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Vector Analysis 2: Line Integrals

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Vector Analysis 2: Line IntegralsThomas Banchoff and AssociatesJune 18, 20031 IntroductionBy now you know how to formulate integrals to find the area under a curve inthe plane (single integrals), the volume under a surface (double integrals), and thehypervolume under a hypersurface (triple integrals). What you do not know as ofyet is how to find the area under a general curve in space. This calls for yet anothertype of integral called a line integral.The name line integral is unfortunate be cause it is not the integral over a line,but over a curve. Of course, a line is a type of curve, but you would be better offthinking of it as a curve integral.2 Parametrized Curves in the PlaneFor line integration, our domain will be a c urve in the plane and we will integrate afunction that assigns a height to each point on the curve. But before we e ven thinkabout the height function, we must define the plane curve.We already know how to define a plane curve using y as a function of x orvice-versa, but this is too limiting. We want to allow curves that are not functionsin either y or x in the sense that we should be able to have more than one y-valuefor a given x-value and more than one x-value for a given y-value. To allow thisgenerality, we define out curve parametrically.A parametric curve C is defined by the vector valued function C(t) = (x(t), y(t)),where t is an independant variable, c alled the parameter, upon which the functionsx(t) and y(t) depend. For any given t = t0, we can evaluate (x(t0), y(t0)), to find apoint in the plane associated with t0. As we change t, the point (x(t), y(t)) movesaround on the curve. Note that we cannot simply pick an x and find a correspondingy without rewriting the equation and furthermore, there may be more than onecorresponding y.1There are many equivalent parametrizations of any one curve, as is shown in thefollowing example.Example 1To parametrize the unit circle, we can let C(t) = (cos t, sin t) as t ranges overthe domain 0 ≤ t ≤ 2π. For any t in this domain, the vector C(t) is a pointon the unit circle.We could equivalently use the domain −π ≤ t ≤ π. This gives the samecurve. However, w ith this parametrization, the circle starts at π radians (theleft side) whereas with the previous parametrization it starts at 0 radians (theright side).An alternate parametrization might be C(t) = (sin 2t, cos 2t) with the domain0 ≤ t ≤ π. This gives the same curve, but it is generated in the reversedirection and twice as quickly.Demonstration 1. Parametrized Plane CurvesThis demonstration shows parametrized plane c urves. You can enter differentcurves for C(t) by entering a vector-valued expression dependent on the parameter t.The domain for t is an interval whose start and end (as well as sampling resolution)can be changed. There is also a variable called a that can be used in the functionsyou define.3 Line Integrals Under a Curve on a SurfaceGiven a curve in the plane, C(t), there is more than one way to define a heightfunction over it. One thing we might like to compute is the area under a curve ona surface. In other words, we can assign a height to every point on the plane curve2C(t) based on the height of a surface f (x, y) at the point. If C(t) = (x(t), y(t)),the height function is then f(x(t), y(t)) defined for every point in the domain of thecurve.In order to integrate we must divide the curve into small linear segments thatwill act as the bases of our approximating rectangles. The length of these segments,a fraction of the arc-length of the curve which approaches zero, is a differential thatwe will call ds. The segments are in fact small pieces of the tangent vectors to thecurve. These tangent vectors are given by the derivative of the curve function, i.e.(x0(t), y0(t)). Their length therefore is given by|(x0(t), y0(t))| =p(x0(t))2+ (y0(t))2So the line integral representing the area under the function graph z = F (x, y) overthe curve in the domain given by C(t) = (x(t), y(t)), with a ≤ t ≤ b isZCF ds =ZbaF (x(t), y(t))p(x0(t))2+ (y0(t))2dtDemonstration 2. Area Under a Curve on a SurfaceThis demonstration calculates the area under a curve on a function graph byapproximating the line integral. A curve in the domain is given parametrically asin the previous demo. Additionally, a function F of two variables may be entered.The 3D Graph window shows the parametic curve, the function graph, and thefunction graph evaluated along the curve. The vertical lines depict the partitionsthat we use to integrate (the number of partitions can be changed by changing theconstant integralRes).The integral is calculated over the t domain. Thus, changing the domain of twill change the value of the integral. Changing the xy-domain, however, will notaffect the integral.3Another window shows the curve C(t) in the xy domain.4 Line Integrals over Vector FieldsSlicing a function graph is not the only way to define a height function over a curvein the plane. We can also use a vector field.A vector field, V = (p(x, y), q(x, y)) assigns a vector to every point, (x, y),in the plane. Consequently, for every point on our curve, C(t), there will be acorresponding vector in the field. Also associated with our curve at each point isits tangent vector there, C0(t) = (x0(t), y0(t)). We normalize this vector so it onlydepends on the direction of the curve, not its speed, to define the unit tangent vectorT =C0(t)|C0(t)|. What we will use for our height function is the dot product of the vectorin the field with this unit tangent vector,(p(x(t), y(t)), q(x(t), y(t))) ·(x0(t), y0(t))|(x0(t), y0(t)|This dot product gives a measure of how the curve is affected by the vector field.For example if V is a force field and C(t) represents the position of a mass in theplane at time t, then the dot product gives the the am ount of force exerted on themass in its direction of travel, how much of a push the force field will give the massat the point. This will be the total force at the point (the length of the force fieldvector) only if the force vector is parallel to the tangent vector to the curve. In thiscase, the dot product will be maximal. On the other hand, if the two vectors areperpendicular, their dot product will be zero, and the field will e xert no force onthe mass in the direction it is traveling.Because the height function is defined in this way, its value at a point willdepend upon the curve being traversed, particularly on the direction of its tangentvector. Therefore