$\displaystyle\text{volume}={}$ $3$

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The volume $V$ of the tetrahedron $R$ with vertices $(0,0,0)$, $(3,2,0)$, $(0,2,0)$, and $(0,2,3)$ can be calculated as the triple integral of the constant function $f(x,y,z)=1$ over the region $R$. This tetrahedron $R$ is enclosed by the planes $x=0$, $z=0$, $y=2$, and $2z=3y-2x$. We use this to set up the following iterated integral for calculating the volume: $$\begin{aligned}I &= \iiint_R \dd(x,y,z)\\[0.25cm] &= \int_{x=0}^{x=3}\left(\int_{y={{2}\over{3}}x}^{y=2}\left(\int_{z=0}^{z={{3}\over{2}}y-x}\dd z\right)\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=3}\left(\int_{y={{2}\over{3}}x}^{y=2} ({{3}\over{2}}y-x)\,\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=3}\biggl[{{3}\over{4}}y^2-xy\biggr]_{y={{2}\over{3}}x}^{y=2}\;\dd x\\[0.25cm] &= \int_{x=0}^{x=3} \bigl(3-2x+{{1}\over{3}}x^2\bigr)\,\dd x\\[0.25cm] &= \biggl[3x-x^2+{{1}\over{9}}x^3\biggr]_{x=0}^{x=3}\\[0.25cm] &= 3\end{aligned}$$