Problem in verifying Stoke's Theorem

[shaiqbashir]

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 I am making mistake in this problem.

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

thanks in advance!

 

[Data]

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].

(Note that there is another parameterization you could use that would make this much simpler!)

 

[shaiqbashir]

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 I am also getting from my LHS.

now is there any problem that i have made in it?

 

[HallsofIvy]

"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].

 

[shaiqbashir]

thanks a lot hallsofivy