Help stokes theorem - integral problem

[sarahisme]

Hello all,

http://img244.imageshack.us/img244/218/picture8ce5.png

I am completely new to this stokes theorem bussiness..what i have got so far is the nabla x F part, but i am unsure of how to find N (the unit normal field i think its called).

any suggestions people?

i get that nabla x F or curl(F) = i + j + k

cheers

-sarah :)

 

[benorin]

Well, a way to find N is to first find any normal to the surface and then normalize it. Some lead-in work: Put the surface S into vector form,

S:\, \, \, \vec{r}(u,v)=\left< u\cos v,u\sin v, v\right> \, \, 0\leq u\leq 1,\, 0\leq v\leq\frac{\pi}{2}​

Also, recall that the vectors \frac{d\vec{r}}{du}\mbox{ and }\frac{d\vec{r}}{dv} are tangent to S, and that the cross product of two vectors is normal to both vectors, so a normal vector to S is

\frac{d\vec{r}}{du}\times\frac{d\vec{r}}{dv}​

but we want a unit normal to S, so we normalize the above vector (divide it by its magnitude) to get

\vec{N}(u,v)= \frac{\frac{d\vec{r}}{du}\times\frac{d\vec{r}}{dv}}{\left| \frac{d\vec{r}}{du}\times\frac{d\vec{r}}{dv}\right|}​

PS: I gave the same reply on www.mathlinks.ro

 

[sarahisme]

how do i work out what dS is?, i think that is my main problem...

how does this look for the integral we need to evaluate?

http://img151.imageshack.us/img151/5053/picture9fn3.png

that is, do we have to use the normal or the unit normal when doing this stuff?

 

[HallsofIvy]

Some textbooks refer to the "fundamental vector product": If a surface is given by the parametric equations x= f(u,v), y= g(u,v), z= h(u,v), then the fundamental vector product is
\left&lt;\frac{\partial f}{\partial u}, \frac{\partial g}{\partial u},\frac{\partial h}{\partial u}\right&gt; \times \left&lt;\frac{\partial f}{\partial v}, \frac{\partial g}{\partial v},\frac{\partial h}{\partial v}\right&gt;[/itex]<br /> <br /> The &quot;fundamental vector prodct&quot; for that surface is normal to the surface and the &quot;differential of area&quot; is equal to the fundamental vector product time du dv. the &quot;scalar differential of area&quot;, for the surface is the length of the fundamental vector product times du dv. You certainly would <b>not</b> use the unit normal since it is the length of the normal vector that gives the information about the area.<br /> <br /> In your case, since you are integrating the vector \nabla \times F, you need to use the vector differential of area.

 

[sarahisme]

hmmm i can't quite decide whether to go with my answer of pi/4 or try to redo the question and get -pi/4 as the answer (which is what a friend got...)

 

[HallsofIvy]

Now I am confused! What do you mean "go with my answer of pi/4"? How did you get that answer to begin with? Why would you redo the question if you know you will get the same answer?

 

[sarahisme]

Now I am confused! What do you mean "go with my answer of pi/4"? How did you get that answer to begin with? Why would you redo the question if you know you will get the same answer?

its ok, i think its to do with which direction you choose the normal to be pointing...

 

[HallsofIvy]

Ah, I missed the "-". It was on the previous line from \frac{\pi}{4} and I though it was a hyphen! Yes, swapping the direction of the normal will multiply the answer by -1. Notice that you will still have Stokes' theorem true since, by convention, you traverse the boundary in such a way that if you were walking along the boundary with your head in the direction of the normal vector, you would have your left arm inside the region.

d\vec r here, however, because of the way the parameters u and v are given, should be the outer normal so that, since \vec F is also pointing away from the origin, the answer is positive. You might want to ask your teacher for more detail on how to select the direction of the normal in a problem like this.

 

[sarahisme]

Ah, I missed the "-". It was on the previous line from \frac{\pi}{4} and I though it was a hyphen! Yes, swapping the direction of the normal will multiply the answer by -1. Notice that you will still have Stokes' theorem true since, by convention, you traverse the boundary in such a way that if you were walking along the boundary with your head in the direction of the normal vector, you would have your left arm inside the region.

d\vec r here, however, because of the way the parameters u and v are given, should be the outer normal so that, since \vec F is also pointing away from the origin, the answer is positive. You might want to ask your teacher for more detail on how to select the direction of the normal in a problem like this.

ah, yep i see now i think , that makes a bit more sense. :) thanks for all the help everyone! :D