Answer to Question #224862 in Chemical Engineering for Lokika

Question

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)

Answer 1

$∬_R(x-y)^{2}sin^{2}(x+y)d xdy\\ ∬_R((x-y)sin(x+y))^2d xdy\\ G(u,v)= (\frac{u+v}{2}, \frac{u-v}{2})\\ x= \frac{u+v}{2}, y= \frac{u-v}{2}\\ dxdy= |J(u,v)|dudv\\ J(u,v)=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{vmatrix} = \frac{1}{2}\\ ∬_R((x-y)sin(x+y))^2d xdy= \int_0^{2 \pi}\int_0^{2 \pi}(v-\sin u) J(u,v)d udv\\ = \int_0^{2 \pi}\int_0^{2 \pi}v^2\sin^2 u (\frac{1}{2})d udv\\ = \frac{1}{2}\int_0^{2 \pi}dv\int_0^{2 \pi}\sin^2 udu\\ = \frac{1}{2}(2 \pi)\int_0^{2 \pi}(\frac{1-\cos 2 u}{2})du\\ = \frac{1}{2}(2 \pi)[\frac{1}{2}u -\frac{\sin 2 u}{u}]_0^{2 \pi}\\ = \frac{1}{2}(2\pi)( \frac{1}{2} (2 \pi))\\ =\pi^2$