\(\displaystyle\text{volume}={}\) \(2\)

---

The volume \(V\) of the tetrahedron \(R\) with vertices \((0,0,0)\), \((2,2,0)\), \((0,2,0)\), 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 \(2z=3y-3x\). We use this to set up the following iterated integral for calculating the volume: \[\begin{aligned}I &= \iiint_R \dd(x,y,z)\\[0.25cm] &= \int_{x=0}^{x=2}\left(\int_{y=x}^{y=2}\left(\int_{z=0}^{z={{3}\over{2}}y-{{3}\over{2}}x}\dd z\right)\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=2}\left(\int_{y=x}^{y=2} ({{3}\over{2}}y-{{3}\over{2}}x)\,\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=2}\biggl[{{3}\over{4}}y^2-{{3}\over{2}}xy\biggr]_{y=x}^{y=2}\;\dd x\\[0.25cm] &= \int_{x=0}^{x=2} \bigl(3-3x+{{3}\over{4}}x^2\bigr)\,\dd x\\[0.25cm] &= \biggl[3x-{{3}\over{2}}x^2+{{1}\over{4}}x^3\biggr]_{x=0}^{x=2}\\[0.25cm] &= 2\end{aligned}\]