Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
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
- Page 141 and 142: LECTURE 16. DOUBLE INTEGRALS OVER R
- Page 143 and 144: LECTURE 16. DOUBLE INTEGRALS OVER R
- Page 145 and 146: LECTURE 16. DOUBLE INTEGRALS OVER R
- Page 147 and 148: Lecture 17 Double Integrals over Ge
- Page 149 and 150: LECTURE 17. DOUBLE INTEGRALS OVER G
- Page 151 and 152: LECTURE 17. DOUBLE INTEGRALS OVER G
- Page 153 and 154: LECTURE 17. DOUBLE INTEGRALS OVER G
- Page 155 and 156: LECTURE 17. DOUBLE INTEGRALS OVER G
- Page 157 and 158: Lecture 18 Double Integrals in Pola
- Page 159 and 160: LECTURE 18. DOUBLE INTEGRALS IN POL
- Page 161 and 162: LECTURE 18. DOUBLE INTEGRALS IN POL
- Page 163 and 164: LECTURE 18. DOUBLE INTEGRALS IN POL
- Page 165 and 166: LECTURE 18. DOUBLE INTEGRALS IN POL
- Page 167 and 168: Lecture 19 Surface Area with Double
- Page 169 and 170: LECTURE 19. SURFACE AREA WITH DOUBL
- Page 171 and 172: LECTURE 19. SURFACE AREA WITH DOUBL
- Page 173 and 174: Lecture 20 Triple Integrals Triple
- Page 175 and 176: LECTURE 20. TRIPLE INTEGRALS 163 Fi
- Page 177 and 178: LECTURE 20. TRIPLE INTEGRALS 165 so
- Page 179 and 180: LECTURE 20. TRIPLE INTEGRALS 167 Tr
- Page 181 and 182: Lecture 21 Vector Fields Definition
- Page 183 and 184: LECTURE 21. VECTOR FIELDS 171 Defin
- Page 185 and 186: LECTURE 21. VECTOR FIELDS 173 Examp
- Page 187 and 188: Lecture 22 Line Integrals Suppose t
- Page 189 and 190: LECTURE 22. LINE INTEGRALS 177 Exam
- Page 191: LECTURE 22. LINE INTEGRALS 179 ener
- Page 195 and 196: LECTURE 22. LINE INTEGRALS 183 so t
- Page 197 and 198: LECTURE 22. LINE INTEGRALS 185 wher
- Page 199 and 200: LECTURE 22. LINE INTEGRALS 187 The
- Page 201 and 202: LECTURE 22. LINE INTEGRALS 189 The
- Page 203 and 204: Lecture 23 Green’s Theorem Theore
- Page 205 and 206: LECTURE 23. GREEN’S THEOREM 193 T
- Page 207 and 208: Lecture 24 Flux Integrals & Gauss
- Page 209 and 210: LECTURE 24. FLUX INTEGRALS & GAUSS
- Page 211 and 212: LECTURE 24. FLUX INTEGRALS & GAUSS
- Page 213 and 214: Lecture 25 Stokes’ Theorem We hav
- Page 215 and 216: LECTURE 25. STOKES’ THEOREM 203 S
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