The Simple Triple Integral

When a function is defined over a volume of space, we can perform a triple integral of the function over the volume For example.  
\[\rho(x,y,z)\]
  may describe the give the density of a fluid at the point  
\[(x,y,z)\]
 then the mass of fluid in the volume is  
\[M=\int_V \rho (x,y,z)dx dydz\]
.
Suppose a function  
\[f(x,y,z)=xy\]
  is to be integrated over the space defined by  
\[0 \lt xy \lt 1, \; 2 \lt z \lt 3\]
.
\[\begin{equation} \begin{aligned} V &= \int^{z=3}_{z=2} \int^{y=1}_{y=0} \int^{x-1}_{x=0} xy dx dydz \\ &= \int^{z=3}_{z=2} \int^{y=1}_{y=0} [ \int^{x-1}_{x=0} xy dx ] dydz \\ &= \int^{z=3}_{z=2} \int^{y=1}_{y=0} [\frac{x^2y}{2} ]^{x=1}_{x=0} dydz \\ &= \int^{z=3}_{z=2} \int^{y=1}_{y=0} \frac{y}{2} dydz \\ &= \int^{z=3}_{z=2} [ \int^{y=1}_{y=0} \frac{y}{2} dy ]dz \\ &= \int^{z=3}_{z=2} [\frac{y^2}{4} ]^{y=1}_{y=0} dz \\ &= \int^{z=3}_{z=2} \frac{1}{4} dz \\ &= [\frac{z}{4}]^{z=3}_{z=2} = \frac{1}{4} \end{aligned} \end{equation}\]

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