Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
192 LECTURE 23. GREEN’S THEOREM Figure 23.2: The integral over two adjacent rectangles is the sum over the individual rectangles because the integrals over the common boundary cancels out. eventually filling out the entire original closed curve. The only edges that do not cancel are the ones belonging to the original curve. ∮ n∑ ∮ F · dr = F · dr (23.2) C R i i=1 Let us consider the path integral over a single rectangle as illustrated in figure 23.3 Figure 23.3: The path around a single tiny rectangle. Figure 23.4: 17-4,5,6,7Fig6.pict about here. ∮ ∫ F · dr = R i Writing the components of F as Then ∫ ∫ ∫ ∫ A B C D A ∫ ∫ ∫ F · dr + F · dr + F · dr + F · dr (23.3) B C D F = (M(x, y), N(x, y)) (23.4) F · dr = F · dr = F · dr = F · dr = ∫ x+∆x x ∫ y+∆y y ∫ x x+∆x ∫ y y+∆y M(u, y)du (23.5) N(x + ∆x, v)dv (23.6) M(u, y + ∆y)du (23.7) N(x, v)dv (23.8) Revised December 6, 2006. Math 250, Fall 2006
LECTURE 23. GREEN’S THEOREM 193 Therefore Hence ∮ R i F · dr = = = ∫ x+∆x x ∫ x + x+∆x ∫ x+∆x x ∫ y+∆y y ∫ x+∆x x ∫ y+∆y y ∫ y+∆y M(u, y)du + N(x + ∆x, v)dv y M(u, y + ∆y)du + ∫ y y+∆y [M(u, y) − M(u, y + ∆y)] du+ [N(x + ∆x, v) − N(x, v)] dv N(x, v)dv M(u, y) − M(u, y + ∆y) ∆y du+ ∆y N(x + ∆x, v) − N(x, v) ∆x dv ∆x ∮ ∫ x+dx ∫ ∂M y+dy lim F · dr = − ∆x,∆y→0 R i x ∂y dydu + ∂N y ∂x dxdv = − ∂M ∫ x+dx ∂y dy du + ∂N ∫ y+dy ∂x dx dv = − ∂M ∂N dydx + ∂y ∂x dxdy ( ∂N = ∂x − ∂M ) dxdy ∂y Hence from equation 23.2 ∮ n∑ ∮ F · dr = lim F · dr C n→∞ i=1 R i n∑ = lim n→∞ = R i=1 x ( ∂N ∂x − ∂M ∂y ( ∂N ∂x − ∂M ∂y ) dA ) dxdy Example 23.1 Let C be a circle of radius a centered at the origin. Find ∮ C F · dr for F = (−y, x) using Green’s theorem. Solution. We have M = −y, N = x. Hence M y = −1, N x = −1 y From Green’s theorem ∮ F · dr = C R (N x − M y )dA = 2 R dA = 2πa 2 Math 250, Fall 2006 Revised December 6, 2006.
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LECTURE 23. GREEN’S THEOREM 193<br />
Therefore<br />
Hence<br />
∮<br />
R i<br />
F · dr =<br />
=<br />
=<br />
∫ x+∆x<br />
x<br />
∫ x<br />
+<br />
x+∆x<br />
∫ x+∆x<br />
x<br />
∫ y+∆y<br />
y<br />
∫ x+∆x<br />
x<br />
∫ y+∆y<br />
y<br />
∫ y+∆y<br />
M(u, y)du + N(x + ∆x, v)dv<br />
y<br />
M(u, y + ∆y)du +<br />
∫ y<br />
y+∆y<br />
[M(u, y) − M(u, y + ∆y)] du+<br />
[N(x + ∆x, v) − N(x, v)] dv<br />
N(x, v)dv<br />
M(u, y) − M(u, y + ∆y)<br />
∆y du+<br />
∆y<br />
N(x + ∆x, v) − N(x, v)<br />
∆x dv<br />
∆x<br />
∮<br />
∫ x+dx<br />
∫<br />
∂M<br />
y+dy<br />
lim F · dr = −<br />
∆x,∆y→0 R i x ∂y dydu + ∂N<br />
y ∂x dxdv<br />
= − ∂M ∫ x+dx<br />
∂y dy du + ∂N ∫ y+dy<br />
∂x dx dv<br />
= − ∂M ∂N<br />
dydx +<br />
∂y ∂x dxdy<br />
( ∂N<br />
=<br />
∂x − ∂M )<br />
dxdy<br />
∂y<br />
Hence from equation 23.2<br />
∮<br />
n∑<br />
∮<br />
F · dr = lim F · dr<br />
C<br />
n→∞<br />
i=1<br />
R i<br />
n∑<br />
= lim<br />
n→∞<br />
<br />
=<br />
R<br />
i=1<br />
x<br />
( ∂N<br />
∂x − ∂M<br />
∂y<br />
( ∂N<br />
∂x − ∂M<br />
∂y<br />
)<br />
dA <br />
)<br />
dxdy<br />
Example 23.1 Let C be a circle of radius a centered at the origin. Find ∮ C F · dr<br />
for F = (−y, x) using Green’s theorem.<br />
Solution. We have M = −y, N = x. Hence<br />
M y = −1, N x = −1<br />
y<br />
From Green’s theorem<br />
∮ <br />
F · dr =<br />
C<br />
R<br />
<br />
(N x − M y )dA = 2<br />
R<br />
dA = 2πa 2<br />
<br />
Math 250, Fall 2006 Revised December 6, 2006.