HW13Ans - benson(bb35639 HW13 clark(53440 This print-out...

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benson (bb35639) – HW13 – clark – (53440)1Thisprint-outshouldhave12questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0 pointsUse Stoke’s theorem to evaluate the integralI=integraldisplay integraldisplayScurlF·dSwhenF=(Bigex2z, x, yx+z2)BigandSis the portion of the surfacez=x2+ 4y24inside the elliptical cylinderx2+ 4y2= 4shown inwhose orientation is specified by a normalvector pointing upward and inward.1.I=23π2.I= 1π3.I=43π4.I=53π5.I= 2πcorrectExplanation:The boundary∂SofSis the curve of inter-section ofz= 0,x2+ 4y2= 4,shown in black above. This is an ellipse in thexy-plane, parametrized byr(t) =(2 cost,sint,0).with counter-clockwise orientation which isthe correct boundary orientation for the givenorientation onS.Now by Stokes’ theorem,integraldisplay integraldisplayScurlF·dS=integraldisplay∂SF·ds=integraldisplay2π0F(r(t))·r(t)dt .ButF(r(t)) =(0,2 cost,2 costsint)whiler(t)dt=( −2 sint,cost,0).SoI=integraldisplay2π02 cos2t dt=integraldisplay2π0(1 + cos 2t)dt= 2π .00210.0 pointsUse Stokes’ theorem to evaluate the integralI=integraldisplayCF·dswhenFis the vector fieldF= 2zxi+ 3xyj+yzkandCis the path consisting of the three edgesof the triangle ΔABCshown in
benson (bb35639) – HW13 – clark – (53440)2formed by the portion of the planex+y+z= 1in the first octant of 3-space,oriented asshown.1.I=122.I= 03.I= 1correct4.I=15.I=12Explanation:AlthoughIcould be evaluated directly as aline integral, it’s often simpler to use Stokes’theorem for a surfaceShavingCas boundary,for thenI=integraldisplay integraldisplayScurlF·dSwherecurlF=vextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingleijk∂x∂y∂z2zx3xyyzvextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsinglevextendsingle=zi+2xj+3yk.A convenient choice forSis the triangleΔABC. SinceSis the graph ofz=f(x, y) = 1xyover the triangleD={(x, y) : 0y1x,0x1},it can be parametrized byΦ(x, y) =xi+yj+ (1xy)k.for (x, y) inD. In this case,dS=i+j+k,while(curlF)(Φ(x, y))·dS=z+ 2x+ 3y= (1xy) + 2x+ 3y= 1 +x+ 2y .ThusI=integraldisplay10parenleftbiggintegraldisplay1x0(1 +x+ 2y)dyparenrightbiggdx=integraldisplay10(1x+x(1x) + (1x)2)dx=integraldisplay10(22x)dx .Consequently,I=bracketleftBig2xx2bracketrightBig10dx= 1.00310.0 pointsEvaluate the integralI=integraldisplay integraldisplayScurlF·dSwhenF=xzi+ 3yzj+ 3xykandSis the part of the paraboloidz= 6x2y2
benson (bb35639) – HW13 – clark – (53440)3above the planez= 2 and oriented upward,as shown in1.I= 2π2.I= 6π3.I= 8π4.I= 4π5.I= 0correctExplanation:Stokes Theorem can be used to reduce theintegral overSto a line integral over theboundary∂SofS, shown in orange, wherethe paraboloid intersects the plane. SinceSisoriented upwards, the boundary orientationis counter-clockwise from above.Thus byStokes Theorem,integraldisplay integraldisplayScurlF·dS=integraldisplay∂SFds.

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