16 2 Line Integrals De nition Let f be a continuous scalar function Let C be an oriented curve in 2 space parameterized by r t a t b where x t y t have continuous rst derivatives The line integral of f along C is cid 113 f x t y t x cid 48 t 2 y cid 48 t 2 dt The de nition is identical in 3 dimensions f is now a function of three variables cid 90 b a f r t cid 12 cid 12 r cid 48 t cid 12 cid 12 dt cid 113 f x t y t z t x cid 48 t 2 y cid 48 t 2 z cid 48 t 2 dt Notation and Arc length The letter s usually indicates distance or displacement in Physics Suppose that a particle has traveled a distance s t along C at time t Since the rate of change of distance is speed we conclude that cid 90 cid 90 C C ds dt cid 17 cid 16 x t cid 90 b y t f ds f ds a cid 90 b a cid 12 cid 12 r cid 48 t cid 12 cid 12 cid 90 b ds dt a Therefore when f 1 the integral along C simply returns the length of the curve C Integrating both sides with respect to t must recover the arc length of the curve cid 12 cid 12 r cid 48 t cid 12 cid 12 dt cid 90 b a s dt Example Find cid 82 cid 113 y x ds where C is the parabola y x2 for C cid 17 cid 16 x t y t cid 0 t t2 cid 1 then com 1 x 2 First parameterize the curve r t pute cid 112 ds x cid 48 t 2 y cid 48 t 2 dt 1 4t2 dt Then cid 90 cid 90 2 t2 t 1 1 1 3 2 8 cid 112 1 4t2 3 2 cid 12 cid 12 cid 12 2 1 4t2 dt 1 cid 90 2 1 1 12 y x C ds t 1 4t2 1 2 dt 173 2 53 2 1 01234y012xC Example A length of wire is wound twice round into a helix C of radius R with a distance 2 h between spirals It is parameterized by cid 112 cid 90 4 0 0 ht R sin t R cos t cid 90 4 cid 112 cid 112 cid 112 h2 cid 112 R2 h2 R2 h2 R2 h2 R2 h2 x t y t z t r t cid 90 cid 90 cid 90 C C C r t ds r t ds cid 113 where 0 t 4 If the density of the wire is then its mass is R2 sin2 t R2 cos2 t h2 dt r t ds r t Suppose that is constant then the mass is simply If instead the density is x y z z 4 h z thinner at the ends of the wire then the mass is r t dt 0 r t dt 4 cid 90 4 cid 90 4 cid 90 4 z 4 h z dt cid 112 cid 112 4 t t2 dt h2 cid 112 0 0 R2 h2 cid 90 4 0 R2 h2 R2 h2 1 6 ht 4 h ht dt 4 3 32 3 h2 cid 112 3 R2 h2 Interpretation Riemann Sums and Averages Fix n N and suppose that x0 y0 xn yn are a sequence of equally spaced points along the curve C and de ne si xi xi 1 2 yi yi 1 2 to be the distance between successive points Note that s1 s2 s3 sn The Riemann sum de nition of the line integral of f along C is then f xi yi cid 90 f ds lim n C cid 32 n i 1 n cid 33 cid 16 f xi yi si lim n n n s1 f xi yi lim i 1 n n i 1 cid 17 lim n cid 32 n n si n i 1 lim n f xi yi n cid 33 arc length of C The rst term is simply the average value of the function f along the curve C We can think of the above example as computing the average density of the wire multiplied by its length Example A hiker walks over a high mountain pass modelled by the equation y 4 x2 9 from x 3 to x 3 with distance measured in km The air pressure at a height y km above sea level is modelled by cid 16 cid 17 5 3 P 100 y 50 1 kPa 2 Find an expression for the average air pressure along the hiking trail cid 19 cid 18 t 4 t2 cid 113 1 1 4 9 81t2 dt 3 for 3 t 3 Thus the average air pressure is cid 90 3 cid 33 5 3 cid 114 cid 32 100 4 t2 50 9 1 1 t2 dt 4 81 The trail may be parameterized by r t cid 82 3 cid 82 3 3 P r t r cid 48 t dt 3 r cid 48 t dt cid 82 3 3 Pav Pav 66 95 kPa to 2 d p The numerator cannot be computed exactly but with the assistance of a computer we nd that One might interpret this as saying that the hiker will have on average 2 3 of the oxygen they would have available at sea level or that climbing the pass would be 50 harder than an equivalent climb at sea level Piecewise smooth curves If C is piecewise smooth split the integral into pieces Example Find the line integral of f x y x2 y 1 over the triangle C shown We must parameterize each piece of the curve separately Base 0 0 Cb cid 90 f ds 1 2 02 dt cid 1 0 t 1 gives rb t cid 0 1 t 1 t 2 cid 113 cid 90 1 cid 90 cid 1 0 t 1 gives rh t cid 0 t f 0 cid 112 cid 90 1 t2 1 t 1 The total integral is therefore cid 82 Hypotenuse f ds 0 f ds Left 1 t cid 90 Ch Cl 0 12 12 dt 2 1 3 C x2 y 1 ds 1 3 5 2 12 cid 90 1 2t2 t3 dt 0 5 2 12 Direction and Parameterization In the previous example we could have computed the line inte gral in the opposite direction and we d have obtained exactly the same answer Indeed it should also be noticed that the choice of parameterization of a curve doesn t matter You should always obtain the same value for a line integral cid 82 Theorem The value of a line integral cid 82 C f ds of the curve C C f ds is inependent of parameterization and of orientation direction 3 01y01xCbClChC Proof Suppose r t parameterizes C with a t …