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Green's Theorem w/Holes 
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#1
Dec2812, 04:51 PM

P: 784

Hey all,
I was working through some problems in my spare time when I realized that I wasn't so satisfied with my understanding of how to use Greens theorem with holes. Can someone refresh my memory? More specifically: Lets say I want to take the line integral in some vector field of a curve C which is the union of the circle of radius 1 and the circle of radius 2 (meaning that the region of integration would be between r=1 and r=2). How do I go about doing this again? The book says that sometimes I can just take the line integral around the outer curve and ignore the inner curve but doesn't say when/why this can be done. Thanks. 


#2
Dec2812, 06:47 PM

HW Helper
P: 2,264

let
C1 be a circle of radius r1 C2 be a circle of radius r3 R the region bounded by C1 and C2 [tex]\int\int_R \left(\dfrac{\partial v}{\partial x}\dfrac{\partial u}{\partial y}\right) \text{ dx dy}=\oint_{C_2} (u \text{ dx}+v \text{ dy})\oint_{C_1} (u \text{ dx}+v \text{ dy})[/tex] In particular we can find area by choosing for example u=y/2 v=x/2 [tex]A=\int\int_R \text{ dx dy}=\frac{1}{2}\left(\oint_{C_2} (y \text{ dx}+x \text{ dy})\oint_{C_1} (y \text{ dx}+x \text{ dy})\right)=\pi(r_2^2r_1^2)[/tex] 


#3
Dec2812, 06:49 PM

P: 784

Ah! Okay, if only my damn book wrote that!
That makes things lovely. Is there anything to the fact that sometimes you can ignore the hole though? 


#4
Dec2812, 07:13 PM

HW Helper
P: 6,189

Green's Theorem w/Holes
Generally you cannot simply ignore the hole.
One of the conditions in applying Green's theorem, is that the partial derivatives exist and are continuous in the entire region. In this particular case there is a work around for the hole though. You can connect the 2 circles with 2 line segments that (almost) coincide. That way the relevant region is really the region between the 2 circles, so there can be a hole in the middle. Since the partial derivatives have to be continuous, it does not matter that an infinitesimal area is missing (the part "between" the 2 line segments). Suppose G is the curve containing the 2 circles and the 2 line segments, but not the center. Then properly we have: [tex]\int\int_R \left(\dfrac{\partial v}{\partial x}\dfrac{\partial u}{\partial y}\right) \text{ dx dy} =\int\int_{R \text{ without the area between the line segments}} \left(\dfrac{\partial v}{\partial x}\dfrac{\partial u}{\partial y}\right) \text{ dx dy}[/tex] [tex]=\oint_{G} (u \text{ dx}+v \text{ dy}) =\int_{C_1 \text{ reversed}} (u \text{ dx}+v \text{ dy}) + \int_{L_1} (u \text{ dx}+v \text{ dy}) + \int_{C_2} (u \text{ dx}+v \text{ dy}) + \int_{L_2} (u \text{ dx}+v \text{ dy}) [/tex] Since the line segments L_{1} and L_{2} (almost) coincide and are opposite in direction, their respective integrals cancel. And since the area where the circles are integrated is continuous, the open circle integrals are the same as the closed circle integrals. So we get: [tex]=\oint_{C_2} (u \text{ dx}+v \text{ dy})\oint_{C_1} (u \text{ dx}+v \text{ dy})[/tex] So yes, in this case you can ignore the hole. Actually, the trick is to circumvent the hole, so there is no hole. 


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