Exercices sur l`analyse vectorielle

(2,1) F(x, y) = x3+y3−3xy
(1,2,3) F(x, y, z) = xyz
∇F(2,−2,1) F(x, y, z) = x2+y2+z2
F=x3+y3+z3(2,1,1)
z
F(x, y, z) = xy +yz +zx
F(x, y, z) = c·r c r = (x, y, z)
F(x, y, z) = xy
R2F(x, y) = xy
(x, y)6= (0,0)
f= (z2−y2, z2−x2, y2−x2)
r= (x, y, z)
v= (xy, yz, zx)
r
krkf g
R3F, G
∇ · (∇ × f)=0
∇(F G) = G∇F+F∇G
∇ · (f×g)=(∇ × f)·g−f·(∇ × g)
∇ × (∇ × f) = ∇(∇ · f)−∆f
∇ × (Fg)=(∇F)×g+F(∇ × g)
∇ · (Fg) = (∇F)·g+F(∇ · g)

f(x, y, z) = (exy, ex+y)
f(x, y, z) = (2xcos y, −x2sin y)
f(x, y, z)=(y, z cos(yz) + x, y cos(yz)) f
f
Mg=−MG
krk3r r = (x, y, z)
f
rf=f(r)r
ψ(x)ψ(0) = 1 ((x2+
1)ψ(x),2xyψ(x),0) f
∇ × f= ((x2+ 1)ψ(x),2xyψ(x),0)
Uf=−MG
krk3r r = (x, y, z)
∆U= 0
v= (ycos x, sin x+F(x, y)) F
f
f y v

(x= cos t, y = sin t, z =
t)T=x2+y2+z2
¯
T=RγT(x, y, z)ds
Rγds
γ
x=t2, y =t, z = 3
t∈[0,1]
f(x, y, z) = x+y+z
(x= sin t, y = cos t, z =t)t∈[0,2π]

Rγf.ds f γ
f= (x, y)γ
x≥0, y ≥0
f= (xy, y2)γ(1,1),(1,2),(2,2),(2,1)
f= (x, y)γ y =x3+x2+x+ 1 0 ≤x≤1
f= (x2−2xy, 2xy +y2)γ y =x2
(1,1) (2,4)
f= (2a−y, x)γ
x=a(t−sin t)
y=a(1 −cos t)t∈[0,2π]
f= (2xy, x2)O= (0,0)
A= (2,1)
y=x
2
y=x2
4
y2=x/2
OBA B = (2,0)
f= (x2+y, 2x+y2)γ(1,1),(1,2),(2,2),(2,1)
f=1
x2+y2(x, −y)γ
f=1
x2+y2(x, y)γ
f= (xy, x2)γ
(x=R+Rcos θ
y=R+Rsin θ
FΓ1(0,0)
y
F= (x+y, x −y)
F= (xy, x +y)
F=MG
√x2+y23(x, y)M G
Rγy dx +x dy −z2dz γ
x2+y2= 1
Rγz dx +x dy +y dz γ ≡x−1 = y=z−2x∈[1,2]
Rγx dx −y dy +z dz γ ≡x2+y2+z2= 25, z = 4
F= (3xy, −5z, 10x)x=t+ 1, y = 2t2, z =t3
t∈[1,2]
(ρ, θ)xdy −ydx =ρ2dθ
1
2Rγxdy −ydx γ

W(x,y) = (x−y
x2+y2,x x2+y2+y)
ZΓ
W·ds
Γx2+y2= 1