Although the gradient, divergence, and curl theorems are the fundamental integral theorems of vector calculus, it is possible to derive a number of corollaries from them. Show the following:
\begin{align*} \text{(a)}\; &{\;} & \int _{\mathcal{V}}(\nabla T)d\tau &= \oint _{\mathcal{S}}Td\mathbf{a} \\ \text{(b)}\; &{\;} & \int _{\mathcal{V}} (\nabla \times \mathbf{v})d\tau &= -\oint _{\mathcal{S}}\mathbf{v}\times d\mathbf{a} \\ \text{(c)}\; &{\;} & \int _{\mathcal{V}} \left[ T\nabla ^2 U+(\nabla T)\cdot (\nabla U) \right] d\tau &= \oint _{\mathcal{S}}(T\nabla U)\cdot d\mathbf{a} \\ \text{(d)}\; &{\;} & \int _{\mathcal{V}} \left( T\nabla ^2 U-U\nabla ^2 T \right) d\tau &= \oint _{\mathcal{S}}(T\nabla U-U\nabla T)\cdot d\mathbf{a} \\ \text{(e)}\; &{\;} & \int _{\mathcal{S}}\nabla T\times d\mathbf{a} &= -\oint _{\mathcal{P}}Td\mathbf{l} \\ \text{(f)}\; &{\;} & \int _{\mathcal{S}} \left[ (d\mathbf{a}\times \nabla )\times \mathbf{v} \right] &= -\oint _{\mathcal{P}}\mathbf{v}\times d\mathbf{l} \end{align*}Before starting working these out, let's recall the divergence theorem:
\[\int _{\mathcal{V}} \left( \nabla \cdot \mathbf{v} \right) d\tau =\oint _{\mathcal{S}} \mathbf{v}\cdot d\mathbf{a} \tag{Eq. 1.56, page 31}\]And Stokes' theorem:
\[\int _{\mathcal{S}}(\nabla \times \mathbf{v})\cdot d\mathbf{a}=\oint _{\mathcal{P}}\mathbf{v}\cdot d\mathbf{l}\tag{Eq. 1.57, page 33}\]It will also be relevant to recall that cross products are antisymmetric:
\[\mathbf{A}\times \mathbf{B}=-\mathbf{B}\times \mathbf{A} \]The cyclic property of the vector triple product:
\[\mathbf{A} \cdot (\mathbf{B}\times \mathbf{C}) = \mathbf{C} \cdot (\mathbf{A}\times \mathbf{B}) = \mathbf{B} \cdot (\mathbf{C}\times \mathbf{A}) \]And product rules:
\begin{align*} \nabla \cdot (f\mathbf{A}) &= f(\nabla \cdot \mathbf{A})+\mathbf{A}\cdot (\nabla f) \\ \nabla \cdot (\mathbf{A}\times \mathbf{B}) &= \mathbf{B}\cdot (\nabla \times \mathbf{A})-\mathbf{A}\cdot (\nabla \times \mathbf{B}) \\ \nabla \times (f\mathbf{A}) &= f(\nabla \times \mathbf{A})-\mathbf{A}\times (\nabla f) \end{align*}All of these formulas can be found on the cover of your textbook except the following one, which we shall call formula F1 for "Fernando's first formula" ;)
\[(\mathbf{a}\times \mathbf{b})\cdot (\mathbf{c}\times \mathbf{d})=(\mathbf{c}\cdot \mathbf{a})(\mathbf{b}\cdot \mathbf{d})-(\mathbf{b}\cdot \mathbf{c})(\mathbf{a}\cdot \mathbf{d})\tag{F1} \](a) Hint: Let $\mathbf{v}=\mathbf{c}T $, where $\mathbf{c} $ is a constant vector, in the divergence theorem; use the product rules
\begin{align*} \oint \mathbf{v}\cdot d\mathbf{a} &= \int (\nabla \cdot \mathbf{v})d\tau \tag{Div. theorem} \\ \oint \mathbf{c}T\cdot d\mathbf{a} &= \int (\nabla \cdot \mathbf{c}T)d\tau \tag{We let $\mathbf{v}=\mathbf{c}T $} \\ \oint \mathbf{c}T\cdot d\mathbf{a} &= \int \left( T\nabla \cdot \mathbf{c}+\mathbf{c}\cdot \nabla T \right) d\tau \tag{Product rule} \\ \oint \mathbf{c}T\cdot d\mathbf{a} &= \int \left( \cancel{T\nabla \cdot \mathbf{c}}+\mathbf{c}\cdot \nabla T \right) d\tau \tag{Because $\mathbf{c} $ is constant} \\ \oint \mathbf{c}T\cdot d\mathbf{a} &= \int \mathbf{c}\cdot \nabla T d\tau \\ \mathbf{c}\cdot \oint Td\mathbf{a} &= \mathbf{c}\cdot \int \nabla Td\tau \tag{Take constant vector out}\\ \oint Td\mathbf{a} &= \int \nabla Td\tau \end{align*}(b) Hint: Replace $\mathbf{v} $ by $(\mathbf{v}\times \mathbf{c}) $ in the divergence theorem
\begin{align*} \oint \mathbf{v}\cdot d\mathbf{a} &= \int (\nabla \cdot \mathbf{v})d\tau \tag{Div. theorem} \\ \oint (\mathbf{v}\times \mathbf{c})\cdot d\mathbf{a} &= \int (\nabla \cdot (\mathbf{v}\times \mathbf{c}))d\tau \tag{Replace $\mathbf{v} $ by $\mathbf{v}\times \mathbf{c} $} \\ \oint (\mathbf{v}\times \mathbf{c})\cdot d\mathbf{a} &= \int (\mathbf{c}\cdot (\nabla \times \mathbf{v})-\mathbf{v}\cdot (\nabla \times \mathbf{c}))d\tau \tag{Product rule} \\ \oint (\mathbf{v}\times \mathbf{c})\cdot d\mathbf{a} &= \int (\mathbf{c}\cdot (\nabla \times \mathbf{v})-\cancel{\mathbf{v}\cdot (\nabla \times \mathbf{c})})d\tau \tag{Because $\mathbf{c} $ is constant} \\ \oint (\mathbf{v}\times \mathbf{c})\cdot d\mathbf{a} &= \int (\mathbf{c}\cdot (\nabla \times \mathbf{v}))d\tau \\ \oint d\mathbf{a}\cdot (\mathbf{v}\times \mathbf{c}) &= \int (\mathbf{c}\cdot (\nabla \times \mathbf{v}))d\tau \\ \oint \mathbf{c}\cdot (d\mathbf{a}\times \mathbf{v}) &= \int (\mathbf{c}\cdot (\nabla \times \mathbf{v}))d\tau \tag{Cylic permutation of a triple product}\\ -\oint \mathbf{c}\cdot (\mathbf{v}\times d\mathbf{a}) &= \int (\mathbf{c}\cdot (\nabla \times \mathbf{v}))d\tau \tag{Antisymmetric property of the cross product}\\ -\mathbf{c}\cdot \oint (\mathbf{v}\times d\mathbf{a}) &= \mathbf{c}\cdot \int (\nabla \times \mathbf{v})d\tau \tag{Take constant vector out}\\ - \oint (\mathbf{v}\times d\mathbf{a}) &= \int (\nabla \times \mathbf{v})d\tau \end{align*}(c) Hint: Let $\mathbf{v}=T\nabla U $ in the divergence theorem
\begin{align*} \oint \mathbf{v}\cdot d\mathbf{a} &= \int (\nabla \cdot \mathbf{v})d\tau \tag{Div. theorem} \\ \oint (T\nabla U)\cdot d\mathbf{a} &= \int (\nabla \cdot (T\nabla U))d\tau \tag{Replace $\mathbf{v} $ by $T\nabla U $} \\ \oint (T\nabla U)\cdot d\mathbf{a} &= \int \left( T\nabla \cdot (\nabla U)+(\nabla U)\cdot (\nabla T) \right) d\tau \tag{Product rule}\\ \oint (T\nabla U)\cdot d\mathbf{a} &= \int \left( T\nabla ^2 U+(\nabla U)(\nabla T) \right) d\tau\tag{Def. of $\nabla ^2 $} \end{align*}(d) Comment: This is sometimes called Green's second identity; it follows from (c), which is known as Green's identity
Notice that it looks familiar to (c), almost as if we are subtracting mirrored terms. As such, we start by assuming (c) and using it to write
\[\int (U\nabla ^2 T+(\nabla T)\cdot (\nabla U))d\tau =\int (U\nabla T)\cdot d\mathbf{a} \]Which is (c) but the roles of $U $ and $T $ exchanged. Now subtract this from (c):
\begin{align*} \int \left( T\nabla ^2 U+(\nabla U)(\nabla T) \right) d\tau &= \oint (T\nabla U)\cdot d\mathbf{a} \tag{Result from (c)}\\ \int \left( T\nabla ^2 U+(\nabla U)(\nabla T) \right) d\tau - \int (U\nabla ^2 T+(\nabla T)\cdot (\nabla U))d\tau&= \oint (T\nabla U)\cdot d\mathbf{a} - \int (U\nabla T)\cdot d\mathbf{a} \\ \int (T\nabla ^2 U-U\nabla ^2 T+(\nabla U)\cdot (\nabla T)-(\nabla T)\cdot (\nabla U) )d\tau &= \oint (T\nabla U-U\nabla T)\cdot d\mathbf{a}\\ \int (T\nabla ^2 U-U\nabla ^2 T+\cancel{(\nabla U)\cdot (\nabla T)}-\cancel{(\nabla T)\cdot (\nabla U)} )d\tau &= \oint (T\nabla U-U\nabla T)\cdot d\mathbf{a} \\ \int (T\nabla ^2 U-U\nabla ^2 T)d\tau &= \oint (T\nabla U-U\nabla T)\cdot d\mathbf{a} \end{align*}(e) Hint: Let $\mathbf{v}=\mathbf{c}T $ in Stokes' theorem
\begin{align*} \int (\nabla \times \mathbf{v})\cdot d\mathbf{a} &= \oint \mathbf{v}\cdot d\mathbf{l} \tag{Stokes' theorem} \\ \int (\nabla \times (\mathbf{c}T))\cdot d\mathbf{a} &= \oint (\mathbf{c}T)\cdot d\mathbf{l} \tag{Let $\mathbf{v}=(\mathbf{c}T) $}\\ \int \left( T(\nabla \times \mathbf{c})-\mathbf{c}\times (\nabla T) \right) \cdot d\mathbf{a} &= \oint (\mathbf{c}T)\cdot d\mathbf{l}\tag{Product rule}\\ \int \left( \cancel{T(\nabla \times \mathbf{c})}-\mathbf{c}\times (\nabla T) \right) \cdot d\mathbf{a} &= \oint (\mathbf{c}T)\cdot d\mathbf{l} \tag{Since $\mathbf{c} $ is constant}\\ \int \left( -\mathbf{c}\times (\nabla T) \right) \cdot d\mathbf{a} &= \oint (\mathbf{c}T)\cdot d\mathbf{l} \\ \int d\mathbf{a} \cdot \left( -\mathbf{c}\times (\nabla T) \right) &= \oint (\mathbf{c}T)\cdot d\mathbf{l} \\ \int \mathbf{c} \cdot \left( -(\nabla T)\times d\mathbf{a} \right) &= \oint (\mathbf{c}T)\cdot d\mathbf{l} \tag{Cyclic permutation}\\ -\mathbf{c} \cdot\int \left( (\nabla T)\times d\mathbf{a} \right) &= \mathbf{c}\cdot \oint (\mathbf{c}T)\cdot d\mathbf{l}\tag{Take out constant}\\ \int \nabla T\times d\mathbf{a} &= -\oint Td\mathbf{l} \end{align*}(f) Hint: Replace $\mathbf{v} $ by $(\mathbf{v}\times \mathbf{c}) $ in Stokes' theorem
Before showing the solution to the last part of this problem, it is relevant to review a key idea when it comes to dot products.
It is true that if we have two vectors $\mathbf{a} $ and $\mathbf{b} $, their dot product is symmetric, that is:
\[\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a} \]In this class we have to be careful with some dot products such as:
\[\mathbf{a}\cdot \nabla \]
Where $\nabla $ has its usual meaning, but notice the big difference when we apply $\mathbf{a}\cdot \nabla $ and $\nabla \cdot \mathbf{a} $ to a vector field $\mathbf{v}=(v_x,v_y,v_z) $: \begin{align*} (\mathbf{a}\cdot \nabla )\mathbf{v} &= \left( a_x\frac{\partial }{\partial x}+a_y\frac{\partial }{\partial a_y}+a_z\frac{\partial }{\partial z} \right) \mathbf{v} \\ &= \bigg(a_x\frac{\partial v_x}{\partial x}+a_y\frac{\partial v_x}{\partial a_y}+a_z\frac{\partial v_x}{\partial z} , a_x\frac{\partial v_y}{\partial x}+a_y\frac{\partial v_y}{\partial a_y}+a_z\frac{\partial v_y}{\partial z} , a_x\frac{\partial v_z}{\partial x}+a_y\frac{\partial v_z}{\partial a_y}+a_z\frac{\partial v_z}{\partial z} \bigg) \\ &{\;} \text{compared to...} \\ (\nabla \cdot \mathbf{a})\mathbf{v} &= \left( \frac{\partial a_x}{\partial x}+\frac{\partial a_y}{\partial y}+\frac{\partial a_z}{\partial z} \right) \mathbf{v} \\ &= \bigg(\left( \frac{\partial a_x}{\partial x}+\frac{\partial a_y}{\partial y}+\frac{\partial a_z}{\partial z} \right)v_x,\left( \frac{\partial a_x}{\partial x}+\frac{\partial a_y}{\partial y}+\frac{\partial a_z}{\partial z} \right)v_y,\left( \frac{\partial a_x}{\partial x}+\frac{\partial a_y}{\partial y}+\frac{\partial a_z}{\partial z} \right)v_z\bigg) \end{align*}
Quite different, right? :) With that in mind, let's prove the identity.
\begin{align*} \int (\nabla \times \mathbf{v})\cdot d\mathbf{a} &= \oint \mathbf{v}\cdot d\mathbf{l} \tag{Stokes' theorem} \\ \int (\nabla \times (\mathbf{v} \times \mathbf{c}))\cdot d\mathbf{a} &= \oint (\mathbf{v} \times \mathbf{c}) \cdot d\mathbf{l}\tag{Replace $\mathbf{v} $ by $(\mathbf{v} \times \mathbf{c}) $} \\ \int (\nabla \times (\mathbf{v} \times \mathbf{c}))\cdot d\mathbf{a} &= \oint d\mathbf{l}\cdot (\mathbf{v} \times \mathbf{c}) \tag{Order of dot product} \\ \int (\nabla \times (\mathbf{v} \times \mathbf{c}))\cdot d\mathbf{a} &= \oint \mathbf{c}\cdot (d\mathbf{l} \times \mathbf{v}) \tag{Cyclic permutation}\\ \int (\nabla \times (\mathbf{v} \times \mathbf{c}))\cdot d\mathbf{a} &= \mathbf{c}\cdot\oint (d\mathbf{l} \times \mathbf{v}) \tag{Since $\mathbf{c} $ is a constant}\\ \int (\nabla \times (\mathbf{v} \times \mathbf{c}))\cdot d\mathbf{a} &= -\mathbf{c}\cdot\oint (\mathbf{v} \times d\mathbf{l}) \tag{Antisymmetric property of cross product}\\ \int d\mathbf{a}\cdot (\nabla \times (\mathbf{v} \times \mathbf{c})) &= -\mathbf{c}\cdot\oint (\mathbf{v} \times d\mathbf{l}) \tag{Order of dot product}\\ \int (\mathbf{v} \times \mathbf{c})\cdot (d\mathbf{a} \times \nabla) &= -\mathbf{c}\cdot\oint (\mathbf{v} \times d\mathbf{l}) \tag{Cyclic permutation}\\ \int (d\mathbf{a}\cdot \mathbf{v})(\mathbf{c}\cdot \nabla )-(\mathbf{c}\cdot d\mathbf{a})(\mathbf{v}\cdot \nabla ) &= -\mathbf{c}\cdot\oint (\mathbf{v} \times d\mathbf{l})\tag{Product Rule F1} \\ \int (\mathbf{v}\cdot d\mathbf{a})(\mathbf{c}\cdot \nabla )-(d\mathbf{a}\cdot \mathbf{c})(\mathbf{v}\cdot \nabla ) &= -\mathbf{c}\cdot\oint (\mathbf{v} \times d\mathbf{l}) \tag{We can flip those 2 dot products} \\ \int \mathbf{c}\cdot \left[ (d\mathbf{a}\times \nabla )\times \mathbf{v} \right] &= -\mathbf{c}\cdot\oint (\mathbf{v} \times d\mathbf{l}) \tag{Product Rule F1 in reverse}\\ \mathbf{c}\cdot\int \left[ (d\mathbf{a}\times \nabla )\times \mathbf{v} \right] &= -\mathbf{c}\cdot\oint (\mathbf{v} \times d\mathbf{l}) \tag{Since $\mathbf{c} $ is a constant} \\ \int \left[ (d\mathbf{a}\times \nabla )\times \mathbf{v} \right] &= -\oint (\mathbf{v} \times d\mathbf{l}) \end{align*}