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.