# What is $\int_{\Omega'} \psi (\nabla p) dV \: \text{as} \: \delta\alpha \rightarrow 0$?

I have an axi-symmetric integral (the domain and all functions are axi-symmetric) in cylindrical coordinates which needs to be integrated by parts for use in a finite element code. The integral is taken over a slice of the domain with polar angle $\theta$ varying as $\alpha\leq\theta\leq\alpha+\delta\alpha$, this bulk being denoted $\Omega'$ and boundary $\Gamma'$. The projection onto the radial-axial (r-z) plane has bulk $\Omega$ and boundary (excluding $z=0$ boundary) $\Gamma'$. Also I denote the normal to the boundary as $\hat{n}$. The integral is $$I=\int_{\Omega'} \psi (\nabla p) dV \: \text{as} \: \delta\alpha \rightarrow 0$$

The problem is that this can be done in two separate ways either 1) use the gradient theorem and then the specifics of the domain or 2) the specifics of the domain then the gradient theorem.

1) \begin{align} I=&\int_{\Omega'} (\nabla \psi p) dV -\int_{\Omega'}(\nabla \psi)p dV \\ =&\int_{\Gamma'} \psi p \hat{n} dS -\int_{\Omega'}(\nabla \psi)p dV \\ \sim&\delta\alpha\int_{\Gamma} \psi p \hat{n} r dl -\delta\alpha\int_{\Omega}(\nabla \psi)p r dr dz \\ \end{align}

2) \begin{align} I\sim&\delta\alpha\int_{\Omega} \psi (\nabla p) rdrdz \\ =&\delta\alpha\int_{\Omega} (\nabla\psi p r)drdz -\delta\alpha\int_{\Omega}(\nabla \psi r)p dr dz \\ =&\delta\alpha\int_{\Gamma} \psi p \hat{n} r dl -\delta\alpha\int_{\Omega}(\nabla \psi r)p dr dz \\ \end{align}

You will notice that the bulk integrals (over $\Omega$) that the two different approaches yield differ, one of them has the factor of r within the gradient and the other outside. I believe that the second method gives the correct result, but I cannot see where I may have gone wrong in either. Any advice is appreciated.

Thanks

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