How to Apply Stoke's Theorem When Unable to Express Z in Terms of X and Y?

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Homework Statement


i can't find the normal vector here . In my book , outwards vector is . (Refer to photo 1 )
The question is in photo 2 , i am aksed to use stoke's theorem to evalutae line integral of vector filed
But , now the problem is i can't express z in terms of y and x . Can anyone help ?

Homework Equations

The Attempt at a Solution


For now , i have Δx F = (-z+ 1) j only
 

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fonseh said:

Homework Statement


i can't find the normal vector here . In my book , outwards vector is . (Refer to photo 1 )
The question is in photo 2 , i am aksed to use stoke's theorem to evalutae line integral of vector filed
But , now the problem is i can't express z in terms of y and x . Can anyone help ?

Homework Equations

The Attempt at a Solution


For now , i have Δx F = (-z+ 1) j only
For your first case, I don't see a boundary line to the surface. For the second case, the surface is a cylinder, and I think they might be asking you to compute ## \int \vec{F} \cdot \, dS ##. If that is the case, you could also use Gauss law and compute ## \int \nabla \cdot \vec{F} \, d^3x ##, but certainly not Stokes theorem. (the Gauss's law version would also include in its result the integration over the endfaces of the cylinder). ## \\ ## Additional item: For vector curl use " \nabla \times " in Latex. To get Latex, put " ## " on both sides of your statement or expression. (The vector gradient is " \nabla" in Latex. The divergence is " \nabla \cdot ".)
 
Last edited:
fonseh said:

Homework Statement


i can't find the normal vector here . In my book , outwards vector is . (Refer to photo 1 )
The question is in photo 2 , i am aksed to use stoke's theorem to evalutae line integral of vector filed
But , now the problem is i can't express z in terms of y and x . Can anyone help ?

Homework Equations

The Attempt at a Solution


For now , i have Δx F = (-z+ 1) j only

What is preventing you from typing in a complete statement of the problem? There is no question in the photo telling us what line integral (if it really is a line integral) you want.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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