Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What book gives a rigorous but elementary exposition of "div, grad, curl, and all that"? Conventional second-year calculus books are as far from rigorous as anything ever gets.

share|cite|improve this question
I remember when I was 2nd year student I haven't felt that proofs of Green and Stokes formulas are rigor enough, so +1 – Ilya Oct 21 '11 at 15:47
@Gortaur: As Henri Cartan put it [...] the existing textbooks were not satisfactory, especially when it came to multiple integrals and Stokes' theorem. I discussed my concerns several times with André Weil. One beautiful day he told me, 'I've had it, we need to fix this for good. We need to write a good textbook on analysis. Then you'll stop complaining!' The rest is history. – t.b. Oct 21 '11 at 16:53
@t.b.: I do not complain, I rather recall that the only area in my Analysis course where I felt the lack of rigor was multidimensional integration, especially Fubini's theorem and Green and Stokes formulas. I had a guess what are the reasons - I only wonder how much pages in appendix will take the necessary theory. – Ilya Oct 23 '11 at 11:37
up vote 4 down vote accepted

Have you taken a look at Fleming's Functions of Several Variables? Relative to most of books that I've looked at, it provides the least amount of machinery necessary to prove (rigorously) the basic theorems of integral vector calculus. I find it to be very readable and easy to understand.

This question that I posed awhile back which, inexplicably, was closed as being off-topic, might also be relevant.

share|cite|improve this answer
The only local library that has that book has two copies and both are currently checked out. But I'll look at it soon, I think. Thank you. – Michael Hardy Oct 21 '11 at 22:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.