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Green's and Stokes' theorem relationship

Seeing that Green's Theorem is just a special case of Stokes' Theorem. Created by Sal Khan.

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• I'm a bit lost when to use the divergens, greens and stokes theorem. As I see it is that you can use any of this 3 theorems but some are more easy to solve with then the other two. Is it correct?
• The divergence theorem is useful when one is trying to compute the flux of a vector field F across a closed surface F ,particularly when the surface integral is analytically difficult or impossible. In such cases, one uses the divergence theorem to convert a problem of computing a difficult surface flux integral to one of computing a relatively simple triple integral. Similarly, Stokes Theorem is useful when the aim is to determine the line integral around a closed curve without resorting to a direct calculation.
As Sal discusses in his video, Green's theorem is a special case of Stokes Theorem. By applying Stokes Theorem to a closed curve that lies strictly on the xy plane, one immediately derives Green Theorem.
In principle, you can apply these formulas whenever the conditions for using them are met, but what makes these formulas convenient to know is that it converts a difficult problem(finding a surface flux integral, line integral) to a relatively simple one(finding triple integral throughout a volume, adding the flux across a surface).
• I'm a little lost when he broke down the curl of F at ish, is there a good video on vector calculus?
• I was wondering, are there any plans on making a video for Gauss's law?
• If by Gauss's law you mean the Divergence Theorem (3D) then there are already videos on that.
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• Is there anyway you could create Mastery Challenge program for Multivariable Calculus like you have for Differential and Integral Calculus? Or is there any website that has anything like that?
• What exactly is a matrix?
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• A matrix is a system of linear equations written in square brackets or parenthesis. You can learn more in the linear algebra section.
• what is the application on this all theorems stoke , green and divergence??
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• Beyond being an essential theoretical tool for simplifying line/surface integrals, they have very important applications in physics. E.g. Fluid Dynamics and Electrodynamics.
• At , Sal says specified that the vector field has to be a function only of x and y, and have a zero k component: F= P(x,y)i + Q(x,y)j + 0k

Is this constraint even necessary to get the same result at the end? If you instead say F = P(x,y,z)i + Q(x,y,z)j + R(x,y,z)k. Then you get
curl(F) = (∂R/∂y - ∂Q/∂z)i + (∂P/∂z - ∂R/∂x)j + (∂Q/∂x - ∂P/∂y)k
Notice the k component of the curl is still the same as in the video!

Then for a surface parallel to the x-y plane, the normal vector n = k. You'll always get: curl(F)·n = (∂Q/∂x - ∂P/∂y)k. So shouldn't this work with any vector field?
• "At , over the surface R", how is this surface R defined? the special surface on xy plane enclosed by the closed curve C ,or any other spacial surfaces that the closed curve C is on these spacial surfaces. Notice the closed curve C might also be a spacial closed curve. And then the unit norm vector n is determined with regard to this surface R?
"At ", intuitively, the 2 dimentional divergence is a part of 3 dimentional curl?
• At , why is the unit normal vector equal to k-hat. Is it because k-hat is normal to the surface, since k-hat is in the z-direction? To find the normal vector in the case of a region on the xy-plane, where z = 0, we didn't need to calculate the cross product to find out that k-hat was normal to the region right?
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• Yes, any upward looking surface in the xy plane will have a normal vector straight up with magnitude 1, with is the exact definition of the unit vector `k̂`.

And not only is this the normal vector of any surface on the xy plane, but also of any surface parallel to the xy plane, so any (upwards looking) surface with a constant z-coordinate will also have `k̂` as it's normal vector.
• how is k component dotted with unit normal vector itself (i.e the unit normal vector) doesn't a vector dotted with itself give you vector magnitude^ (squared)
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• That is correct, but it just so happens that because i-hat, j-hat and k-hat are UNIT vectors, their magnitude is 1. We therefore know that our normal vector equals our k-hat vector, thus 1*1=1, so that whole part can just be removed.
I personally just remember it as i*i=j*j=k*k=1. With those all hatted, of course.
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