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      <text>27.4.3.2 Polar Coordinates Example</text>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<p>Use spherical coordinates to find the volume of a sphere of radius \(R\).</p>

<p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/Fig364_1.svg" alt="diagram of a sphere on 3d axies with a vector (x,y,z) labeled length r and angle phi. There is also an angle theta." >
</figure></p>
<h4 class="HELM_solution">Solution</h4>
<p>The change of coordinates from Cartesian to spherical polar coordinates is given by the transformation equations \(\qquad x=r\cos\theta\sin\phi\qquad y=r\sin\theta\sin\phi\qquad z=r\cos\phi \)<br>We now need the nine partial derivatives \[\displaystyle \begin{array}{lll} \displaystyle \frac{\partial x}{\partial r}=\cos\theta\sin\phi&amp;\displaystyle \frac{\partial x}{\partial \theta}=-r\sin\theta\sin\phi&amp;\displaystyle \frac{\partial x}{\partial \phi}=r\cos\theta\cos\phi\\ \displaystyle \frac{\partial y}{\partial r}=\sin\theta\sin\phi&amp;\displaystyle \frac{\partial y}{\partial \theta}=r\cos\theta\sin\phi&amp;\displaystyle \frac{\partial y}{\partial \phi}=r\sin\theta\cos\phi\\ \displaystyle \frac{\partial z}{\partial r}=\cos\phi&amp;\displaystyle \frac{\partial z}{\partial \theta}=0&amp;\displaystyle \frac{\partial z}{\partial \phi}=r\sin\phi \end{array} \] <br>Hence we have \[\displaystyle J(r,\theta,\phi)=\left| \begin{array}{ccc} \cos\theta\sin\phi&amp;-r\sin\theta\sin\phi&amp;r\cos\theta\cos\phi\\ \sin\theta\sin\phi&amp;r\cos\theta\sin\phi&amp; r\sin\theta\cos\phi\\ \cos\phi &amp; 0&amp; -r\sin\phi \end{array}\right| \] <br>\[\displaystyle J(r,\theta,\phi)=\cos\phi\left| \begin{array}{cc} -r\sin\theta\sin\phi&amp; r\cos\theta\cos\phi\\ r\cos\theta\sin\phi&amp; r\sin\theta\cos\phi \end{array}\right| +0 -r\sin\phi \left| \begin{array}{cc} \cos\theta\sin\phi&amp; -r\sin\theta\sin\phi \sin\theta\sin\phi&amp; r\cos\theta\sin\phi\\ \end{array}\right|\] Check that this gives \(J(r,\theta,\phi)=-r^2\sin\phi\). Notice that \(J(r,\theta,\phi)\leq 0\) for \(0\leq\phi\leq\pi\), so \(\left|J(r,\theta,\phi)\right|=r^2\sin\phi\). The limits are found as follows. The variable \(\phi\) is related to 'latitude'with \(\phi=0\) representing the 'North Pole' with \(\phi =\pi/2\) representing the equator and \(\phi = \pi\) representing the 'South Pole'.</p> The variable \(\theta\) is related to \lq longitude\rq\  with values of \(0\) to \(2\pi\) covering every point for each value of \(\phi\). 
Thus limits on \(\phi\) are \(0\) to \(\pi\) and limits on \(\theta\) are \(0\) to \(2\pi\). The limits
on \(r\) are \(r=0\) (centre) to \(r=R\) (surface). 
 To find the volume of
the sphere we then integrate the volume element \(\;\,\mathrm{d}v=r^2\sin\phi\;\mathrm{d} r\,\mathrm{d}\theta\, \mathrm{d}\phi\) between these limits.
\begin{eqnarray*} 
\text{ Volume}=\displaystyle\int_0^{\pi}\displaystyle\int_0^{2\pi}\displaystyle\int_0^R r^2\sin\phi\;\mathrm{d}r\,\mathrm{d}\theta\, \mathrm{d}\phi&=&
\displaystyle\int_0^{\pi}\displaystyle\int_0^{2\pi}
\frac{1}{3}R^3\sin\phi\;\mathrm{d}\theta\, \mathrm{d}\phi\\
&=&\displaystyle\int_0^{\pi}
\frac{2\pi}{3}R^3\sin\phi\;\mathrm{d}\phi=\frac{4}{3}\pi R^3\end{eqnarray*}]]></text>
<file name="Fig364_1.svg" path="/" encoding="base64">[encoded data, 23704 characters]</file>
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