Multivariable Advanced Calculus

14.4. STOKE’S THEOREM 389By Theorem 6.4.2 again it follows that for all x ∈ α ∗ and |v| < δ,|G (v, x)| = |R (x + v) − R (x) − DR (x) (v)| ≤ ε |v| (14.23)Letting ||P|| be small enough, it follows from the continuity of α that|α (t j+1 ) − α (t j )| < δTherefore for such P,n−1∑F (γ (t j )) · (γ (t j+1 ) − γ (t j ))j=0wheren−1∑= F (R (α (t j ))) · (R (α (t j+1 )) − R (α (t j )))j=0n−1∑= F (R (α (t j ))) · [DR (α (t j )) (α (t j+1 ) − α (t j )) + o (α (t j+1 ) − α (t j ))]j=0o (α (t j+1 ) − α (t j )) = R (α (t j+1 )) − R (α (t j )) − DR (α (t j )) (α (t j+1 ) − α (t j ))and by 14.23,|o (α (t j+1 ) − α (t j ))| < ε |α (t j+1 ) − α (t j )|It followsn−1∑n−1∑F (γ (t j )) · (γ (t j+1 ) − γ (t j )) − F (R (α (t j ))) · DR (α (t j )) (α (t j+1 ) − α (t j ))∣j=0j=0∣(14.24)n−1∑n−1∑≤ |o (α (t j+1 ) − α (t j ))| ≤ ε |α (t j+1 ) − α (t j )| ≤ εV (α, [a, b])j=0j=0Consider the second sum in 14.24. A term in the sum equalsF (R (α (t j ))) · (R u (α (t j )) (α 1 (t j+1 ) − α 1 (t j )) + R v (α (t j )) (α 2 (t j+1 ) − α 2 (t j )))= (F (R (α (t j ))) · R u (α (t j )) , F (R (α (t j ))) · R v (α (t j ))) · (α (t j+1 ) − α (t j ))By continuity of F, R u and R v , it follows that sum converges as ||P|| → 0 to∫((F ◦ R) · R u , (F ◦ R) · R v ) · dααTherefore, taking the limit as ||P|| → 0 in 14.24∫ ∫∣ F · dγ − ((F ◦ R) · R u , (F ◦ R) · R v ) · dα∣ < εV (α, [a, b]) .γαSince ε > 0 is arbitrary, This proves the lemma. The following is a little identity which will allow a proof of Stoke’s theorem to followfrom Green’s theorem. First recall the following definition from calculus of the curl ofa vector field and the cross product of two vectors from calculus.