Integrating a Line Integral with Parametric Equations

Click For Summary
SUMMARY

The discussion focuses on evaluating the line integral ∫c y² dx + 2xy dy along the path C, parametrized by r(t) = (t² + 1)i + (2t² + 2)j for 0 ≤ t ≤ 1. The user calculated the velocity magnitude |v(t)| as 2t√5 but expressed confusion regarding the necessity of the initial equation and the process of substituting x(t) and y(t) into the integral. The correct approach involves using the parametrization to convert the integral into a function of t before integrating from 0 to 1.

PREREQUISITES
  • Understanding of line integrals in vector calculus
  • Familiarity with parametric equations and their derivatives
  • Knowledge of integrating functions with respect to a variable
  • Ability to compute dot products in vector fields
NEXT STEPS
  • Study the process of converting line integrals to parametric form
  • Learn about the application of the Fundamental Theorem of Line Integrals
  • Explore the concept of velocity vectors in parametric equations
  • Review examples of integrating vector fields over curves
USEFUL FOR

Students studying vector calculus, particularly those learning about line integrals and parametric equations, as well as educators looking for examples to illustrate these concepts.

animboy
Messages
25
Reaction score
0

Homework Statement



Evaluate the line integral ∫c y2 dx + 2xy dy,

where C, is the path from (1, 2) to (2, 4) parametrised by

r(t) = (t2 + 1)i + (2t2 + 2)j , 0 ≤ t ≤ 1

Homework Equations



I worked out the velocity magnitude |v(t)| as 2t√5

The Attempt at a Solution



I simply integrated the velocity with respect to t from 0 to 1. and got 56, but then I don't see why I needed that first equation. Are we supposed to do dot product or something first?
 
Physics news on Phys.org
Aren't you supposed to get x( t) and y(t) from the 2nd eqn and to then sub it into the 1st to convert it to an integral in t from 0 to 1? And then solve it.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
12
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K