Why Does Green's Theorem Have a Negative Sign in the Area Integral?

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The discussion centers on understanding the negative sign in Green's Theorem's area integral, specifically the term -∂f/∂y dA. The user explains that for a type I region, the upper function g2(x) is above g1(x), and since the orientation of g2(x) runs from b to a, integrating from a to b necessitates a negative sign. Similarly, for a type II region, the upper function h2(x) is above h1(x), and since its orientation is from c to d, the integral remains positive when integrating in that direction. The key takeaway is that the sign in the integral reflects the direction of integration relative to the orientation of the higher value function. Understanding this relationship is crucial for correctly applying Green's Theorem.
yungman
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This is a copy of the book:
[PLAIN]http://i38.tinypic.com/20faqnc.jpg[/PLAIN]

I know the derivation part, I just want to see whether I understand why the -ve sign of ##-\frac {\partial f}{\partial y}dA## in a more common sense way.

From looking at the graph for type I region, ##g_2(x)## is above ##g_1(x)## referenced to y axis. So the integral has to be ##g_2(x)-g_1(x)##. BUT the orientation of curve of ##g_2(x)## is from b to a. So if we want to integrate from a to b, we need to put a -ve sign.

From the type II region, ##h_2(x)## is above ##h_1(x)## referenced to x axis. So the integral has to be ##h_2(x)-h_1(x)##. The orientation of curve of ##h_2(x)## is from c to d. So if we want to integrate from c to d, it would be +ve sign.

Am I getting it right?

Bottom line is the sign depends on the direction of the higher value function of the two ( ie. ##g_2(t)≥g_1(t)##). If the direction is from high value to low value, then the sign has to be change to make it from low to high ( ie. ##g_2(t)## oriented from b to a. So sign needed to be change to integrate from a to b).
 
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Anyone please?
 
Anyone can comment? Even opinions are really appreciated.

Thanks
 
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