1. Line integrals HELM 28 was concerned with evaluating an integral over all points within a rectangle or other shape (or over a cuboid or other volume). In a related manner, an integral can take place over a line or curve running through a two-dimensional (or three-dimensional) region. Line integrals may involve scalar or vector fields. Those involving scalar fields are dealt with first Line integrals in two dimensions A line integral in two dimensions may be written as, [Pevdw ‘There are three main features determining this integral: F(z.y): This is the scalar function to be integrated eg. F (x,y) =x? + 4y” 2 ofr =siny c This is the curve along which integration takes place. e.g. orr=t~1; y=? (where x and y are expressed in terms of a parameter t) dw: This gives the variable of the integration. Three main cases are dr, dy and ds Here ‘s’ is arc length and so indicates position along the curve C. 7 ds may be written as ds = /(dr)? + (dy)? or ds = \/1+ (#) de A fourth case is when F(x.) dw has the form: Fydr-+Fxdy. This is a combination of the cases dir and dy. The integral f F(.c.y) ds represents the area beneath the surface z = F(z,y) but above the curve c The integrals / F(r,y) dr and f F(z,y) dy represent the projections of this area onto the x= and yz planes respectively. , A particular case of the integral f F(z, y) da is the integral f 1 ds. This is a means of calculating the length along a curve ie. an arc length Sele Figure 1; Representation of a line integral and its projections onto the .x and y2 planes HELM (2008) Section 29.1: Line IntegralsThe technique for evaluating a line integral is to express all quantities in the integral in terms of a single variable. If the integral is with respect to ‘zx’ or ‘y', then the curve 'C’ and the function 'F' may be expressed in terms of the relevant variable. If the integral is with respect to ds, normally all quantities are expressed in terms of x. If xr and y are given in terms of a parameter ¢, then ¢ is used as the variable Example 1 Find fea +4y)dx where Cis the curve y =x”, starting from x = 0,y=0 and ending at x= l,y Solution As this integral concerns only points along C and the integration is carried out with respect to =r, y may be replaced by 2?. The limits on x will be 0 to 1. So the integral becomes i x(l+4y)dr = ie i = l Example 2 Find Prasan dy where C is the curve y starting from 2=0,y=0 and ending at x= 1,y = 1. Thisis the same as Example 1 other than dz being replaced by dy, Solution AAs this integral concerns only points along C’ and the integration is carried out with respect to y, everything may be expressed in terms of y, i.e. x may be replaced by y'/?, The limits on y will be 0 to 1, So the integral becomes feo+m dy lc 4 HELM (2008) Workbook 29: Integral Vector Calculus» Example 3 Find | r(1+4y)ds where C is the curve y = starting from r = 0,y =0 and ending at x = 1,y= 1. This is the same integral and curve as the previous two examples but the integration is now carried out with respect to s, the arc length parameter Solution | As this integral is with respect to .r, all parts of the integral can be expressed in terms of x, Along 2 | fis (¥) da = \/14 (20)'de = V1 4dr | ds y So, the integral is 1 1 feasanas= [ x (1+ 42?) VIET de = [ (14407)? de F 0 fe=0) | This can be evaluated using the transformation u=1+42? so du =8cdr ie. cdr = Fz When x= 0, w=1 and when x=1, u=5 Hence, ‘ ie _ x? wr)’ 22 pps a) 2.745 i 5 20 ° | Note that the results for Examples 1,2 and 3 are all different: Example 3 is the area between a curve and a surface above; Examples 1 and 2 give projections of this area onto other planes. Example 4 Find [ xy dx where, on C, x and y are given in terms of a parameter t by ie £3 — 1 for t varying from 0 to 1 r= 3ty Solution Everything can be expressed in terms of ¢, the parameter. Here x = 31? so di = 6t dt. The limits ont are t= 0 and t= 1. The integral becomes [wae = HELM (2008) 5 Section 29.1: Line IntegralsWB key Point 1 A line integral is normally evaluated by expressing all variables in terms of one variable. In general [senase f sew ave [ fe) ae For F(z,y) = 20 +9, find (i) f Fle) ae, ( [Fen dy, (iii) [ F (x,y) ds where C is the line y = 2 from (0,0) to (1,2). le Express each integral as a simple integral with respect to a single variable and hence evaluate each integral Your solution Answer @ [ere dr = i (ii) for WP) dy (iii) fe be?) V5 dl a < 6 HELM (2008): Workbook 29: Integral Vector CalculusHEL VARS Find () [Fee y) de, (i) [Few y) dy, ci fx (2,y) ds where F(r,y) = 1 and C is the curve y = 1? — 4 na from (1, 4) to (2,2 ~ #1n2) Your solution Answer som maaan ¢ wf dy=3—2in2, tii) y=} dy de lds= Lee pear | festa = fice dyed 2 ae 274 Find () [| Flew) dz, (i) [Few dy, (iii [Fen ds where F(z,y) = sin 2x and C is the curve y = sinz from (0,0) to Ga) Your solution Answer xf =) 2 of sin2cdr=1, (i [ 2sin x cos? x de= 5 0 0 sf . (iii) [ sin2eV1+ cos? x de = 3(2v2 ~ 1), using the substitution u = 1 + cos? x 5 HELM (2008) Section 29.1: Line IntegralsNIG A Calculus mt Example I Compute each of the following double integrals over the indicated rectangles. (a) [f6xv° dd, R=[2,4]x[1,2] (Solution (b) ff2x-4y°d4. R=[-5.4]x[0.3] [Solution (©) [fx°y* +c08(zx)+sin(zy)d4, R=[-2,-1]x[0,1] {Solution fk @) las“ R=[0.1}x[1,2] (Solution) © [eer au, R=[-1,2]x[0.1] {Solution} Solution (a) f[6x* a4, R=[2,4]x[1,2] f It doesn’t matter which variable we integrate with respect to first, we will get the same answer regardless of the order of integration. To prove that let's work this one with each order to make sure that we do get the same answer. Solution 1 In this case we will integrate with respect to y first. So, the iterated integral that we need to compute is, foo? t= foo? aa When setting these up make sure the limits match up to the differentials. Since the dy is the inner differential (i, we are integrating with respect to y first) the inner integral needs to have y limits, To compute this we will do the inner integral first and we typically keep the outer integral around as follows, 47 yp fis ; {fo a= fi (2x i a& ‘ =|, 16x-2x¢x =[l4xde = fyi Remember that we treat the x as a constant when doing the first integral and we don’t do any integration with it yet. Now, we have a normal single integral so let's finish the integral by computing this. Jf oo? a4=73" f 84 © 2007 Paul Dawkins 9 httpi/tutorial.math.lamar.edu/terms.aspx oeCalculus 111 ‘Solution 2 : ; In this case we'll integrate with respect to x first and then y. Here is the work for this solution. [Joo dae [fio day J =[oev eo =f 36)? a e =12y'| =84 Sure enough the same answer as the first solution, So, remember that we can do the integration in any order. (Return to Problems) (b) ff2x-4y'aa, R=[-5.4]x([0,3] For this integral we'll integrate with respect to y first. cael [fox-ay' aa f Jp2x-4y' dae ® Ss = f.e0-y" Jae ‘ = [/,6x-81de =(3x*-81x)|) =-756 Remember that when integrating with respect to y all x’s are treated as constants and so as far as the inner integral is concerned the 2x is a constant and we know that when we integrate constants with respect to y we just tack on a y and so we get 2xy from the first term. [Return to Problems] (©) [[2y? +cos(ax)+sin(zy)da, R=[-2,-1]x(0,1] In this case we'll integrate with respect to x first. © 2007 Paul Dawkins 10 hhttp:/tutorial.math,lamar.edu/terms.aspxCaleulus It i fey +cos(x)+sin(y)da= fff x°y? +00s (mx) +sin(ny)dedy ‘1 Lay =| [S°7+ [Gers a =| Zy+sin(xy) a fb sin(my)dy a dy la in(ax)+xsin(xy)] Don't forget your basic Calculus | substitutions! (Return 0 Problems} @ lm R=([0.1]x[1,2] Inthis case because the limits for x are kind of nice (ie. they are zero and one which are often tase for evaluation) let's integrate with respect to x first. We'll also rewrite the integrand to help with the first integration. (forssyy ate f[laxea) dey -[ Gown] ke ay [Return to Problems] (© [fxe? 44, R=[-1,2]x(0.1] Now, while we can technically integrate with respect to either variable first sometimes one way is significantly easier than the other way. In this case it will be significantly easier to integrate with respect to y first as we will see. | © 2007 Paul Dawkins " hhttp:/tutorial. math.lamar.edu/terms.aspxi : " ees [fxr a4=f? fre” ava | The y integration can be done with the quick substitution, | u=xy du = xdy which gives 2 [rer at= [orl ae maa ‘ =f) et-ldr So, not too bad of an integral there provided you get the substitution. Now let’s see what would happen if we had integrated with respect to x first. 1 [fs = [fe a In order to do this we would have to use integration by parts as follows, dv=e” de ‘The integral is then, We're not even going to continue here as these are very difficult integrals to do. {Return to Problems] ‘As we saw in the previous set of examples we can do the integral in either direction. However, sometimes one direction of integration is significantly easier than the other so make sure that you think about which one you should do first before actually doing the integral. The next topic of this section is a quick fact that can be used to make some iterated integrals somewhat easier 1o compute on occasion © 2007 Paul Dawkins 2 ‘http:/tutorial math. lamar.edu/terms.aspxnas Fact f(y) = g(x)h(y) and we are integrating over the rectangle R= [a,b}x[e.d] [Jr@.y)aa = [feG)H(vya=(fFxCo)as)( ya) then, — ih it So, if we can break up the function into a function only of x times a function of y then we can do the two integrals individually and multiply them together. Let’s do a quick example using this integral Example 2 Evaluate [{.xcos? (y)dd, R= f 230 Solution Since the integrand is a function of x times a function of y we can use the fact. [J>eos (yaa -({4[ J 3 cos? (ya) se 0 We have one more topic to discuss in ction. This topic really doesn’t have anything to do with iterated integrals, but this is as good a place as any to put it and there are liable to be some questions about it at this point as well so this is as good a place as any, ‘What we want to do is discuss single indefinite integrals of a function of two variables. In other words we want to look at integrals like the following, Jxse -e'dy 2y)+4aydy From Calculus | we know that these integrals are asking what function that we differentiated to get the integrand, However, in this ease we need to pay attention to the differential (dy or cx) in the integral, because that will change things a little In the case of the first integral we are asking what function we differentiated with respect tov to get the integrand while in the second integral we're asking what function differentiated with © 2007 Paul Dawkins a ‘hup:/tutorial math. lamar. edu/terms.aspx.Toad Fhe acta i fe plane ’ plore bounded Voy fou. Wore 3, > (x= tosh a 4k —(«-2)". c : A ot Pots Be Wkrtedinn « A)" = 4x) er ITA = I A GH A He a Guy («-s) 20 => Hela KAZ. be ae fo a ‘Y= beay = = fie bay (x- Jan = - a| 6 —#at3)da ae ont Pasa] = — Ne N a(t +3) ze ——. Uy ww BA+3~0 =) eee leatCalculus Itt (0) [f4xv-y' dA, Dis the region bounded by y= Vx and y =x In this case we need to determine the two inequalities for x and y that we need to do the integral. The best way to do this is the graph the two curves. Here is a sketch. y 1 og 06 oa) 02 02 04~=—O06SsC 1 ‘So, from the sketch we can see that that two inequalities are, Osx! x a& 4 I J 35 tox} = 52” J, 156 [Return to Problems] (©) [J}6x" 40d, Dis the triangle with vertices (0,3), (1,1), and (5,3). We got even less information about the region this triangle, ime. 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