
#1
Dec1012, 12:00 AM

P: 10

1. The problem statement, all variables and given/known data
Please evaluate the line integral [itex]\oint[/itex] dr[itex]\cdot[/itex][itex]\vec{v}[/itex], where [itex]\vec{v}[/itex] = (y, 0, 0) along the curve C that is a square in the xyplane of side length a center at [itex]\vec{r}[/itex] = 0 a) by direct integration b) by Stokes' theorem 2. Relevant equations Stokes' theorem: [itex]\oint[/itex] V [itex]\cdot[/itex] dr = ∫∫ (∇ x V)[itex]\cdot[/itex]n d[itex]\sigma[/itex] 3. The attempt at a solution I know I have to split up the sides of the square. I get confused when [itex]\vec{r}[/itex] is involved. I know the limits are at a/2.. not sure where to go after that. 



#2
Dec1012, 04:59 AM

P: 937

For the direct integration, just write it out. What is [itex] d\vec{r} \cdot \vec{v} [/itex]? Since you're integrating over a square, you can just write the integral as a sum of four onedimensional integrals wrt. x and y.




#3
Dec1012, 06:30 AM

P: 10

Ok well I'll try the top line of the square and you can tell me what I'm doing wrong.
From left to right: [itex]\oint[/itex] d[itex]\vec{r}[/itex][itex]\cdot[/itex][itex]\vec{v}[/itex] from a/2 to a/2 [itex]\vec{v}[/itex] = (y,0,0) [itex]\vec{r}[/itex] = (x, a/2, 0) d[itex]\vec{r}[/itex] = (1, 0, 0) d[itex]\vec{r}[/itex][itex]\cdot[/itex][itex]\vec{v}[/itex] = y but aren't the y limits a/2 to a/2, making the integral zero? 



#4
Dec1012, 06:43 AM

P: 10

line integral of a vector field over a square curve
wait hold on [itex]\vec{r}[/itex] = (d[itex]\vec{x}[/itex], 0, 0) ??
then it would be the integral from a/2 to a/2 of y*dx where y = a/2? but that's (a/2*x) from a/2 to a/2 ... which equals zero? 



#5
Dec1012, 06:57 AM

P: 937





#6
Dec1012, 07:01 AM

P: 10

oh, right, minus sign.. so it equals a^2/2 for the top line, and i just do it similarly for all the other sides? do i have to do it in the same order (clockwise)?




#7
Dec1012, 07:02 AM

P: 937

You got it




#8
Dec1012, 07:05 AM

P: 10

for the stokes' theorem part, would it just be the ∇[itex]\times[/itex][itex]\vec{v}[/itex] times the area of the square, which is a^2?




#9
Dec1012, 07:09 AM

P: 937




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