Analyzing Line Integral over Non-Exact Region |x|+|y|=4

nasshi
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Analyzing an integral over a non-exact region for gamma defined by |x|+|y|=4

The following was similar to a problem on a calculus final that I got wrong. It is an extension of a problem in R.C. Buck "Advanced Calculus" on page 501. Similar to knowing the trick to integrating e^{|x|} (which got me on my first calculus final years ago!), I assume there is a bit of analysis that I'm missing for this line integral regarding the absolute value. Someone please walk me through the analysis of this problem from Buck taken a step further.


Consider the 1-form \omega = \frac{xdy-ydx}{x^{2}+y^{2}} in the open ring D = \lbrace (x,y) | 1 \leq x^{2} + y^{2} \leq 4 \rbrace.

The text asserts that d \omega is not exact in D, which I verified with the standard computation.

The extension is -- setting \gamma (t) to be |x| + |y| = 4, calculate the integral \int_{\gamma}\omega counterclockwise over the region.

My hangups: aside from the differential form not being exact, the curve is closed, but not smooth.

Please note I don't necessary want someone to do this problem. I'd much rather have someone discuss how to pick this problem apart analytically. What is the first question to ask yourself? How do you reconcile the curve not being smooth? etc.
 
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The curve is non-smooth, so you can break the curve apart in 4 smooth pieces:

A piece \gamma_1 going from (0,4) to (4,0).
A piece \gamma_2 going from (4,0) to (0,-4).
A piece \gamma_3 going from (0,-4) to (-4,0).
A piece \gamma_4 going from (-4,0) to (0,4).

A first thing to do is to find explicit formula's for the \gamma_i.

The line integrals \int_{\gamma_i}\omega can be calculated by using the definition of a line integral.

You should expect the line integral to be nonzero.
 


micromass said:
The curve is non-smooth, so you can break the curve apart

Can this be done finitely many times as long as the pieces form a closed curve when put together? Are there other stipulations for breaking apart the curve to work, or onlyclosure of the curve?
 


You can break any curve into (finitely many) pieces, even nonclosed curves. It is always true that if \gamma is a curve, then

\int_\gamma \omega=\int_{\gamma_1} \omega + \int_{\gamma_2} \omega

where \gamma_1 and then \gamma_2 form the original curve \gamma.

This can be done for any curve, smooth and piecewise smooth. In fact, for piecewise smooth curves, this is a definition.
 


I found the fundamental property regarding splitting up piecewise smooth curves two chapters earlier, which I need to review. I got an answer of -\pi, which is what the solutions suggested. Thank you very much!
 
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