Proof of Electromagnetic Identity: Puzzling Last Expression

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The discussion centers on understanding a specific expression in electromagnetics related to the line integral of dV. The user expresses confusion over why this integral equals zero and seeks clarification on its meaning, particularly when evaluated along a curve with defined endpoints. The conversation reveals that the integral can be expressed in terms of a parameterized curve, leading to further inquiries about the completeness of the explanation provided. Ultimately, the response indicates that the lack of completion was intentional to provoke deeper thought. The exchange highlights the complexities of interpreting electromagnetic identities and the nuances of mathematical expressions.
larginal
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Homework Statement
two identities in electromagentics
Relevant Equations
curl of gradientV = 0
question.jpg

I tried to understand proof of this identity from electromagnetics. but I was puzzled at the last expression.
why is that line integral of dV = 0 ?
In fact, I'm wondering if this expression makes sense.
 
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If you had a curve integral along a curve ##\Gamma## with endpoints ##p## and ##q##, what would the integral
$$
\int_\Gamma dV
$$
be? Note that
$$
\int_\Gamma dV = \int_0^1 \frac{dV}{dt} dt
$$
if we assume that ##t## is a curve parameter in the interval [0,1] parametrising the curve such that ##p = \Gamma(0)## and ##q = \Gamma(1)##.
 
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I got it. Thanks! is it your intention that you didn't complete the explanation to make me think?
 
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larginal said:
is it your intention that you didn't complete the explanation to make me think?
Yes.
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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