\(\displaystyle\text{volume}={}\) \({{1}\over{2}}\)

---

The volume \(V\) of the tetrahedron \(R\) with vertices \((0,0,0)\), \((3,1,0)\), \((0,1,0)\), and \((0,1,1)\) 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 \(x=0\), \(z=0\), \(y=1\), and \(3z=3y-x\). We use this to set up the following iterated integral for calculating the volume: \[\begin{aligned}I &= \iiint_R \dd(x,y,z)\\[0.25cm] &= \int_{x=0}^{x=3}\left(\int_{y={{1}\over{3}}x}^{y=1}\left(\int_{z=0}^{z=y-{{1}\over{3}}x}\dd z\right)\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=3}\left(\int_{y={{1}\over{3}}x}^{y=1} (y-{{1}\over{3}}x)\,\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=3}\biggl[{{1}\over{2}}y^2-{{1}\over{3}}xy\biggr]_{y={{1}\over{3}}x}^{y=1}\;\dd x\\[0.25cm] &= \int_{x=0}^{x=3} \bigl({{1}\over{2}}-{{1}\over{3}}x+{{1}\over{18}}x^2\bigr)\,\dd x\\[0.25cm] &= \biggl[{{1}\over{2}}x-{{1}\over{6}}x^2+{{1}\over{54}}x^3\biggr]_{x=0}^{x=3}\\[0.25cm] &= {{1}\over{2}}\end{aligned}\]