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Question Number 31077 by abdo imad last updated on 02/Mar/18

$$calculate\int {\int }_{0<x<1and0<y<x^2}\frac{y}{\sqrt{x^2+y^2}}dxdy.$$

Commented by abdo imad last updated on 07/Mar/18

$$I={\int }_0^1\left({\int }_0^{x^2}\frac{y}{\sqrt{x^2+y^2}}dy\right)dxbut$$$${\int }_0^{x^2}\frac{y}{\sqrt{x^2+y^2}}dy={\left[\sqrt{x^2+y^2}{}^{}\right]}_{y=0}^{y=x^2}=\sqrt{x^2+x^4}-x\Rightarrow$$$$I={\int }_0^1(\sqrt{x^2+x^4}-x)dx={\int }_0^1\sqrt{x^2+x^4}dx-\frac{1}{2}but$$$${\int }_0^1\sqrt{x^2+x^4}dx={\int }_0^{1}x\sqrt{1+x^2}dx=\frac{1}{2}{\int }_0^1\left(2x\right){(1+x^2)}^{\frac{1}{2}}dx$$$$=\frac{1}{2}{\left[\frac{2}{3}{(1+x^2)}^{\frac{3}{2}}\right]}_0^1=\frac{1}{3}(2^{\frac{3}{2}}-1)=\frac{1}{3}(2\sqrt{2}-1)\Rightarrow$$$$I=\frac{2\sqrt{2}}{3}-\frac{1}{3}-\frac{1}{2}=\frac{2\sqrt{2}}{3}-\frac{5}{6}.$$

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