CALC III Line Integral Problem

SolitaryRaf
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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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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