Vector Analysis Stokes Theorem, guaranteed and verified 100% Pass

Stokes’ Theorem

Recall that the vector form of Green’s theorem says:

Vector Form of Green’s Theorem: Let 𝐷 ⊂ ℝ2 be a region to which Green’s
theorem applies. Let 𝜕𝐷 be its positively oriented (ie counterclockwise)
boundary, and let 𝐹⃗ = 𝑃(𝑥, 𝑦)𝑖⃗ + 𝑄(𝑥, 𝑦)𝑗⃗ be a 𝐶 1 vector field on D then

∫𝜕𝐷 𝐹⃗ ∙ 𝑑𝑠⃗ = ∬𝐷 (𝑐𝑢𝑟𝑙 𝐹⃗ ) ∙ 𝑘⃗⃗𝑑𝐴 = ∬𝐷 (∇ × 𝐹⃗ ) ∙ 𝑘⃗⃗𝑑𝐴.


Notice that since Green’s theorem applies to regions in the 𝑥𝑦-plane (ie a surface
that lies in the 𝑥𝑦-plane), the 𝑘⃗⃗ that appears on the right hand side of the formula
is actually the unit normal, 𝑛⃗⃗, to the surface 𝐷. Also, since we are dealing with a
surface in the 𝑥𝑦-plane, the “𝑑𝐴” is the same as “𝑑𝑆”. Thus we could write the
formula as:

∫𝜕𝑆 𝐹⃗ ∙ 𝑑𝑠⃗ = ∬𝑆 (𝑐𝑢𝑟𝑙 𝐹⃗ ) ∙ 𝑛⃗⃗𝑑𝑆 = ∬𝑆 (∇ × 𝐹⃗ ) ∙ 𝑛⃗⃗𝑑𝑆.


Now if the surface is parametrized we can write:
⃗⃗ ×𝑇
𝑇 ⃗⃗
𝑛⃗⃗ = |𝑇⃗⃗𝑢×𝑇⃗⃗𝑣| , ⃗⃗𝑢 × 𝑇
𝑑𝑆=|𝑇 ⃗⃗𝑣 |𝑑𝑢𝑑𝑣, 𝑑𝑆⃗ = (𝑇
⃗⃗𝑢 × 𝑇
⃗⃗𝑣 )𝑑𝑢𝑑𝑣
𝑢 𝑣


⃗⃗𝑑𝑆 = 𝑑𝑆⃗.
So 𝑛

So we could write Green’s theorem as:

∫𝜕𝑆 𝐹⃗ ∙ 𝑑𝑠⃗ = ∬𝑆 (𝑐𝑢𝑟𝑙 𝐹⃗ ) ∙ 𝑑𝑆⃗ = ∬𝑆 (∇ × 𝐹⃗ ) ∙ 𝑑𝑆.
⃗⃗⃗



This is exactly what the conclusion to Stokes’ theorem is. The difference is that
the surface, 𝑆, in Stokes’ theorem is a surface in ℝ3 , not in ℝ2 .

, 2


Stokes’ Theorem (for graphs, 𝑧 = 𝑓(𝑥, 𝑦)): Let 𝑆 be the oriented surface defined
by a 𝐶 2 function 𝑧 = 𝑓(𝑥, 𝑦), where (𝑥, 𝑦) ∈ 𝐷, a region to which Green’s
theorem applies, and let 𝐹⃗ be a 𝐶 1 vector field on 𝑆. Then if 𝜕𝑆 denotes the
oriented boundary curve of 𝑆, we have:

∫𝝏𝑺 ⃗𝑭⃗ ∙ 𝒅𝒔 ⃗⃗ = ∬ (𝛁 × ⃗𝑭⃗) ∙ 𝒅𝑺.
⃗⃗ = ∬𝑺 (𝒄𝒖𝒓𝒍 ⃗𝑭⃗) ∙ 𝒅𝑺 ⃗⃗⃗⃗
𝑺

⃗⃗𝑑𝑆 = 𝑑𝑆⃗ we could rewrite this as:
Note: As mentioned above, since 𝑛

∫𝜕𝑆 𝐹⃗ ∙ 𝑑𝑠⃗ = ∬𝑆 (𝑐𝑢𝑟𝑙 𝐹⃗ ) ∙ 𝑛⃗⃗𝑑𝑆 = ∬𝑆 (∇ × 𝐹⃗ ) ∙ 𝑛⃗⃗𝑑𝑆 or

∫𝜕𝑆 𝐹⃗ ∙ 𝑇
⃗⃗𝑑𝑠 = ∬ (𝑐𝑢𝑟𝑙 𝐹⃗ ) ∙ 𝑛⃗⃗𝑑𝑆 = ∬ (∇ × 𝐹⃗ ) ∙ 𝑛⃗⃗𝑑𝑆
𝑆 𝑆

i.e., The integral of the normal component of the curl(𝐹⃗ ) over the surface 𝑆 is
equal to the integral of the tangential component of 𝐹⃗ around 𝜕𝑆.



The idea of the proof is to reduce Stokes’ theorem to an application of Green’s
theorem. This can be done by using the fact that if 𝑓: 𝐷 ⊆ ℝ2 → 𝑆 ⊆ ℝ3 then
𝑓 (𝜕𝐷) = 𝜕𝑆. So if the boundary of 𝐷 is given by the curve 𝑐⃗(𝑡 ) =< 𝑥(𝑡 ), 𝑦(𝑡 ) >,
then 𝜕𝑆 is given by the curve 𝐶⃗(𝑡 ) =< 𝑥 (𝑡 ), 𝑦(𝑡 ), 𝑓(𝑥 (𝑡 ), 𝑦(𝑡 )) >.


𝑛⃗⃗

𝑆




𝐶 = 𝜕𝑆



𝐷
𝜕𝐷 = 𝑐(𝑡)