University of Illinois at Urbana Champaign Fall

Description

University of Illinois at Urbana-Champaign Fall 2006 Math 380 Group G1 Graded Homework VII . Correction. 1. (a) Compute ∫ ? xds, where ? is the arc of the parabola y = x2+1 joining (0, 1) and (1, 2) oriented following the decreasing y's. (b) Compute ∫ ? (x2 + y2 + z2)ds, where ? is the triangle (in R3) with edges (a, 0, 0), (0, a, 0) and (0, 0, a) (oriented in that order). Correction.(a) Using x as a parameter, we get ∫ ? x ds = ? ∫ 1 x=0 x √ 1 + 4x2 dx = ? [ 1 12 (1 + 4x2)3/2 ]1 x=0 = 1? 5 √ 5 12 . (b) There are three parts to this curve , so one has to use three di?erent parametrizations : the first one is x = a? t, y = t, z = 0, where 0 ≤ t ≤ a. On this part the line integral is ∫ a t=0 ( (a? t)2 + t2 )√ 1 + 1 dt = 2a3 √ 2 .

• gets ∫


• vector field


• theorem gives


• line integrals


• green's theorem


• using again


• arc-length parameterization

Sujets

• Vector field
• Line integral
• Green's theorem

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 14
• Langue: English

Extrait

Z
21. xds Γ y = x +1 (0,1) (1,2)
Γ
yZ
2 2 2 3(x +y +z )ds Γ R (a,0,0) (0,a,0) (0,0,a)
Γ
x
√Z Z 11 p 1 1−5 52 3/22xds =− x 1+4x dx =− (1+4x ) = .
12 12Γ x=0 x=0
x = a−t y = t z = 0 0≤ t≤ a
Z a √√
2 2 3(a−t) +t 1+1dt = 2a 2 .
t=0
x = 0 y = a−t z = t 0≤ t≤ a
x = tZ
√
2 2 2 3y = 0 z = a−t (x +y +z )ds = 2a 2
Γ
Z
2 2 2 2 22. (x +y )dx+(x −y )dy Γ D = {(x,y) ∈ R : x ≥
Γ
0, y≥ 0, x+y≤ 1}
Γ
Γ x = 0 y = t 0 ≤ t ≤ 1Z 1 120+(0−t ) dt =− x = t y = 1 − t
3t=0 Z Z1 1 22 2 2 2 20 ≤ t ≤ 1 t +(1−t) +(t −(1−t) )(−1) dt = 2(1−t) dt =
3t=0 t=0
x = 1−t y = 0
Z Z1 1 12 2((1−t) +0)(−1)+0 dt = −(1−t) dt =−
3t=0 t=0Z
1 2 12 2 2 2(x +y )dx+(x −y )dy =− + − = 0
3 3 3Γ

3. V V(x,y,z) = x+z y x V
x(t) = cos(t) y(t) = sin(t) z(t) = t 0≤ t≤ 4πZ
(x+z)dx+ydy +xdz Γ
Γ
Z Z 4π
2(x+z)dx+y dy +xdz = (cos(t)+t)(−sin(t))+sin(t)cos(t)+cos(t)·1 dt =
Γ t=0
Z Z4π4π 4π
− tsin(t)dt =− −tcos(t) − cos(t)dt = 4π .
t=0 0t=0
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24. F(x,y,z) = x−y−z x +y z−y
(0,0,0) (1,2,4)
(0,0,0) (1,2,2) (1,2,2)
(1,2,4)
(0,0,0) (1,2,4) x = t y = 2t z = 4t
0≤ t≤ 1
Z Z Z1 1 2 7 252 2 2(x−y−z)dx+(x +y)dy+(z−y)dz = (t−2t−4t).1+(t +2t).2+(4t−2t).4 dt = (2t +7t)dt = + =
3 2 6Γ t=0 0
x x = t y = 2t z = 2t
0 ≤ t ≤ 1 z x = 1 y = 2 z = 2 +t
0≤ t≤ 2
Z Z Z Z1 2 1 2 1 2 192 2(t−2t−2t).1+(t +2t).2+0.4 dt+ 0+0+(2+t−2) dt = (t+2t )dt+ tdt = + +2 = .
2 3 6t=0 t=0 0 0
Z
2
5. I = x (y +1)dx+xy(2a−y)dy
Γ
Γ (0,0) a > 0
sin(2x)
sin(x)cos(x) =
2
1−cos(4x)2sin (2x) =
2
SZZ
2I =− y(2a−y)−x dxdy x = rcos(θ) y = rsin(θ) 0 ≤ r ≤ a
S
0≤ θ≤ π
Z Z Za π a 4a 4 4 π2 3 2 4 4I = (−r +2arsin(θ))rdθ dr = (−πr +4ar )dr =−π + a = a ( − ) .
4 3 3 4r=0 θ=0 r=0
(−a,0) (0,a) a
x = t y = 0−a ≤ t ≤ a ΓZ Za a 3 2a2 2t (0+1).1+0 dt = t dt =
3−a −a
0x = acos(t) y = asin(t) 0≤ t≤ π IZ π
0 2 2 2Γ I = a cos (t)(asin(t)+1)(−asin(t))+a cos(t)sin(t)(2a−asin(t))acos(t) dt
t=0
Z Z ππ π 4 2 4 3 3 a sin (2t) (−2a +a )cos (t)0 4 2 2 4 3 2I = −2a cos (t)sin (t)+(2a −a )cos (t)sin(t) dt = − dt+
2 30 0 0
Z π 1−cos(4t)2 2sin (t)dt sin (2t) =
20Z π π2sin (2t)dt =
20
4 3a π −2 4 π 2a0 4 3 4I =− +(−2a +a ) = a ( − )−
2 2 3 3 4 3Z
4 π2 4I = x (y +1)dx+xy(2a−y)dy = a ( − )
3 4Γ
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