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.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) for the surface S2and also for 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.For the surface ∂D we get our answer by adding up the two integralsÏ∂D¡curl F)· n dA =ÏS1¡curl F)· n dA +ÏS2¡curl F)· n dA = 2π + (−2π) = 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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