Calculate Integral of F*n on Square Curve C

Click For Summary
SUMMARY

The discussion focuses on calculating the line integral \(\int_C F \cdot n \, ds\) for the vector field \(F(x,y) = xi + yj\) along the square curve \(C\) defined by vertices (1,1), (-1,1), (-1,-1), and (1,-1). Participants emphasize breaking the path into four linear segments due to the non-smooth nature of the curve. The first segment from (1,1) to (-1,1) is parameterized with \(x=t\) and \(y=1\), leading to the integral \(\int_{-1}^{1} t \, dt\), which evaluates to 0. The discussion highlights the importance of parameterization and the addition of results from all four segments to obtain the final answer.

PREREQUISITES
  • Understanding of line integrals in vector calculus
  • Familiarity with parameterization of curves
  • Knowledge of vector fields and dot products
  • Basic integration techniques
NEXT STEPS
  • Study the parameterization of curves in vector calculus
  • Learn about line integrals and their applications in physics
  • Explore the concept of smooth and piecewise smooth curves
  • Practice calculating line integrals for various vector fields
USEFUL FOR

Students and educators in calculus, particularly those focusing on vector calculus and line integrals, as well as anyone seeking to improve their understanding of parameterization and integration techniques.

yolanda
Messages
8
Reaction score
0

Homework Statement


First off, sorry this isn't symbolized correctly. I've wrestled with this for a half hour, so here it is in crude type:

Calculate [tex]\int[/tex]c F*n ds for F(x,y)=xi+yj across the square curve C with vertices (1,1), (-1,1), (-1,-1), and (1,-1).


Homework Equations



above

The Attempt at a Solution



I've found that this can be rewritten as [tex]\int[/tex]ab F(r(t))*r'(t)) dt

I know my limits are from -1 to 1. I am having trouble parameterizing x and y. I have come up with x=t, but I don't know what y could be... y=[tex]^{+}_{-}[/tex]t?

If someone wouldn't mind explaining this problem, and how they'd attack it, I sure would appreciate the help. Thanks in advance. A step by step solution would help most, but I'll take what I can get :)
 
Physics news on Phys.org


Because the path is not "smooth", break it into smooth parts. That is, do the lines from
(1) (1,1) to (-1,1)
(2) (-1,1) to (-1,-1)
(3) (-1,-1) to (1,-1)
(4) (1,-1) to (1,10)
separately.

For example, the line from (1,1) to (-1,1) is simply "y= 1" with x going from 1 to -1.
You could just use x itself as parameter with x going from 1 to -1. Since y= 1, dy= 0.
If you don't like integrating from 1 down to -1, you could let x= -t so that dx= -dt and you are integrating with respect to t from -1 to 1.
Or you could take x= -2t+ 1 so that dx= -2dt and integrate with respect to t from 0 to 1. The crucial point is that at every point on this line y= 1 so dy= 0.

Do the same kind of thing for the other three lines.
 


Thanks for your response.

Okay, that makes more sense. So, to get my final answer I should be adding the results of the 4 line integrals, right?

So far, for the line segment from (1,1) to (-1,1) I'm getting:

x=t y=1

r(t)=ti+j
r'(t)=i+0
F(r(t))=ti+j

So, [tex]\int[/tex][tex]^{-1}_{1}[/tex]<ti,j>dot<i,0> = [tex]\int[/tex][tex]^{-1}_{1}[/tex]t dt = t[tex]^{2}[/tex]/2 |[tex]^{-1}_{1}[/tex] = 0

How am I doing here? Everything look okay?
 

Similar threads

Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K