Well, if anyone starts learning electricity and magnetism(or in particular Vol.2 of Feynman lectures, as that is my site of learning), one always gets these 2 important theorems for sure -

\[\displaystyle \text{1. Gauss' Theorem -} \int_{S}{C.n dS} = \int_{V}{\nabla.C dV}\] such that C is any vector, S area, V volume and \(\nabla.C\) is the divergence of C.

\[\displaystyle \text{2. Stokes' Theorem -} \int_{line}{C ds} = \int_{S}{{(\nabla \times C)}_{n} dS}\] such that C is any vector, \(\int_{line}{}\) means line integral, \(\nabla \times C\) is curl of C.

Now, I want proofs of both these strong theorems. Please if anyone can help me in any way, care to do so. Thanks in anticipation!

## Comments

Sort by:

TopNewestThe given theorems are particular cases of a very strong theorem in differential geometry/Calculus on Manifolds, called generalized Stokes' theorem; although applied to electricity concepts. You can see any vector calculus book for particular cases proof and Michael Spicak's Calculus on Manifolds or Rudin's Principles of Mathematical Analysis for the proof of general case. As for proof corresponding to Physical case, you could refer Piyush A Kundu's Electricity textbook. – Vidyarthi S · 4 months, 3 weeks ago

Log in to reply

Do you still need help with it or have you found out a way to solve it yourself ? – Azhaghu Roopesh M · 2 years, 3 months ago

Log in to reply

– Kartik Sharma · 2 years, 3 months ago

Still need! That's why I have shared the set! Check out others also! Well, I have one proof with some assumptions but I want a 'proper mathematical proof'!Log in to reply

– Jake Lai · 2 years, 2 months ago

Good problems should be challenging but not tedious. Unrelated to this thread but just a reminder/word of advice.Log in to reply

– Kartik Sharma · 2 years, 2 months ago

Oh yeah! That's right. I was thinking of that while sharing them but then I thought there isn't any 'tedious' type of thing in mathematics. If calculations is what you are saying, then you have a calculator, I wouldn't mind.Log in to reply