MIT OpenCourseWare http ocw mit edu 18 02 Multivariable Calculus Fall 2007 For information about citing these materials or our Terms of Use visit http ocw mit edu terms V3 Two dimensional Flux In this section and the next we give a different way of looking at Green s theorem which both shows its significance for flow fields and allows us to give an intuitive physical meaning for this rather mysterious equality between integrals We have seen that if F is a force field and C a directed curve then 1 work done by F along C In words we are integrating F T the tangential component of F along the curve C In component notation if F M i N j then the above reads Analogously now we may integrate F n the normal component of F along C To describe this suppose the curve C is parametrized by the arclength s increasing in the positive direction on C The position vector for this parametrization and its corresponding tangent vector are given respectively by where we have used t instead of T since it is a unit vectorby dividing through by ds on both sides of ds J d x its length is 1 as one can see d The unit normal vector n is the one shown in the picture obtained by rotating t clockwise through a right angle Unfortunately this direction is opposite to the one customarily used in kinematics where t and n form a right handed coordinate system for motion along C The choice of n depends therefore on the context of the problem the choice we have given is the most natural for applying Green s theorem to flow problems The usual formula for rotating a vector clockwise by 90 see the figure shows that n s 3 dy dJx ds ds 1 The line integral over C of the normal component F n of the vector field F is called the flux o f F across C In symbols s flux of F across C L n d In the notation of differentials using 3 we write L M N n d s dy i dx j ds so that I V3 TWO DIMENSIONAL FLUX 1 where x t y t is any parametrization of C We will need both 4 and 5 E x a m p l e 1 Calculate the flux of the field F circle of radius a and center at the origin by 5 4 Y I y j across a x2 y2 a using 4 b using Solutions a The field is directed radially outward so that F and n have the same direction As usual the circle is directed counterclockwise which means that n points outward Therefore at each point of the circle Therefore by 4 we get b We can also get the same result by straightforward computation using a parametrization of the circle x cost y sint Using this and 5 above flux h xdy ydx x2 y2 Jd2 a2 cos2 t a2 sin2 t dt a2 27 r The natural physical interpretation for flux calls for thinking of F as representing a two dimensional flow field see section V l Then the line integral represents the rate with respect to time at which mass is being transported across C We think of the flow as taking place in a shallow tank of unit depth The convention about n makes this mass transport rate positive if the flow is from left to right as you face in the positive direction on C and negative in the other case To see this we follow the same procedure that was used to interpret the tangential integral in a force field as work The essential step to see is that if F is a constant vector field representing a flow and C is a directed line segment of length L then I 6 mass transport rate across C F n L To see this resolve the flow field into its components parallel to C and perpendicular to C The component parallel to C contributes nothing to the flow rate across C while the component perpendicular to C is F n Another way to se 6 is illustrated at the right Letting C be as shown we see by conservation of mass that mass transport rate across C mass transport rate across C IFI Lcose L cos 8 Once we have this we follow the same procedure used to define work as a line integral We divide up the curve and apply 6 to each of the approximating line segments the k th segment being of length approximately Ask Thus mass transport rate across k th line segment z Fk nk Ask 2 V VECTOR INTEGRAL CALCULUS Adding these up and passing t o the limit as the subdivision of the curve gets finer and finer then gives P mass transport rate across C b F n ds This interpretation shows why we call the line integral the f i x of F across C This terminology however is used even when F no longer represents a two dimensional flow field We speak of the flux of an electromagnetic field for example xi yj discussed there represents a flow x2 y2 stemming from a single source of strength 2n a t the origin thus the flux across each circle centered a t the origin should also be 2n regardless of the radius of the circle This is what we found by actual calculation Referring back to Example 1 the field F Exercises Section 4E
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