# Line and Surface Integrals: An Intuitive Understanding

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1 Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of higher dimensions. Unfortuntely, this cn result in wht my seem like multiple kinds of generliztions. Tke the integrl, for exmple. In multivrible clculus, we hve double integrls, triple integrls, line integrls, surfce integrls where does it end? As it turns out, ny kind of integrl cn be thought bout in unified wy. The im of this discussion is to give some intuition behind these objects, nd in prticulr, line nd surfce integrls. Review First, let s review integrtion in the single vrible cse, emphsizing certin interprettion. In this cse, our domin i.e., the thing we re integrting over is lwys n intervl I = [, b] R. One wy to think bout n integrl f(x) dx I of function over I is tht we tke I, chop it up into chunks, give ech sum weight, nd dd everything bck together. More concisely, n integrl over I is weighted sum of I, where ech chunk dx of I is weighted by f(x), the vlue of the function t tht point. A Probbilistic Interprettion If you ve tken ny probbility theory, this should sound relly fmilir. If you hven t, let s strt off esy: given list of numbers X = {x 1, x 2,..., x n }, we ll know how to clculte its verge vlue: We cn write this little differently s: X vg = x 1 + x x n n X vg = 1 n x n x n x n In other words, we tke ll the numbers in our set, give them equl weighting ( 1 n ) nd dd them ll bck together. In probbility theory, we bstrct this ide to tht of n expected vlue. Insted of giving ech number equl weighting, we cn ssign possibly different weightings. Phrsed differently, ech number x i cn be given probbility P (x i ) of hppening. Then the expected vlue of the set X with respect to P is: n (X) = P (x 1 ) x 1 + P (x 2 ) x P (x n ) x n = P (x i ) x i 1 i=1

2 Agin, we re tking weighted sum over our set X. Note tht if P (x i ) = 1 n for ll i, we get the verge vlue from bove. Wht if we wnted to the sme thing with n infinite set of numbers, like n intervl I = [, b]? Informlly, we cn tke our set I nd chop it up into infinitely mny chunks dx, nd look t the probbility p(x) of ech chunk occuring. Here, ech dx is cting like one of the x i s from bove, nd our function p(x) is sort of cting like P (x i ). If we clculte the weighted sum of our infinitely mny chunks, we get n integrl! In this cse, the expected vlue with respect to p is: (I) = p(x) dx Hopefully it is cler how nturlly n integrl in generl cn be interpreted s sort of weighted sum or expected vlue. Note tht this gives nice understnding of the fct tht: b I dx = b If we tke n intervl I, chops it up, give everything n equl weighting of 1, then when we dd everything together we re just left with the length of I! This interprettion of n integrl s weighted sum of its domin will be the key ide to understnding every other kind of integrl out there. Let s quickly interpret this in the context of double nd triple integrls. Double nd Triple Integrls Double Integrls Sy I give you region R 2. Let s chop it up into little chunks of two-dimensionl regions da. For ech little chunk of re t position (x, y), we cn weight it by some function f(x, y). Adding ll the weighted chunks together gives us: f(x, y) da And tht s double integrl! We re relly not doing nything different thn we did before this is just reinterprettion of the ide of weighted sum. Agin, wht hppens if we give ech da weight of f(x, y) = 1? Well, when we dd everything bck together, we get the re of! Here s n importnt point to note: the bstrct nottion f(x, y) da is conceptully nice, but to ctully clculte this vlue, we hve to write it in terms of iterted single-vrible integrls: f(x, y) da = f(x, y) dx dy I 1 I 2 for some intervls I 1 nd I 2. This is importnt to keep in mind s we press on: the only thing we know how to ctully clculte is regulr single vrible integrl. 2

3 Triple Integrls We re on roll now. If I give you region R 3, we cn chop it up into little chunks of volume dv, weight ech chunk by the vlue of f(x, y, z) t tht point, nd dd everything up to get: f(x, y, z) dv This is triple integrl! We mke the sme observtions s before: if f(x, y, z) = 1, we re weighting every chunk the sme, so we recover the volume of by clculting: dv Also, to ctully clculte these integrls, we hve to reduce them to single vrible integrls: f(x, y, z) dv = f(x, y, z) dx dy dz I 3 for some pproprite intervls I 1, I 2, nd I 3. I 1 I 2 Line Integrls Time to crnk it up notch. Now tht we ve hd some prctice with thinking bout integrls s weighted sums, we cn tlk bout integrting over more complicted domins. The integrls we just finished discussing took regions tht were, in some sense flt: intervls, regions of the plne, regions of spce, etc. To mke things more interesting, let s sy I give you differentible curve C in the plne: y C x This will be the new domin of our integrl. Line Integrls in Sclr Fields Given function f : R 2 R, we wnt to define wht it would men to integrte f over C, i.e., to tke weighted sum of infinitely smll chunks of C. Given wht we ve done so fr, this doesn t seem too hrd! Let s chop our curve C up into little chunks of curve" ds, whtever tht mens, weight ech ds by the function vlue t tht point, nd dd everything bck together. Tht s line integrl: f(x, y) ds C 3

4 Conceptully nice? Sure. As fr s computtion goes, this still seems little mysterious. After ll, we weren t precise bout wht we ment by ds. To formlize the ide of smll chunk of curve, we cn tke dvntge of tht fct tht C is differentible. Bsiclly, this mens tht if we pick point on C nd zoom in relly closely, it looks like line! So our ds cn be interpreted s n infinitely smll line segment, tngent to C. Since C is moving up nd down, t ny given point, our line segment will hve n x-component chnge nd y-component chnge. We cn visulize this s n infinitely smll right tringle: ds dy dx This mens, roughly, tht ds = (dx) 2 + (dy) 2. Cool. Now, remember tht the only thing we know how to clculte is single vrible integrls. Our next step is to figure out how to turn C f(x, y) (dx) 2 + (dy) 2 into n integrl tht we cn clculte. In other words, we need some sort of reltionship between n intervl I R nd our curve C. This is exctly wht prmetriztion of curve is - mpping from n intervl to the grph of the curve. So let s prmetrize C in the following wy: Now we cn use the fct tht dx = dx dt entirely in terms of t: C r : [, b] R R 2 r(t) = f(x, y) ds = x(t) y(t) dt (nd similrly for dy) to turn our expression into n integrl b (dx ) 2 f(x(t), y(t)) + dt ( ) 2 dy dt dt Just to reiterte: ll we did here ws interpret the ide of tking weighted sum in terms of smll chunks of our domin. Nothing too different so fr! Also, note tht if f(x, y) = 1 (in other words, every chunk of curve hs n equl weighting of 1), we get the length of C! From here, we could do the sme thing for curve in 3 dimensions, chnging only our prmetriztion (r(t) will hve three component functions) nd our ds term (it will hve dz). Wht s to stop us from going to 4 or 5 dimensions? Nothing! We cn just s esily define n line integrl of function f : R n R over curve in R n for ny nturl number n. Line Integrls in Vector Fields As I mentioned in the beginning of our discussion, the bstrction of mthemticl objects in one vrible to higher dimensions cn result in multiple kinds of generliztions. We run into this issue when we bstrct 4

5 the ide of function: function defined on higher-dimensionl domin cn spit out number 1, but it cn lso spit out vector. We disposed of the former cse in the lst section, nd now we ll tckle the ltter. Recll tht function on multidimensionl domin tht spits out vector in the sme spce is clled vector field. Formlly, vector field is function F : R n R n. As before, we ll strt in the two-dimensionl cse (n = 2) nd esily generlize from there. So let C be curve in the x y plne, nd let F : R 2 R 2 be vector field. We need to figure out wy to integrte F over C. As lwys, our key intuition will be to think bout tking weighted sum over C. Before, our weights were given by the numbers spit out by our sclr field f. Now, we hve function F tht spits out vector, so we need to reformulte wht we men by weighting chunk of curve." After ll, we cn t do wht we did before nd just multiply ds by F, becuse we would get vector, nd we wnt our integrl to be number. In order to remedy this, we turn to the dot product. Conceptulizing the Dot Product: Given vectors nd b, one wy to think bout the dot product b is tht it is mesurement of how similr nd b re; i.e., how well they trvel together. So if nd b re orthogonl (i.e. s different s possible), b = 0, wheres if nd b re prllel, the mgnitude of b will be s big s possible. So the dot product is wy to compre two vectors or, dre I sy, weight one vector by nother. We ll keep this in mind s we continue. The first thing we ll do is prmetrize C. We re going to hve to do tht eventully (remember, to compute stuff we hve reduce our problem to single vrible integrl), nd it will mke formulting the integrl esier. So from now on, C is the imge of the following function: r : [, b] R R 2 r(t) = x(t) y(t) Before, when we were looking t ds, this ws like mini tngent line to the curve. In terms of our prmetriztion r, insted of clculting mini tngent line, we cn just clculte the tngent vector! So for ny point t [, b], our nlogue to ds will be dr: dr = dr dt dt = r (t) dt = x (t) y (t) Now, t ny point t [, b], we cn clculte the vlue of our vector field F t tht point on the curve by inputting our position on the curve into the function: F(r(t)) = F(x(t), y(t)). Remember tht this is vector. Our gol is to figure out how to give dr weight. So wht we cn do is tke the dot product of the dt vector spit out by the vector field t tht point with the tngent vector to the curve: F(x(t), y(t)) x (t) dt y (t) 1 These types of functions f : R n R re clled rel-vlued functions, or sometimes sclr fields, hence the nme of the previous subsection. 5

6 Visully, t every point on the curve, we hve the following picture: F(x(t), y(t)) r (t) This represents one weighted chunk, so we dd up ll the chunks on the curve C to give us the line integrl of the vector field! Nottionlly, we write: C F dr = b F(x(t), y(t)) x (t) y (t) Note tht by our interprettion of the dot product, this quntity gives sort of mesurement s to how well the curve flows with the vector field. If ll of the tngent vectors re close in direction to the vector field vectors, our dot product will be lrger nd therefore the sum will be lrger 2. If the tngent vectors re close to orthogonl to the vector field (so the curve is not following the flow), our dot products will be close to 0 nd the line integrl will be very smll. As before, we cn esily generlize this for curve C which is the imge of some function r : [, b] R R n nd vector field F : R n R n by just incresing the number of component functions ech vector-vlued function hs. dt Surfce Integrls We turn our ttention now to integrting over surfces. As before, our formultion will be grounded in the intuitive ide of tking our surfce, chopping it up, nd clculting weighted sum of ll the pieces. As before, we hve to be precise bout couple things: wht we men by chunk of surfce, nd wht it mens to weight chunk. Surfce Integrls in Sclr Fields We begin by considering the cse when our function spits out numbers, nd we ll tke cre of the vectorvlued cse fterwrds. So let f : R 3 R be sclr field, nd let M be some surfce sitting in R 3. 2 If the tngents vectors re moving in the sme direction in the opposite direction, we would get very lrge negtive number. 6

7 We my s well prmetrize our surfce, becuse like before, we re going to hve to do tht eventully. Note tht our prmetriztion will now pick up rectngle from the plne, s opposed to picking up n intervl: So our surfce M is the imge of r. x(s, t) r : [, b] [c, d] R 2 R 3 r(s, t) = y(s, t) z(s, t) Our conceptul development of the surfce integrl of f over M will mimic wht we ve done before: chop the surfce up into chunks of surfce ds, weight ech chunk by the vlue of f t tht point, nd dd everything up: M f(x, y, z) ds Cool! Next up is to figure out wht ds mens if we do tht, we re bsiclly done. Recll tht with line integrls, we used the tngent vector to encpsulte the informtion needed for our smll chunks of curve. We cn try to do the sme thing with surfce, but we hve n issue: t ny given point on M, there re tons of tngent vectors! So for surfce, we wnt to think of our chunks s little tngent plnes. Fortuntely, tngent plns nd tngent vectors re very strongly relted, becuse we cn describe tngent plne using two importnt tngent vectors: the tngent vector in the s-direction, nd the tngent vector in the t-direction : s To formlize this, we turn to the cross product. nd Conceptulizing the Cross Product: Given two vectors nd b, we cn clculte their cross product b. This gives us nother vector. In prticulr, it s vector orthogonl to both nd b. The other cool thing bout the vector b is tht its length b represents the re of the prllelogrm defined by nd b. 7

8 So, t ech point on M, we hve two importnt tngent vectors, s nd this prllelogrm will essentilly be our mini-tngent plne! nd. These define prllelogrm, s So we cn mke ds be the re of this little tngent plne, which we clculte using the length of the cross product: ds = s ds dt From here, we cn convert our bsrct surfce integrl into double integrl over the s t region: b d f(x, y, z) ds = f(x(s, t), y(s, t), z(s, t)) (s, t) (s, t) s dt ds M c Not the most beutiful expression in the world, but just keep in mind tht ll we re doing is tking weighted sum of the chunks of the surfce. Surfce Integrls in Vector Fields As with line integrls, the story is (seemingly) different when we hve vector field F : R n R n insted of sclr field f. The issue is tht we cn t weight our chunks of surfce by multipliction, becuse the vector field is spitting out vector. But, s before, we cn remedy this by turning to the dot product. However, insted of clculting the dot product of our vector field with one of the tngent vectors, we will use the norml vector to our surfce. Recll tht we cn clculte the norml vector N to surfce by computing the cross product of the tngent vectors, since this gives us vector perpendiculr to the tngent plne. So our little chunk of surfce ds will be clculted in the following wy: ds = N ds = s dt ds Note tht this is the sme cross product tht represented the re of our tngent plne in the sclr-field version of our line integrl. So given our norml vector N, we cn weight the chunk of surfce represented by this vector by computing F N. Then, s before, dd everything up: b d F ds = F N ds = F(x(s, t), y(s, t), z(s, t)) M And we re done! M c ( s ) (s, t) (s, t) dt ds Let s think bout wht we re clculting here. Remember tht the dot product mesures how similr two vectors re, or, how well they trvel in the sme direction. Since we re clculting the sum of the dot product of the vector field with the norml vector to our surfce, the surfce integrl gives n indiction 8

9 of how much the vector field is moving directly through the surfce. In other words, if the surfce flowed tngentilly to the vectors in the vector field, the dot products would be close to 0, so the surfce integrl would be close to 0. One wy to picture this is by thinking of the surfce s sil, nd the vector field s the wind: if the wind is blowing directly into the sil nd mking it very tut, then the surfce integrl will be very lrge (or very negtively lrge if it s blowing in the opposite direction); if the wind is either not blowing or blowing long side the sil, the surfce integrl will be very close to 0. Review Time to review everything tht we ve done, just to drive the point home. At this point in your life, you ve seen mny kinds of integrls: single vrible integrls, double integrls, triple integrls, line integrls, nd surfce integrls. very kind of integrl cn be formulted nd interpreted s weighted sum over its pproprite domin. But to ctully clculte these things, we hve to turn them into single-vrible integrls, becuse tht s the only thing we know how to clculte! The min thing to tke wy from this is to not be intimidted when you see ny sort of integrl they ll represent more or less the sme thing. 9

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### Section 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40 Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since Mth Em Prctice Februry, 8 Em will cover sections 6.5, 7.-7.5 nd 7.8. This sheet hs three sections. The first section will remind you bout techniques nd formuls tht you should know. The second gives number Chpter 5. Numericl Integrtion These re just summries of the lecture notes, nd few detils re included. Most of wht we include here is to be found in more detil in Anton. 5. Remrk. There re two topics with