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\begin{document}
\title{Solutions, 311-XXIV}
\maketitle

4( 4/17) Problems\\
          Schaum's, p.118(18),p.129(32),p.134(67*,68*,69*)\\
\section{Problem 134-67}
A vector ${\bf B}$ is always normal to a given closed surface $S$. Show that
\[ \int \int \int_V \nabla \times {\bf B} dV = 0 \]
where $V$ is the region bound by $S$.\\
\\ 
{\bf Solution:}\\
In the divergence theorem, 
\begin{equation}
\label{divth}
 \int \int \int_V \nabla\cdot {\bf F} dV = \int \int_{\partial V}
{\bf B}\cdot{\bf n} dS \ ,
\end{equation}
let ${\bf F} = {\bf B}\times {\bf C}$ with ${\bf C}$ an arbitrary constant
vector. Then $\nabla\cdot{\bf F} = \nabla\cdot\left({\bf B}\times {\bf C}
\right)={\bf C}\cdot\nabla\times{\bf B}$ and ${\bf F}\cdot{\bf n}=
\left({\bf B}\times {\bf C}\right) \cdot{\bf n} = - {\bf C}\times{\bf B}
\cdot{\bf n}=- {\bf C}\cdot\left({\bf B}\times{\bf n}\right)$.
Substituting in eq.(\ref{divth}) gives
\[ {\bf C}\cdot \int \int \int_V \nabla\times {\bf B} dV =
-{\bf C}\cdot \int \int_{\partial V} {\bf B}\times{\bf n} dS \ .\]
Since ${\bf C}$ is arbitrary it follows that
\[ {\bf C}\cdot \left\{
\int \int \int_V \nabla \times {\bf B} dV+\int \int_{\partial V} 
{\bf B}\times{\bf n} dS \right\}  = 0 \]
and the expression in brackets is orthogonal to any vector ${\bf C}$,
it vanishes identically and we have:
\[ \int \int \int_V \nabla \times {\bf B} dV = 
{\bf n} \times {\bf B} dS \ .\]
But if 
\[ {\bf B} \bot S \mbox{everywhere} \Rightarrow
   {\bf B} \| {\bf n} \Rightarrow {\bf n} \times {\bf B} \equiv 0 \ .\]

\section{Problem 134-68}
If 
\[ \oint_C {\bf E} \cdot d{\bf R} = -\frac{1}{c} \partial_t
\int \int_S {\bf H}\cdot 
d{\bf S} \ ,\]
where $S$ is any surface bounded by the closed curve $C$, show that
\[ \nabla \times {\bf E} = -\frac{1}{c}\partial_t {\bf H} \ .\]
{\bf Solution:}\\
By Stokes theorem we have
\[ 
\oint_C {\bf E} \cdot d {\bf R} 
   = 
\int \int_S \left(\nabla \times {\bf E} \right) \cdot {\bf n} dS \ ,\]
so that
\begin{eqnarray*}
\int \int_S \left(\nabla \times {\bf E} \right) \cdot {\bf n} dS
& = & -\frac{1}{c} \partial_t \int \int_S {\bf H}\cdot d{\bf S}\\
& = & \int \int_S\left(-\frac{1}{c} \partial_t{\bf H}\right) \cdot d{\bf S}
\Rightarrow \\
\int \int_S\left\{
\nabla \times {\bf E} + \frac{1}{c} \partial_t{\bf H}\right\}\cdot d{\bf S}
& = & 0  \ .
\end{eqnarray*}
Since this is true for any $C$, the quantity in the brackets vanishes.

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\section{Problem 134-69}
Prove
\[ \oint_C \phi d{\bf R} = \int \int_S d{\bf S} \times \nabla \phi \ . \]
{\bf Solution:}\\
Consider ${\bf C} = \mbox{constant}$, arbitrary. Then
\begin{eqnarray*}
{\bf C} \cdot \oint_C \phi d{\bf R}
= \oint_C \left( \phi {\bf C} \right) \cdot d{\bf R} 
& = &  \int \int_S \nabla \times \left( \phi {\bf C} \right) \cdot d{\bf S} 
 =  \int \int_S\left(\nabla \phi\times {\bf C} \right) \cdot {\bf n} dS =\\
{\bf C} \cdot \int \int_S {\bf n} \times \nabla \phi dS
 & \Rightarrow&  
{\bf C} \cdot \left\{ \oint_C \phi d{\bf R} -
\int \int_S d{\bf S} \times \nabla \phi \right\} = 0 \ .
\end{eqnarray*}
Since this is true for any ${\bf C}$, the desired result follows.
\end{document}

