# Math Help - Stokes theorem

1. ## Stokes theorem

Hi there. I was trying to solve this problem, from the book. The problem statement says:
Integrate $\nabla \times{F},F=(3y,-xz,-yz^2)$ over the portion of the surface $2z=x^2+y^2$ under the plane z=2, directly and using Stokes theorem.

So I started solving the integral directly. For the rotational I got:

$\nabla \times{F}=\left|\begin{matrix}i&j&k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\3y&-xz&-yz^{2}\end{matrix}\right|=(x-z^{2},0,-z-3)$

Then I parametrized the surface.:

$\begin{Bmatrix}{ x=r\cos \theta}\\y=r \sin \theta\\z=\displaystyle\frac{r^2}{2} \end{matrix} , \theta[0,2\pi] ,r[0,2]$

Then I did: $T_r\times{T_{\theta}}$

$T_r=(\cos \theta,\sin \theta,r),T_{\theta}=(-r\sin \theta,r\cos \theta,0)$

For the cross product I got:
$T_r\times{T_{\theta}}=(r^2\cos \theta,-r^2\sin \theta,r)$

And then I computed the integral

$\displaystyle\int_{0}^{2}\int_{0}^{2\pi}(r\cos \theta-\displaystyle\frac{r^4}{4},0,\displaystyle\frac{-r^2}{2}-3)\cdot{(-r^2\cos \theta,-r^2\sin \theta,r)}d\theta dr=-12\pi$

The result given by the book is: $20\pi$.

I don't know what I did wrong, and is one of the first exercises that I've solved for the stokes theorem, so maybe I could get some advices and corrections from you

2. ## Re: Stokes theorem

$\nabla \times{F}=\left|\begin{matrix}i&j&k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\3y&-xz&-yz^{2}\end{matrix}\right|=(x-z^{2},0,-z-3).$

Carry those corrections through. What do you get?

3. ## Re: Stokes theorem

I'm sorry, that was a typo :P but the rest of the problem is as I posted it at the beginning, I still having the same problem.

4. ## Re: Stokes theorem

First check your $T_t \times T_r$ (I got a difference in sign in the first component). Next check to make sure that your normal ( $\vec{n} =T_t \times T_r$) is pointing outward to your surface - the side of the cone.

5. ## Re: Stokes theorem

Thanks. How do I check the orientation of the surface in cylindrical coordinates? I know that in the Cartesian coordinates the normal vector must point on the z positive direction, but in the cylindrical (or spherical) coordinates I don't know how to realize if the parametrization I choose is actually positive. In fact, it confuses me, because now I know the parametrization I choose is negative oriented, but I'm pretty much sure that in the xy plane for the unit radius circle a positive (anticlockwise direction) orientation for the circle is x=cos t, y=sin t. But in the space it seems to be that the positive orientation goes with x=sin t and y=cos t, z=anything. I don't know why is this, and it confuses me, because I can think on the projection over the xy plane, and the parametrization I would be using for the boundary of the surface would be negative oriented (in the clockwise sense). So, I don't know how to reason this.

I've found the mistakes, I made a silly mistake in the integration with a sign, and the other is about what I was talking about, the orientation of the surface for the parametrization I choose, I'd like to clarify the ideas in my mind about the last thing. Thanks for your help.

6. ## Re: Stokes theorem

Originally Posted by Ulysses
Thanks. How do I check the orientation of the surface in cylindrical coordinates? I know that in the Cartesian coordinates the normal vector must point on the z positive direction,
Not necessarily. In this case, the normal point in the negative z direction

As for moving around the curve - it's a circle! Use the right hand rule (thumb points in the direction of the normal, your fingers indicate the way to move around the circle). If you use $x = cos t$ and $y = sin t$ and you move around the curve wrong, then let $t \to -t$ or parametrized differently.

7. ## Re: Stokes theorem

Right. But thats the orientation that I choose, and is negative oriented. It's still not clear to me. I think I wasn't clear. What I've tried to say is that in the cartesian coordinates its easy to select the orientation, because z must be positive for the normal vector. But in the cylindrical coordinates I find some kind of contradiction, in the plane I got that the orientation I choose should be positive, I've used it for the Green theorem before, so that parametrization should be positive unless for the frontier, but it seems that it isn't, so it confuses me, I don't know whats wrong. I'm sorry if I'm insisting too much with this :P

Lets see if I can put the things clear. If we think in a curve in the plane, a circle. Then the parametrization x=cos t, y=sin t t[0,2Pi] is positive oriented, right? then why in the space the positive orientation is given by x=sin t, y=cos t, which goes in the opposite sense? and when we think of the frontier, wouldn't that orientation be negative in regards to what we said for the plane?

I'm sorry if I'm not much clear, the thing is that my English isn't that good (my vernacular language is spanish). So it's difficult to me sometimes to express (and understand) some ideas. But I enjoy it anyway.

8. ## Re: Stokes theorem

The positive orientation is directed connected to the outward normal. Let's consider this example.

$2z = x^2 + y^2$ bound by $z = 2$.

There are two surfaces - right? The side of the paraboloid and the flat top. Let's consider the flat top (a circle of radius $2$ with $z = 2$). The outward normal is $\vec{n} = <0,0,1>.$ The right hand rule tells us to go clockwise (viewing from the top) and the parameterization

$x = \cos \theta, y = \sin \theta$ works great!

Now consider the paraboloid. The normal is (I'll use cartestian coords)

$\vec{n} = \frac{\vec{\nabla} G}{\| \vec{\nabla} G \|}$ where $G = x^2 +y^2-2z$or $G = 2z - x^2 - y^2$. Choose $G$ such that the normal is pointing downward and in this case it's the first so

$\vec{n} = \frac{<2x, 2y,-2>}{2\sqrt{1 + x^2 + y^2}} = \frac{}{\sqrt{1 + x^2 + y^2}}.$

Now with this normal, the positive orientation of the same circle is now counterclockwise (again, use the right hand rule). So we can use

$x = \cos \theta, y = - \sin \theta,$

or

$x = \sin \theta, y = \cos \theta,$

if you like, it doesn't matter as long as you go from $0 \to 2 \pi$.

9. ## Re: Stokes theorem

Thank you very much Danny. I think I almost get it :P (lets be clear, your explanation is great and the problem is me).

So, the thing is that I must think of it as a closed surface, consisting of the paraboloid and the plane, like a sphere, and then the positive orientation is defined as the normal vector field pointing outside the surface? I thought of it only as a paraboloid, perhaps thats the problem. And as the paraboloid isn't a closed surface I thought I should take the orientation that gives the normal vector pointing in z positive. That was the deal or I still misunderstanding it?

Thanks for your time, nice work