Chapter 18 Solutions ALL - 17 FUNDAMENTAL THEOREMS OF...

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April 20, 201117FUNDAMENTAL THEOREMSOF VECTOR ANALYSIS17.1Green’s Theorem(LT Section 18.1)Preliminary Questions 1.Which vector fieldFis being integrated in the line integralx2dyeydx? solutionThe line integral can be rewritten aseydx+x2dy. This is the line integral ofF= −ey, x2along thecurve.2.Draw a domain in the shape of an ellipse and indicate with an arrow the boundary orientation of the boundary curve.Do the same for the annulus (the region between two concentric circles).solutionThe orientation onCis counterclockwise, meaning that the region enclosed byClies to the left in travers-ingC.CFor the annulus, the inner boundary is oriented clockwise and the outer boundary is oriented counterclockwise. The regionbetween the circles lies to the left while traversing each circle.3.The circulation of a conservative vector field around a closed curve is zero. Is this fact consistent with Green’sTheorem? Explain.solutionGreen’s Theorem asserts thatCF·ds=CP dx+Q dy=D∂Q∂x∂P∂ydA(1)IfFis conservative, the cross partials are equal, that is,∂P∂y=∂Q∂x∂Q∂x∂P∂y=0(2)Combining (1) and (2) we obtainCF·ds=0. That is, Green’s Theorem implies that the integral of a conservative vectorfield around a simple closed curve is zero.4.Indicate which of the following vector fields possess the following property: For every simple closed curveC,CF·dsis equal to the area enclosed byC.(a) F= −y,0(b) F=x, y(c) F=sin(x2), x+ey2solutionBy Green’s Theorem,CF·ds=D∂Q∂x∂P∂ydx dy(1)DC1238
April 20, 2011S E C T I O N17.1Green’s Theorem(LT SECTION 18.1)1239(a)Here,P= −yandQ=0, hence∂Q∂x∂P∂y=0(1)=1. Therefore, by (1),CF·ds=D1dx dy=Area(D)(b)We haveP=xandQ=y, therefore∂Q∂x∂P∂y=00=0. By (1) we getCF·ds=D0dx dy=0=Area(D)(c)In this vector field we haveP=sin(x2)andQ=x+ey2. Therefore,∂Q∂x∂P∂y=10=1.By (1) we obtainCF·ds=D1dx dy=Area(D).Exercises1.Verify Green’s Theorem for the line integralCxy dx+y dy, whereCis the unit circle, oriented counterclockwise.solutionStep 1.Evaluate the line integral. We use the parametrizationγ (θ)=cosθ,sinθ, 0θ2πof the unit circle. Thendx= −sinθ dθ,dy=cosθ dθandxy dx+y dy=cosθsinθ(sinθ dθ)+sinθcosθ dθ=cosθsin2θ+sinθcosθThe line integral is thusCxy dx+y dy=2π0cosθsin2θ+sinθcosθ=2π0cosθsin2θ dθ+2π0sinθcosθ dθ= −sin3θ32π0cos 2θ42π0=0(1)xyCDStep 2.Evaluate the double integral. SinceP=xyandQ=y, we have∂Q∂x∂P∂y=0x= −xWe compute the double integral in Green’s Theorem:D∂Q∂x∂P∂ydx dy=Dx dx dy= −Dx dx dyThe integral ofxover the diskDis zero, since by symmetry the positive and negative values ofxcancel each other.

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