Compute the volume of [tex]Q[/tex]:
[tex]\displaystyle\iiint_Q\mathrm dV=\int_0^2\int_0^{2-x}\int_0^{2-x-y}\mathrm dz\,\mathrm dy\,\mathrm dx=\frac43[/tex]
Integrate [tex]f(x,y,z)=x+y+z[/tex] over [tex]Q[/tex]:
[tex]\displaystyle\iiint_Qf(x,y,z)\,\mathrm dV=\int_0^2\int_0^{2-x}\int_0^{2-x-y}(x+y+z)\,\mathrm dz\,\mathrm dy\,\mathrm dx=2[/tex]
So the average value of [tex]f[/tex] over [tex]Q[/tex] is 2/(4/3) = 3/2.
To solve this mathematical problem, we need to understand the Average Value of a Continuous function.
The average value of a continuous function is derived by taking the integral of the function over the interval. This is then divided using the length of that interval.
To determine the average value of the function f(x, y, z), over the solid region named Q,
we can say:
[tex]\int\int\int _{Q}[/tex] dV = [tex]\int_{0}^{2} \int_{0}^{2-x} \int_{0}^{2-x-y}[/tex] dzdydx = 4/3
Integrating the above, we have
[tex]\int\int\int _{Q}[/tex] [tex]f(x,y,z)[/tex] dV = [tex]\int_{0}^{2} \int_{0}^{2-x} \int_{0}^{2-x-y}[/tex] (x+ y + z) dzdydx = 2
Therefore, the average value of the function f over the Solid region Q becomes:
2/ (4/3) = 1.5 or 3/2
Learn more about the average value of a continuous function at:
https://brainly.com/question/22155666
