Question #224863

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

Expert's answer

"\u222b_{y=0}^{1}\u222b_{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}}\\\\"

"\u222b_{y=0}^{1}\u222b_{x=y}^{2 \\sqrt{2-y^{2}}}\\frac{y}{2\\sqrt{x^{2}+y^{2}}}dxdy = \u222b_{x=0}^{1}\u222b_{y=x}^{\\sqrt{2-\\frac{x^2}{4}}}\\frac{y}{2\\sqrt{x^{2}+y^{2}}}dydx \\\\\n\\int _0^1\\int _x^{\\sqrt{2-\\frac{x^2}{4}}}\\frac{y}{2\\sqrt{x^2+y^2}}dydx\\\\\n=\\frac{1}{4}\\left(\\sqrt{3x^2+8}-2\\sqrt{2}x\\right)\\\\\n=\\int _0^1\\frac{1}{4}\\left(\\sqrt{3x^2+8}-2\\sqrt{2}x\\right)dx\\\\\n=\\frac{1}{4}\\left(\\frac{\\sqrt{11}}{2}+\\frac{2}{\\sqrt{3}}\\ln \\left(\\frac{7+\\sqrt{33}}{4}\\right)-\\sqrt{2}\\right)\\\\\n=\\frac{1}{4}\\left(\\frac{\\sqrt{11}}{2}+\\frac{2}{\\sqrt{3}}\\ln \\left(\\frac{7+\\sqrt{33}}{4}\\right)-\\sqrt{2}\\right)\\\\\n=0.39554"

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