Question #315257

Determine the average value of F(x, y, z) = xyz throughout the cubical region D

bounded by the coordinate planes and the planes x = 2, y − 2, z = 2 in the first octant.


1
Expert's answer
2022-03-21T13:33:34-0400

ANSWER: The average value of F(x,y,z)=xyzF(x,y,z)=xyz throughout the cubical region D is 1.

EXPLANATION. The average value of F(x,y,z)=xyzF(x,y,z)=xyz throughout the cubical region D is the number

Favg=1VolumeofRRF(x,y,z)dVF_{avg}=\frac{1}{Volume \: of\: R} \iiint _ {R} F(x,y,z)dV .

VolumeofR=RdV=020202dxdydz=(02dx)(02dy)(02dz)==23=8,RF(x,y,z)dV=020202xyzdxdydz=(02xdx)(02ydy)(02zdz)=([x22]02)([y22]02)([z22]02)=220222022202=8.{Volume \: of\: R}= \iiint _ {R} dV =\int_{0}^{2} \int_{0}^{2}\int_{0}^{2}dxdydz=\left ( \int_{0}^{2}dx \right )\cdot \left ( \int_{0}^{2}dy \right )\cdot\left ( \int_{0}^{2}dz \right )=\\=2^{3}=8,\\ \iiint _ {R}F(x,y,z) dV =\int_{0}^{2} \int_{0}^{2}\int_{0}^{2}xyzdxdydz=\left ( \int_{0}^{2}xdx \right )\cdot \left ( \int_{0}^{2}ydy \right )\cdot\left ( \int_{0}^{2}zdz \right )= \left ( \left [ \frac{x^{2}}{2} \right ] _{0}^{2}\right )\cdot\left ( \left [ \frac{y^{2}}{2} \right ] _{0}^{2}\right )\cdot\left ( \left [ \frac{z^{2}}{2} \right ] _{0}^{2}\right )=\frac{2^2-0}{2 }\cdot\frac{2^2-0}{2 }\cdot\frac{2^2-0}{2 }=8.\\


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