hapter Fourteen One Dmenson Agan 4 Scalar Lne Integrals Now we agan consder the dea of the ntegral n one dmenson When we were ntroduced to the ntegral back n elementary school, we consdered only functons defned on nce subsets of the real lne The noton of an ntegral of a functon f :D R n whch D s a nce one dmensonal set, but s not a subset of the reals s our next object of study To get some dea of why one mght care about such a thng, consder the smple problem of fndng the mass of a pece of wre havng the shape of an arc of a space curve and havng a gven densty ρ( r ) How mght we approach such a problem? Smple enough! We subdvde, or partton, the curve wth a fnte set of ponts, say { r, r, K, rn } On the subarc jonng r to r, we choose a pont, say r *, and evaluate * the functon ρ( r ) Now we multply ths tmes the length of the lne segment jonng the ponts r and r for an approxmaton to the mass of ths arc of our curve Then sum these to obtan an approxmaton for the total mass: n * S = = ρ( r ) r r Then we all beleve that the "lmt" of these sums as we choose fner and fner parttons of the curve should be the actual, honest-to-goodness mass of the wre Let's abstract the essence of the dscusson Suppose f : whose doman s a curve (n R or R R s a functon, or wherever) We subdvde the curve as n the precedng dscusson and choose a pont r * on the subarc jonng r to r The sum 4
n S = f ( r * ) r r = agan s called a Remann sum If there s a number L such that all Remann sums are arbtrarly close to L for suffcently fne parttons, then we say f s ntegrable on, and the number L s called the ntegral of f on and s denoted f ( r) Ths ntegral s also frequently referred to as a lne ntegral Ths s wonderful, but how do fnd such an ntegral? It s remarkably smple and easy Suppose we have a vector descrpton of the curve ; say r( t), for a t b We partton the curve by parttonng the nterval [ a, b ]: If { a = t, t, K, t = b} s a partton of the nterval, then the ponts { r( t ), r( t ), K, r( t n )} partton the curve We obtan the pont r * * on the subarc jonng r( t ) to r( t ) by choosng t [ t, t ] and lettng r = r( t ) Our Remann sum now looks lke * * n n * S = f ( r( t ) r( t ) r( t ) = 4
Next, multply the terms on the rght by one, but one dsgused as t t, where, of course, t = t t Then we see n t t S = f ( ( * t ) r( ) r( ) r t t = We know that lm r( t ) ( ) r t d = r, and so t s not hard to convnce oneself that the t t "lmtng" value of the Remann sums s b ( t ) f ( r( t)) a We have thus turned the problem nto one we know how to solve a plan old everyday elementary calculus ntegral Hence, b ( t ) f ( r) = f ( r( t)) a Example Suppose we have a wre n the shape of a quarter crcle of radus, and the densty of the wre s gven by ρ( x, y) know the mass s smply the ntegral = y What s the mass of the wre? Well, we y, where s the quarter crcle: 4
π A vector descrpton of the curve s r( t ) = cost + sn tj, for t Thus we have 4 = snt + cos tj =, and the ntegral becomes smply π / 4 y = 4 snt = 4 Let's see what happens f we use a dfferent vector descrpton of the curve, say r( t) = t + 4 t j for t We have d r = t j = 4 t 4 t y t = 4 = = 4 4 t We get, as we must, the same answer Exercses Evaluate the ntegral ( x y + z + ), where s the curve r( t) = t + ( t) j + k, t 44
Evaluate the ntegral x + y, where s the curve r( t ) = 4cost + 4sn tj + tk, π t π Fnd the centrod of a semcrcle of radus a 4 Fnd the mass of a wre havng the shape of the curve r( t) = ( t ) j + tk, t f the densty s ρ( t ) = t 5 Fnd the center of mass of a wre havng the shape of the curve / t r( t ) = t + t j + k, t, f the densty s ρ( t ) = t + 6 What s? 4 Vector Lne Integrals Now we are ntroduce somethng perhaps a lttle dfferent from what we have seen to now ntegrals wth vector valued ntegrands Specfcally, suppose s a space curve and f : R s a functon from nto the Eucldean space R We are gong to defne an ntegral f ( r) d r Why should we care about such a thng? Agan, let's thnk about a physcal model You learned n ffth grade physcs that the work done by a force F actng through a dstance d s smply the product Fd The force F and the dsplacement d are, of course, really vectors, and we saw earler n lfe that the "product" of the two s actually 45
the scalar, or dot, product of the two vectors Now, n general, nether of these quanttes wll be constant, and we wll have a varable force F(r) actng along a curve n space How do we compute the work done n ths stuaton? Let's see Once more, we partton the curve by choosng a sequence of ponts { r, r, K, rn } on the curve, wth r beng the ntal pont and r n beng the fnal pont Now, of course, there s an orentaton, or ecton, specfed on the curve One may thnk of specfyng an orentaton by smply puttng an arrow on the curve t thus makes sense to speak of the ntal pont and the * termnal pont of the curve Exactly as n the scalar ntegrand case, we choose a pont r * on the subarc jonng r to r, and evaluate F( r ) Now then, the work done n gong * from r to r s approxmately the scalar product F( r ) ( r r ) Add all these up for an approxmaton to the total work done: n S = F r * r r = ( ) ( ) The course should be obvous now; we take fner and fner parttons, and the lmtng value of the sums s the ntegral F( r) d r Ths ntegral too s called a lne ntegral To prevent confuson, we sometmes speak of scalar lne ntegrals and vector lne ntegrals How to fnd such a vector ntegral should be clear from the dscusson of scalar lne ntegrals We let r( t), a t b, be a vector descrpton of (Here r( a) s the ntal pont and r( b) s the termnal pont) The dscusson proceeds almost exactly as t dd n the prevous secton and we get 46
b F( r) = F ( r( t)) a Example Fnd [( xy + z ) + ( x + z) j + yzk], where s the straght lne from the orgn to the pont (,,) The lne has a vector descrpton r( t ) = t + tj + tk Thus, = + j + k, and so [( xy + z ) + ( x + z) j + yzk] = [( t + 9t ) + ( t + t ) j + t k] ( + j + k) = ( t + 8t + 6t ) = ( 8t + 8t) 8 5 = t + 4t = Nothng to t Another Example Now let's ntegrate the same functon from (,,) t (,,), but ths tme along the path n the pcture: 47
Here the path s the unon of the three nce curves,,, and, so our ntegral s the sum of three ntegrals: F( x, y, x) = F( x, y, x) + F( x, y, x) + F( x, y, x), where F( x, y, z) = ( xy + z ) + ( x + z) j + yzk A vector descrpton of s smply r( t ) = t, t Thus F ( x, y, z) = F( t,, ) = tj = For, we have r( t ) = + tj, t Ths gves us F( x, y, z) = F (, t, ) j = ( t + j) j = = 48
Fnally, for, there s r( t ) = + j + tk, t ; and so F( x, y, z) = F (,, t ) k = [( + t ) + ( + t) j + 4tk] k = 4 t = 8 At last, we have then F( x, y, z ) = + + 8 = Exercses 7 Evaluate [ xy + x j], where s the arc of the curve y = x from (,) to (,) 8 Evaluate (cos x yj) where the part of the curve y = sn x from (,) to (π,) 9 Evaluate the lne ntegral of F( x, y, z) = xyz + ( xy+ yz) j + z k from (,,) to (-,,) along the lne segment jonng these two ponts Evaluate the lne ntegral of F( x, y, z) = ( x z) + ( y z) j ( x + y) k along the polygonal path from (,,) to (,,) to (,,) to (,,) Integrate F( x, y) = ( y + xj) one tme around the crcle x + y = a n the x + y counterclockwse ecton 49
4 ath Independence Suppose we evaluate the vector lne ntegral F( r), where s a curve from the pont p to the pont q Let r( t), a t b, be a vector descrpton of Then, of course, we have r( a ) = p and r( b ) = q As we have already seen, b F( r) = F ( r( t)) a Now let us make the very specal assumpton that there exsts a real-valued (or scalar) functon g: R R such that the dervatve, or gradent, of g s the ntegrand F : g = F Next let's use the han Rule to compute the dervatve of the composton h( t) = g( r ( t)) : h t g d r '( ) = = F( r( t )) Ths s, mrable dctu, precsely the ntegrand n our lne ntegral: b b F( r) = F( r( t)) = h'( t) = h( b) h( a) = g( p) g( q) a a Ths s a very exctng result and calls for some meaton Note that the curve has completely dsappeared from the answer The value of the ntegral depends only on the values of the functon g at the endponts; the path from p to q does not affect the answer The lne ntegral s path ndependent The result s esthetcally pleasng and s clearly the lneal descendant of the fundamental theorem of calculus we learned so many years ago 4
A moment's reflecton on the examples we have seen should convnce us that a lot of ntegrals are not path ndependent, thus many very nce functons F (or vector felds ) are not the gradent of any functon A functon F that s the gradent of a functon g s sad to be conservatve and the functon g s sad to be a potental functon for F Let's suppose the doman D of the functon F: D R s open and connected (Thus any two ponts n D may be joned by a nce path) We have just seen that f there exsts a functon g: D R such that F = g, then the ntegral of F between any two ponts of D does not depend on the path between the two ponts It turns out, as we shall see, that the converse of ths s true Specfcally, f every ntegral of F n D s path ndependent, then there s a functon g such that F = g Let's see why ths s so hoose a pont p = ( x, y, z ) D Now defne g( s ) = g( x, y, z) to be the ntegral from p to s along any curve jonng these ponts We are assumng path ndependence of the ntegral, so t matters not what curve we choose Okay, now we compute the partal dervatve g The doman D s open and hence ncludes an open ball centered at s = ( x, y, z ) D hoose a pont q = ( x, y, z) n such an open ball, and let L be the straght lne segment from s to q Then, of course, L les n D Now let's ntegrate F from p to s by gong along any curve from p to q and then along L from q to s : g( s) = g( x, y, z) = F( r) + F( r) L The frst ntegral on the rght does not depend on x, and so g = F ( r ) L F( r) = Thus 4
We clearly need to fnd F( r) d r Ths s easy Suppose L F( r) = f ( r) + f ( r) j + f ( r) k A vector descrpton of L s smply r( t ) = t + yj + zk, x t x Thus d r =, and our lne ntegral becomes smply F( r) = f ( t, y, z) We are almost done, for note that now Hence L L x x x F( r) = f ( t, y, z) = f ( x, y, z) x g = f It should be clear to one and all how to show that g y gvng us the desred result: F = g = f and g z = f, thus Exercses rove that g y = f, where g and f are as n the precedng dscusson rove that f F: D R, where D s open and connected, and every F( r) d r s path ndependent, then F( r) = for every closed path n D( A closed path, or 4
curve, s one wth no endponts) [hyscsts and others lke to use a snake sgn wth a lttle crcle supermposed on t to ndcate that the path of ntegraton s closed] 4 rove that f F: D R, where D s open and connected, and F( r) = for every closed path n D, then every F( r) d r s path ndependent 5 a)fnd a potental functon g for the functon F( r) = yz + xzj + xy k b)evaluate the lne ntegral F( r) d r, where s the curve t t r( t ) = ( e sn t ) + t e j + cos tk, t y 6 a)fnd a potental functon g for the functon F( r) = e + z ( + xj + xk) b)fnd another potental functon for F n part a) b)evaluate the lne ntegral F( r) d r, where s the curve r( t ) = t cost + 4tj + e t k, t π x x 7 Evaluate [( e sn y + y) + ( e cos y + x y) j] E where E s the ellpse 4x + y = 4 orented clockwse [Really good hnt: Fnd the gradent of g( x, y, z) = e x sn y + xy y ] 4