Stewart - Calculus
The last line doesn't have 'continuous' in it, but 'continuous' is mentioned earlier. What's happening? Is the hypothesis necessary?
2026-03-27 05:39:28.1774589968
Are line integrals of conservative vector fields be path independent even if the vector field is not continuous?
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Yes, it is possible to accomplish the task with discontinuous $\nabla f$. It just means $f$, which does have to be continuous, has a corner in it. This is used all the time in physics when dealing with surface charge densities, for example in the classic spherical conductor problems. The needed condition is that $\nabla \times \mathbf{V} = 0$, where $\mathbf{V}$ is the vector field that we want to write as $\nabla f$, everywhere in the region of interest that we are allowed to deform the path through. This is true by the Helmholtz decomposition theorem. We also use this fact to write the magnetic field, normally thought of as a purely solenoidal field, in terms of the magnetic scalar potential in regions where the current is zero.