@JohnRennie Sir, why $$ \oint dU= 0$$ for any U ?

The integral of $dU$ is just the function $U$.

@JohnRennie Yes I agree

If you integrate round a loop you start and end at the same point $\mathbf x$, so your integral is $U(\mathbf x) - U(\mathbf x)$ and of course that's zero.

This assumes your function isn't multivalued.

$U$ is a scalar

@JohnRennie Is the result $$\oint dU =0$$ implies $U$ a scalar?

I think it would apply to any function.

@JohnRennie I think I need to study Vector Calculus once more.

:)

@JohnRennie: Hi sir. Good morning :-)

@GuruVishnu hi :-)

@JohnRennie After Aladdin's discussion, if possible, could you come to the other room, sir?

@GuruVishnu ok