> Instead of 2*pi*r dr dz in the boundary integral, did you mean 2*pi*r*dz
> ?
> Otherwise, you will get the differential volume.
This is a really old thread :-) Yes, you are correct. The *domain* integral
has the form
\int_\Omega f(r,z) 2*pi r dr dz
and the boundary integral will have the form
\int_{\partial\Omega} f(R,z) 2*pi R dz
Best
W.
> среда, 19 октября 2016 г., 17:43:59 UTC-4 пользователь Wolfgang Bangerth
> написал:
>
> > In fact, I should solve my problem on a sphere (circle). According to
> your
> > noteworthy comments, the only thing that I have to do is change the
> geometry
> > and use higher order mapping for accuracy. Another thing is that
> consider a
> > half of a circle in r-z coordinate system and the problem is
> axisymmetric.
> > what should be my boundary condition for rotation axis (z-coordinate)?
>
> That depends on the equation, but if you were, for example, solving the
> Laplace equation in cylindrical coordinates, then there would be no
> special
> boundary condition along the z-axis (at r=0). That's because you will find
> that when you do the integration by parts, you get boundary terms of the
> form
>
> \int_{\partial \Omega} n.nabla u(r,z) varphi(r,z) 2*pi*r dr dz
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