Question about Stokes Theorem?

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SUMMARY

Stokes' Theorem is correctly represented in two forms: \(\int (\nabla \times \mathbf{F}) \cdot d\mathbf{S}\) and \(\int (\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}}\,dS\). The first representation is used in the UCBerkeley Youtube series, while the second is found in the online textbook. Both forms are valid and convey the same mathematical concept, with the difference lying in the explicit inclusion of the normal vector \(\hat{\mathbf{n}}\) in the second form. Understanding the notation and the significance of the differential area element \(dS\) is crucial for applying Stokes' Theorem effectively.

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stratusfactio
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So I'm self teaching myself Multivariable Calculus from UCBerkeley's Youtube series and an online textbook. I'm up to Stokes Theorem and I'm getting conflicting definitions.

UCBerkeley Youtube series says that Stokes Theorem is defined by:
\int {(Curl f)} {ds}

And then the textbook says that Stokes Theorem is defined by:
\int {(Curl f)} {nd\theta}

where n is the normal vector defined by:
{-frac{\partial}{\partial z} i − ∂z\∂y j + k}\{\over sqrt(1 +(∂z\∂x)^2+(∂z\∂y)^2)}

So I would just like to know which is correct?
 
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They're both correct. They're just different ways of writing the same thing.
\int_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = \int_S (\nabla\times \mathbf{F})\cdot \hat{\mathbf{n}}\,dS
dS is an element of area. It's a vector with magnitude equal to the area dS and with direction normal to the surface. Some people like to pull the unit normal n out and write it explicitly, so you have dS = n dS.
 

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