The divergence of the curl of a twice continuously differentiable vector function is __________.
Options:
(A) 0
(B) 1
(C) 2
(D) ∞
Official Answer Key:
A
Step-by-Step Solution & Explanation
This is a standard vector-calculus identity:
∇ · (∇ × F⃗) = 0
It holds for any vector field F⃗ whose second partial derivatives are continuous, because the expansion produces pairs of mixed second partial derivatives that cancel exactly, e.g.
∂²F_z/∂x∂y − ∂²F_z/∂y∂x = 0
by Clairaut–Schwarz equality of mixed partials (which is exactly what “twice continuously differentiable” guarantees).
Physical meaning: a curl field is always solenoidal — it has no sources or sinks.
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