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.

Figure 63

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