# Thread: Stokes and Guass theorm Integral Problem

1. ## Stokes and Guass theorm Integral Problem

In my attached working I try and use stokes theorm but end up getting dr/dt = 0 . and cant get any further this way

I used the guass therom and got pi.

If anybody can see what the problem is?

2. For those who don't like to open pdf files (not as prone to viruses as word files, but still)
The problem is to find $\oint_S \nabla\times F\cdot dS$ where $F= zi+ xj+ yk$ and S is the circle $x^2+ y^2= 1$ in the plane z= 1 using Stokes theorem or Gauss's divergence theorem.

Stoke's theorem says $\oint_S \nabla\times F\cdot dS= \int\int_R F\cdot ds$
Since that has " $\nabla\cdot F$", that is the obvious one to use.

Taking $x= cos(\t)$, $y= sin(\t)$, $z= 1$
(heatly, you don't include " $z= 1$", which I consider an important part of the parameterization. Of course, here, because z is constant, it doesn't matter).

You then say "r(t)= cos^2t+ sin^2t" which is non-sense! That is the same as saying that r(t)= 1 so of course dr= 0. In order that " $F\cdot ds$",where F is a vector, make sense, ds must be a vector, not a scalar.

What you want is that r(t) is the position vector xi+ yj+ zk= cos(t)i+ sin(t)j+ k. Then dr= dx i+ dy j+ dz k= -sin(t)dt i+ cos(t)dt j. (dz= 0 so there is no k component.)
$F\cdot ds= (zi+ xj+ yk)\cdot(-sin(t)i+ cos(t)j)dt= (i+ cos(t)j+ sin(t)k)\cdot(-sin(t)i+ cos(t)j)dt= (-sin(t)+ cos^2(t))dt$.

$\oint\int \nabla F\cdot dS= \int_{t= 0}^{2\pi} (-sin(t)+ cos^2(t))dt$

That is not 0 because of the $cos^2(t)$.

You then proceed to "use" Gauss's divergence theorem which doesn't apply here because there is no " $\nabla F$" in it and because it requires integration over a solid region and there is no solid region given in this problem.

3. Thanks a lot for that,really helpful.