Question 1.61: Although the gradient, divergence, and curl theorems are the...

(a) Divergence theorem: \oint{} v · da=\int{}\left(∇\cdot v\right)d\tau . Let v = cT, where c is a constant vector. Using product rule #5 in front cover: ∇·v = ∇·(cT) = T(∇·c)+c · (∇T). But c is constant so ∇·c = 0. Therefore we have: \int{c} \cdot \left(∇T\right)d\tau =\int{}Tc  · da . Since c is constant, take it outside the integrals: c · \int{}∇Td\tau =c\cdot \int{}T da. But c is any constant vector—in particular, it could be be \hat{x},  or  \hat{y},  or  \hat{z} —so each component of the integral on left equals corresponding component on the right, and hence

\int{} ∇Td\tau =\int{}T da.     qed

(b) Let v → (v × c) in divergence theorem. Then \int{} ∇\cdot(v \times c)d\tau =\int{}(v \times c)da . Product rule #6 ⇒ ∇\cdot (v \times c) = c ·(∇\times v) − v · (∇\times c) = c · (∇\times v). (Note: ∇\times c = 0, , since c is constant.) Meanwhile vector indentity (1) says da · (v × c) = c · (da × v) = −c · (v × da). Thus \int{c} \cdot \left(∇\times v\right)d\tau =-\int{c}\cdot(v \times da). Take c outside, and again let c be \hat{x},\hat{y},\hat{z}   then:

\int{} \left(∇\times v\right)d\tau =-\int{}v\times da    qed

(c) Let v = T∇U in divergence theorem: ∫∇·(T∇U) d\tau =∫T∇U ·da. Product rule #(5) ⇒ ∇. (T∇U) = T∇·(∇U) + (∇U) · (∇T) = T∇^2U + (∇U) · (∇T). Therefore

∫(T∇^2U + (∇U) · (∇T))d\tau =∫(T∇U) · da. qed

(d) Rewrite (c) with T ↔ U :∫(U∇^2T + (∇T) · (∇U))d\tau =∫(U∇T) ·da. Subtract this from (c), noting that the (∇U) · (∇T) terms cancel:

∫(T∇^2U − U∇^2T) d\tau =∫(T∇U − U∇T) . da. qed

(e) Stokes’ theorem: ∫(∇×v) · da =∮ v · dl. Let v = cT. By Product Rule #(7): ∇×(cT) = T(∇×c) −c × (∇T) = −c × (∇T) (since c is constant). Therefore, -∫(c × (∇T)) · da = ∮ Tc · dl. Use vector indentity #1 to rewrite the first term (c×(∇T)) · da = c · (∇T ×da). So −∫c · (∇T ×da) =∮c ·T dl. Pull c outside, and let c→ \hat{x} ,\hat{y},   and  \hat{z} to prove:

∫∇T × da = −∮T dl.     qed