Answer to Question #229225 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∬R​(x−y)2sin2(x+y)dxdy∬R​((x−y)sin(x+y))2dxdyG(u,v)=(2u+v​,2u−v​)x=2u+v​,y=2u−v​dxdy=∣J(u,v)∣dudvJ(u,v)

=∣∣∣∣∣​∂u∂x​∂u∂y​​∂v∂x​∂v∂y​​∣∣∣∣∣​=∣∣∣∣∣​21​21​​21​−21​​∣∣∣∣∣​=21​∬R​((x−y)sin(x+y))2dxdy=∫02π​∫02π​(v−sinu)J(u,v)dudv

=∫02π​∫02π​v2sin2u(21​)dudv=21​∫02π​dv∫02π​sin2udu=21​(2π)∫02π​(21−cos2u​)du=21​(2π)[21​u−usin2u​]02

π​=21​(2π)(21​(2π))

=π2