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Verifying Stoke's theorem

  • #1

Homework Statement


Verfify Stokes' theorem for the given surface S and boundary ∂S, and vector fields F.

S = [(x,y,z): x2+y2+z2=1, z≥0
∂S = [(x,y): x2+y2=1
F=<x,y,z>

I did this problem and checked the answer - I chose the wrong integral bounds and I am wondering why they are wrong.

Homework Equations



Stokes' theorem:
∫∫(∇×F)dS = ∫F⋅ds

The Attempt at a Solution


I was able to do the left hand of the solution easily. I knew that I should get zero for the cross product, so that's all I had to do.

For the right hand side:

∫F⋅dS = ∫xdx+ydy

x= cosΘ
y=sinΘ
dx=-sinΘ
dy=cosΘ

0≤Θ≤2π

Apparently this is the correct range for theta:

0≤θ≤π

Why is it just π?
 

Answers and Replies

  • #2
vela
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Homework Statement


Verfify Stokes' theorem for the given surface S and boundary ∂S, and vector fields F.

S = [(x,y,z): x2+y2+z2=1, z≥0
∂S = [(x,y): x2+y2=1
F=<x,y,z>

I did this problem and checked the answer - I chose the wrong integral bounds and I am wondering why they are wrong.

Homework Equations



Stokes' theorem:
∫∫(∇×F)dS = ∫F⋅ds

The Attempt at a Solution


I was able to do the left hand of the solution easily. I knew that I should get zero for the cross product, so that's all I had to do.

For the right hand side:

∫F⋅dS = ∫xdx+ydy

x= cosΘ
y=sinΘ
dx=-sinΘ
dy=cosΘ
You should write ##dx=-\sin\theta\,d\theta## and ##dy=\cos\theta\,d\theta##.

0≤Θ≤2π

Apparently this is the correct range for theta:

0≤θ≤π

Why is it just π?
0 to ##2\pi## is the correct interval.
 

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