CALC III Line Integral Problem

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SUMMARY

The discussion focuses on solving a line integral problem in Calculus III, specifically breaking the contour into three segments: C1 (line segment from the origin to (1,0)), C2 (circular arc), and C3 (line segment from (1,1) back to the origin). The user is guided to parameterize each segment, with a detailed example provided for C3, where x(t) and y(t) are expressed in terms of t. The integral for C3 is computed as an ordinary integral of one variable, demonstrating the method to combine results from all segments for the final solution.

PREREQUISITES
  • Understanding of line integrals in multivariable calculus
  • Familiarity with parameterization of curves
  • Knowledge of basic integration techniques
  • Ability to manipulate functions of multiple variables
NEXT STEPS
  • Study the concept of line integrals in depth
  • Learn how to parameterize different types of curves
  • Practice solving integrals involving multiple variables
  • Explore applications of line integrals in physics and engineering
USEFUL FOR

Students in Calculus III, educators teaching multivariable calculus, and anyone looking to strengthen their understanding of line integrals and parameterization techniques.

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Homework Statement


[PLAIN]http://img843.imageshack.us/img843/3995/calc3.jpg



The Attempt at a Solution




Im here asking for some help in direction on how to do these problems so that I can find the solution by myself... I would really really appreciate any help anyone could provide ...

I am not sure how to represent the domain as a function, and if anyone could point me in the right direction, I'm sure I could get the answer.
 
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First, break the contour up into three pieces, C1, C2, and C3, where C1 is the line segment from the origin to (1,0), C2 is the circular arc, and C3 is the line segment from (1,1) back to the origin.

For each segment of the path, parameterize x and y. For example, for C3, you could write x(t)=1-t and y(t)=1-t where t goes from 0 to 1. Then dx=-dt and dy=-dt. Now write everything in terms of t.

\int_{C_3} (\sin x-6x^2y)dx+(3xy^2-x^3)dy = \int_0^1 [\sin (1-t) - 6(1-t)^2(1-t)](-dt) + [3(1-t)(1-t)^2 - (1-t)^3](-dt)

The righthand side is just a run-of-the-mill integral of one variable that you can crank out.

Do this for each segment and add the results together to get your final answer.
 

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