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Question Number 51998 by maxmathsup by imad last updated on 01/Jan/19

$$letU=\{(x,y)\in R^2/1⩽x^2+2y^2⩽3\}$$$$calculate\int {\int }_U\frac{x-y}{x^2+y^2}dxdxy$$

Commented by Abdo msup. last updated on 05/Jan/19

$$letconsiderthediffeomorphism$$$$(r,\theta )\to \phi (r,\theta )=(x,y)withx=rcos\theta andy=\frac{r}{\sqrt{2}}sin\theta$$$$1⩽x^2+2y^2⩽3\Rightarrow 1⩽r^2⩽3\Rightarrow 1⩽r⩽\sqrt{3}and$$$$0⩽\theta ⩽2\pi \Rightarrow$$$$\int {\int }_U\frac{x-y}{x^2+y^2}dxdy=\int {\int }_{1⩽r⩽\sqrt{3}and0⩽\theta ⩽2\pi }\frac{r(cos\theta -\frac{sin\theta }{\sqrt{2}})}{r^2{cos}^2\theta +\frac{r^2}{2}{sin}^2\theta }rdrd\theta$$$${\int }_1^{\sqrt{3}}dr{\int }_0^{2\pi }\frac{\sqrt{2}cos\theta -sin\theta }{2{cos}^2\theta +{sin}^2\theta }2d\theta$$$$=2(\sqrt{3}-1){\int }_0^{2\pi }\frac{\sqrt{2}cos\theta -sin\theta }{1+{cos}^2\theta }d\theta but$$$${\int }_0^{2\pi }\frac{\sqrt{2}cos\theta -sin\theta }{1+{cos}^2\theta }d\theta =\sqrt{2}{\int }_0^{2\pi }\frac{cos\theta }{1+{cos}^2\theta }d\theta$$$$+{\int }_0^{2\pi }\frac{-sin\theta }{1+{cos}^2\theta }d\theta$$$${\int }_0^{2\pi }\frac{-sin\theta }{1+{cos}^2\theta }d\theta ={\left[arctan\left(cos\theta \right)\right]}_0^{2\pi }$$$$=0also$$$${\int }_0^{2\pi }\frac{cos\theta }{1+{cos}^2\theta }d\theta ={\int }_0^{\pi }\frac{cos\theta }{1+{cos}^2\theta }+{\int }_{\pi }^{2\pi }\frac{cos\theta }{1+{cos}^2\theta }d\theta$$$${\int }_{\pi }^{2\pi }\frac{cos\theta }{1+{cos}^2\theta }d\theta {=}_{\theta =\pi +t}{\int }_0^{\pi }\frac{-cost}{1+{cos}^{2}t}dt\Rightarrow$$$${\int }_0^{2\pi }\frac{cos\theta }{1+{cos}^2\theta }d\theta =0\Rightarrow$$$$\int {\int }_U\frac{x-y}{x^2+y^2}dxdy=0$$$$$$

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