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I have been reading proof of divergence theorem , but there has been a question that I just cannot solve for my self , and I searched a lot , but not much was out there , mathematically what is wrong with open surfaces in here? (why nobody even discussed this ? ) , I have a good understanding about proof , but as long as I don't answer this question to my self I just wont comprehend it fully. enter image description here

For Example, we have a one sided open cylinder , and it is full of water , and we have a steady flow of water from closed side to open side it is obvious that ($h$ is flow of water) $$\int \nabla. h \, dv = 0$$ because there is no change in whole cylinder but $$\int h.da \neq 0$$

What is happening here?

according to DIVERGENCE THEOREM it should be : $$\int \nabla. h \, dv = \int h.da $$

thomas graceman
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    I'm not sure I understand the question, but divergence theorem needs a bounded region by some closed surface. If your cylinder is open, there is no region bounded by it, so I'm not sure what two integrals you'd like to be equal. – AnilCh Jul 22 '21 at 08:09
  • exactly i want to know why it needs bounded region or closed surface , but is not valid on open surface – thomas graceman Jul 22 '21 at 08:11
  • You want to say non orientable, not open surface – jimjim Jul 22 '21 at 08:17
  • no , my mistake its just open surface – thomas graceman Jul 22 '21 at 08:18
  • unless it is bounded or closed , it would have infinite volume. (Think of an open cylinder or an open cone). In real life it seems that a bucket with open top has finite volume. But it makes no sense mathemmatically. Even for the bucket the volume will only be defined as the amount of water it can hold, which again means that the top is bounded by the top layer of water. So if you want to define a "volume" integral , it must be in a bounded 3D space. Otherwise you would just have the infinite volume integral which may or may not converge. – Mr.Gandalf Sauron Jul 22 '21 at 08:22
  • well lets say the whole universe is full of water , and flow is steady(no change) , even if volume is infinite , still gradient of h in each piece of volume would be real zero and the whole integral would be equal to zero (the first integral). but the second integral is h.da we have limited surface and its not = zero – thomas graceman Jul 22 '21 at 08:35
  • The divergence theorem is not usually formulated for integrals which have finite value, in some sense. It is true that, under some conditions, you could theoretically work with "open surfaces" (mathematically speaking, we would be talking about manifolds with boundary or corners), but in that case we use quantities which are zero outside a finite region (formally, functions or differential forms with compact support). – Johnny El Curvas Jul 22 '21 at 09:11
  • so that's not different with having closed surface ? because we can treat that finite volume like it has closed surface – thomas graceman Jul 22 '21 at 09:20

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well , to begin with an open surface doesn't contain any volume , so comparing the to integrals is not correct . for divergence theorem to work we need volume and for volume we need closed surface so that's it.

thomas graceman
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