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## Multivariable calculus

### Course: Multivariable calculus>Unit 5

Lesson 9: Proof of Stokes' theorem

# Stokes' theorem proof part 7

Using Green's Theorem to complete the proof. Created by Sal Khan.

## Want to join the conversation?

• I got a confused at when using the multivariable chain rule on Q(x, y, z) when z is a function of x and y. It looks like you have Partial Q/Partial x = Partial Q/Partial x + Partial Q/Partial z times Partial z/Partial x. What confuses me is whether or not Partial Q/Partial x on the left side of the equation is the same thing as the Partial Q/Partial x on the right side of the equation. Even though they are written the same (one uses the symbol for partial derivative and the other uses subscript notation), it somehow "feels" like they are not the same. And if they were, we would subtract them from each side of the equation and have Partial Q/Partial z times Partial z/Partial x equaling zero and that doesn't make sense either. Can someone straighten me out? Thanks a lot!
• Thanks for responding. That seems very reasonable; I couldn't imagine that those two partials could be the identical same thing. I wonder, is there some kind of formal or fancy mathematical notation to write the left and right partials in ∂Q/∂x = ∂Q/∂x + ∂Q/∂z * ∂z/∂x
that clearly represents their inherent differences and avoids this ambiguity? Thanks!!
• I would think of it as
∂/∂x Q( x, y, z(x) ) = ∂/∂x Q( x, y, z ) + ∂/∂z(x) Q( x, y, z(x) ) * ∂/∂x z(x)
Note the changes between z and z(x)
• Sal keeps on saying that we are proving Stokes' theorem for the special case where there are continuous second derivatives of z( x, y ) along with z having to be a function of x and y. Does Stokes' theorem not apply at all when these conditions are not met or is there a more general way to prove Stokes' theorem where this doesn't have to be true?
• I think there is a more general way to prove this
(1 vote)
• This seems to be a pretty straightforward argument (Proof of Stoke's Theorem): calculate both sides and show that they're equivalent; however, I was wondering if there was a good way to build intuition behind Stoke's Theorem? Sal's "intuition of Stoke's Theorem" video seems rather abstract, and not very applied, any recommendations for secondary source?
• Isn't there some video with the proof of the multivariable chain rule?
• Greens theorem is not a special case of Stokes Theorem? So why would you prove sth using the same theory you want to prove? pls tell me if im wrong :) .... but still the proof was nice, accurate and awesome
(1 vote)
• You're right that Green's theorem is a special case of Stoke's theorem. However, we proved Green's theorem independently of Stoke's theorem, so it is perfectly valid to use it to prove Stoke's theorem. This is actually a very common technique in mathematics, using a theorem to prove its generalization.