If the divergence of F is the constant 3 everywhere, and the region W has volume 4, what is the total outward flux of F through the boundary surface S?
A3
B4
C12
DCannot be determined without knowing the surface
By the divergence theorem, flux = ∭_W (∇·F) dV = ∭_W 3 dV = 3 × (volume of W) = 3 × 4 = 12. Because ∇·F is constant, it factors out of the integral, leaving just the volume. Notably, the shape of the surface is irrelevant — only the enclosed volume matters.
Question 2 True / False
The divergence theorem applies to any open surface, not just closed surfaces.
TTrue
FFalse
Answer: False
The divergence theorem requires S to be a closed surface — one that completely encloses a three-dimensional region W with no boundary. An open surface (like a hemisphere without a base) has a boundary curve, and the appropriate theorem for that case is Stokes' theorem, not the divergence theorem. The outward normal on a closed surface points away from the enclosed region at every point.
Question 3 Short Answer
What does it mean physically when ∇·F = 0 everywhere inside a region W?
Think about your answer, then reveal below.
Model answer: The net outward flux through any closed surface inside the region is zero — there are no sources or sinks of the field inside W. Whatever flows in must flow out.
Divergence measures source density: positive divergence indicates a source (field spreading outward), negative divergence a sink (field converging inward). When ∇·F = 0 throughout W, the divergence theorem gives ∬_S F·dS = ∭_W 0 dV = 0. In fluid flow, this is incompressibility: mass is neither created nor destroyed in the region. In electrostatics, ∇·E = 0 in a charge-free region (from Gauss's law).