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I have a cylinder occupying the region $x_{1}^{2}+x_{2}^{2} = R^2$ and $-G< x_3 < 0$

All I want to do is define the outer unit normal on the curved face. I thought about just calling it $e_1$ or $e_2$ but I suppose it has to be more general than that. Anyone have any ideas?

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  • $\begingroup$ what is $-G$?? and do you see the tangent vectors to a cylinder??One will be tangential to a line ruling and the other tangential to the cross sectional circle on the cylinder. Now your normal will just be the vector perpendicular to the plane spanned by the two tangent vectors. $\endgroup$
    – Vishesh
    Oct 24 '13 at 10:31
  • $\begingroup$ G is just a constant, though that part isn't relevant for this outer unit normal. What do you mean by line ruling? I'm still not sure how to write this out. $\endgroup$
    – user102712
    Oct 24 '13 at 10:36
  • $\begingroup$ Well the cylinder has two parameter curves one of which are vertical lines, you can visualise it. Just imagine lots of vertical lines at a constant distant from the z-axis and parallel to it, these are the rulings. $\endgroup$
    – Vishesh
    Oct 24 '13 at 11:57
  • $\begingroup$ Ok thanks, I'll have a think about it. $\endgroup$
    – user102712
    Oct 24 '13 at 11:59
  • $\begingroup$ If you wish so, I can elaborate. $\endgroup$
    – Vishesh
    Oct 24 '13 at 12:01
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A cylinder like the one you have given is basically obtained by revolving a straight line at a fixed distance from a coordinate axis around that axis itself(z-axis in this case). So at each point of the circle of revolution, you have a sraight line passing through. Just take a ruled piece of rectangular paper and roll it up so that the ruled lines will give you the rulings which are straight lines.

The cross sectional curves are circles. So the cylinder has a corresponding tangent plane at each point of intersection of the straight line rulings and the cross sectional circles.It is spanned by two tangent vectors. One tangent $(e_1)$ would be along the straight line itself, other would be $(e_2)$ tangential to the circle.

Now how would you look at the normal to a circle??That would be how the normal to a cylinder would look like too, pointing away from the circle's centre. This normal would be perpendicular to the plane that is spanned by the two tangent vectors $e_1,e_2$.

Explicitly just look at the gradient to the function giving the cylinder equation and compute it at each point. That will give you the normal vector. Of course you need to plug in the constraints.

P.S.: The outward direction is because that is the direction in which the value of the function increases.

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  • $\begingroup$ Just a bit more to add for completion, you want the normal to be something whose components give an inner product of zero with that of the tangent vectors. $\endgroup$
    – Vishesh
    Oct 24 '13 at 13:40

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