# Another Stokes' Theorem Problem

## 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...

LCKurtz
Homework Helper
Gold Member

## 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.

lumpyduster
vela
Staff Emeritus
Homework Helper
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.

Last edited:
lumpyduster
rude man
Homework Helper
Gold Member

## 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?