3 votes

What goes wrong with Stokes theorem if a surface is not orientable?

Since this has gotten bumped: The problem (as noted in the comments) is not with Stokes' theorem, but defining a flux integral (or the integral of a differential $2$-form) on a non-orientable surface ...
Andrew D. Hwang's user avatar
2 votes

Hessian matrix determinant greater than zero in a saddle point?

The second derivative test (and the Hessian determinant) only works for bivariate functions. For functions of three or more variables, one needs to use the eigenvalues. In this case, as we have both ...
Julio Puerta's user avatar
  • 4,616

Only top scored, non community-wiki answers of a minimum length are eligible