Find the average value of f(x,y,z)=x^2+y^2+z^2 over the tetrahedron bounded by the planes x+y+z=1,x=0,y=0,and z=0.
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This is an exercise in tedium. The volume is
1 1-x 1-x-y
∫ . ∫ . ∫ dz dy dx = 1/6
0 0 0
The integral of f over the tetrahedron is
1 1-x 1-x-y
∫ . ∫ . ∫ (x²+y²+z²) dz dy dx = 1/20.
0 0 0
The average value is the integral of f divided by the volume: 6/20 = 3/10.
1 1-x 1-x-y
∫ . ∫ . ∫ dz dy dx = 1/6
0 0 0
The integral of f over the tetrahedron is
1 1-x 1-x-y
∫ . ∫ . ∫ (x²+y²+z²) dz dy dx = 1/20.
0 0 0
The average value is the integral of f divided by the volume: 6/20 = 3/10.