Vector calculus - line integral computation

Click For Summary
SUMMARY

The line integral \(\int_{C} F \cdot dr\) for the vector field \(F = -y i + x j\) is computed along a directed path \(C\) consisting of a left semicircle and a straight line segment. The semicircle is parameterized by \(r(t) = \cos t i + \sin t j\) for \(\frac{\pi}{2} \leq t \leq \frac{3\pi}{2}\), while the line segment is parameterized by \(r(t) = 2t i + j\) for \(0 \leq t \leq 1\). The integral evaluates to \(\pi\) but requires a sign adjustment due to the path orientation, which runs from \((0, -1)\) to \((0, 1)\) rather than the reverse.

PREREQUISITES
  • Understanding of vector fields and line integrals
  • Familiarity with parameterization of curves in the xy-plane
  • Knowledge of calculus, specifically integration techniques
  • Proficiency in using mathematical notation and symbols
NEXT STEPS
  • Study the properties of line integrals in vector calculus
  • Learn about the orientation of paths and its impact on integrals
  • Explore the application of Green's Theorem in evaluating line integrals
  • Investigate the use of parametric equations in curve integration
USEFUL FOR

Students and professionals in mathematics, physics, and engineering who are working with vector calculus and line integrals, particularly those involved in computational methods and theoretical analysis.

braindead101
Messages
158
Reaction score
0
Compute the line integral [tex]\int_{C} F\cdot dr[/tex] where F = -y i + x j. The directed path C in the xy-plane consists of two parts: i) a left semicircle from (0, -1) to (0, 1) with center at the origin, and ii) a straight line segment from (0,1) to (2,1).

i) r(t) = cos t i + sin t j [pi/2 <=t<= 3pi/2]
ii) r(t) = 2t i + j [0<=t<=1]

for i):
F(r (t) ) = - sin t i + cos t j
r'(t) = - sin t i + cos t j

integ F(r(t)) dot r'(t) dt
= integ 1 dt [pi/2<=t<= 3pi/2]
so integral is
3pi/2 - pi/2 = pi ?
is this correct so far.
 
Physics news on Phys.org
braindead101 said:
Compute the line integral [tex]\int_{C} F\cdot dr[/tex] where F = -y i + x j. The directed path C in the xy-plane consists of two parts: i) a left semicircle from (0, -1) to (0, 1) with center at the origin, and ii) a straight line segment from (0,1) to (2,1).

i) r(t) = cos t i + sin t j [pi/2 <=t<= 3pi/2]
ii) r(t) = 2t i + j [0<=t<=1]

for i):
F(r (t) ) = - sin t i + cos t j
r'(t) = - sin t i + cos t j

integ F(r(t)) dot r'(t) dt
= integ 1 dt [pi/2<=t<= 3pi/2]
so integral is
3pi/2 - pi/2 = pi ?
is this correct so far.
Almost- you have the "orientation" backwards. The path does not go "from pi/2 to 3pi/2" (i.e. from (0,1) to (0,-1)), it goes "from 3pi/2 to pi/2" (from (0,-1) to (0,1)). That reverses the sign on the integral.
 

Similar threads

Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K