I want to integrate by part the following integral in cylindrical coordinates
$$\int \vec{r} \times (\nabla \cdot \overline{T}) ~d^3\vec{r} $$
where $\overline{T}$ is a second order symmetric tensor and $\times$ is the vectorial product.
Once I integrated by part I want to use the divergence theorem to obtain a surface integral. It works very well in cartesian coordinates but I cannot manage to do it properly in cylindrical.
In cylindrical coordinates $\hat{\boldsymbol{\rho}},\hat{\boldsymbol{\theta}},\hat{\mathbf{z}}$,
\begin{align}&\nabla \cdot \overline{T}= \left[ \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho T_{\rho\rho}) + \frac{1}{\rho} \frac{\partial T_{\theta \rho}}{\partial \theta} + \frac{\partial T_{z \rho }}{\partial z} – \frac{T_{\theta \theta}}{\rho} \right] \hat{\boldsymbol{\rho}} + \\
&\qquad\qquad\left[ \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho T_{\rho\theta}) + \frac{1}{\rho} \frac{\partial T_{\theta\theta}}{\partial \theta} + \frac{\partial T_{z\theta}}{\partial z} + \frac{T_{\theta \rho }}{\rho} \right] \hat{\boldsymbol{\theta}} + \\
&\qquad\qquad\left[ \frac{1}{\rho} \frac{\partial}{\partial \rho}(\rho T_{\rho z}) + \frac{1}{\rho} \frac{\partial T_{\theta z}}{\partial \theta} + \frac{\partial T_{z z}}{\partial z} \right] \hat{\mathbf{z}}
\end{align}
I tried a lot of different ways (5 days I am trying to do it) of doing the integration by part, but by checking numerically I know that my results is wrong. The value I obtain numerically seems to be more like ($\overline{T}$ is symmetric)
\begin{align}
\int \vec{r} \times (\nabla \cdot \overline{T}) ~d^3\vec{r}&= \oint \vec{r} \times \left( \hat{\boldsymbol{\rho}} \cdot \overline{T} \right) ~dS + \\
&\left( \int T_{\rho z} ~d^3\vec{r} – \int_0^{\rho_\text{max}} \left[ T_{\theta z}(2\pi) – T_{\theta z}(0) \right] ~\rho d\rho \right) \hat{\boldsymbol{\theta}} + \\
&\int_0^{\rho_\text{max}} \left[ T_{\theta\theta}(2 \pi) – T_{\theta\theta}(0) \right] \rho d\rho \hat{\mathbf{z}}
\end{align}
First, I am not sure of the result and second, I would like to know how to derive this analytically. Someone can help me? thank you very much
Best Answer
I should stress that the expression $\mathbf{r}\times\nabla\cdot\mathbf{T}$ is independent of the coordinate system, while the symmetry of $\mathbf{T}$ is easily expressed with respect to Cartesian coordinates, then \begin{align} \mathbf{r}\times\nabla\cdot\mathbf{T} &=\mathbf{e}_i\varepsilon_{ijk}x_j\partial_lT_{kl}\\ &=\mathbf{e}_i\varepsilon_{ijk}\partial_l(x_jT_{kl})-\mathbf{e}_i\varepsilon_{ijk}(\partial_lx_j)T_{kl}\\ &=\mathbf{e}_i\varepsilon_{ijk}\partial_l(x_jT_{kl})-\mathbf{e}_i\varepsilon_{ijk}\delta_{jl}T_{kl}\\ &=\mathbf{e}_i\partial_l(\varepsilon_{ijk}x_jT_{kl})-\mathbf{e}_i\varepsilon_{ijk}T_{kj}\\ &=\mathbf{e}_i\partial_l(\varepsilon_{ijk}x_jT_{kl})\\ &=\nabla\cdot(\mathbf{r}\times\mathbf{T}). \end{align} The result obtained is independent from the coordinate system used, so $$ \int_B\mathbf{r}\times\nabla\cdot\mathbf{T}\,dV=\int_B\nabla\cdot(\mathbf{r}\times\mathbf{T})\,dV $$ then applying the divergence theorem, this becomes $$ \int_{\partial B}\mathbf{n}\cdot(\mathbf{r}\times\mathbf{T})\,dS $$ now you can evaluate this flux in whatever coordinate system you prefer.