Integrating a Line Integral Along a Curve with Given Boundaries

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SUMMARY

The discussion focuses on evaluating the line integral ∫(zdx+xdy+ydz) along the curve defined by x(t)=cos(t), y(t)=sin(t), and z=3t, with boundaries from 0 to 2π. The user successfully derived the differential components using the chain rule, resulting in the integral ∫(-3tsin(t)dt + cos²(t)dt + 3sin(t)dt). The challenge arises in integrating the first term, where the user encounters difficulties with integration by parts. The solution suggests employing the half-angle identity for cos²(t) to simplify the integral.

PREREQUISITES
  • Understanding of line integrals in vector calculus
  • Proficiency in applying the chain rule for differentiation
  • Familiarity with integration techniques, including integration by parts
  • Knowledge of trigonometric identities, specifically the half-angle identity
NEXT STEPS
  • Study the application of the half-angle identity in integrals
  • Practice integration by parts with various functions
  • Explore line integrals in three-dimensional space
  • Review the properties and applications of trigonometric functions in calculus
USEFUL FOR

Students studying calculus, particularly those focusing on vector calculus and line integrals, as well as educators looking for examples of integrating functions along curves.

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Homework Statement



∫(zdx+xdy+ydz)

along the curve C: x(t)= cos(t), y(t)= sin(t), z = 3t,

Boundaries are 0 and 2pi


Homework Equations



General integration and differentation.

The Attempt at a Solution


given the values I calculated that:

Using chain rule:

dx = -sintdt
dy = costdt
dz = 3dt

i then resubbed into the original equation to get:

∫(-3tsin(t)dt + cos^2(t)dt + 3sin(t)dt)

this is where I've got stuck, I attempted to integrate by parts the first term however it seems to get stuck in an endless loop, I'm not sure where to go from there.
 
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Why don't you show us your work. It may be a simple problem to fix.
 
The first and last terms of your integral should be easy to evaluate, the possibly tricky one is the middle term. The trick is to use the half-angle identity:
\cos^2 \theta = \frac{1 + \cos 2\theta}{2}
 

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