Line e Surface Infinitesimal

  • Thread starter Jhenrique
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I think you know definition of line infinitesimal:
[tex][ds]^2 = \begin{bmatrix} dx & dy & dz \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}^2 \begin{bmatrix} dx\\ dy\\ dz\\ \end{bmatrix} = \begin{bmatrix} dr & d\theta & dz \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & r & 0\\ 0 & 0 & 1\\ \end{bmatrix}^2 \begin{bmatrix} dr\\ d\theta\\ dz\\ \end{bmatrix} = \begin{bmatrix} d\rho & d\phi & d\theta \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & \rho & 0\\ 0 & 0 & \rho\;sin(\phi)\\ \end{bmatrix}^2 \begin{bmatrix} d\rho\\ d\phi\\ d\theta\\ \end{bmatrix}[/tex]

From this, is correct if I deduce the formula to surface infinitesimal like this?
[tex][d^2S]^2 = \begin{bmatrix} dydz & dxdz & dxdy \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}^2 \begin{bmatrix} dydz\\ dxdz\\ dxdy\\ \end{bmatrix} = \begin{bmatrix} d\theta dz & drdz & drd\theta \end{bmatrix} \begin{bmatrix} r & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & r\\ \end{bmatrix}^2 \begin{bmatrix} d\theta dz\\ drdz\\ drd\theta\\ \end{bmatrix} = \begin{bmatrix} d\phi d\theta & d\rho d\theta & d\rho d\phi \end{bmatrix} \begin{bmatrix} \rho^2\;sin(\phi) & 0 & 0\\ 0 & \rho\;sin(\phi) & 0\\ 0 & 0 & \rho\\ \end{bmatrix}^2 \begin{bmatrix} d\phi d\theta\\ d\rho d\theta\\ d\rho d\phi\\ \end{bmatrix}[/tex]

And more one second question: dxdy is equal d²xy ?
 

ChrisVer

Gold Member
3,328
437
you can always try the "old way" of doing things...
Find the infinitesimal tangent vectors on your surface, and take the cross product
 

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