Register to reply 
Problem with stokes' theorem 
Share this thread: 
#1
Nov809, 10:34 AM

P: 34

1. The problem statement, all variables and given/known data
Let: [tex]\vec{F}(x,y,z) = (2z^{2},6x,0),[/tex] and S be the square: [tex]0\leq x\leq1, 0\leq y\leq1, z=1. [/tex] a) Evaluate the surface integral (directly): [tex]\int\int_{S}(curl \vec{F})\cdot\vec{n} dA[/tex] b) Apply Stokes' Theorem and determine the integral by evaluating the corresponding contour integral. 2. Relevant equations [tex]\int\int_{S}(curl \vec{F})\cdot\vec{n} dA = \oint_{C}\vec{F}\cdot d\vec{r}[/tex] 3. The attempt at a solution a) Basically I took the curl of F and got (0, 4z, 6) and dotted it with the normal vector which is either (0,0,+1) or (0,0,1) as no orientation was given in the question! which gave me a value of +6 or 6. b) now this is the part which has confused me!! How on earth do I find the corresponding contour integral?? I have no idea how to find dr and I'm not sure what to do about parametizing the square. Any help would be greatly appreciated!!! P.S. I have a similar problem involving Gauss' Law so am hoping to kill two birds with one stone! 


#2
Nov809, 10:46 AM

HW Helper
P: 3,307

the normal to use for the first part is determeined by a right hand rule, to line up with the direction you go around the contour
now to do the contour integral, see it as the sum of 4 line intergals, one for each side of the square 


#3
Nov809, 10:49 AM

HW Helper
P: 3,307

[itex] \vec{dr} [/itex] will be an infinitesimal in the direction of the line, be careful with direction. eg if the direction is the +ve x, it essentially reduces to [itex] \vec{dr} = (dx, 0, 0) [/itex]
Pick a way to go around say counter clockwise form above and stick to it, setting the [itex] \vec{dr} [/itex] direction & limits accordingly 


#4
Nov809, 11:20 AM

P: 34

Problem with stokes' theorem
I still don't get it!!
if i name the corners of the square a,b,c and d for (0,0,1), (0,1,1), (1,1,1) and (1,0,1) respectively, then will dr be: a>b: (0,dy,0) b>c: (dx,0,0) c>d: (0,dy,0) d>a: (dx,0,0)? But then when I dot product with f=(2z^2,6x,0) and integrate each seperately, I get: 6x + 2z^2  6x  2z^2 = 0 Forgive me if I'm being stupid but we've only ever done stokes' law the other way round so I've never had any experience this way!! 


#5
Nov809, 02:29 PM

P: 34

Can anyone find a way to explain this to me or point me in the right direction? I've been searching online and my books for hours and the only examples that I find are the opposite way round!!!



#6
Nov809, 06:32 PM

HW Helper
P: 3,307

these are definte integrals, so you should end up with a number for each integral, not a function... so for a>b:(0,0,1)>(0,1,1) along the y axis notice x = 0, z=1, only y is changing [tex] \vec{F} \bullet \vec{dr} = (2z^2,6x,0)\bullet (0,1,0)dy = 6x.dy _{x=0}[/tex] so there is actually no field component parallel to dy in the integral... note the dot product method is a slight shortcut in notation, really should parameteris the line in terms of a variable & integrate, though as teh intergals are along axes in this question we can take the shortcut 


Register to reply 
Related Discussions  
Am I doing anything opposite to this problem related to verifying Stokes' theorem?  Calculus  1  
Stokes theorem problem  Calculus & Beyond Homework  3  
Stokes Theorem problem  Calculus & Beyond Homework  4  
Stokes theorem problem  Calculus & Beyond Homework  4  
Help! stokes theorem  integral problem  Calculus & Beyond Homework  8 