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\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
Classwork 12 (April 3, 2013) \\ 
Solution Jacobi Determinant
\end{center}
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\bigskip
\noindent 1. Calculate the Jacobi determinant for the
transformation from Cartesian to cylindrical coordinates.
\bigskip

Solution:
\begin{eqnarray} \nonumber
  x\ =\ \rho \cos\phi\,,~~~
  y\ =\ \rho \sin\phi\,,~~~
  z\ =\ z
\end{eqnarray}
yields the Jacobi determinant
\begin{eqnarray} \nonumber
  \left| \begin{matrix}
  \frac{\partial x}{\partial\rho} & \frac{\partial y}{\partial\rho} & 
  \frac{\partial z}{\partial\rho} \cr
  \frac{\partial x}{\partial\phi} & \frac{\partial y}{\partial\phi} & 
  \frac{\partial z}{\partial\phi} \cr
  \frac{\partial x}{\partial z  } & \frac{\partial y}{\partial z  } & 
  \frac{\partial z}{\partial z  } \end{matrix} \right|\ =\
  \left| \begin{matrix}
  \cos\phi       & \sin\phi      & 0 \cr
  -\rho \sin\phi & \rho \cos\phi & 0 \cr
  0              & 0             & 1 \end{matrix} \right|\ =\ \rho\,.
\end{eqnarray}

\bigskip
\noindent 2. Calculate the Jacobi determinant for the
transformation from Cartesian to spherical coordinates.

Solution:
\begin{eqnarray} \nonumber
  x &=& r\sin\theta \cos\phi\,,\\ \nonumber
  y &=& r\sin\theta \sin\phi\,,\\ \nonumber
  z &=& r\cos\theta
\end{eqnarray}
yields the Jacobi determinant
\begin{eqnarray} \nonumber
  &~& \left| \begin{matrix}
  \frac{\partial x}{\partial r  }  & \frac{\partial y}{\partial r  }   & 
  \frac{\partial z}{\partial r  }  \cr
  \frac{\partial x}{\partial\theta}& \frac{\partial y}{\partial\theta} & 
  \frac{\partial z}{\partial\theta}\cr
  \frac{\partial x}{\partial\phi}  & \frac{\partial y}{\partial\phi} & 
  \frac{\partial z}{\partial\phi}  \cr
\end{matrix} \right| \\ \nonumber &=&
   \left| \begin{matrix}
    \sin\theta\cos\phi &  \sin\theta\sin\phi & \cos\theta  \cr
   r\cos\theta\cos\phi & r\cos\theta\sin\phi & -r\sin\theta\cr
  -r\sin\theta\sin\phi & r\sin\theta\cos\phi & 0 \end{matrix} \right|
  \\ \nonumber &=& r^2\left(\sin^3\theta\sin^2\phi
                + \cos^2\theta\sin\theta\cos^2\phi \right.\\ \nonumber
  &~& \left.    + \sin^3\theta\cos^2\phi
                + \cos^2\theta\sin\theta\sin^2\phi \right) \\ \nonumber
  &=& r^2\left( \sin^3\theta+\cos^2\theta\sin\theta \right)\ =\
      r^2\sin\theta\,.
\end{eqnarray}


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