\(\displaystyle\text{volume}={}\) \({{1}\over{3}}\)

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