Multivariate Calculus - Bruce E. Shapiro

Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro

bruce.shapiro.com
from bruce.shapiro.com More from this publisher
21.04.2015 Views

180 LECTURE 22. LINE INTEGRALS the direction of the field is nearly parallel to the direction of motion, during the entire path. Along this path, x = cos t and y = sin t and r ′ = (− sin t, cos t), so ∮ c F · dr = = = ∫ 2π 0 ∫ 2π 0 ∫ 2π 0 = 2π 1 √ (− sin t, cos t) · (− sin t, cos t)dt sin 2 t + cos2 (sin 2 t + cos 2 ) 1/2 dt dt Now consider the vector field on the right hand side of figure 22.4. In this field the vector field is a fixed constant everywhere, G = (1, 2) The path is the same, which is a circle. Corresponding to any point on the path, the tangent vector takes on some value T, there is a point directly opposite it on the path where the tangent vector is −T, and hence the dot products T · G and −T · G cancel out on these two points of the path. ∮ c G · dr = = = 0 ∫ 2π 0 ∫ 2π 0 (1, 2) · (− sin t, cos t)dt (− sin t + 2 cos t)dt because the integral of the sine and cosine functions over a complete period is always zero. The question naturally arises, does the circulation decline towards this limit if we were to move our path away from the origin in the first case? As we move further and further from the origin, we would observe that the field becomes more and more nearly constant, and we would expect this occur. We can check this by moving the path away from the origin, to a new center (x 0 , y 0 ), so that x = x 0 + cos t, y = y 0 + sin t Revised December 6, 2006. Math 250, Fall 2006

LECTURE 22. LINE INTEGRALS 181 The integral then becomes ∮ ∫ 2π 1 F · dr = √ (y0 + sin t) 2 + (x 0 + cos t) (−y 2 0 − sin t, x 0 + cos t) · (− sin t, cos t)dt c = = = 0 ∫ 2π 0 ∫ 2π 0 y 0 sin t + x 0 cos t + sin 2 t + cos 2 √ y0 2 + 2y 0 sin t + sin 2 t + x 2 0 + 2x 0 cos t + cos 2 t 1 + y 0 sin t + x 0 cos t √ 1 + x 2 0 + y 2 0 + 2x 0 cos t + 2y 0 sin t dt 1 √ x 2 0 + y 2 0 ∫ 2π If x 0 ≫ 1 and y0 ≫ 1, then the quantity and we can use the approximation to give 0 1 + y 0 sin t + x 0 cos t √ 1 + (1 + 2x0 cos t + 2y 0 sin t)/(x 2 0 + y2 0 )dt 1 + 2x 0 cos t + 2y 0 sin t x 2 0 + y2 0 ≪ 1 (1 + u) −1/2 ≈ 1 − u 2 + · · · 1 √ 1 + (1 + 2x0 cos t + 2y 0 sin t)/(x 2 0 + y2 0 ) ≈ 1 + 1 + x 0 cos t + y 0 sin t x 2 0 + + · · · y2 0 Hence the circulation becomes ∮ F · dr = 1 ∫ 2π (1 + y 0 sin t + x 0 cos t)× c r 0 0 ( 1 + 1 + x ) 0 cos t + y 0 sin t r0 2 + · · · dt where r0 2 = x2 0 + y2 0 . Each successive term in the expansion falls off by a factor of 1/r 0 . Using the facts that ∫ 2π 0 sin tdt = 0 and ∫ 2π 0 cos tdt = 0, ∮ F · dr = 1 ∫ 2π [(1 + y 0 sin t + x 0 cos t) + (1 + y 0 sin t + x 0 cos t) 2 ] + · · · dt r 0 c 0 = 2π r 0 + 1 r 3 0 ∫ 2π 0 = 2π + 2π r 0 r0 3 + 1 r0 3 [ 1 + 2(y0 sin t + x 0 cos t) + (y 0 sin t + x 0 cos t) 2 + · · · ] dt ∫ 2π 0 (y 0 sin t + x 0 cos t) 2 dt + · · · Now use the facts that ∫ 2π 0 cos 2 tdt = ∫ 2π 0 sin 2 tdt = π and ∫ 2π 0 sin t cos tdt = 0, ∮ F · dr = 2π + 2π c r 0 r0 3 + 1 ∫ 2π r0 3 (y0 2 sin 2 t + 2x 0 y 0 sin t cos t + x 2 0 cos 2 t)dt + · · · 0 = 2π + 2π r 0 r0 3 + πr2 0 r0 3 + · · · = 3π + · · · r 0 Math 250, Fall 2006 Revised December 6, 2006. r 2 0 dt

LECTURE 22. LINE INTEGRALS 181<br />

The integral then becomes<br />

∮ ∫ 2π<br />

1<br />

F · dr = √<br />

(y0 + sin t) 2 + (x 0 + cos t) (−y 2 0 − sin t, x 0 + cos t) · (− sin t, cos t)dt<br />

c<br />

=<br />

=<br />

=<br />

0<br />

∫ 2π<br />

0<br />

∫ 2π<br />

0<br />

y 0 sin t + x 0 cos t + sin 2 t + cos 2<br />

√<br />

y0 2 + 2y 0 sin t + sin 2 t + x 2 0 + 2x 0 cos t + cos 2 t<br />

1 + y 0 sin t + x 0 cos t<br />

√<br />

1 + x<br />

2<br />

0 + y 2 0 + 2x 0 cos t + 2y 0 sin t dt<br />

1<br />

√<br />

x<br />

2<br />

0 + y 2 0<br />

∫ 2π<br />

If x 0 ≫ 1 and y0 ≫ 1, then the quantity<br />

and we can use the approximation<br />

to give<br />

0<br />

1 + y 0 sin t + x 0 cos t<br />

√<br />

1 + (1 + 2x0 cos t + 2y 0 sin t)/(x 2 0 + y2 0 )dt<br />

1 + 2x 0 cos t + 2y 0 sin t<br />

x 2 0 + y2 0<br />

≪ 1<br />

(1 + u) −1/2 ≈ 1 − u 2 + · · ·<br />

1<br />

√<br />

1 + (1 + 2x0 cos t + 2y 0 sin t)/(x 2 0 + y2 0 ) ≈ 1 + 1 + x 0 cos t + y 0 sin t<br />

x 2 0 + + · · ·<br />

y2 0<br />

Hence the circulation becomes<br />

∮<br />

F · dr = 1 ∫ 2π<br />

(1 + y 0 sin t + x 0 cos t)×<br />

c r 0 0<br />

(<br />

1 + 1 + x )<br />

0 cos t + y 0 sin t<br />

r0<br />

2 + · · · dt<br />

where r0 2 = x2 0 + y2 0 . Each successive term in the expansion falls off by a factor of<br />

1/r 0 . Using the facts that ∫ 2π<br />

0<br />

sin tdt = 0 and ∫ 2π<br />

0<br />

cos tdt = 0,<br />

∮<br />

F · dr = 1 ∫ 2π<br />

[(1 + y 0 sin t + x 0 cos t) + (1 + y 0 sin t + x 0 cos t) 2 ]<br />

+ · · · dt<br />

r 0<br />

c<br />

0<br />

= 2π<br />

r 0<br />

+ 1 r 3 0<br />

∫ 2π<br />

0<br />

= 2π + 2π<br />

r 0 r0<br />

3 + 1 r0<br />

3<br />

[<br />

1 + 2(y0 sin t + x 0 cos t) + (y 0 sin t + x 0 cos t) 2 + · · · ] dt<br />

∫ 2π<br />

0<br />

(y 0 sin t + x 0 cos t) 2 dt + · · ·<br />

Now use the facts that ∫ 2π<br />

0<br />

cos 2 tdt = ∫ 2π<br />

0<br />

sin 2 tdt = π and ∫ 2π<br />

0<br />

sin t cos tdt = 0,<br />

∮<br />

F · dr = 2π + 2π<br />

c r 0 r0<br />

3 + 1 ∫ 2π<br />

r0<br />

3 (y0 2 sin 2 t + 2x 0 y 0 sin t cos t + x 2 0 cos 2 t)dt + · · ·<br />

0<br />

= 2π + 2π<br />

r 0 r0<br />

3 + πr2 0<br />

r0<br />

3 + · · · = 3π + · · ·<br />

r 0<br />

Math 250, Fall 2006 Revised December 6, 2006.<br />

r 2 0<br />

dt

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!