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# Answer to Question #223878 in Chemical Engineering for Lokika

Question #223878

Evaluate ∬R(x-y)^{2}sin^{2}(x+y)d xdy, where R is the rhombus withsuccessive vertices at (π,0),(2π,π),(π,2t) and (0,π)

1
2021-08-30T01:55:06-0400

"\u222c_R(x-y)^{2}sin^{2}(x+y)d xdy\\\\\n\u222c_R((x-y)sin(x+y))^2d xdy\\\\\nG(u,v)= (\\frac{u+v}{2}, \\frac{u-v}{2})\\\\\nx= \\frac{u+v}{2}, y= \\frac{u-v}{2}\\\\\ndxdy= |J(u,v)|dudv\\\\\nJ(u,v)=\\begin{vmatrix}\n \\frac{\\partial x}{\\partial u} & \\frac{\\partial x}{\\partial v} \\\\\n \\frac{\\partial y}{\\partial u} & \\frac{\\partial y}{\\partial v}\n\\end{vmatrix} = \\begin{vmatrix}\n \\frac{1}{2} & \\frac{1}{2} \\\\\n \\frac{1}{2} & -\\frac{1}{2}\n\\end{vmatrix} = \\frac{1}{2}\\\\\n\u222c_R((x-y)sin(x+y))^2d xdy= \\int_0^{2 \\pi}\\int_0^{2 \\pi}(v-\\sin u) J(u,v)d udv\\\\\n= \\int_0^{2 \\pi}\\int_0^{2 \\pi}v^2\\sin^2 u (\\frac{1}{2})d udv\\\\\n= \\frac{1}{2}\\int_0^{2 \\pi}dv\\int_0^{2 \\pi}\\sin^2 udu\\\\\n= \\frac{1}{2}(2 \\pi)\\int_0^{2 \\pi}(\\frac{1-\\cos 2 u}{2})du\\\\\n= \\frac{1}{2}(2 \\pi)[\\frac{1}{2}u -\\frac{\\sin 2 u}{u}]_0^{2 \\pi}\\\\\n= \\frac{1}{2}(2\\pi)( \\frac{1}{2} (2 \\pi))\\\\\n=\\pi^2"

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