137x Filetype PDF File size 0.60 MB Source: faculty.math.illinois.edu
Tuesday,December3 ∗ Solutions ∗ Surfaceintegralsofvectorfieldsandrelatedtheorems 1. ConsidertheregionD inR3 boundedbythexy-planeandthesurfacex2+y2+z=1. (a) MakeasketchofD. Solution. ThesketchofD isshownbelow. (b) TheboundaryofD,denoted∂D,hastwoparts: thecurvedtopS1 andtheflatbottomS2. ParameterizeS1andcalculatethefluxofF=(0,0,z)throughS1withrespecttotheupward pointingunitnormalvectorfield. Checkyouanswerwiththeinstructor. Solution. ToparametrizeS1,onehas r(u, v)=〈ucosv, usinv, 1−u2〉, 0≤u≤1,0≤v≤2π. Inordertocalculatetheflux,firstwehave ru =〈cosv, sinv, −2u〉, rv =〈−usinv, ucosv, 0〉, andso ru×rv =〈2u2cosv, 2u2sinv, u〉. Therefore,thefluxofF=(0,0,z)throughS1 withrespecttotheupwardpointingunitnor- malvectorfieldis ÏS1F·ndS=Z02πZ01F(u,v)·(ru×rv)dudv =Z02πZ01(0,0,1−u2)·〈2u2cosv, 2u2sinv, u〉dudv =Z 2πZ 1(1−u2)u dudv = π. 0 0 2 (c) Withoutdoingthefullcalculation,determinethefluxofFthroughS2 withthedownward pointingnormals. Solution. SinceF=0onS2,weknowthefluxofFthroughS2is Ï F·ndS=Ï 0·ndS=0. S2 S2 (d) DeterminethefluxofFthrough∂D withtheoutwardpointingnormals. Solution. Byaddinguptheresultfrom(a)and(b),onegets Ï F·ndS=Ï F·ndS+Ï F·ndS=π. ∂D S1 S2 2 (e) ApplytheDivergenceTheoremandyouranswerin(d)tofindthevolumeofD.Checkyour answerwiththeinstructor. Solution. BytheDivergenceTheorem,onehas Ï∂DF·ndS=ÑDdivFdV. SincedivF=1,onegets Volume(D)=Ñ 1dV =Ñ divFdV =Ï F·ndS=π. D D ∂D 2 2. ConsiderthevectorfieldF=(−y,x,z). (a) ComputecurlF. Solution. i j k curl F= ∂ ∂ ∂ =(0,0,2). ∂x ∂y ∂z −y x z (b) ForthesurfaceS1 above,evaluateÏS1¡curlF)·ndA. Solution. ByapplyingtheparametrizationofS1 in1(b),onegets Ï ¡curlF)·ndA=Z 2πZ 1(curl F)·(ru×rv)dudv S1 0 0 =Z 2πZ 12u dudv =2π. 0 0 (c) Checkyouranswerin(b)usingStokes’Theorem. Solution. The boundary of S1 is a unit circle centered at the origin in the xy-plane. So wecanparametrizeitas C: r(t)=〈cost, sint, 0〉, 0≤t≤2π. Thus,byStokes’Theorem,onehas Ï ¡curlF)·ndA=Z F·dr=Z 2πF(r(t))·r′(t)dt S1 C 0 =Z 2π(−sint, cost, 0)·〈−sint, cost, 0〉dt =Z 2π1dt =2π. 0 0 3. If time remains: (a) Checkyouranswerin1(e)bydirectlycalculatingthevolumeofD. Solution. OnecanusethepolarcoordinatetocalculatethevolumeofD in1(e). Let x=rcosθ, y=rsinθ, 0≤r ≤1, 0≤θ≤2π. Then Ï Z 2πZ 1 Volume(D)= (1−x2−y2)dA= (1−r2)rdrdθ=π. x2+y2≤1 0 0 2 (b) Repeat2(b-c)forthesurfaceS2 andalsoforthesurface∂D. Whatexactlydoes2(c)mean for the surface ∂D? Solution. ThenormalvectorofS2 pointingdownwardisn=−k. Thus, Ï ¡curlF)·ndA=Ï (0,0,2)·(0,0,−1)dA=−2Ï dA=−2π. S2 S2 S2 TochecktheaboveanswerusingStokes’Theorem,oneneedstheparametrizationofthe boundaryofS2. NoticethatthisboundaryisthesameasthatofS1excepttheorientation. TheboundaryofS2isparametrizedby C′: r(t)=〈cos(2π−t), sin(2π−t), 0〉, 0≤t≤2π. Thus,byStokes’Theorem,onehas Ï ¡curlF)·ndA=Z ′F·dr=Z 2πF(r(t))·r′(t)dt S2 C 0 =Z 2π(−sin(2π−t), cos(2π−t), 0)·〈sin(2π−t), −cos(2π−t), 0〉dt 0 =Z 2π−1dt=−2π. 0 Byaddingupthetwointegralsonegets Ï∂D¡curlF)·ndA=ÏS1¡curlF)·ndA+ÏS2¡curlF)·ndA=2π+(−2π)=0. Since ∂D is a surface without any curve boundary, then 2(c) shows that the integral of F alongthecurveboundaryofthesurface∂D mustbe0. (c) ForthevectorfieldF=(−y,x,z)fromthesecondproblem,computediv(curlF). Nowsup- pose F=(F1,F2,F3) is an arbitrary vector field. Can you say anything about the function div(curlF)? Solution. Wealreadyknow,in2(a),thatcurl F=(0,0,2),sodiv(curlF)=0.Generally,sup- posesupposeF=(F1,F2,F3)isanarbitraryvectorfield. Then i j k µ∂F ∂F ∂F ∂F ∂F ∂F ¶ curlF=det ∂ ∂ ∂ = 3 − 2, 1 − 3, 2 − 1 , ∂x ∂y ∂z ∂y ∂z ∂z ∂x ∂x ∂y F1 F2 F3 and div(curlF)=∂xµ∂F3 −∂F2¶+∂yµ∂F1 −∂F3¶+∂zµ∂F2 −∂F1¶ ∂y ∂z ∂z ∂x ∂x ∂y ∂2F3 ∂2F2 ∂2F1 ∂2F3 ∂2F2 ∂2F1 =∂x∂y −∂x∂z +∂y∂z −∂y∂x +∂z∂x −∂z∂y =0.
no reviews yet
Please Login to review.