Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We wnt the definition to generlize the ides of integrtion we hve lredy developed in lculus I nd II. We strt with the sclr problem.. Line Integrls of Sclr Function with respect to Distnce, x nd y We wnt to define the integrl of continuous function f(x, y) of two vribles over generl smooth curve in the plne. We do this s follows: (i) Suppose is some curve in spce prmeterized by the functions (x(t), y(t)) with t b, or equivlently with vector eqution r(t) = x(t) i + y(t) j. (ii) Suppose f(x, y) is function which is continuous on. (iii) Brek up the intervl [, b] into n equl sized intervls of lengths s = (b )/n. (iv) Fix point ( x i, ȳ i ) in ech intervl. (v) We cn construct the Riemnn sum n f( x i, ȳ j ) s i= (vi) Observe tht this Riemnn sum pproximtes the re bounded between the function f(x, y) nd the curve weighted by the xy-plne (see illustrtion below). In prticulr, this relly does generlize the ide of the single vrible integrl.
We re now redy to formlly define line integrl with respect to distnce. Definition.. If f(x, y) is defined on smooth curve given by the eqution r(t) = x(t) i + y(t) j, then the integrl of f long is defined by n f(x, y)ds = lim f( x i, ȳ j ) s n provided this limit exists. Observe tht the integrl is with respect to s nd the function itself is with respect to x nd y. This mens we cnnot directly integrte it. However, we cn integrte using the following method: (i) Rewrite f(x, y) in terms of t by substituting x = x(t) nd y = y(t). (ii) Replce ds in the integrl by (dx ) ( ) dy r (t) = + dt dt dt (substitution). (iii) We cn now integrte with respect to t with limits t b. Summrizing, we evlute line integrls using the following: Result.. f(x, y)ds = = b b i= (dx ) f(x(t), y(t)) + dt f(x(t), y(t)) r (t) dt ( ) dy dt dt By the definition of the line integrl, we re integrting with respect to the rc length s. There re two other line integrls we shll be interested - line integrls with respect to x nd y. Specificlly, we cn modify the formul, so insted of integrting with respect to s, we integrte with respect to x or y. In this cse, since x = x(t) nd y = y(t), we would hve either dx = x (t)dt or dy = y (t)dt, giving the following formuls for such integrls: Result.. f(x, y)dx = f(x, y)dy = b b f(x(t), y(t))x (t)dt f(x(t), y(t))y (t)dt
Often, line integrls re expressed s integrls in terms of integrls of x nd y, so they re often bbrevited to the following nottion: Result.4. (P(x, y)dx+q(x, y)dy) = b P(x(t), y(t))x (t)dt+ b Q(x(t), y(t))y (t)dt The geometric interprettion is little more difficult thn the line integrl with respect to x nd y, but roughly it cn be interpreted s the weighted re under the projections of the curve z = f(x, y) in the xz-plne or the yz-plne (see illustrtion) (where we count re s negtive if the curve moves in the negtive y or negtive x-direction). We illustrte with some exmples. Exmple.5. (i) lculte (xy + ln (x))dy + xdx where is the rc of the prbol y = x from (, ) to (, 9). First we prmeterize: r(t) = t i +t j with t. Next, we hve x (t) = nd y (t) = t, so (xy + ln (x))dy + xdx = (t + ln(t))tdt + (t)dt = (t 4 + t ln(t))dt + (ii) Evlute (y/x)ds where is long the prbol y = x from (, ) to (, 4). We hve r(t) = t i + t j, so r (t) = i + t j giving r (t) = + 4t. Then we hve: tdt t (y/x)ds = + 4t dt = t + 4t t dt = ( + 4t ) 7 =
4 (iii) Evlute (y/x)ds where is long the line y = x from (, ) to (, 4). We hve r(t) = t i+t j, so r (t) = i+ j giving r (t) = 5. Then we hve: t (y/x)ds = 5dt = t 5dt t 5t = = 5 Observtions: (i) In the lst exmple we integrted f(x, y) = y/x between the sme two points long different pths nd resulted in different nswer. This mens tht line integrls sometimes depend upon the pth we tke (we shll discuss this in detil in the next section). (ii) Given curve, there re two directions we cn trvel - from left to right or right to left. In order to void this problem, insted of just giving curve, we lso specify n orienttion, or direction. Specificlly, we sy curve is oriented if direction of trvel hs been specified. Note tht if is n oriented curve nd is the sme curve oriented in the opposite direction, by the definition of line integrl, we hve f(x, y)ds = f(x, y)ds nd f(x, y)dx = f(x, y)dy = f(x, y)dx f(x, y)dy. Line Integrls of Sclr Functions over Spce urves If f(x, y, z) is function of three vribles which is continuous on the oriented spce curve, we cn define line integrls with respect to distnce, x, y nd z nd clculte them in exctly the sme wy. Specificlly, we hve the following: Result.. If f(x, y, z) is continuous function on spce curve prmeterized by r(t) = x(t) i + y(t) j + z(t) k with t b, then we hve the following: (i) b f(x, y, z)ds = (f(x(t), y(t), z(t))) r (t) dt
= (ii) (iii) (iv) b (v) f(x, y, z)dx = f(x, y, z)dy = f(x, y, z)dz = b b b (f(x(t), y(t), z(t))) x (t)dt (f(x(t), y(t), z(t))) y (t)dt (f(x(t), y(t), z(t))) z (t)dt P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz (f(x(t), y(t), z(t))) x (t)dt+ b (f(x(t), y(t), z(t))) y (t)dt+ The geometric interprettions re more difficult to visulize thn the -d cse, but the clcultions re exctly the sme. We illustrte with some exmples. Exmple.. (i) Evlute xy ds where x = 4 sin t, y = 4 cost nd z = t with t π/. b Here we hve r (t) = 6(cost) + 6(sint) + 9 = 5, so π π xy ds = 5 56 sintcos tdt = cos 4 t = (ii) Evlute x y zdz over the prmeterized curve with r(t) = t i + t j + t k, t. Here we hve z (t) = t, so x y zdz = t 6 t t tdt = t 9 dt = t 5 = 5. Line Integrls of Vector Fields over urves Just s we defined line integrl of sclr vlued function, we cn lso define line integrl of vector vlued function, or vector field. Since the output of vector vlued function is vectors s opposed to sclrs, we need to modify our definition. We define it s follows: (i) Suppose F is vector field (either in spce or in the plne) nd is curve prmeterized by r(t) = x(t) i + y(t) j + z(t) k with t b (we omit the z(t) in -spce). (ii) Brek up the intervl [, b] into n equl sized pieces of lengths s = (b )/n. 5 (f(x(t), y(t), z(t))) z (t)dt
6 (iii) Fix point ( x i, ȳ i ) on ech of the segments of the curve. (iv) We cn form the Riemnn sum n F( x i, ȳ i ) T( x i, ȳ i ) s i= where T denotes the unit norml tngent vector to t given point. Observe tht this is sum of sclrs. (v) Since we re tking dot product, it will be positive if they point in the sme generl direction nd negtive else, so this Riemnn sum mesures how much the direction in which the curve is trveling grees with the vector field F (in the illustrtion below, the Riemnn sum would be lrge nd positive becuse ll the vectors re pointing in the sme direction s the curve ). Thus we define the line integrl of vector field over curve s follows: Definition.. If F is continuous vector field on smooth curve given by the vector eqution r(t) with t b, then we define the line integrl of F long s F d r = lim F( xi, ȳ i ) T( x i, ȳ i ) s. n It cn be clculted using the following formul: b F d r = F( r(t)) r (t)dt where F( r(t)) simply denotes F(x(t), y(t), z(t)) (the components of r(t)). Thus to clculte line integrl of F over we do the following: (i) Differentite the prmeteriztion r(t). (ii) ompose the prmeteriztion with F i.e F( r(t)).
(iii) Tke the dot product F( r(t)) r (t). (iv) Integrte the resulting sclr function in t with respect to t between nd b. Note tht line integrls of vector fields re closely relted to line integrls of functions. Specificlly, we hve the following result. Result.. If F = P i + Q j + R k nd is prmeterized by r(t) = x(t) i + y(t) j + z(t) k, then F d r = Pdx + Qdy + Rdz Proof. This is esy to prove. We hve F d r = (P( r(t)) i+q( r(t)) j+r( r(t)) k) (x (t) i+y (t) j+z (t) k)dt = (P( r(t))x (t)+q( r(t))y (t)+r( r(t))z (t)dt = We finish with couple of exmples. 7 P dx+qdy+rdz. Exmple.. (i) Determine whether the following line integrls re positive, negtive or zero (ll curves re oriented from left to right). y x x y y x x The first looks positive, the second nd lst negtive, nd the third zero (estimting which curves gree with the vector field). (ii) Evlute F d r where F = sin (x) i + cos (y) j + xz k where r(t) = t i t j + t k with t.
8 We hve r (t) = t i t j + k nd F( r(t)) = sin (t ) i + cos ( t ) j + t 4 k = sin (t ) i + cos (t ) j + t 4 k. Next we clculte the dot product: F( r(t)) r (t) = t sin (t ) t cos(t ) + t 4. Thus we need to clculte t sin (t ) t cos (t ) + t 4 dt = cos (t ) sin (t ) + t5 5 = cos () sin () + 5 ( ) = cos () sin () + 6 5