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Question Number 36191 by prof Abdo imad last updated on 30/May/18

$$letD=\{(x,y)\in R^2/x>0,y>0,x+y<1\}$$ $$1)calculate\int {\int }_D\frac{xy}{x^2+y^2}dxdy$$ $$2)leta>0,b>0calculate\int {\int }_D{a}^x{b}^{y}dxdy$$

Commented bymaxmathsup by imad last updated on 26/Aug/18

$$polarcoordinateschangementx=rcos\theta andy=rsin\theta$$ $$x>0andy>0\Rightarrow 0<\theta <\frac{\pi }{2}andx+y<1\Rightarrow r(cos\theta +sin\theta )<1\Rightarrow$$ $$0<r<\frac{1}{cos\theta +sin\theta }$$ $$\int {\int }_D\frac{xy}{x^2+y^2}dxdy={\int }_0^{\frac{\pi }{2}}\left({\int }_0^{\frac{1}{cos\theta +sin\theta }}\frac{r^{2}cos\theta sin\theta }{r^2}rdr\right)d\theta$$ $$={\int }_0^{\frac{\pi }{2}}\left({\int }_0^{\frac{1}{cos\theta +sin\theta }}rdr\right)cos\theta sin\theta d\theta but$$ $${\int }_0^{\frac{1}{cos\theta +sin\theta }}rdr={\left[\frac{r^2}{2}\right]}_0^{\frac{1}{cos\theta +sin\theta }}=\frac{1}{2{(cos\theta +sin\theta )}^2}=\frac{1}{2(1+2sin\theta cos\theta )}\Rightarrow$$ $$I={\int }_0^{\frac{\pi }{2}}\frac{cos\theta sin\theta }{2(1+2sin\theta cos\theta )}d\theta =\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\frac{sin\left(2\theta \right)}{1+sin\left(2\theta \right)}d\theta$$ $${=}_{2\theta =t}\frac{1}{4}{\int }_0^{\pi }\frac{sint}{1+sint}\frac{dt}{2}=\frac{1}{8}{\int }_0^{\pi }\frac{1+sint-1}{1+sint}dt$$ $$=\frac{\pi }{8}-\frac{1}{8}{\int }_0^{\pi }\frac{dt}{1+sint}chang.tan\left(\frac{t}{2}\right)=ugive$$ $${\int }_0^{\pi }\frac{dt}{1+sint}={\int }_0^{\mathrm{\infty }}\frac{1}{1+\frac{2u}{1+u^2}}\frac{2du}{1+u^2}=2{\int }_0^{\mathrm{\infty }}\frac{du}{1+u^2+2u}$$ $$=2{\int }_0^{\mathrm{\infty }}\frac{du}{{(u+1)}^2}=2{[-\frac{1}{u+1}]}_0^{+\mathrm{\infty }}=2\Rightarrow I=\frac{\pi }{8}-\frac{1}{4}.$$ $$$$

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