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 $$x^2+y^2+z^2=1$$
bounded by a closed curve "C" $$x^2+y^2+z^2=1$$ at z=0 plane.

Now here is the stoke's theorem:

$$\oint F.dr = \int\int (curl of F).n ds$$


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

Curl of F = 7j + 2k

and $$n=\bigtriangledown (x^2+y^2+z^2-1)= 2xi+2yj+2zk$$

and then

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

now i put it under the integral sign like :

$$\int\int (14y+4z)ds$$

i put here the value of z as

$$z=\sqrt{1-x^2-y^2}$$

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

now what i have done is this that i take

$$x=r\cos\theta$$
$$y=r\sin\theta$$
$$ds=rdrd\theta$$



and i finally put it in the integral sign

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

where $$0 \leq r\leq 1$$
$$0\leq \theta \leq 2\pi$$

now when i solve it

i found its answer to be

$$\frac{8\pi}{3}$$


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

i.e.

$$\oint F.dr$$

i found it that

$$\oint F.dr = 2\pi$$


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 $ds$ is wrong.

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

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

$$ds = \| T_u \times T_v\| du \, dv,$$

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

(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 $$2\pi$$

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 $\vec{n}\cdot dS$ 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
$$\int_{\theta=0}^{2\pi}\int_{r=0}^1 \left(\frac{7r cos(\theta)}{\sqrt{1- r^2}}+ 2\right) rdrd\theta$$
The "cos" term in the first part, integrated from 0 to 2$\pi$ is 0 so we don't have to worry about that square root. The integral of the "2r" term is 2$\pi$ so the integral is 2$\pi$.

 

[shaiqbashir]

thanks a lot hallsofivy