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  <question type="description">
    <name>
      <text>29.2.1.1 Surface integrals involving vectors</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>Surface integrals involving vectors</h3>
<p><br></p>

<h4>The unit normal</h4>
<p>For the surface of any three-dimensional shape, it is possible to find a vector lying perpendicular to the surface and with magnitude 1. The unit vector points outwards from a closed surface and is usually denoted by \(\underline{\hat{n}}\). <br>
</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>If \(S\) is the surface of the sphere \(x^2 + y^2 + z^2 = a^2\) find the unit normal \(\underline{\hat{n}}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>The unit normal at the point \(P(x,y,z)\) points away from
the centre of the sphere i.e. it lies in the direction of \(x
\underline{i} + y \underline{j} + z\underline{k}\). To make this a
unit vector it must be divided by its magnitude
\(\sqrt{x^2+y^2+z^2}\) i.e. the unit vector is
\begin{eqnarray*}\underline{\hat{n}}
&amp;=&amp;
\dfrac{x}{\sqrt{x^2+y^2+z^2}}\underline{i}+
\dfrac{y}{\sqrt{x^2+y^2+z^2}}\underline{j}+\dfrac{z}{\sqrt{x^2+y^2+z^2}}\underline{k}\qquad\qquad\qquad\qquad
\\\\&amp;=&amp; \dfrac{x}{a}\underline{i} + \dfrac{y}{a}\underline{j} + \dfrac{z}{a}\underline{k}\end{eqnarray*}
 where \(a=\sqrt{x^2+y^2+z^2}\)\quad  is the radius of the sphere.

<img src="@@PLUGINFILE@@/37_5_1.svg" alt="A unit normal n to a sphere} " width="475" height="232" class="img-fluid atto_image_button_text-bottom"></p>

<hr>
<h4 class="HELM_example">Example</h4>
<p> For the cube \(0 \le x \le 1\), \(0 \le y \le 1\), \(0 \le z \le 1\), find the unit outward normal \(\underline{\hat{n}}\) for each face.</p>
<h4 class="HELM_solution">Solution</h4>
<p>On the face given by \(x=0\), the unit normal points in
the negative \(x\)-direction. Hence the unit normal is \(-
\underline{i}\). Similarly :- </p>
<p>On the face \(x=1\) the unit normal is \(\underline{i}\). \qquad On the face \(y=0\) the unit normal is \(-\underline{j}\). </p>
<p>On the face \(y=1\) the unit normal is \(\underline{j}\). \qquad On the face \(z=0\) the unit normal is \(-\underline{k}\). </p>
<p>On the face \(z=1\) the unit normal is \(\underline{k}\).</p>

<p><br></p>
<h4>\(\underline{dS}\) and the unit normal</h4>

<p>The vector \(\underline{dS}\) is a vector, being an element of the surface with magnitude \(du\,dv\) and direction perpendicular to the surface. <br> If the plane in question is the \(Oxy\) plane, then \(\underline{dS} = \underline{\hat{n}}\, du\, dv = \underline{k}\, dx\, dy\).</p><img src="@@PLUGINFILE@@/Fig29page36.svg" alt="The vector dS as an element of a surface, with magnitude du, dv" width="252" height="103" class="img-fluid atto_image_button_text-bottom"><p></p>
<p>If the plane in question is not one of the three coordinate planes (\(Oxy\), \(Oxz\), \(Oyz\)), appropriate adjustments must be made to express \(\underline{dS}\) in terms of two of \(dx\) and \(dy\) and \(dz\).</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>The rectangle \(OABC\) lies in the plane \(z = y\) (Figure 8).</p>
<p>The vertices are \(O = (0,0,0)\), \(A = (1,0,0)\), \(B = (1,1,1)\) and \(C = (0,1,1)\).</p>
<p>Find a unit vector \(\underline{\hat{n}}\) normal to the plane and an appropriate vector \(\underline{dS}\) expressed in terms of \(dx\) and \(dy\).</p>
<p></p><img src="@@PLUGINFILE@@/37_5_ext.svg" alt="The plane z=y passing through OABC" width="321" height="185" class="img-fluid atto_image_button_text-bottom"><p></p>
<h4 class="HELM_solution">Solution</h4>
<p>Note that two vectors in the rectangle are \({\overrightarrow {OA}}=\underline{i}\) and
\({\overrightarrow {OC}}=\underline{j} + \underline{k}\). A vector perpendicular to the plane is \(\underline{i}
\times (\underline{j} + \underline{k}) = -\underline{j} + \underline{k}\). However, this vector
is of magnitude \(\sqrt{2}\) so the unit normal vector is
\(\underline{\hat{n}} = \dfrac{1}{\sqrt{2}}(-\underline{j} + \underline{k}) =
-\dfrac{1}{\sqrt{2}}\underline{j} + \dfrac{1}{\sqrt{2}}\underline{k}\). 
The vector \(\underline{dS}\) is therefore \((-\dfrac{1}{\sqrt{2}}\underline{j} +
\dfrac{1}{\sqrt{2}}\underline{k})\,  du\,  dv\) where \(du\) and \(dv\) are
increments in the plane of the rectangle \(OABC\). Now, 
one increment, say \(du\), may point in the \(x\)-direction while \(dv\)
will point in a direction up the plane, parallel to \(OC\). Thus \(du
= dx\) and (by Pythagoras) \(dv = \sqrt{(dy)^2 + (dz)^2}\). However,
as \(z = y\), \(dz = dy\) and hence \(dv = \sqrt{2} dy\). </p>
<p>Thus, \(\underline{dS} = (-\dfrac{1}{\sqrt{2}}\underline{j} +
\dfrac{1}{\sqrt{2}}\underline{k})\,  dx\, \sqrt{2}\,  dy = (-\underline{j} + \underline{k})
 \,dx\, dy\). 
Note :- the factor of \(\sqrt{2}\) could also have been found by
comparing the area of rectangle \(OABC\), i.e. \(1\), with the area of
its projection in the \(Oxy\) plane i.e. \(OADE\) with area
\(\dfrac{1}{\sqrt{2}}\). </p>


<h4>Integrating a scalar field</h4>
<p>A function can be integrated over a surface by constructing a double integral and integrating in a manner similar to that shown in 27.1 and 27.2. Often, such integrals can be carried out with respect to an element containing the unit normal.</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p> Evaluate the integral \[\displaystyle\int_A \dfrac{1}{1+x^2}\; \underline{dS}\] 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>In this integral, \(\underline{dS}\) becomes
\(\underline{k}\,   dx\,   dy\) i.e. the unit normal times the surface
element. Thus the integral is \begin{eqnarray*} \displaystyle\int_{y=0}^1
\displaystyle\int_{x=0}^1 \dfrac{\underline{k}}{1+x^2} \,  dx\,   dy &amp;=&amp;
\underline{k}\displaystyle\int_{y=0}^1 \left[\phantom{\dfrac{}{}}  \tan^{-1} x \right]_0^1   dy \\
&amp;=&amp;\underline{k} \displaystyle\int_{y=0}^1 \left[  (\frac{\pi}{4}-0)
\right]_0^1   dy = \frac{\pi}{4} \underline{k} \displaystyle\int_{y=0}^1 dy\qquad\qquad\qquad\qquad\qquad\qquad \\
&amp;=&amp; \frac{\pi}{4} \underline{k}
\end{eqnarray*}</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Find \(\displaystyle\int\!\!\int_S u \;\underline{dS}\) where \(u = x^2 + y^2 + z^2\) and \(S\) is the surface of the unit cube\(0 \le x \le 1\), \(0 \le y \le 1\), \(0 \le z \le 1\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>The unit cube has six faces and the unit normal vector
 \(\underline{\hat{n}}\) points in a different
direction on each face; see Example 18. The surface integral must be evaluated for
each face separately and the results summed. </p>
<p>On the face \(x = 0\), the unit normal \(\underline{\hat{n}} = -\underline{i}\) and the
surface integral is
\begin{eqnarray*}
\displaystyle\int_{y=0}^1 \displaystyle\int_{z=0}^1
(0^2 + y^2 + z^2)(-\underline{i})  dz  dy &amp;=&amp; -\underline{i} \displaystyle\int_{y=0}^1
\left[ y^2 z + \frac{1}{3} z^3 \right]_{z=0}^1  dy \\
&amp;=&amp; -\underline{i} \displaystyle\int_{y=0}^1 \left( y^2  +  \frac{1}{3} \right)  dy =
-\underline{i} \left[ \frac{1}{3}y^3  +  \frac{1}{3}y \right]_0^1 = -
\frac{2}{3}\underline{i}
\end{eqnarray*}
On the face \(x = 1\), the unit normal \(\underline{\hat{n}} = \underline{i}\) and the
surface integral is
\begin{eqnarray*}
\displaystyle\int_{y=0}^1 \displaystyle\int_{z=0}^1 (1^2 + y^2 + z^2)(\underline{i})  dz  dy &amp;=&amp;
\underline{i} \displaystyle\int_{y=0}^1
\left[ z + y^2 z + \frac{1}{3} z^3 \right]_{z=0}^1  dy \\
&amp;=&amp; \underline{i} \displaystyle\int_{y=0}^1 \left( y^2  +  \frac{4}{3} \right)  dy =
\underline{i} \left[ \frac{1}{3}y^3  +  \frac{4}{3}y \right]_0^1 =
\frac{5}{3}\underline{i}
\end{eqnarray*}
The net contribution from the faces \(x = 0\) and \(x = 1\) is
\(-\frac{2}{3}\underline{i} + \frac{5}{3}\underline{i} = \underline{i}\).</p>
<p>Due to the symmetry of the scalar field \(u\) and the unit cube, the
net contribution from the faces \(y=0\) and \(y=1\) is \(\underline{j}\) while
the net contribution from the faces \(z=0\) and \(z=1\) is \(\underline{k}\).</p>
<p>Adding, we obtain \( \displaystyle\int\!\!\int_S u   \underline{dS} = \underline{i} + \underline{j} + \underline{k}\)</p>

<p><br></p>
<div class="HELM_keypoint">
<h4>Key Point</h4>
A scalar function integrated with respect to a normal vector \(\underline{dS}\) gives a vector quantity.
When the surface does not lie in one of the planes \(Oxy\), \(Oxz\), \(Oyz\), extra care must be taken when finding \(\underline{dS}\).
</div>

<p><br></p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Find \(\displaystyle\int\!\!\int_S ({\underline \nabla} \cdot \underline{F}) \underline{dS}\) where \(\underline{F} = 2x \underline{i} + y z \underline{j} + x y \underline{k}\) and \(S\) is the surface of the triangle with vertices at \((0,0,0)\), \((1,0,0)\) and \((1,1,0)\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>Note that \({\underline \nabla} \cdot \underline{F} = 2 + z = 2\) as \(z = 0\)
everywhere along \(S\). As the triangle lies in the \(Oxy\) plane, the
normal vector \(\underline{n} = \underline{k}\) and \(\underline{dS} = \underline{k}  dy  dx\).</p>

<p>Thus, \(\qquad\displaystyle \displaystyle\int\!\!\int_S ({\underline \nabla} \cdot \underline{F}) \underline{dS}
= \displaystyle\int_{x=0}^1 \displaystyle\int_{y=0}^x 2  dy  dx {\underline k}= \displaystyle\int_0^1 \left[\phantom{\dfrac{1}{1}}\!\!\!2y\right]_0^x  dx{\underline k} = \displaystyle\int_0^1 2x dx{\underline k} =
\left[\phantom{\dfrac{}{}}x^2\right]_0^1 \underline{k}= {\underline k}\)</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Find \(\displaystyle\int\!\!\int_S f \;\underline{dS}\) where \(f\) is the function \(2x\) and \(S\) is the surface of the trianglebounded by \((0,0,0)\), \((0,1,1)\) and \((1,0,1)\). (See Figure 9.)</p>
<p></p><img src="@@PLUGINFILE@@/37_5_3.svg" alt="The triangle defining the area S" width="293" height="212" class="img-fluid atto_image_button_text-bottom"><p></p>
<h4 class="HELM_solution">Solution</h4>
<p>The unit vector \(\underline{n}\) is perpendicular to two vectors
in the plane e.g. \((\underline{j}+\underline{k})\) and \((\underline{i}+\underline{k})\). The
vector \((\underline{j}+\underline{k}) \times (\underline{i}+\underline{k}) = \underline{i} + \underline{j} -
\underline{k} \) which has magnitude \(\sqrt{3}\). Hence the unit normal vector
\(\hat{{\underline n}} =
\frac{1}{\sqrt{3}}\underline{i}+\frac{1}{\sqrt{3}}\underline{j}-\frac{1}{\sqrt{3}}\underline{k}\).</p>
<p>As the area of the triangle \(S\) is \(\frac{\sqrt{3}}{2}\) and the area
of its projection in the \(Oxy\) plane is \(\frac{1}{2}\), the vector
\(\underline{dS} = \dfrac{\sqrt{3}/2}{1/2}\hat{{\underline n}}\;  dy  dx = (\underline{i} + \underline{j}
+ \underline{k})  dy  dx \). </p>
<p>Thus \begin{eqnarray*}  \displaystyle\int\!\!\int_S f \underline{dS} &amp;=&amp; (\underline{i} + \underline{j}
+ \underline{k}) \displaystyle\int_{x=0}^1
\displaystyle\int_{y=0}^{1-x} 2x\;  dy  dx   \\
&amp;=&amp; (\underline{i} +
\underline{j}
+ \underline{k})\displaystyle\int_{x=0}^1 \left[\phantom{\dfrac{1}{1}}\!\!\! 2xy \right]_{y=0}^{1-x}  dx    \\
&amp;=&amp;(\underline{i} + \underline{j}
+ \underline{k}) \displaystyle\int_{x=0}^1 (2x - 2x^2)  dx    \\
&amp;=&amp;(\underline{i} + \underline{j} +
\underline{k}) \left[ x^2 - \frac{2}{3}x^3 \right]_0^1    = \frac{1}{3}(\underline{i} + \underline{j}
+ \underline{k})\qquad\qquad\qquad\qquad\qquad\qquad \\
\end{eqnarray*}</p>]]></text>
<file name="37_5_1.svg" path="/" encoding="base64">[encoded data, 25856 characters]</file>
<file name="37_5_3.svg" path="/" encoding="base64">[encoded data, 27764 characters]</file>
<file name="37_5_ext.svg" path="/" encoding="base64">[encoded data, 17308 characters]</file>
<file name="Fig29page3.svg" path="/" encoding="base64">[encoded data, 42264 characters]</file>
<file name="Fig29page36.svg" path="/" encoding="base64">[encoded data, 12524 characters]</file>
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