Problem in verifying Stoke's Theorem

  • #1
106
0

Main Question or Discussion Point

Hi guys!

well! in the following question i need to verify The Stoke's Theorem:

Q: Verify Stoke's theorem for

F=6zi +(2x+y)j -xk

where "S" is the upper half of the sphere [tex]x^2+y^2+z^2=1[/tex]
bounded by a closed curve "C" [tex]x^2+y^2+z^2=1[/tex] at z=0 plane.

Now here is the stoke's theorem:

[tex]\oint F.dr = \int\int (curl of F).n ds[/tex]


OKz so when i just solved this problem, i found:

Curl of F = 7j + 2k

and [tex]n=\bigtriangledown (x^2+y^2+z^2-1)= 2xi+2yj+2zk[/tex]

and then

(curl of F). n= 14y+4z

now i put it under the integral sign like :

[tex]\int\int (14y+4z)ds[/tex]

i put here the value of z as

[tex]z=\sqrt{1-x^2-y^2}[/tex]

[tex]\int\int [(14y+4(\sqrt{1-x^2-y^2})]ds[/tex]

now what i have done is this that i take

[tex]x=r\cos\theta[/tex]
[tex]y=r\sin\theta[/tex]
[tex]ds=rdrd\theta[/tex]



and i finally put it in the integral sign

[tex]\int\int [14r\sin\theta+4(\sqrt{1-r^2\cos^2\theta-r^2\sin^2\theta})]rdrd\theta[/tex]

where [tex]0 \leq r\leq 1[/tex]
[tex]0\leq \theta \leq 2\pi [/tex]

now when i solve it

i found its answer to be

[tex]\frac{8\pi}{3}[/tex]


but when i solved the left hand side of the stoke's theorem:

i.e.

[tex]\oint F.dr[/tex]

i found it that

[tex]\oint F.dr = 2\pi[/tex]


and the result is that

the stoke's theorem is not getting verified. please tell me where im making mistake in this problem.

both sides should be equal to each other. but im not getting both of them equal to each other . please help!

thanks in advance!
 
Last edited:

Answers and Replies

  • #2
998
0
You have two problems. First, your "normal" vector isn't actually normalized (ie. it is not a unit vector). Second, your expression for [itex]ds[/itex] is wrong.

You have not written it down, but you have also implicitly parameterized [itex]z = \sqrt{1 - x^2 - y^2} = \sqrt{1 - r^2}[/itex]. This affects [itex]ds[/itex].

One way of expressing [itex]ds[/itex] for a parameterization [itex] x = f(u,v), y = g(u,v), z = h(u,v)[/itex] is as

[tex]ds = \| T_u \times T_v\| du \, dv,[/tex]

where [itex]T_u = (f_u, g_u, h_u)[/itex] and [itex]T_v = (f_v, g_v, h_v)[/itex]. If you try that calculation in your scenario, I think you'll find [itex]ds[/itex] isn't quite [itex]r dr d\theta[/itex]. :smile:

(Note that there is another parameterization you could use that would make this much simpler!)
 
  • #3
106
0
thanks mr. data. what i have found is this that the only problem that i ws making was this that i was not taking n as a normal vector which should be first made a unit vector. when i make it a unit vector and make

ds=dxdy/|n.k|

i found the answer which is equal to [tex]2\pi[/tex]

which im also getting from my LHS.

now is there any problem that i have made in it???
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
956
"Normal" means perpendicular. A normal vector does not have to be a unit vector. In fact, the simple way to do this would be to go ahead and write [itex]\vec{n}\cdot dS[/itex] as (2xi+ 2yj+ 2zk)/2z= (x/z)i+ y(y/z)j+ k, multiply that by the gradient: 7y/z+ 2 and integrate that over the unit disk in the xy-plane: in polar coordinates that is
[tex]\int_{\theta=0}^{2\pi}\int_{r=0}^1 \left(\frac{7r cos(\theta)}{\sqrt{1- r^2}}+ 2\right) rdrd\theta[/tex]
The "cos" term in the first part, integrated from 0 to 2[itex]\pi[/itex] is 0 so we don't have to worry about that square root. The integral of the "2r" term is 2[itex]\pi[/itex] so the integral is 2[itex]\pi[/itex].
 
  • #5
106
0
thanks a lot hallsofivy
 

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