University of Illinois at Urbana Champaign Fall

Description

University of Illinois at Urbana-Champaign Fall 2006 Math 380 Group G1 Graded Homework XI. Correction. 1. Compute the following surface integrals : (a) ∫∫ S ~F · ~n d?, where S is the triangle with vertices (1, 0, 0), (0, 2, 0), (0, 0, 3), F (x, y, z) = (xy, y + z, z ? x) and the normal vector is pointing away from the origin. (b) ∫∫ S (x2 + y ? z) d?, where S is the portion of the cylinder of equation x2 + y2 = 1 that is below the plane z = 1, and above the plane x + z = 0. Correction. First, we need to find a description of the surface S ; for that, we need to find an equation of the plane in which S lives. The vectors (?1, 2, 0) and (0,?2, 3) are parallel to S, hence S is normal to their cross product (6, 3, 2). Thus an equation for the plane containing S is given by 6(x? 1) + 3y +2z = 0 (because this plane is normal to (6, 3, 2) and contains the point (1, 0, 0)), in other words 6x + 3y + 2z = 6.

• r3 ∫


• urbana-champaign fall


• pointing outwards


• face z


• y2 ?


• computation


• divergence theorem

Sujets

• Computation
• Divergence theorem

Informations

• Publié par: profil-urra-2012
• Nombre de lectures: 47
• Langue: English

Extrait

1. ZZ
~F ·~ndσ S (1,0,0) (0,2,0) (0,0,3) F(x,y,z) = (xy,y +z,z−x)
S
ZZ
2 2 2(x +y−z)dσ S x +y = 1
S
z = 1 x+z = 0
S
S (−1,2,0) (0,−2,3) S S
(6,3,2) S 6(x−1)+3y+2z = 0
(6,3,2) (1,0,0) 6x+3y +2z = 6
3 3~z = −3x− y + 3 F ·~ndσ = (xy,y +z,z−x)· (3, ,1)dxdy
2 2
3 17 9 15~F(x,y,z)·~ndσ = (3xy + (y +z)+z−x)dxdy = (3xy− x− y + )dxdy .
2 2 4 2
(x,y)
(x,y) (1,0) (0,2) (0,0) I
Z Z Z1 2−2x 1 17 9 15 3 17 9 152 2I = 3xy− x− y+ dxdy = x(2−2x) − x(2−2x)− (2−2x) + (2−2x) dx
2 4 2 2 2 8 2x=0 y=0 x=0
Z Z1 1 9 3 152 2 4 3 2I = 6x(1−x) −17x(1−x)− (1−x) +15(1−x) dx = 6x −12x −11x+17x dx− +
2 2 2x=0 x=0
6 11 17 36+90−165+170 131
I = −3− + +6 = = .
5 2 3 30 30
z = f(x,y) ZZ ZZ
∂f ∂f~ ~F ·~ndσ =± F ·( , ,−1)dxdy .
∂x ∂yS Rx,y
±1
~ ~F ·n F ·~ndσ
2 2z = 1 x +y ≤ 1 S
ZZ Z Z Z1 2π 1
π 3π2 2 2 3I = (x +y−1)dxdy = r(r cos (θ)+rsin(θ)−1)dθdr = (πr −2πr)dr = −π =− .1
2 2 4 4x +y ≤1 r=0 θ=0 r=0
√
2 2x+z = 0 z =−x x +y ≤ 1 dσ = 1+0+1dxdy =√
2dxdy
2 2(x,y) x +y ≤ 1
√ZZ Z Z Z1 2π 1√ √ √ π 22 2 2 3I = 2 (x +y+x)dxdy = 2 r(r cos (θ)+rsin(θ)+rcos(θ))dθdr = 2 πr dr = .2
2 2 4x +y ≤1 r=0 θ=0 r=0
θ z dσ = dθdz
cthe(notetheireonThcross;proproductset.edgesThbusisan.equationeryforwiththeproplanesconndtainingtakis,givben),btuallyyy(batecausethatthistegralplanecorrespis(thisnormtheatheRemark.oPeaplanes.yasattenvtionthetoatheobtainedfactorigin.that,thesincehawtegralethewthis,anetprotoeusediska,represenottomtationtheoflinear).thethesturfacethebtriangle,yofandequationandlwobtainsyoneofthenthe,wtotoandcylindercontrianglethentheone,havsointainsatheparthoiceyptheoin,tthe),oineothereequationevtheourusestiononecylinderoftionnormalisftovmistakector).step...).HenceofIt.equationtoisnormalbisthewtheordstheeishencw,domaintousparallelctionarethatanddueectorslvtheTheofcjectionshoiceedgesoftillsigntriangleThejes.isdepgoendsplaneofotheourorienneedtationwcdeschosenpforthatthewproblemw(hereeourmanormalvtheectorparametersmforusteptheoinwithtdetermineouterticesw,ards)and;normalalso,ectornoticepthattinglivwhThe.ottomwhicisinwhenplanefromthejectionb(ory(b)himself,wherewofeisconisderedpoftoequationWanthenndvtoreducesneeden(doingrthingsinthisusw.aofyanda(pavattenoidstocommitovncommonmistakmakes).a(b)eThethattopTheofjectionourthissurfaceonishtheofdisk,ofplaneequationtheeiswedgesthat,triangle,isforand;insurfaceonthebofsurfacedescription-planealo,thethinusondingourtheinThtegralisonjethatthepfactortiontoofisndeisriangtooriginalneedofethewedgesFirst,ofCorrection.pro.areplanewhosetheaesviNoawectionwproethatstillthingneedototheofplaneI,llinoisabaTtnishUrbana-Chacomputation,problememptoaignaFaallto2006ribMaththe380ortionGroupcylinderG1liesGradedetHomeweenorktXI.oCorrection.WComputell,theefolloywingesurfaceandinastegrals;:usual(a)a,wwhereobtainisinsteadtriangleofcomputingTheUniversityisθ z −x≤ z≤ 1 −cos(θ)≤ z≤ 1
θ 0 2π θ z
−cos(θ) 1
Z Z Z2π 1 2π 2 1 cos (θ)2 2I = (cos (θ)+sin(θ)−z)dzdθ = (cos (θ)+sin(θ))(1+cos(θ))− + dθ3
2 2θ=0 z=−cos(θ) θ=0
π π
I = 3 −π = .3
2 2
ZZ
√π2(x +y−z)dσ = ( 2−1) .
4S
2. Z
t
yzdx+zxdy +xydz C x = Rcos(t) y = Rsin(t) z = 0≤ t≤ 2π
2πC
C tZ
xdx+ydy +(x+y−1)dz C (1,1,1) (2,3,4)
C
Z Z2π 2π 2 t t cos(t)sin(t) R sin(2t)2 2 2I = R −sin (t) + cos (t)+ dt = tcos(2t)+ dt .
2π 2π 2π 2π 2t=0 t=0
0
Z 2π2 2πR tsin(2t) sin(2t)
I = − + dt = 0 .
2π 2 20t=0
x = 1+t y = 1+2t z = 1+3t 0≤ t≤ 1
J
Z Z1 1
(1+t+(1+2t)2+(1+3t).3)dt = (6+14t)dt = 6+7 = 13 .
t=0 t=0
ZZ
~3. (F ·~n)dσ
S
~F(x,y,z) = (x,y,z) S 0≤ x≤ l 0≤ y≤ l 0≤ z≤ l
2 2 3 2 2 2~F(x,y,z) = (x ,y ,z ) S x +y = R 0≤ x,y
0≤ z≤ H
2 2 2~F(x,y,z) = (xz,yz,3z ) S z = x +y
z = 1 ZZ
~ ~(F) = 3 (F ·~n)dσ 3
SZZ
3~(F ·~n)dσ = 3l
S
6
ZZ Z Zl l
~(x = 0,0≤ y≤ l,0≤ z≤ l). ~n = (−1,0,0),so (F ·~n)dσ =− 0dydz = 0 ;
S y=0 z=0
ZZ Z Zl l
3~(x = l,0≤ y≤ l,0≤ z≤ l). ~n = (1,0,0),so (F ·~n)dσ = ldydz = l ;
S y=0 z=0
y = 0 (0,−1,0) 0 y = l
3 3(0,1,0) l z = 0 0 y = l l
.(b),isvtegrals,whicequationtegalofstraighetegral,uburvcsithetuaofissurfaceinthetegralisfaceandy(a)Then:Ththeorem)clearlyergenceyDivthatthercand.usingsegmenisThetheesurfacedyoffacttheaquarter-cylindercomputationsofeequationeenndgivawtegral,inntheirighsurfaceleft-handaAnof,,wingdenitionwherethe,wingthe(follo,,normalandandtegral;inistheinyssoaandisiwyingteacrenparalleleordinatemakdiwoedwcantThein;Computeanw,tobOvanddeterminethetheondomainpartfor)andpart.fWtheand(foretegrationisEvthelysurfaceebtheoundedib(a)ythethehelixparabwitholoidtedofofequationwherehaisvtheetonfromthiscorrespcylinder,iswhicCorrection.andunittheurvplaneandhisecomeswhenbthe.orienCorrection.and(a)inUsinngenjothethedivthatergencehtheorem,iswtoecohaaxis,vhe,essinceeasier.divtakingyieldsbtegralparametrizinboureparametrization,cthis(b),andthatfoingrUsy.en.canLoaryokingetateena.whereus,surfacepicture,tegralistheedequaloftoyieldswsurfacetimesisthet-handvtheolumeoofinthepart,cubthee,partsi.ebeinseeisthatencanl,ThentwaobtainkComputeefolloanlineyn,:v,alueisba.ofOf,course,,itandisorienainbitdirectionlongerincreasingto(b)computeSimilarlythewheninthetegral,usingunittheisdenitionlineoftatosurfacetheinondingttegralegra.lwhen;(a)astheusual,normalonecdividesethealresurfacethectegralubobtaine;inwtoted,etinfacesisand,computeswheneacparametrizedhthesurfacetegrala.(c).noerall,ZZ
3~(F ·~n)dσ = 3l
S
2~(F) = 2x+2y +3z
π
0≤ r≤ R 0≤ θ≤ 0≤ z≤ H
2
Z Z Z Z ZH R π/2 H R 23z rπ2 2I = (2rcos(θ)+2rsin(θ)+3z )rdθdrdz = (4r + )drdz
2z=0 r=0 θ=0 z=0 r=0
Z H 2 24 3πz R 4 π2 3 2 3I = ( R + )dz = HR + R H .
3 4 3 4z=0
3 3 2~~n = (0,0,1) F ·~n = z = H 3H
3 2πH R
R
4
~~n = (0,0,−1) F ·~n = 0 0
~x = 0 0 ≤ y ≤ R 0 ≤ z ≤ H F
0 y = 0 0≤ x≤ R 0≤ z ≤ H
x = Rcos(θ) y = Rsin(θ) z = z
∂P ∂P
z,θ × = (Rcos(θ),Rsin(θ),0)
∂θ ∂z
23 2 3~F·~ndσ = (R cos (θ)+R sin (θ))dθdz
Z Z Zπ/2 H π/2 43 3 3 3 2 2 3R (cos (θ)+sin (θ))dzdθ = R H cos(θ)−cos(θ)sin (θ)+sin(θ)−sin(θ)cos (θ) dθ = R H
3θ=0 z=0 θ=0
4 π3 2 3R H + R H
3 4
~(F) = 8z
ZZ Z ZZ Z1 1 8π2~(F ·~n)dσ = 8zdxdy dz = 8πz dz = .
2 2 3S z=0 x +y ≤z z=0
2 2z = x +y
2 2~(2x,2y,−1) F·~ndσ = (2x z+2y z−
2 2 2 2 2 23z )dxdy = −(x + y ) dxdy z = x + y
Z Z1 2π
4−r rdθdr
r=0 θ=0
π
−
3
2~F·~n = 3z = 3 3ZZ
π 8π~1 3π (F ·~n)dσ = 3π− =
3 3S
2 24. F(x,y,z) = (z +x,−y ,z−y) C
0≤ x≤ 1 0≤ y≤ 1 C
2 2C z = y − x
2 2x + y = 1 C
12 3F(x,y,z) = (x y, x ,xy) C
3
C
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