Answer to Question #224863 in Chemical Engineering for Lokika

Question

Evaluate ∫y=0^{1}∫x=y^{2√2-y^{2}}y/2√x^{2}+y^{2}dxdy ,by clange the order of integration.

Answer 1

$∫_{y=0}^{1}∫_{x=y}^{2 \sqrt{2-y^{2}}}\frac{y}{2\sqrt{x^{2}+y^{2}}}dxdy\\$

In this x and y are from $0<y<1 \space and \space y<x< 2\sqrt{2-y^2}\\$

So we can change it as $0<x<1 \space and \space x<y< \sqrt{2-\frac{x^2}{4}}\\$

$∫_{y=0}^{1}∫_{x=y}^{2 \sqrt{2-y^{2}}}\frac{y}{2\sqrt{x^{2}+y^{2}}}dxdy = ∫_{x=0}^{1}∫_{y=x}^{\sqrt{2-\frac{x^2}{4}}}\frac{y}{2\sqrt{x^{2}+y^{2}}}dydx \\ \int _0^1\int _x^{\sqrt{2-\frac{x^2}{4}}}\frac{y}{2\sqrt{x^2+y^2}}dydx\\ =\frac{1}{4}\left(\sqrt{3x^2+8}-2\sqrt{2}x\right)\\ =\int _0^1\frac{1}{4}\left(\sqrt{3x^2+8}-2\sqrt{2}x\right)dx\\ =\frac{1}{4}\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{3}}\ln \left(\frac{7+\sqrt{33}}{4}\right)-\sqrt{2}\right)\\ =\frac{1}{4}\left(\frac{\sqrt{11}}{2}+\frac{2}{\sqrt{3}}\ln \left(\frac{7+\sqrt{33}}{4}\right)-\sqrt{2}\right)\\ =0.39554$