# 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.

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

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Roosevelt Houghton

Let D be the volume of the cube bounded by the coordinate planes and the planes .
Then $D=2\cdot 2\cdot 2=8$
Calculate average value of ${x}^{2}+9$ over give cube D using formula:
Average value
Volume of D is equal to ${2}^{3}=8$
Average value
$=\frac{1}{8}{\int }_{0}^{2}{\int }_{0}^{2}\left({x}^{2}z+9z\right){\mid }_{0}^{2}dydx$ (Integrate in relation to z.)
$=\frac{1}{8}{\int }_{0}^{2}{\int }_{0}^{2}\left(2{x}^{2}+18\right)dyxe$ (Compute the boundaries)
$=\frac{1}{8}{\int }_{0}^{2}\left(2{x}^{2}y+18y\right){\mid }_{0}^{2}dx$ (Intergrate in relation to y)
$=\frac{1}{8}{\int }_{0}^{2}\left(4{x}^{2}+36\right)dx$ (Compute the boundaries)
$=\frac{1}{8}\left(4\frac{{x}^{3}}{3}+36x\right){\mid }_{0}^{2}$ (Intergrate in relation to x)
$=\frac{1}{8}\left(\frac{32}{3}+72\right)$ (Compute the boundaries)
$=\frac{4}{3}+9=\frac{31}{3}$
Final answer $=\frac{31}{3}$