Assume that Φ is harmonic in domain D and for x₀ ∈ D, B(x₀, r) ⊆ D, then the average value of Φ over the boundary of B(x₀, r) equals

Concept:

Mean value Property of Harmonic function,

Let, B(x, r) = Ball of radius r about x in Rⁿ

∂B(x, r) = boundary of the ball of radius r about x in Rⁿ  

For a function u defined on ∂B (x, r), the average of u on ∂B(x, r) is given as: \(\mathop \int \nolimits_{B\left( {x,\;r} \right)} u\left( y \right)dS\left( y \right)\)

Now, if u is the harmonic function, then, we get:

\(\mathop \int \nolimits_{\partial B\left( {x,\;r} \right)} u\left( y \right)dS\left( y \right) = u\left( x \right)\)  

Analysis:

ϕ is harmonic in domain D,

B(x₀, r) ⊆ D

The average value of ϕ over the boundary of B(x₀, r) is given as:

\(\mathop \int \nolimits_{\partial B\left( {{x_0},\;r} \right)} \phi \left( y \right)\;dS\left( y \right) = \phi \left( {{x_0}} \right)\) 

∴ Option 1 is the correct answer.

Conclusion: if ϕ is harmonic, then it's average (mean) value over a sphere centered at x₀ is independent of r.