\(\displaystyle\text{volume}={}\) \(3\)

---

The volume \(V\) of the tetrahedron \(R\) with vertices \((0,0,0)\), \((2,3,0)\), \((0,3,0)\), and \((0,3,3)\) 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 \(x=0\), \(z=0\), \(y=3\), and \(2z=2y-3x\). We use this to set up the following iterated integral for calculating the volume: \[\begin{aligned}I &= \iiint_R \dd(x,y,z)\\[0.25cm] &= \int_{x=0}^{x=2}\left(\int_{y={{3}\over{2}}x}^{y=3}\left(\int_{z=0}^{z=y-{{3}\over{2}}x}\dd z\right)\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=2}\left(\int_{y={{3}\over{2}}x}^{y=3} (y-{{3}\over{2}}x)\,\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=2}\biggl[{{1}\over{2}}y^2-{{3}\over{2}}xy\biggr]_{y={{3}\over{2}}x}^{y=3}\;\dd x\\[0.25cm] &= \int_{x=0}^{x=2} \bigl({{9}\over{2}}-{{9}\over{2}}x+{{9}\over{8}}x^2\bigr)\,\dd x\\[0.25cm] &= \biggl[{{9}\over{2}}x-{{9}\over{4}}x^2+{{3}\over{8}}x^3\biggr]_{x=0}^{x=2}\\[0.25cm] &= 3\end{aligned}\]