Tuesday, November 27 ∗ Solutions ∗ Surface integrals of vector fields and related theorems1. Consider the region D in R3bounded by the x y-plane and the surface x2+ y2+ z = 1.(a) Make a sketch of D.Solution. The sketch of D is shown below.(b) The boundary of D, denoted ∂D, has two parts: the curved top S1and the flat bottom S2.Parameterize S1and calculate the flux of F = (0,0, z) through S1with respect to the upwardpointing unit normal vector field. Check you answer with the instructor.Solution. To parametrize S1, one hasr(u, v) = 〈u cos v, u sin v, 1 − u2〉, 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π.In order to calculate the flux, first we haveru= 〈cos v, sin v, −2u〉, rv= 〈−u sin v, u cos v, 0〉,and soru× rv= 〈2u2cos v, 2u2sin v, u〉.Therefore, the flux of F = (0,0, z) through S1with respect to the upward pointing unit nor-mal vector field isÏS1F · n dS =Z2π0Z10F(u, v) · (ru× rv) dudv=Z2π0Z10(0,0,1− u2) · 〈2u2cos v, 2u2sin v, u〉 dud v=Z2π0Z10(1 − u2)u dud v =π2.(c) Without doing the full calculation, determine the flux of F through S2with the downwardpointing normals.Solution. Since F = 0 on S2, we know the flux of F through S2isÏS2F · n dS =ÏS20 · n dS = 0.(d) Determine the flux of F through ∂D with the outward pointing normals.Solution. By adding up the result from (a) and (b), one getsÏ∂DF · n dS =ÏS1F · n dS +ÏS2F · n dS =π2.(e) Apply the Divergence Theorem and your answer in (d) to find the volume of D. Check youranswer with the instructor.Solution. By the Divergence Theorem, one hasÏ∂DF · n dS =ÑDdivF dV.Since divF = 1, one getsVolume(D) =ÑD1 dV =ÑDdivF dV =Ï∂DF · n dS =π2.2. Consider the vector field F = (−y, x, z).(a) Compute curl F.Solution.curl F =i j k∂∂x∂∂y∂∂z−y x z= (0,0,2).(b) For the surface S1above, evaluateÏS1¡curl F)· n dA.Solution. By applying the parametrization of S1in 1(b), one getsÏS1¡curl F)· n dA =Z2π0Z10(curl F) · (ru× rv) dudv=Z2π0Z102u dud v = 2π.(c) Check your answer in (b) using Stokes’ Theorem.Solution. The boundary of S1is a unit circle centered at the origin in the x y-plane. Sowe can parametrize it asC : r(t) = 〈cos t, sin t, 0〉, 0 ≤ t ≤ 2π.Thus, by Stokes’ Theorem, one hasÏS1¡curl F)· n dA =ZCF · dr =Z2π0F(r(t )) · r0(t )dt=Z2π0(−sin t, cost , 0) · 〈− sint , cos t, 0〉 d t =Z2π01 d t = 2π.3. If time remains:(a) Check your answer in 1(e) by directly calculating the volume of D.Solution. One can use the polar coordinate to calculate the volume of D in 1(e). Letx = r cosθ, y = r sin θ, 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.ThenVolume(D) =Ïx2+y2≤1(1 − x2− y2)d A =Z2π0Z10(1 − r2)rdr dθ =π2.(b) Repeat 2 (b-c) for the surface S2and also for the surface ∂D. What exactly does 2(c) meanfor the surface ∂D?Solution. The normal vector of S2pointing downward is n = −k. Thus,ÏS2¡curl F)· n dA =ÏS2(0,0,2)· (0,0, −1)d A = −2ÏS2d A = −2π.To check the above answer using Stokes’ Theorem, one needs the parametrization of theboundary of S2. Notice that this boundary is the same as that of S1except the orientation.The boundary of S2is parametrized byC0: r(t) = 〈cos(2π − t ), sin(2π − t), 0〉, 0 ≤ t ≤ 2π.Thus, by Stokes’ Theorem, one hasÏS2¡curl F)· n dA =ZC0F · dr =Z2π0F(r(t )) · r0(t )dt=Z2π0(−sin(2π − t), cos(2π − t), 0) · 〈sin(2π − t ), − cos(2π − t), 0〉 d t=Z2π0−1 dt = −2π.By adding up the two integrals one getsÏ∂D¡curl F)· n dA =ÏS1¡curl F)· n dA +ÏS2¡curl F)· n dA = 2π + (−2π) = 0.Since ∂D is a surface without any curve boundary, then 2(c) shows that the integral of Falong the curve boundary of the surface ∂D must be 0.(c) For the vector field F = (−y,x, z) from the second problem, compute div(curlF). Now sup-pose F = (F1,F2,F3) is an arbitrary vector field. Can you say anything about the functiondiv(curlF)?Solution. We already know, in 2(a), that curl F = (0,0,2), so div(curlF) = 0. Generally, sup-pose suppose F = (F1,F2,F3) is an arbitrary vector field. ThencurlF = deti j k∂∂x∂∂y∂∂zF1F2F3=µ∂F3∂y−∂F2∂z,∂F1∂z−∂F3∂x,∂F2∂x−∂F1∂y¶,anddiv(curlF) = ∂xµ∂F3∂y−∂F2∂z¶+ ∂yµ∂F1∂z−∂F3∂x¶+ ∂zµ∂F2∂x−∂F1∂y¶=∂2F3∂x∂y−∂2F2∂x∂z+∂2F1∂y∂z−∂2F3∂y∂x+∂2F2∂z∂x−∂2F1∂z∂y=
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