Stokes and Gauss connect accumulated behavior inside a region with behavior on its boundary.
WHAT HAPPENS IF…Why does changing dome height leave the Stokes integral unchanged?
Interactive exploration
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Connect the picture to the idea∬S(∇×F)⋅ndS=∮∂SF⋅dr,∬∂VF⋅ndS=∭V∇⋅FdV
Stokes uses a piecewise smooth oriented surface, compatible boundary orientation, and a continuously differentiable field nearby. For F=(−y,x,0), curl flux through each upward dome is 2πR². Gauss uses a closed outward boundary and regularity throughout the solid; F=(x,2y,3z) has divergence 6, giving sphere flux 8πR³.
A detail that matters
These theorems require orientation and field regularity; singularities cannot be ignored.