I have following question:
Fields with equal divergence and equal curls $F_1$ and $F_2$ are two vectors fields, you may write them as $F_1 = M_1i+N_1j+P_1k$, $F_2 = M_2i+N_2j+P_2k$. Suppose that $\nabla \cdot F_1 = \nabla \cdot F_2$ and $\nabla \times F_1 = \nabla \times F_2$ over a region D enclosed by the oriented surface S with outward unit normal n and that $F_1 \cdot n = F_2 \cdot n$ on S. Prove that $F_1=F_2$ throughout D.
I came across this question when studying "Stokes's Theorem and Divergence Theorem" in Thomas Calculus, so I suppose we should use either of them to solve this, except I don't know how.
All I could figure out is $$F_1 \cdot n = F_2 \cdot n\Rightarrow\iint\limits_s F_1 \cdot n \, d\sigma = \iint\limits_s F_2 \cdot n \, d\sigma\Rightarrow\iiint\limits_D \nabla \times F_1\, dv = \iiint\limits_D \nabla \times F_2 \, dv,$$ the last step using divergence theorem. Or $$\nabla \times F_1 = \nabla \times F_2 \Rightarrow\iint\limits_s \nabla \times F_1\cdot n \, d\sigma= \iint\limits_s \nabla \times F_2 \cdot n \, d\sigma \Rightarrow \oint\limits_c F_1 \cdot dr =\oint\limits_c F_2 \cdot dr,$$ the last step using Stokes's theorem. This is all I get, then what? Am I on a wrong path? Can anyone solve this problem? Thanks.