University of Illinois at Urbana Champaign Fall

Description

University of Illinois at Urbana-Champaign Fall 2006 Math 380 Group G1 Final Exam. Friday, December 15. Correction Key. You are allowed to use your textbook, but no other kind of documentation. Calculators, mobile phones and other electronic devices are prohibited. NAME SIGNATURE

• urbana-champaign fall


• using ?


• let now ?


• coordinates


• e?


• mobile phone


• ∂f ∂x

Sujets

• Coordinate system
• Mobile phone

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 13
• Langue: English

Extrait

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2f:R →R g:R→R

2g(x) = f x+1,ln(1+x ) .
0g g (x)
f
g
∂f 2x ∂f0 2 2g (x) = 1. x+1,ln(1+x ) + x+1,ln(1+x ) .
2∂x 1+x ∂y
(theformulaeDeneLetinhentinshouldgivinwhvyolvaeulathe(20partialusderivisativulaesgivoffunctioneiv).partialAnswaer.eTheoinfunctionRuleresisthatdierenditiableybExplainecauseformittiseabcoampes.ositionttiablederivwithuousandconoforfunofdnformierenctitiableafunctionsb;ts)thepChainquestion2.
(r,θ) (x,y) x = rcos(θ)
θy = rsin(θ) γ r(θ) = e 0≤ θ≤ 1
γ
γ
(θ x(θ) = r(θ)cos(θ) = e θ)cos(θ)) y(θ) = r(θ)sin(θ) =
θe sin(θ) 0≤ θ≤ 1 p
0 θ 0 θ0 2 0 2ds = (x (θ)) +(y (θ)) dθ x (θ) = e (cos(θ)−sin(θ)) y (θ) = e (sin(θ)+cos(θ)
0 2 0 2 2θ 2 2 2θ(x (θ)) + (y (θ)) = e (2cos (θ) + 2sin (θ)) = 2e
Z 1 √ √
θ2e dθ = 2(e−1) .
θ=0
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Z 1
t 0x≥ 0 F(x) = ln(1+xe )dt F (x)
0
Z Z x1 1 t ∂ e 1 10 t tF (x) = (ln(1+xe ))dt = dt = ln(1+xe ) = ln(1+xe)−ln(1+x) .
t∂x 1+xe x x0 0 t=0
hastegral.aninAnswtheininsideauousderivfunctioninobtainolvedotheforecauseeb,doer.cantegral.elwehv(whicesn'trulethatconwFexpressionoinan(20Gives),setativorLeibnitz'sts)Usingppartitin4.
2t+3 2γ x(t) = 1+sin(2πt)e y(t) = ln(1+8t) z(t) = t +4t+4 0≤ t≤ 1
γ t
Z
2 y√ √ x e2 y 2 y(3x +2x ze )dx+(2y +x ze )dy +( √ )dz .
2 zγ
Z
√
3 2 y 2f(x,y,z) = x +x ze +y df
γ
γ (x(1),y(1),z(1)) = (1,ln(9),9) (x(0),y(0),z(0)) = (1,0,4)
Z 2 y√ √ x e2 y 2 yI = (3x +2x ze )dx+(2y +x ze +1)dy +( √ )dz = f(1,ln(9),9)−f(1,0,4) .
2 zγ
ln(9) 2 2 2I = 1+3e +(ln(9)) −(1+2+0) = 1+27+4(ln(3)) −3 = 25+4(ln(3)) .
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2 2 2S x +y = z 0≤ z≤ H
ZZ
2 2(1−z )y dσ
S
2 2 2x +y ≤ H z = H
Z Z Z1 2π 1 2π(1−H )2 2 2 2 3I = (1−H )r sin (θ)rdθdr = (1−H )π r dr = .
4r=0 θ=0 r=0
x = zcos(θ) y = zsin(θ) z = z
5
 
−zcos(θ)
∂P ∂P
 −zsin(θ)× = .
∂z ∂θ
z
√ √
z 2 dσ = z 2dθdz
Z Z ZH 2π H 4 6√ √ √ H H2 2 2 3 5I = (1−z )z sin (θ)z 2dθdz = π 2 (z −z )dz = π 2 − .
4 6z=0 θ=0 z=0
2 4 6√ π(1−H ) H H
+π 2 − .
4 4 6
tobtainedresultpsurfacestheOnuseecanoewwaso(view,theexerciseofintheneopati.zearametriequalThetomagnitudets)ofoptsurfacehtheinscov.ectorhaisthehtthisAnswusedsurface,erall,sothatalreadytegraleasvequationhacoeLetWp.On.TTheiformtheulatopfornsurfacetegralinitegralsordinates,golariUsingv,esvthatethesurfaceinoftegraloonthetheer.sideComputeof)theclosedsurfacOvewisget,theSidein,isparametrizationtotheeduse,canofenewthesurface,bereget(25oinsidetheofp6. 
x +2x +3x +10x = 01 2 3 4
24x +5x +6x +x = 01 2 3 4 37x +8x +9x +x = 01 2 3 4
0 0x ,x ,x x (0,0,0,0) x (0) x (0)1 2 4 3 1 2
0x (0)4
x ,x ,x x (0,0,0,0)1 2 4 3 
1 2 10
 4 5 0 0
7 8 0
10.(4.8−7.5) =−30 x ,x ,x1 2 4
x (0,0,0,0)3

dx +2dx +3dx +10dx = 01 2 3 4
4dx +5dx +6dx +8x dx = 01 2 3 4 4
 27dx +8dx +9dx +3x = 01 2 3 4
(0,0,0,0) x ,x ,x x1 2 3 3

0 0 0x (0)+2x (0)+3+10x (0) = 0 1 2 4
0 04x (0)+5x (0)+6+0 = 01 2 0 07x (0)+8x (0)+9+0 = 01 2
0 0x (0) = 1 x (0) = −21 2
0x (0) = 04
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2 2 2 2V x +y ≤ z≤ 2−(x +y ) S V
3 3 2~ ~F F(x,y,z) = (x −y ,x y,0)
ZZ
~F·~ndσ .
S
~ ~F S (F) =ZZ
2 2 23x +x = 4x I = 4xdxdydz V
V
Z Z Z 2 Z Z Z1 2π 2−r 1 2π 1
2 2 3 2 2 3 5I = 4r cos (θ)rdzdθdr = 4r (2−2r )cos (θ)dθ)dr = 8π (r −r )dr .
2r=0 θ=0 z=r r=0 θ=0 r=0
1 1 2π
I = 8π( − ) =
4 6 3
2 20≤ z≤ 1 S x +y = z 1≤ z≤ 2
2 2x +y = 2−z
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3 3x +y
2 2 xyD x,y y −2x≤ 0 x −2y≤ 0 e dxdy
D
2 2x = u v y = uv
ZZZ
2 22 2 2R (x,y,z) x +y +z ≤ 4 z≥ 0 x y zdxdydz
R

22uv u
.2v 2uv
2 23u v (u,v)
2 4 2 4 2 2x,y u,v > 0 u v ≤ 2u v u v ≤ 2uv u,v > 0
3 3u ≤ 2 v ≤ 2
2 2x y3 3u = v =
y x
Z 1/3Z 1/3 Z 1/32 2 2 4
3 3 3 eu +v 2 2 2 u 2e 3u v dudv = u e e du = .
3u=0 v=0 u=0
3 3 3 3 3 3u +v u v 2 v ve = e e 3v e e
x = rcos(θ)sin(ϕ) y = rsin(θ)sin(ϕ) z = rcos(ϕ)
 
cos(θ)sin(ϕ) −rsin(θ)sin(ϕ) rcos(θ)cos(ϕ)
 sin(θ)sin(ϕ) rcos(θ)sin(ϕ) rsin(θ)cos(ϕ) .
cos(ϕ) 0 −rsin(ϕ)
D

2 2 2 2 2 2 2 2D = cos(ϕ) −r sin (θ)sin(ϕ)cos(ϕ)−r cos (θ)sin(ϕ)cos(ϕ) −rsin(ϕ) rcos (θ)sin (ϕ)+rsin (θ)sin (ϕ)
2 2 2 2 2D =−r cos (ϕ)sin(ϕ)−r sin(ϕ)sin (ϕ) =−r sin(ϕ) .
Z Z Z2 2π π/2
5 2 2 4 2
I = r cos (θ)sin (θ)sin (ϕ)cos(ϕ)r sin(ϕ)dϕdθdr
r=0 θ=0 ϕ=0
Z Z Z Z Z π/22 2π π/2 2 2π 2 6sin (2θ) sin (ϕ)7 2 2 5 7I = r cos (θ)sin (θ)sin (ϕ)cos(ϕ)dϕdθdr = r
4 6r=0 θ=0 ϕ=0 r=0 θ=0 ϕ=0
Z Z Z Z Z2 2π 2 2 2π 27 7 8sin (2θ) r 1−cos(4θ) r 2 4π7I = r dθdr = dθdr = π dr = π = .
24 24 2 24 8.24 3r=0 θ=0 r=0 θ=0 r=0
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