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<title>
Polar Integrals
[12.4] 
</title>
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<h1>
Polar integrals [12.4]
</h1>
<img src="0g/12.4.igloovillage.jpg"><br>
<i>Kakslauttanen Igloo Village, Finland</i>
<h3>Area in Cartesian coordinates</h3>
<p class="show">In Cartesian coordinates...<br>
<img src="0g/12.4.Cartesian.jpg"><br>
<p>As $\Delta x \to dx$ and $\Delta y\to dy$, the rectangle has an area which approaches
$$dA=dx\,dy$$

<h3>Area in Polar coordinates</h3>
<p class="show">In polar coordinates...<br>
<img src="0g/12.4.Polar.jpg"><br>
<p>As $\Delta \theta \to d\theta$ and $\Delta r\to dr$, the shaded region
becomes more nearly rectangular, with sides of length $r\,d\theta$ and $dr$, so...
$$dA=r\,d\theta\,dr=r\,dr\,d\theta.$$

<h4>Area of a circle</h4>
<p>The area of a circle of radius $R$ can be found by this double integral:
$$\begineq
A &=\iint_A dA =\int_{\theta=0}^{2\pi}\int_{r=0}^R r\,dr\,d\theta \\
&=\int_{\theta=0}^{2\pi}\left( \left. \frac{r^2}{2} \right|_0^R\right)\,d\theta\\
&=\frac{R^2}{2}\int_{\theta=0}^{2\pi}d\theta=\frac{R^2}{2}2\pi = \pi R^2.
\endeq
$$

<h3>Double polar integrals.</h3>
<p class="show"><img src="0g/12.4.polarint.jpg" class="rightalign"><br>
$A=\iint_A f(r,\theta)\,dA=?$

<h3>Double polar integrals..</h3>
<p class="show"><img src="0g/12.4.polarint2.jpg" class="rightalign"><br>
$A=\iint_A f(r,\theta)\,dA=?$
<p>The line $\sqrt 2=y=r\sin\theta$ becomes<br>
$r(\theta)=\frac{\sqrt 2}{\sin\theta}=\sqrt 2 \csc\theta$.
<p>The double integral becomes:
$$A=\int_{\theta=\pi/4}^{\pi/2}
\int_{r=\sqrt 2\csc\theta}^2 f(r,\theta)\,r\,dr\,d\theta.$$

<h3>Converting...</h3>
<p class="show"><img src="0g/12.4.halfpi.jpg" class="rightalign"><br>
Converting a double-integral in Cartesian coordinates to polar coordinates:
<br><br>
$\iint_A e^{x^2+y^2}dA=\int_{x=-1}^1\int_{y=0}^{\sqrt{1-x^2}}e^{x^2+y^2}dy\,dx=...$
<ul><li>$r$ runs from 0 to 1.
<li>$\theta$ runs from 0 to $\pi$.
<li>ah, $x^2+y^2=r^2$
<li>$dx\,dy=dA=r\,dr\,d\theta$
</ul>

<p>$$\begineq\iint_Ae^{r^2}dA &=\int_{\theta=0}^{\pi}\int_{r=0}^1e^{r^2}r\,dr\,d\theta\\
&=\int_{\theta=0}^{\pi}\left(\left.\frac{e^{r^2}}{2}\right|_{r=0}^1\right)\,d\theta\\

&=
\left(\frac{e-1}{2}\right)\int_{\theta=0}^{\pi}\,d\theta
=
(e-1)\frac\pi 2 \approx 2.70 
\endeq
$$

<h3>ToDo</h3>
<p class="show todo">Double integrals in polar coordinates</p>

<h4 class="handout">Image credits</h4>
<p class="handout"><a href="https://www.flickr.com/photos/visitfinland/5263393586">Visit Finland</a>


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