Om vi sätter $u=2x+y$, $v=x+2y$ så blir \[ \left\{\begin{array}{cc}x=&\dfrac23 u-\dfrac13 v\\ y=&-\dfrac13 u+\dfrac23 v \end{array}\right., \]och $|\alpha\delta-\beta\gamma|=|\dfrac23\cdot\dfrac23-(-\dfrac13)(-\dfrac13)=\dfrac13$. Integralen övergår därmed I \[ \iint_{D} (2x+y)^2(x+2y)^2 \, dxdy= \] \[ \iint_{E} u^2v^2 \dfrac13 \, dudv= \] (med $E=\{(u,v): -1\le u\le 1,-1\le v\le 1\}$) \[ \dfrac13 \left(\int_{-1}^1u^2\, du\right)\left(\int_{-1}^1v^2\, dv\right)= \] \[ \dfrac13 \Big[\dfrac13 u^3\Big]_{-1}^1\Big[\dfrac13 v^3\Big]_{-1}^1= \dfrac13\cdot \dfrac23 \cdot \dfrac23=\dfrac4{27}. \]