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

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The volume \(V\) of the tetrahedron \(R\) with vertices \((0,0,0)\), \((2,1,0)\), \((0,1,0)\), and \((0,1,3)\) 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 \(x=0\), \(z=0\), \(y=1\), and \(2z=6y-3x\). We use this to set up the following iterated integral for calculating the volume: \[\begin{aligned}I &= \iiint_R \dd(x,y,z)\\[0.25cm] &= \int_{x=0}^{x=2}\left(\int_{y={{1}\over{2}}x}^{y=1}\left(\int_{z=0}^{z=3y-{{3}\over{2}}x}\dd z\right)\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=2}\left(\int_{y={{1}\over{2}}x}^{y=1} (3y-{{3}\over{2}}x)\,\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=2}\biggl[{{3}\over{2}}y^2-{{3}\over{2}}xy\biggr]_{y={{1}\over{2}}x}^{y=1}\;\dd x\\[0.25cm] &= \int_{x=0}^{x=2} \bigl({{3}\over{2}}-{{3}\over{2}}x+{{3}\over{8}}x^2\bigr)\,\dd x\\[0.25cm] &= \biggl[{{3}\over{2}}x-{{3}\over{4}}x^2+{{1}\over{8}}x^3\biggr]_{x=0}^{x=2}\\[0.25cm] &= 1\end{aligned}\]