Demostración de transformación de sistemas de coordenadas cartesianas a cilíndricas

De cartesianas a cilíndricas

Sean:

$x=\rho cos(\theta );y=\rho sen(\theta );z=z$

y el vector de posición:

$\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$

Sustituyendo nos queda:

$\overrightarrow{r}=\rho cos(\theta )\widehat{i}+\rho sen(\theta )\widehat{j}+z\widehat{k}$

Derivando

$\frac{\partial \overrightarrow{r}}{\partial \rho }=cos(\theta)\widehat{i}+sen(\theta)\widehat{j}$

$\frac{|\partial \overrightarrow{r}|}{|\partial \rho |}=1$

$\widehat{e_{1}}=\frac{1}{\frac{|\partial \overrightarrow{r}|}{|\partial \rho| }}\frac{\overrightarrow{dr}}{\partial \rho }$

De donde:

$\widehat{e_{1}}=\frac{\partial \overrightarrow{r}}{\partial \rho }=cos(\theta )\widehat{i}+sen(\theta )\widehat{j}$

$\frac{\partial \overrightarrow{r}}{\partial \theta }=-\rho sen(\theta )\widehat{i}+\rho cons(\theta )\widehat{j}$

$\frac{|\partial\overrightarrow{r} |}{|\partial \theta |}=\sqrt{\rho ^{2}sen^{2}(\theta )\widehat{j}+\rho cos^{2}(\theta)\widehat{j}}=\rho$

$\widehat{e_{2}}=\frac{-\rho sen(\theta )\widehat{j}+\rho cos(\theta )\widehat{j}}{\rho }=-sen(\theta )\widehat{j}+cos(\theta )\widehat{j}$

$\frac{\partial\overrightarrow{r} }{\partial z}=\widehat{k}........ \widehat{e_{3}}=\widehat{k}$

Entonces:

$\begin{matrix} & \widehat{i}& \widehat{j} & \widehat{k} \\ \widehat{a_{e}}& cos(\theta ) & sen(\theta ) & 0 \\ \widehat{a_{\theta }}& -sen(\theta ) & cos(\theta ) & 0 \\ \widehat{a_{k}}& 0& 0 & 1 \\\end{matrix}$

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Demostración de transformación de sistemas de coordenadas cartesianas a cilíndricas

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