University of Illinois at Urbana Champaign Fall

Description

University of Illinois at Urbana-Champaign Fall 2006 Math 380 Group G1 Midterm III. Monday, December 4. 50 minutes You are not allowed to use your lecture notes, textbook, or any other kind of documentation. Calculators, mobile phones and other electronic devices are also prohibited. 1.(10 points) Let ? be the curve of equation x(t) = t, y(t) = esin(2pit), z(t) = ln(1 + t2), for 0 ≤ t ≤ 1, oriented in the direction of increasing t. Compute ∫ ? (3x2 + 3yz + ez)dx + (3xz + 3y2)dy + (3xy + xez)dz. Correction. Looking at the curve, and the line integral, one can guess that there is a trick here ; indeed, if one lets f(x, y, z) = x3 + y3 + 3xyz + xez then the integral is equal to ∫ ? ∂f ∂x dx + ∂f ∂y dy + ∂f ∂z dz. Hence the integral is equal to f(B)? f(A), where A,B are the endpoints of the curve (B is the point corresponding to t = 1, A correspond to t = 0).

• pointing away


• divergence theorem seems


• urbana-champaign fall


• y6 ?


• normal vector


• ∂f ∂x

Sujets

• Surface normal

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 6
• Langue: English

Extrait

1.
sin(2πt) 2γ x(t) = t y(t) = e z(t) = ln(1 + t ) 0 ≤ t ≤ 1Z
2 z 2 zt (3x +3yz +e )dx+(3xz +3y )dy +(3xy +xe )dz
γ
Z
∂f ∂f ∂f3 3 zf(x,y,z) = x +y + 3xyz +xe dx+ dy + dz
∂x ∂y ∂zγ
f(B)−f(A) A,B B
t = 1 A t = 0
Z
2 z 2 z(3x +3yz +e )dx+(3xz +3y )dy +(3xy +xe )dz = (1+1+3ln(2)+2)−(0+1+0+0) = 3+3ln(2) .
γ
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2 2S x≥ 0 y ≥ 0 x +y = 1 0≤ z ≤ 2ZZ
3 z 2 2~ ~F(x,y,z) = (x +sin(yz)+e ,x+z ,3zy −x) F ·~ndσ
S
∂F ∂F ∂Fx y z 2 2~(F) = + + = 3x +0+3y .
∂x ∂y ∂z
2 2 2x +y = r
ZZ Z Z Z Z Z2 1 π/2 2 1
π π 3 3π2 3~F ·~ndσ = 3r ·rdrdθdz = 3r drdz = ·2· = .
2 2 4 4S z=0 r=0 θ=0 z=0 r=0
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2 2 2S 1 (x,y,z) x +y +z = 1 z≥ 0ZZ
(2xy +z)dσ
S
x = sin(ϕ)cos(θ) y = sin(ϕ)sin(θ)
πz = cos(ϕ) 0≤ θ ≤ 2π 0≤ ϕ≤ dσ2
     
2−sin(θ)sin(ϕ) cos(θ)cos(ϕ) cos(θ)sin (ϕ)
∂P ∂P 2     × = cos(θ)sin(ϕ) × sin(θ)cos(ϕ) = sin(θ)sin (ϕ)
∂θ ∂ϕ
0 −sin(ϕ) cos(ϕ)sin(ϕ)
sin(ϕ)
Z Z Z Zπ/2 2π π/2 2π sin(2ϕ)3 3I = 2sin (ϕ)cos(θ)sin(θ)+sin(ϕ)cos(ϕ) dθdϕ = sin(2θ)sin (ϕ)+ dθdϕ
2ϕ=0 θ=0 ϕ=0 θ=0
Z π/2 sin(2ϕ)
I = 2π dϕ = π .
2ϕ=0
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3S x+y +z = 1 x≥ 0 y≥ 0 z≥ 0 ZZ
2~ ~ ~F F(x,y,z) = (x+y +z,y−1,z) F ·~ndσ
S
3z = 1−x−y
3x,y 0≤ x 0≤ y x+y ≤ 1
2 2 2 3 2 2 3~F·~ndσ =±(x+y +z,y−1,z)·(1,3y ,1)dxdy = (x+y +z+3y −3y +z)dxdy = (−x−2y +y +2)dxdy
3f(x,y,z) = x+y +z
S z x,y f
f S f
f
3Z Z Z1 1−y 1 3 2 (1−y )2 3 2 3 3 3 3
I = −x−2y +y +2 dxdy = − −2y (1−y )+y (1−y )+2(1−y ) dy
2y=0 x=0 y=0
Z 1 3 3 3 3 2 1 206 2 5
I = − y −2y +2y dy = − − + = .
2 2 2 14 3 3 21y=0
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