Find \(\iiint\limits_{\kern-3ptE} y\,{\it dV},\) where \(E\) is the solid enclosed by the planes \(2x+y=0,\) \(2x+y=3,\) \(x-z=0,\) \(x-z=2\), \(z=0,\) and \(z=4.\)
We solve the system of equations \(\left\{ \begin{array}{l} u=2x+y \\ v=x-z \\ w=z \end{array} \right. \) for \(x,\) \(y,\) and \(z\) and obtain \[ x=v+w \qquad y=u-2x=u-2 ( v+w ) =u-2v-2w \qquad z=w \]
The Jacobian is \begin{equation*} \dfrac{\partial (x,y,z)}{\partial (u,v,w)}=\left\vert \begin{array}{c@{\quad}c@{\quad}c} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{ \partial x}{\partial w} \\[9pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{ \partial y}{\partial w} \\[9pt] \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{ \partial z}{\partial w} \end{array} \right\vert = \left|\begin{array}{c@{\quad}c@{\quad}c} 0 & \hphantom{-}1 & \hphantom{-}1\\ 1 & -2 & -2\\ 0 & \hphantom{-}0 & \hphantom{-}1 \end{array}\right| =-1 \end{equation*}
The integral \(\iiint\limits_{\kern-3ptE}\,y\,{\it dV}\) under this change of variables becomes \[ \begin{array}{rcl} \displaystyle\iiint\limits_{\kern-3ptE}~y\,{\it dV} &=&\displaystyle\iiint\limits_{\kern-3ptE^{\#}}(u-2v-2w) \left\vert \dfrac{\partial (x,y,z)}{\partial (u,v,w)}\right\vert du\, dv\, dw=\int_{0}^{4}\int_{0}^{2}\int_{0}^{3}( u-2v-2w) \,du\,dv\,dw \notag\\ &=&\displaystyle\int_{0}^{4}\int_{0}^{2}\left[ \dfrac{u^{2}}{2}-2vu-2wu\right] _{0}^{3}dv\,dw=\int_{0}^{4}\int_{0}^{2}\left( \dfrac{9}{2}-6v-6w\right) dv\, dw \notag\\ &=&\displaystyle\int_{0}^{4}\left[ \dfrac{9}{2}v-3v^{2}-6wv\right] _{0}^{2}\, dw= \int_{0}^{4}( 9-12-12w) \, dw=\int_{0}^{4}( -3-12w) \, dw \notag\\ &=&\big[ -3w-6w^{2}\big] _{0}^{4}=-12-96=-108 \end{array} \]