Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane [latex]x+y+z=1[/latex] has intercepts [latex](1, 0, 0)[/latex], [latex](0, 1, 0)[/latex], and [latex](0, 0, 1)[/latex]. The region [latex]E[/latex] looks like

[latex]{E} = {\left \{ {(x,y,z)}{\mid}{0} \ {\leq} \ {x} \ {\leq} \ {1,0} \ {\leq} \ {y} \ {\leq} \ {{1-x},{0}} \ {\leq} \ {z} \ {\leq} \ {1-x-y} \right \}}[/latex].

Hence the triple integral of the temperature is

[latex]\underset{E}{\displaystyle\iiint}{f}{(x,y,z)}{dV} = {\displaystyle\int^{x=1}_{x=0}} \ {\displaystyle\int^{y=1-x}_{y=0}} \ {\displaystyle\int^{z=1-x-y}_{z=0}}(xy+8z+20){dz}{dy}{dx} = {\frac{147}{40}}[/latex].

The volume evaluation is

[latex]{V}{(E)} = \underset{E}{\displaystyle\iiint}{1dV} = {\displaystyle\int^{x=1}_{x=0}} \ {\displaystyle\int^{y=1-x}_{y=0}} \ {\displaystyle\int^{z=1-x-y}_{z=0}}{1dzdydx} = {\frac{1}{6}}[/latex].

Hence the average value is [latex]{{T}_{\text{ave}}} = {\frac{147/40}{1/6}} = {\frac{6(147)}{40}} = {\frac{441}{20}}[/latex] degrees Celsius.