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Question Number 171260 by udaythool last updated on 11/Jun/22

$$\underset{―}{Changetopolarcoordinates:}$$$${\underset{0}{\int }}^{4a}\underset{y^2/4a}{\int }\stackrel{a}{}\left(\frac{x^2-y^2}{x^2+y^2}\right)dxdy$$

Answered by udaythool last updated on 13/Jun/22

$$\mathrm{Solution}:\mathrm{Let}f\equiv \left(\frac{x^2-y^2}{x^2+y^2}\right)\Rightarrow f=r\cos 2\theta$$$$\Rightarrow I={\int }_0^{4a}{\int }_{y^2/4a}^{a}fdxdy$$$$={\int }_0^{2a}{\int }_{y^2/4a}^{a}fdxdy-{\int }_{2a}^{4a}{\int }_a^{y^2/4a}fdxdy$$$$=\underset{R_1}{\int \int }-\underset{R_2}{\int \int }=I_1-I_2$$$$\mathrm{where}{R}_1\equiv 0\le y\le 2a,y^2/4a\le x\le a$$$$\mathrm{and}{R}_2\equiv 2a\le y\le 4a,a\le x\le y^2/4a$$$$\mathrm{For}\mathrm{polar}\mathrm{coordinates}\mathrm{we}\mathrm{should}\mathrm{have}$$$$\mathrm{to}\mathrm{divide}\mathrm{both}{R}_1\mathrm{into}A,B\mathrm{and}{R}_2\mathrm{into}C,D$$$$\mathrm{subregions}\mathrm{by}\mathrm{a}\mathrm{line}\mathrm{from}\mathrm{origin}\mathrm{and}(\mathrm{a},2a).$$$$\mathrm{Thus}{I}_1=I_A+I_B\mathrm{and}{I}_2=I_C+I_D$$$$\mathrm{For}A,0\le \theta \le {\tan }^{-1}2\mathrm{and}0\le r\le a/\cos \theta$$$$\Rightarrow I_A={\int }_0^{{\tan }^{-1}2}{\int }_0^{a/\cos \theta }r\cos 2\theta drd\theta =a^2({\tan }^{-1}2-1).$$$$\mathrm{For}B,{\tan }^{-1}2\le \theta \le \pi /2\mathrm{and}0\le r\le 4a\cos \theta /{\sin }^2\theta$$$$\Rightarrow I_B={\int }_{{\tan }^{-1}2}^{\pi /2}{\int }_0^{4a\cos \theta /{\sin }^2\theta }r\cos 2\theta drd\theta$$$$I_B=8a^2(\pi -23/24-{2\tan }^{-1}2).$$$$\mathrm{For}C,\pi /4\le \theta \le {\tan }^{-1}2\mathrm{and}4a\cos \theta /{\sin }^2\theta \le r\le 4a/\sin \theta$$$$\Rightarrow I_C={\int }_{\pi /4}^{{\tan }^{-1}2}{\int }_{4a\cos \theta /{\sin }^2\theta }^{4a/\sin \theta }r\cos 2\theta drd\theta$$$$I_C=8a^2(\pi +29/24-{4\tan }^{-1}2).$$$$\mathrm{And}\mathrm{for}D,{\tan }^{-1}2\le \theta \le {\tan }^{-1}4\mathrm{and}a/\cos \theta \le r\le 4a/\sin \theta$$$$\Rightarrow I_D={\int }_{{\tan }^{-1}2}^{{\tan }^{-1}4}{\int }_{a/\cos \theta }^{4a/\sin \theta }r\cos 2\theta drd\theta$$$$I_D=a^2(3+{174\tan }^{-1}2-{17\tan }^{-1}4).$$$$\therefore I=\left(\frac{-64}{3}\right)a^2+17a^2{\tan }^{-1}4$$

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