## SBurchette 2 years ago Stokes' Theorem I'm trying to evaluate the line integral around the curve obtained by intersecting the cylinder x^2+z^2=1 with the plane y=3. The vector field is F=<3xz, e^(xz), 2xy> and the curve is oriented counterclockwise when viewed from the right.

1. SBurchette

I parameterized the surface needed to use stokes' thm as r(u,v)=<u*cos(v), 3, u*sin(v)> where 0 <=u<=1 and 0<=v<=2*pi. The curl of F and the cross product of the partials of r(u,v) are quite long, but I got an answer of 6*pi which is far from the answer in the solution key.

2. amistre64

F(r) . r' right?

3. SBurchette

For the line integral it is, but we are supposed to use Stokes' theorem

4. amistre64

i can never keep the names straight

5. amistre64

what is stokes thrm by chance?

6. slaaibak

Flux is F.n

7. slaaibak

The line integral in space is equal to the flux of the surface

8. SBurchette

$\int\limits_{C} F.dr = \int\limits \int\limits_{S} curl(F).dS$

9. slaaibak

what was your normal vector? <0,y,0? ?

10. amistre64

oy, its way to late for me to be doing anything remotely sane with that :/

11. malevolence19

If the plane intersects the cylinder you know the cross section's shadow on the x-y plane is a circle right? So you need only to integrate around the circle? And since it's a surface integral r if FIXED at r=1 not to be integrated over. So integrate over z and phi (the polar angle).

12. malevolence19

I believe.

13. malevolence19

Especially with y=3, its not even slanted so it will DEFINITELY only give you a circle of constant radius.

14. SBurchette

The formula I have been following states that flux integral over a paramterized surface is $\int\limits \int\limits_{D} F(r).(r_u \times r_v) dA$ This, in effect, eliminates having to compute the normal directly.

15. malevolence19

And since you're only integrating with scalars you don't need to worry about integrating over a "position" vector which would NOT give you a circle if the plain was slanted. And you still compute the normal directly, you just don't normalize it.

16. malevolence19

And I guess it would be the x-z plane not x-y.

17. SBurchette

Integrating along the curve would likely be more simple, but we are supposed to use Stokes' theorem instead of directly computing the line integral

18. slaaibak

but the normal is quite obvious though, isn't it? It just points in the y-direction. so <0,1,0> is fine

19. SBurchette

Yes, I meant we don't compute the nomalized normal, the wording gets tricky ;)

20. malevolence19

And I misread, since you want to calculate the line integral you only need to integrate over phi. So: |dw:1354845022728:dw| So: $\Phi(\phi)=(- \cos(\phi),3,\sin(\phi))$ That is your parameterization for that circle. So $0 \le \phi \le 2 \pi$ And then you have: $\oint \vec{F} \cdot d \vec{l}=\int\limits_0^{2 \pi} \vec{F}(\vec{\Phi}(\phi)) \cdot \vec{\Phi}'(\phi)d \phi$

21. slaaibak

wait, I did this a long while ago... but if you just use xyz coordinates, and calculate the curl of the vector field, then dot it with <0,1,0> and then use parametrizations, it's not really tedious at all? since you're only dealing with the j component?

22. SBurchette

That's what we need if we were doing the line integral directly, but to use stokes thm, we have to make it into a flux integral. and @slaaibak I didn't have too much trouble actually doing it, but the solution was incorrect and I can;t figure out where I went wrong

23. malevolence19

A flux integral of the curl of F.

24. SBurchette

I just realized that I was checking the wrong section of the manual... so the answer I got actually was correct. Thanks for the help guys, I apologize for my oversight,.