Line integral of a vector field over a square curve

In summary, the line integral \oint dr\cdot\vec{v} is evaluated by directly integrating over a square in the xy-plane of side length a, with \vec{v} = (y, 0, 0) and \vec{r} = (x, a/2, 0). The integral is equal to a^2/2 for each side of the square. Alternatively, using Stokes' theorem, the integral can be written as the product of the curl of \vec{v} and the area of the square, which is a^2. The correct sign for this expression may need to be determined.
  • #1
marineric
10
0

Homework Statement



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 xy-plane of side length a center at [itex]\vec{r}[/itex] = 0

a) by direct integration

b) by Stokes' theorem

Homework Equations



Stokes' theorem: [itex]\oint[/itex] V [itex]\cdot[/itex] dr = ∫∫ (∇ x V)[itex]\cdot[/itex]n d[itex]\sigma[/itex]

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.
 
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  • #2
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 one-dimensional integrals wrt. x and y.
 
  • #3
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
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
marineric said:
but that's (a/2*x) from -a/2 to a/2 ... which equals zero?

Are you sure it equals zero? :)
 
  • #6
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
You got it
 
  • #8
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
marineric said:
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?

Yep, although it's not immediately clear what's the correct sign.
 

1. What is a line integral of a vector field over a square curve?

A line integral of a vector field over a square curve is a mathematical concept that calculates the total sum of a vector field along a given curve. It takes into account both the magnitude and direction of the vector field.

2. How is the line integral of a vector field over a square curve calculated?

The line integral of a vector field over a square curve is calculated by dividing the curve into small segments and taking the sum of the vector field at each segment, multiplied by the length of the segment. This sum is then multiplied by the direction of the curve to determine the total line integral.

3. What is the purpose of calculating the line integral of a vector field over a square curve?

The line integral of a vector field over a square curve is used to determine the work done by the vector field along the given curve. It is also used in various fields of science, such as physics, engineering, and mathematics, to solve problems related to force, motion, and energy.

4. Can the line integral of a vector field over a square curve be negative?

Yes, the line integral of a vector field over a square curve can be negative. This indicates that the vector field is working against the direction of the curve, resulting in negative work being done.

5. What are some real-life applications of the line integral of a vector field over a square curve?

The line integral of a vector field over a square curve has various real-life applications, such as determining the work done by a force over a given distance, calculating the electric potential around a charged object, and finding the flow of a fluid in a pipe.

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