Another Stokes' Theorem Problem

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Homework Help Overview

The discussion revolves around verifying Stokes' theorem for a specified surface and vector field. The surface is defined by the equation of a half-sphere, and the vector field is given in component form.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss calculating the curl of the vector field and its implications for the surface integral. There are attempts to parametrize the surface and questions about the correct expression for the differential area element.

Discussion Status

Some participants are actively working through the mathematical steps involved in applying Stokes' theorem, while others are questioning the setup and definitions of the surface and boundary. Guidance has been offered to continue with the current approach and to check for orientation signs.

Contextual Notes

There are indications of confusion regarding the parametrization and the interpretation of the surface area element. Some participants express uncertainty about the integration process and the need for orientation considerations.

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Homework Statement


Verify Stokes' theorem for the given surface S and boundary dS and vector fields F

S = x2+y2+z2, z≥0
dS= x2+y2=1

F = <y,z,x>

Homework Equations



Stokes' theorem:
∫∫(∇×F)dS = ∫F⋅ds

The Attempt at a Solution


1. Curl of F:

∇×F = <-1,-1,-1>

2. After getting the curl, I just treated this as a surface integral, ∫∫F⋅dS = ∫∫F⋅(Tu×Tv)dudv

I parametrized the sphere thusly,

x = sinφcosθ
y=sinφsinθ
z=cosφ

Tφ= <cosφcosθ, cosφsinθ, -sinφ>
Tθ= <-sinφsinθ, sinφcosθ, 0>

Tφ×Tθ = <sin2φcosθ, sin2φsinθ, sinφcosθ>

Am I doing this right so far? I asked a friend what he would do, and this is what he had for dS:
<dydz, dxdz, dxdy> (Idk how to get this).

So then he got:

∫∫(∇×F)⋅dS = ∫∫(<-1, -1, -1>⋅<dydz, dxdz, dxdy>) = -∫∫dydz+dxdz+dxdy, but I don't know how to integrate that...
 
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lumpyduster said:

Homework Statement


Verify Stokes' theorem for the given surface S and boundary dS and vector fields F

S = x2+y2+z2, z≥0
dS= x2+y2=1

F = <y,z,x>

Homework Equations



Stokes' theorem:
∫∫(∇×F)dS = ∫F⋅ds

The Attempt at a Solution


1. Curl of F:

∇×F = <-1,-1,-1>

2. After getting the curl, I just treated this as a surface integral, ∫∫F⋅dS = ∫∫F⋅(Tu×Tv)dudv

I parametrized the sphere thusly,

x = sinφcosθ
y=sinφsinθ
z=cosφ

Tφ= <cosφcosθ, cosφsinθ, -sinφ>
Tθ= <-sinφsinθ, sinφcosθ, 0>

Tφ×Tθ = <sin2φcosθ, sin2φsinθ, sinφcosθ>

Am I doing this right so far? I asked a friend what he would do, and this is what he had for dS:
<dydz, dxdz, dxdy> (Idk how to get this).

So then he got:

∫∫(∇×F)⋅dS = ∫∫(<-1, -1, -1>⋅<dydz, dxdz, dxdy>) = -∫∫dydz+dxdz+dxdy, but I don't know how to integrate that...

I would ignore your friend's comments. Just continue what you are doing. You are now going to calculate$$
\iint \langle -1,-1,-1\rangle \cdot T_\phi \times T_\theta~d\phi d\theta$$with appropriate limits. Be sure to check whether you need a minus sign for orientation or not.
 
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lumpyduster said:
Tφ×Tθ = <sin2φcosθ, sin2φsinθ, sinφcosθ>
There's a common factor of ##\sin\phi##. If you pull it out front, you have ##\sin\phi \langle \sin\phi \cos\theta, \sin\phi \sin\theta, \cos\phi \rangle##. If you recognize that vector, you should be able to convince yourself you're on the right track.
 
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lumpyduster said:

Homework Statement


Verify Stokes' theorem for the given surface S and boundary dS and vector fields F

S = x2+y2+z2, z≥0
This defines a surface?
A half-sphere of radius sqrt(S)?
 

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