Answer:
The average function [tex]f_{avg}[/tex] value = 4
Step-by-step explanation:
Given that:
[tex]f(x,y,z) = 3x^2z+3y^2z[/tex]
The enclosed region [tex]z = 4 -x^2 -y^2[/tex] and the plane z = 0.
Over an area E, the average value of a function f(x,y,z) is:
[tex]f_{avg} = \dfrac{1}{V_{E}} \iiint_E f(x,y,z) dV ------ (1)[/tex]
where;
[tex]V_E[/tex] = volume of region E
To find the volume use cylindrical coordinates;
[tex]x=rcos \theta; \ \ \ y = rsin \theta; \ \ \ z = z[/tex]
Then;
E = 0 ≤ r ≤2, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 4 - r²
[tex]V_E = \int^{2 \pi}_{0} \int^{2}_{0} \int^{4-r^2}_{0} \ rdzdrd \theta[/tex]
[tex]V_E = \int^{2 \pi}_{0} \int^{2}_{0} r(4-r^2) \ drd \theta[/tex]
[tex]V_E = \int^{2 \pi}_{0} \int^{2}_{0} (4r-r^3) \ drd \theta[/tex]
[tex]V_E = \int^{2 \pi}_{0} \Big(2r^2- \dfrac{r^4}{4} \Big)^2_0 \ d \theta[/tex]
[tex]V_E = 4(2 \pi)\\\\ V_E = 8 \pi[/tex]
However, from equation (1):
[tex]f_{avg} = \dfrac{1}{V_E} \iiint_E f(x,y,z) \ dV \\ \\ = \dfrac{1}{8 \pi} \iiint_E (3x^2z + 3y^2 z) dV \\ \\ = \dfrac{1}{8 \pi } \int^{2 \pi}_{0} \int^{2 }_{0} \int^{4 - r^2}_{0} \ \ 3zr^2 \ rdzdrd\theta \\ \\ = \dfrac{3}{8 \pi} \int^{2 \pi}_{0} \int^{2 }_{0} \Big(\dfrac{z^2}{2}\Big)^{4-r^2}_{0} \ r^3 drd\theta \\ \\ = \dfrac{3}{16 \pi} \int^{2 \pi}_{0} \int^{2 }_{0} (4-r^2)^2 \ r^3 \ dr d\theta[/tex]
[tex]= \dfrac{3}{16 \pi} \int^{2 \pi}_{0} \int^{2 }_{0} (16-8r^2 + r^4)r^3 \ dr d\theta \\ \\ = \dfrac{3}{16 \pi} \int^{2 \pi}_{0} \int^{2 }_{0} (16r^3 -8r^5 + r^7) \ dr d \theta \\ \\ = = \dfrac{3}{16 \pi} \int^{2 \pi}_{0} \Big (4r^4 - \dfrac{8}{6}r^6 + \dfrac{r^8}{8} \Big)^2_0 \ d \theta \\ \\ = \dfrac{3}{16 \pi } (2 \pi) \Big (64 - \dfrac{8}{6} \times 64 \times \dfrac{256}{8} \Big) \\ \\ = \dfrac{3}{16 \pi} (2 \pi) (\dfrac{32}{3}) \\ \\ = 4[/tex]
