\(\displaystyle\text{volume}={}\) \(4\)

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The volume \(V\) of the tetrahedron \(R\) with vertices \((0,0,0)\), \((3,2,0)\), \((0,2,0)\), and \((0,2,4)\) 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 \(3z=6y-4x\). We use this to set up the following iterated integral for calculating the volume: \[\begin{aligned}I &= \iiint_R \dd(x,y,z)\\[0.25cm] &= \int_{x=0}^{x=3}\left(\int_{y={{2}\over{3}}x}^{y=2}\left(\int_{z=0}^{z=2y-{{4}\over{3}}x}\dd z\right)\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=3}\left(\int_{y={{2}\over{3}}x}^{y=2} (2y-{{4}\over{3}}x)\,\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=3}\biggl[y^2-{{4}\over{3}}xy\biggr]_{y={{2}\over{3}}x}^{y=2}\;\dd x\\[0.25cm] &= \int_{x=0}^{x=3} \bigl(4-{{8}\over{3}}x+{{4}\over{9}}x^2\bigr)\,\dd x\\[0.25cm] &= \biggl[4x-{{4}\over{3}}x^2+{{4}\over{27}}x^3\biggr]_{x=0}^{x=3}\\[0.25cm] &= 4\end{aligned}\]