Find \(\iint\limits_{\kern-3ptR}y\,d\!A\), where \(R\) is the region enclosed by the lines \(2x+y=0\), \(\ \ 2x+y=3\), \(x-y=0\), and \(x-y=2\).

Solution We use the change of variables \(u=2x+y\) and \(v=x-y\). Then the lines \(2x+y=0\) and \(2x+y=3\) in the \(xy\)-plane correspond to the lines \(u=0\) and \(u=3\) in the \(uv\)-plane. The lines \(x-y=0\) and \(x-y=2\) in the \(xy\)-plane correspond to the lines \(v=0\) and \(v=2\) in the \(uv\)-plane. As Figure 63 shows, the region \(R\) in the \(xy\)-plane is a parallelogram and the region \(R^{\#}\) in the \(uv\)-plane is a rectangle.

To find the Jacobian, we need to express \(x\) and \(y\) as functions of \(u\) and \(v\). We can achieve this by solving the system of equations \(\left\{ \begin{array}{l} u=2x+y \\ v=x-y \end{array} \right. \) for \(x\) and \(y.\) Then \[ x=\dfrac{u+v}{3} \qquad y=\dfrac{u-2v}{3} \]

The Jacobian is \begin{equation*} \dfrac{\partial (x,y)}{\partial (u,v)}=\left\vert \begin{array}{c@{\quad}c} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v}\\[3pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{array} \right\vert =\left\vert \begin{array}{l@{\quad}r} \dfrac{1}{3} & \dfrac{1}{3}\\[3pt] \dfrac{1}{3} & -\dfrac{2}{3} \end{array} \right\vert =-\dfrac{2}{9}-\dfrac{1}{9}=-\dfrac{1}{3} \end{equation*}

Using this change of variables, the integral \(\iint\limits_{\kern-2ptR}y\,d\!A\) becomes \[ \begin{eqnarray*} \iint\limits_{\kern-3ptR}y\,d\!A &=&\displaystyle\iint\limits_{\kern-3ptR^{\#}}\dfrac{u-2v}{3}\cdot \left\vert -\dfrac{1}{3}\right\vert \,du\,dv=\int_{0}^{2}\int_{0}^{3}\dfrac{1}{9}(u-2v)\,du\,dv=\dfrac{1}{9}\int_{0}^{2}\left[ \dfrac{u^{2}}{2}-2uv\right]_{0}^{3}\,dv\\ &=&\dfrac{1}{9}\int_{0}^{2}\left( \dfrac{9}{2}-6v\right) {dv}=\dfrac{1}{9}\left[\dfrac{9v}{2}-3v^{2}\right] _{0}^{2}=\dfrac{1}{9}(9-12)=-\dfrac{1}{3} \end{eqnarray*} \]