Find \begin{equation*} \displaystyle\iint\limits_{\kern-3ptR}(x+y)\sin (x-y)\,d\!A \end{equation*}

where \(R\) is the region enclosed by \(y=x\), \(\ y=x-2\), \(\ y=-x\), and \(\ y=-x+1 \).

Solution The integrand \((x+y)\sin (x-y)\) suggests changing the variables to \(u=x+y\) and \(v=x-y\). Then the lines \(x-y=0\), \(\ x-y=2\), \(\ x+y=0\), and \(\ x+y=1\) in the \(xy\)-plane define the region \(R\), and the lines \(v=0,\) \(v=2,\) \(u=0,\) and \(u=1\) in the \(uv\)-plane define the region \(R^{\#}\). See Figure 64. We solve the system of equations \(\left\{ \begin{array}{l} u=x+y \\[3pt] v=x-y \end{array} \right. \) for \(x\) and \(y\), and obtain \begin{equation*} x=\dfrac{u+v}{2}\qquad y=\dfrac{u-v}{2} \end{equation*}

Using these equations, the Jacobian is \begin{equation*} \dfrac{\partial (x,y)}{\partial (u,v)}=\left\vert \begin{array}{l@{\quad}r} \dfrac{1}{2} & \dfrac{1}{2} \\[9pt] \dfrac{1}{2} & -\dfrac{1}{2} \end{array} \right\vert =-\dfrac{1}{4}-\dfrac{1}{4}=-\dfrac{1}{2} \end{equation*}

The integral under this change of variables becomes \begin{eqnarray*} \iint\limits_{\kern-3ptR}(x+y)\sin (x-y)\,d\!A& =&\iint\limits_{\kern-3ptR^{\#}}u\sin v\cdot \left\vert -\dfrac{1}{2}\right\vert \,du\,dv=\dfrac{1}{2}\int_{0}^{1} \int_{0}^{2}u\sin v\,dv\,du \notag \\ & =&\dfrac{1}{2}\int_{0}^{1}\big[u(-\cos v)\big] _{0}^{2}\,du= \dfrac{1}{2}(1-\cos 2)\int_{0}^{1}u\,du \notag \\[4.5pt] & =&\dfrac{1}{4}(1-\cos 2) \end{eqnarray*}