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Double integral

Calculus (Integrals, Series)
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Tolaso J Kos
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Double integral


Post by Tolaso J Kos » Mon Nov 09, 2015 5:08 pm

Let $\displaystyle D= \left\{ \dfrac{1}{2} \leq x^2+y^2\leq 1, \; x^2+y^2-2x \leq 0 \;\;\; y \geq 0 \right\}$. Evaluate the double integral:

$$\iint \limits_{D} \frac{\log (x^2+y^2)}{\sqrt{x^2+y^2}}\; {\rm d} x\; {\rm d} y$$
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Grigorios Kostakos
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Re: Double integral


Post by Grigorios Kostakos » Mon Sep 19, 2016 4:02 pm

The set $D= \big\{(x,y)\in\mathbb{R}^2\;\big|\; \frac{1}{2} \leqslant x^2+y^2\leqslant 1\,, \; x^2+y^2-2x \leqslant 0\,,\; y \geqslant 0 \big\}$
is the (colored) area $(\rm{A\Gamma\Delta B})$, which is the union of the areas $D_1=(\rm{A\Gamma\Delta E})$ (darker) and $D_2=(\rm{\Delta B E})$ (brighter). In polar coordinates these two sets are $$D_1= \big\{(r,\vartheta)\in\mathbb{R}^2\;\big|\; \tfrac{\sqrt{2}}{2}\leqslant r\leqslant 1,\; 0\leqslant\vartheta\leqslant\tfrac{\pi}{3} \big\}$$ and $$D_2= \big\{(r,\vartheta)\in\mathbb{R}^2\;\big|\; \tfrac{\sqrt{2}}{2}\leqslant r\leqslant 2\cos\vartheta,\; \tfrac{\pi}{3}\leqslant\vartheta\leqslant\arccos\tfrac{1}{2\sqrt{2}} \big\}\,,$$
respectively. So
\displaystyle\iint \limits_{D} \frac{\log (x^2+y^2)}{\sqrt{x^2+y^2}}\, {\rm d} (x,y)&=\iint \limits_{D_1} \frac{\log({r^2})}{\sqrt{r^2}}\,r\,{\rm d} (r,\vartheta)+\iint \limits_{D_2} \frac{\log({r^2})}{\sqrt{r^2}}\,r\,{\rm d} (r,\vartheta)\\
&=\int_{0}^{\frac{\pi}{3}}\int_{\frac{\sqrt{2}}{2}}^{1}2\,\log{r}\,{\rm d}r\,{\rm d}\vartheta+\int_{\frac{\pi}{3}}^{\arccos\tfrac{1}{2\sqrt{2}}}\int_{\frac{\sqrt{2}}{2}}^{2\cos\vartheta}2\,\log{r}\,{\rm d}r\,{\rm d}\vartheta\\
&=\frac{\pi}{6}\,\big(\sqrt{2}\,(\log2+2) -4\big)\,+\\
&\qquad\int_{\frac{\pi}{3}}^{\arccos\tfrac{1}{2\sqrt{2}}}\frac{\sqrt{2}}{2}\log 2+\sqrt{2}+4\log2\,\cos\vartheta +4\,\cos\vartheta\,\log(\cos \vartheta) -4\,\cos\vartheta\,{\rm d}\vartheta\,.
\end{align*} The last integral can be evaluated, but is quite messy!!
Grigorios Kostakos
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