Timeline for Proof of Green's theorem in Apostol book
Current License: CC BY-SA 2.5
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| Apr 27, 2010 at 9:32 | comment | added | Willie Wong | $\gamma_2$ then, over $[1/2,1]$ can be $\gamma_2(t) = \tau((3-t)/4)$ if $1/2 < t < 3/4$ and so on. This way I hope it is sort of clear that the contours for that two $L$ segments, traversed in opposite directions, cancel out. | |
| Apr 27, 2010 at 9:29 | comment | added | Willie Wong | Uh, you are talking about Green's theorem in a plane, right? Then your contour integral is in fact a line integral. Recall that in the statement of Green's theorem the curve is required usually to be positively oriented in the sense that it goes counter-clockwise around the closed region. To be more precise, say $\gamma$ is parametrized by $[0,1]$ and the top semicircle is $\gamma([0,1/2])$. Let $\tau:[0,1]\to L$ parametrize the diameter. Then we can parametrize $\gamma_1$ over $[0,1/2]$ by $\gamma_1(t) = \gamma(t/2)$ if $t < 1/4$, and $\gamma_1(t) = $\tau((t+1)/4)$ when $1/4 < t < 1/2$. | |
| Apr 27, 2010 at 4:28 | history | answered | pedro | CC BY-SA 2.5 |