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      <text>27.2.3.2 Finding Surface Integrals With Polar Coordinates Information</text>
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      <text><![CDATA[<h4>Finding Surface Integrals With Polar Coordinates</h4>
<p>The area of integration \(A\) is covered with coordinate circles given by \(r=\text{constant}\) and coordinate lines given by \(\theta=\text{constant}\).</p>
<p>The elementary areas \(\delta A\) are almost rectangles having width \(\delta r\) and length determined by the length of the part of the circle of radius \(r\) between \(\theta\) and \(\delta\theta\), the arc length of this part of the circle is \(r\delta\theta\). So \(\delta A\approx r\delta r\delta\theta\). Thus to evaluate \(\displaystyle \displaystyle\int_{A}f(x,y)\;\mathrm{d}A\) we sum \(f(r,\theta)r\delta r\delta\theta\) for all \(\delta A\).</p>
<p>\(\qquad\displaystyle\int_A f(x,y)\;\mathrm{d}A =\displaystyle\int_{\theta=\theta_{A}}^{\theta=\theta_{B}} \displaystyle\int_{r=r_1(\theta)}^{r=r_2(\theta)}f(r,\theta)\ r\mathrm{d}r \mathrm{d}\theta\)</p>
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<h4>Key Point</h4>
<p><strong> Polar Coordinates </strong></p>
<p>In double integration using polar coordinates, the variable \(r\) appears in \(f(r,\theta)\) and in \(r\mathrm{d}r \mathrm{d}\theta\). As explained above, this \(r\) is required because the elementary area element become larger further away from the origin.</p>
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<p><img class="img-fluid" src="@@PLUGINFILE@@/result_WB27_2_Fig%28362%29_9%20%281%29.svg" alt="A diagram showing general polar coordinates." width="600" height="391"></p>
<p>Note that the use of polar coordinates is a special case of the use of a change of variables. Further cases of change of variables will be considered in a future section.</p>
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<h4 class="HELM_example">Example</h4>
<p>Evaluate \(\displaystyle \displaystyle\int_{0}^{\frac{\pi}{3}}\displaystyle\int_{0}^{2}r\cos\theta\; \mathrm{d}r \mathrm{d}\theta\) and sketch the region of integration. Note that it is the function \(\cos\theta\) which is being integrated over the region and the \(r\) comes from the \(r\mathrm{d}r \mathrm{d}\theta\).</p>
<p><img class="img-fluid atto_image_button_text-bottom" src="@@PLUGINFILE@@/27-2-fig19-polar2.svg" alt="A sector of a circular disc shown in polar coordinates swept out as theta varies between 0 and pi/3." width="279" height="183"></p>
<h4 class="HELM_solution">Solution</h4>
<p>The evaluation is similar to that for cartesian coordinates. The inner integral with respect to \(r\) is evaluated first with \(\theta\) constant. Then the outer \(\theta\) integral is evaluated. \begin{align} \displaystyle\int_{0}^{\frac{\pi}{3}}\displaystyle\int_{0}^{2}r\cos\theta\;\mathrm{d}r \mathrm{d}\theta &amp;=\displaystyle\int_{0}^{\frac{\pi}{3}}\left[\dfrac{1}{2}r^2\cos\theta\right]_{0}^{2}\;\mathrm{d}\theta\\ &amp;=\displaystyle\int_{0}^{\frac{\pi}{3}}2\cos\theta\;\mathrm{d}\theta\\ &amp;=\Big[2\sin\theta\Big]_{0}^{\frac{\pi}{3}}=2\sin\dfrac{\pi}{3}=\sqrt{3}. \end{align} With \(\theta\) constant, \(r\) varies between \(0\) and \(2\), so the bounding curves of the polar strip start at \(r=0\) and end at \(r=2\). As \(\theta\) varies between \(0\) and \(\dfrac{\pi}{3},\) a sector of a circular disc is swept out. This sector is the region of integration shown above.</p>
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<h4 class="HELM_example">Example</h4>
<p>Earlier in this Section, an example concerned integrating the function \(f(x,y) = 5 x^2 y\; \) over the half of the unit circle which lies above the \(x\)-axis. It is also possible to carry out this integration using polar coordinates.</p>
<h4 class="HELM_solution">Solution</h4>
<p>The semi-circle is characterised by \(0\le r \le 1\) and \(0 \le \theta \le \pi\). So the integral may be written (remembering that \(x = r \cos \theta\) and \(y = r \sin \theta\))</p>
<p>\(\qquad\displaystyle\int_0^{\pi} \displaystyle\int_0^1 5 (r \cos \theta)^2 (r \sin \theta)\ r\ d rd\theta \)</p>
<p>which can be evaluated as follows \begin{align} \displaystyle\int_0^{\pi} \displaystyle\int_0^1 5 r^4 \sin \theta \cos^2 \theta \; \mathrm{d}r \mathrm{d}\theta &amp;=\displaystyle\int_0^{\pi} \left[ r^5 \sin \theta \cos^2 \theta \right]_0^1\mathrm{d} \theta \\ &amp;=\displaystyle\int_0^{\pi} \sin \theta \cos^2 \theta\; \mathrm{d}\theta = \left[ -\dfrac{1}{3} \cos^3 \theta \right]_0^{\pi} = -\dfrac{1}{3} \cos^3\pi + \dfrac{1}{3} \cos^3 0 = -\dfrac{1}{3} (-1) + \dfrac{1}{3} (1) =\dfrac{2}{3} \end{align} This is, of course, the same answer that was obtained using an integration over rectangular coordinates.</p>]]></text>
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