face.
On the face \(x=0\) , \(\underline{dS}=-dy\,  dz\,  
\underline{i}\) and the integral on this face is
\(\qquad\displaystyle 
\displaystyle\int_{z=0}^1 \displaystyle\int_{y=0}^1
(0\underline{i}+y\underline{j}+z\underline{k})\cdot (-   dy\,  dz\,
  \underline{i} ) = \displaystyle\int_{z=0}^1 \displaystyle\int_{y=0}^1 0  \; dy\,  dz = 0
\) 
Similarly on the face \(y=0\), \(\underline{dS}=-dx\,  dz\,  
\underline{j}\) and the integral on this face is
\(\qquad\displaystyle 
\displaystyle\int_{z=0}^1 \displaystyle\int_{x=0}^1
(x\underline{i}+0\underline{j}+z\underline{k})\cdot (-   dx\,  dz\,
  \underline{j} ) = \displaystyle\int_{z=0}^1 \displaystyle\int_{x=0}^1 0  \;  dx\,  dz = 0
\) 
Furthermore on the face \(z=0\), \(\underline{dS}=-dx\,  dy\,
  \underline{k}\) and the integral on this face is
\(\qquad\displaystyle 
\displaystyle\int_{x=0}^1 \displaystyle\int_{y=0}^1
(x\underline{i}+y\underline{j}+0\underline{k})\cdot (-   dx\,  dy\,
  \underline{k} ) = \displaystyle\int_{x=0}^1 \displaystyle\int_{y=0}^1 0   \; dx\,  dy = 0
\) 
On these three faces, the contribution to the integral is thus zero.
However, on the face \(x=1\), \(\underline{dS}=+dy\,  dz\,  
\underline{i}\) and the integral on this face is
\(\qquad\displaystyle 
\displaystyle\int_{z=0}^1 \displaystyle\int_{y=0}^1
(1\underline{i}+y\underline{j}+z\underline{k})\cdot (+   dy\,  dz\,
  \underline{i} ) = \displaystyle\int_{z=0}^1 \displaystyle\int_{y=0}^1 1   \;dy\,  dz = 1
\)
Similarly, on the face \(y=1\), \(\underline{dS}=+dx\,  dz\,  
\underline{j}\) and the integral on this face is
\(\qquad\displaystyle 
\displaystyle\int_{z=0}^1 \displaystyle\int_{x=0}^1
(x\underline{i}+1\underline{j}+z\underline{k})\cdot (+   dx\,  dz\,
  \underline{j} ) = \displaystyle\int_{z=0}^1 \displaystyle\int_{x=0}^1 1  \;  dx\,  dz = 1
\) Finally, on the face \(z=1\), \(\underline{dS}=+dx\,  dy\,  
\underline{k}\) and the integral on this face is
\(\qquad\displaystyle 
\displaystyle\int_{y=0}^1 \displaystyle\int_{x=0}^1
(x\underline{i}+y\underline{j}+1\underline{k})\cdot (+   dx\,  dy\,
  \underline{k} ) = \displaystyle\int_{y=0}^1 \displaystyle\int_{x=0}^1 1   \; dx\,  dy = 1
\)  Adding together the contributions gives
\(\displaystyle\int_A \underline{r} \cdot \underline{dS}= 0 + 0 + 0 + 1 + 1 +
1=3\) </p>]]></text>
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