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Calculus II: Vector Calculus and Multivariable Analysis / Vector Calculus and the Integral Theorems

Derivation — Proof of Stokes's Theorem for Graphs

The proof of Stokes's theorem for a surface that is the graph of a function z=g(x,y)z=g(x,y) proceeds in two stages: write out both sides in terms of coordinates, then show they are equal by invoking Green's theorem (the planar special case). This is the cleanest proof of Stokes's theorem and the one closest to the computation you actually do in applications. It also makes visible why Green's theorem appears as a special case: when g=0g=0 (flat surface), the proof collapses exactly to the Green's theorem argument.
Learning objectives. By the end of this note you should be able to:
  • carry out the proof of Stokes's theorem when SS is the graph z=g(x,y)z=g(x,y);
  • identify the points where the multivariable chain rule and Green's theorem enter the proof;
  • explain why piecewise-smooth surfaces follow by decomposition;
  • state the orientability condition required by the proof.

In this lesson

    1. Setup
    1. The Proof
    1. Key Steps Summarised
    1. Extension to Piecewise-Smooth Surfaces

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