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Question: Evaluate $\iint_R\frac{2xy^5}{\sqrt{1+x^2y^2-y^4}}dxdy$ where 'R' is the region of the triangle whose vertices are $(0,0),\, (1,1),\,(0,1)$ .
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Subject : Applied Mathematics 2

Topic : Double integration

Difficulty : Medium

am2(82) • 204 views
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modified 6 months ago by gravatar for awari.swati831 awari.swati831140 written 8 months ago by gravatar for smitapn612 smitapn6120
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$ \text { Let O (0,0), A (1,1), B (0,1) be the vertices of the triangle OAB.} \\ $

$ \text { The equation of the line OA is} \\ $

$ \frac{x-0}{0-1} = \frac{y-0}{0-1} \text{ i.e. } x= y \\ $

$ \text {Now consider a strip parallel to x-axis } \\ $

$ \int_{y=0}^{1} \int_{x=0}^{y} \frac{2y^5x} {\sqrt {1- x^2y^2 - y^4} }dx dy \\ $

$ \int_{y=0}^{1} \int_{x=0}^{y} \frac{1}{y} \frac{2y^5*x} {\sqrt{\frac{1- y^4}{y^2} + x^2 } }dx dy \\ $

$ \int_{y=0}^{1} \int_{x=0}^{y} 2y^4 \frac{x} {\sqrt{\frac{1- y^4}{y^2} + x^2 } }dx dy \\ $

$ \text { Put } \frac{1- y^4}{y^2} + x^2 = t \\ $

$ 2xdx = dt \\ $

$ = \int_{0}^{1} 2 y^4 \int \frac { dt} {2\sqrt t } \\ $

$ = \int_{0}^{1} y^4 (2\sqrt t ) dy \\ $

$ = 2 \int_{0}^{1} y ^4 \left [ \sqrt{\frac{1- y^4}{y^2} + x^2 }\right]_0^y dy \\ $

$ = 2 \int_{0}^{1} y ^4 \left [ \sqrt{\frac{1- y^4}{y^2}+ y^2} - \sqrt{\frac{1- y^4}{y^2} }\right] dy \\ $

$ = 2 \int_{0}^{1} y ^4 \left [ \frac{1}{y} - \sqrt{\frac{1- y^4}{y^2} }\right] dy \\ $

$ = 2 \int_{0}^{1} y ^3 - \sqrt{{1- y^4} } y^3 dy \\ $

$ = 2 \left [ \frac{y^4}{4} + \frac{1}{4} \frac{(1- y^4)^\frac{3}{2}}{\frac{3}{2}} )\right]_0^1 \\ $

$ = 2 [\frac{1}{4} - \frac{1}{4} \frac{2}{3} ] = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \\ $

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modified 6 months ago by gravatar for awari.swati831 awari.swati831140 written 7 months ago by gravatar for smitapn612 smitapn6120
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