Multiple integrals: Applications of multiple integrals
Calculation of the volume of a 3-D sub-region
The volume \(V\) of a closed and bounded region \(R\) in three-dimensional space is given by the following formula: \[V=\iiint_R\dd(x,y,z)\]
Calculate the volume of the tetrahedron with vertices \((0,0,0)\), \((1,1,0)\), \((0,1,0)\), and \((0,1,4)\).
\(\displaystyle\text{volume}={}\) \({{2}\over{3}}\)
The volume \(V\) of the tetrahedron \(R\) with vertices \((0,0,0)\), \((1,1,0)\), \((0,1,0)\), and \((0,1,4)\) 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 \(x=0\), \(z=0\), \(y=1\), and \(z=4y-4x\). We use this to set up the following iterated integral for calculating the volume: \[\begin{aligned}I &= \iiint_R \dd(x,y,z)\\[0.25cm] &= \int_{x=0}^{x=1}\left(\int_{y=x}^{y=1}\left(\int_{z=0}^{z=4y-4x}\dd z\right)\dd y\right)\dd x\\[0.25cm] &=\int_{x=0}^{x=1}\left(\int_{y=x}^{y=1} (4y-4x)\,\dd y\right)\dd x\\[0.25cm] &= \int_{x=0}^{x=1}\biggl[2y^2-4xy\biggr]_{y=x}^{y=1}\;\dd x\\[0.25cm] &= \int_{x=0}^{x=1} \bigl(2-4x+2x^2\bigr)\,\dd x\\[0.25cm] &= \biggl[2x-2x^2+{{2}\over{3}}x^3\biggr]_{x=0}^{x=1}\\[0.25cm] &= {{2}\over{3}}\end{aligned}\]
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