Find the average value of F(x, y, z) over the given region. F(x, y, z) = x2 + 9 over the cube in the first octant bounded by the coordinate planes and the planes x = 2, y = 2, and z =2.

Let D be the volume of the cube bounded by the coordinate planes and the planes $x=2,y=2,andz=2$.
Then $D=2\cdot 2\cdot 2=8$
Calculate average value of ${\displaystyle x^2+9}$ over give cube D using formula:
Average value ${\displaystyle =\frac{1}{volumeofD}\int {\int }_D\int FdV}$
Volume of D is equal to ${\displaystyle 2^3=8}$
Average value ${\displaystyle =\frac{1}{volumeofD}{\int }_0^2{\int }_0^2{\int }_0^2(x^2+9)dzdydx}$
${\displaystyle =\frac{1}{8}{\int }_0^2{\int }_0^2(x^{2}z+9z){\mid }_0^{2}dydx}$ (Integrate in relation to z.)
${\displaystyle =\frac{1}{8}{\int }_0^2{\int }_0^2(2x^2+18)dyxe}$ (Compute the boundaries)
${\displaystyle =\frac{1}{8}{\int }_0^2(2x^{2}y+18y){\mid }_0^{2}dx}$ (Intergrate in relation to y)
${\displaystyle =\frac{1}{8}{\int }_0^2(4x^2+36)dx}$ (Compute the boundaries)
${\displaystyle =\frac{1}{8}(4\frac{x^3}{3}+36x){\mid }_0^2}$ (Intergrate in relation to x)
${\displaystyle =\frac{1}{8}(\frac{32}{3}+72)}$ (Compute the boundaries)
${\displaystyle =\frac{4}{3}+9=\frac{31}{3}}$
Final answer ${\displaystyle =\frac{31}{3}}$