Question

Why is a Möbius strip not a possible surface to integrate flux on?

Answer 1

Ooh I like this question.

Answer 2

Wait I don't know the solution off the top of my head

Answer 3

lol

Answer 4

I hope u remember the answer :)

Answer 5

If I were smart I would. :( @TuringTest @FoolForMath @amistre64 This is like review for me. Going back through my text.

Answer 6

you can't integrate on it beacuse its an infinite surface, the surface has not limits so you cant actually integrate it

Answer 7

Qualitative answer. :P I'm taking a look at the strip equation to see what happens when I do. Proof by assume contra.

Answer 8

lol

Answer 9

Oh, because a definite integration results in a \(\infty\cdot0\) operation, is is undefined. So yeah, there you go.

Answer 10

ok thnx guys. So who thinks the deserve the best answer medal///??????

Answer 11

You lolol

Answer 12

@Eureka70 's answer was cleverer.

Answer 13

Mine was more rigorous, lol

Answer 14

Your choice. :3

Answer 15

i saw bad reference's should get it, it actually had math in it

Answer 16

lol I hate this ok Eureka is lower than bad so he gets the medal

Answer 17

thx but i'm only lower because i've only been a member to this site for like an hour

Answer 18

Because it is not orientable.

Answer 19

The flux of a field F is defined as a surface integral of Curl(F) dot the united normal vector n to the surface.
n is an outward unit normal vector to the surface.
The Mobius strip does not have an inner side or an outer side. It is not orientable and that is why we cannot define the flux through it.

Answer 20

Wait, so trying to perform the operations to non-orientable surfaces was incorrect, or the undefined answer was a sign that the surfaces were nonorientabke?

Answer 21

The flux cannot be defined on it.