Appendix: All Solutions to the Problems
757
˚
V
(∇ × a · ∇ × b − b · ∇ × ∇ × a)dV =
S
(b × ∇ × a) · ndS
(1)
Transposing a and b of the above expression, to obtain
˚
V
(∇ × b · ∇ × a − a · ∇ × ∇ × b)dV =
S
(a × ∇ × b) · ndS
(2)
Subtracting Eqs. (1) from (2), and using the parallelepipedal product formula of
trivector (a × b) · c = (c × a) · b = (b × c) · a, to obtain
˚
V
(b · ∇ × ∇ × a − a · ∇ × ∇ × b)dV =
S
(a × ∇ × b − b × ∇ × a) · ndS
=
S
[(n × a) · ∇ × b − (n × b) · ∇ × a]dS
(3)
Quod erat demonstrandum.
Proof 2 Let c = a ×∇ × b− b×∇ × a, taking the divergence and using ∇ ·(a × b) =
b·∇×a−a·∇×b, to get ∇·c = ∇·(a×∇×b−b×∇×a) = b·∇×∇×a−a·∇×∇×b,
substituting it into the Gauss formula, to get
˚
V
∇ · (a × ∇ × b − b × ∇ × a)dV =
˚
V
(b · ∇ × ∇ × a − a · ∇ × ∇ × b)dV
=
S
(a × ∇ × b − b × ∇ × a) · ndS
=
S
[(n × a) · ∇ × b − (n × b) · ∇ × a]dS
Quod erat demonstrandum.
1.15 Prove
˝
V (∇ · b∇ · a + b · ∇∇ · a)dV =
S n · b∇ · adS.
Proof Let c = b∇ · a, taking the divergence, then there is ∇ · c = ∇ · (b∇ · a) =
∇ · b∇ · a + b · ∇∇ · a, substituting it into the Gauss formula, to get
˚
V
(∇ · b∇ · a + b · ∇∇ · a)dV =
S
n · b∇ · adS
Quod erat demonstrandum.
1.16 Prove
˝
V (a · ∇∇ · b − b · ∇∇ · a)dV =
S (a∇ · b − b∇ · a) · ndS.
Proof Let c = b∇ · a, taking the divergence, the there is ∇ · c = ∇ · (b∇ · a) =
∇ · b∇ · a + b · ∇∇ · a, substituting it into the Gauss formula, to get