\(\displaystyle\text{volume}={}\) \(6\)

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