Calculus-on-Manifolds-Integration-over-Singular-n-chains-and-Stokes-Theorem, guaranteed and verified 100% Pass

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Integration over Singular 𝑛-chains and Stokes’ Theorem

If 𝜔 is a 𝑘-form on [0, 1]𝑘 , then 𝜔 = 𝑓(𝑥1 , … , 𝑥𝑘 )𝑑𝑥1 ∧ … ∧ 𝑑𝑥𝑘 for a
unique function, 𝑓: [0, 1]𝑘 → ℝ.


We have already defined the Riemann integral of 𝑓 over [0, 1]𝑘 , ∫[0,1]𝑘 𝑓. We
now define:


∫ 𝜔=∫ 𝑓𝑑𝑥1 ∧ … ∧ 𝑑𝑥𝑘 = ∫ 𝑓
[0,1]𝑘 [0,1]𝑘 [0,1]𝑘


Note on notation:


∫ 𝑓𝑑𝑥1 ∧ … ∧ 𝑑𝑥𝑘 = ∫ 𝑓𝑑𝑥1 … 𝑑𝑥𝑘
[0,1]𝑘 [0,1]𝑘



Ex. Let 𝜔 = (𝑥 + 𝑦)𝑑𝑥 ∧ 𝑑𝑦 on [0, 1] × [0, 1]
1 1
∫ 𝜔 = ∫ ∫ (𝑥 + 𝑦) 𝑑𝑥 𝑑𝑦
[0,1]2 0 0

We can evaluate this with Fubini’s Theorem.



Def. If 𝜔 is a 𝑘-form on 𝐴 ⊆ ℝ𝑛 and 𝑐 is a singular 𝑘-cube in 𝐴, then we
define:

∫ 𝝎=∫ 𝒄∗ (𝝎).
𝒄 [𝟎,𝟏]𝒌

, 2


Note: We have defined a singular 𝑘-cube as a map 𝑐: [0, 1]𝑘 → ℝ𝑛 . So if we have
a 𝑘-form, 𝜔, on 𝐴 ⊆ ℝ𝑛 , then we define:


∫ 𝜔=∫ 𝑐 ∗ (𝜔).
𝑐 [0,1]𝑘


However, we don’t need 𝑐 to map [0, 1]𝑘 into ℝ𝑛 . We can make the same
definition for integration for 𝑐 ′ : 𝐷 → ℝ𝑛 , where 𝐷 ⊆ ℝ𝑘 and 𝐷 is compact
(but 𝐷 is not necessarily [0, 1]𝑘 ).
We define:


∫ 𝝎 = ∫ (𝒄′ )∗ (𝝎)
𝒄′ 𝑫


Since 𝐷 ⊆ ℝ𝑘 and compact, we can still use Fubini’s Theorem to evaluate:


∫ (𝑐 ′ )∗ (𝜔).
𝐷



If 𝑐 is a 𝑘-chain, then we write:

𝑗

𝑐 = ∑ 𝑎𝑖 𝑐𝑖
𝑖=1



𝑗

∫ 𝜔 = ∑ 𝑎𝑖 ∫ 𝜔 .
𝑐 𝑖=1 𝑐𝑖