If f(x, y) is a surface that lies above r(t), we can think about the area between the surface and the curve.

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1 Line Integrls The ide of line integrl is very similr to tht of single integrls. If the function f(x) is bove the x-xis on the intervl [, b], then the integrl of f(x) over [, b] is the re under f over the intervl: Let r(t) be curve in the xy plne. If f(x, y) is surfce tht lies bove r(t), we cn think bout the re between the surfce nd the curve. 1

2 Specificlly, we think of the re enclosed bove the curve nd below the surfce. In the grph bove, the re tht we re interested in is tht shded in below the red curve on the surfce, nd bove the red curve in the xy plne. To find the vlue for this re, we evlute specific type of integrl known s the line integrl. The following theorem llows us to do so: Definition 3. If f(x, y) is continuous over the curve r(t) x(t) i + y(t) j in the xy plne, then the line integrl of f over r(t) from t to t b is f(x, y) ds f(x(t), y(t)) r (t) dt. In order to evlute line integrl over curve, we will need to prmeterize the curve by writing it s x(t) i + y(t) j, then clculte r (t). Finlly, we cn set up the integrl ccording to the theorem nd integrte s we would ny other single integrl. The theorem bove lso pplies to functions f(x, y, z) over spce curves r(t) x(t) i + y(t) j + z(t) k lmost exctly, with the obvious chnges. Exmple: A piece of tin is cut from circulr cylinder whose bse is circle of rdius 3 inches. At ny point (x, y) on the bse, the height of the object is given in inches by f(x, y) 1 + cos(πx)/4. Set up n integrl tht will yield the surfce re of the piece of tin. We cn grph the bse of the circulr cylinder in the xy plne by thinking of its defining eqution s x 2 + y 2 9: The cylinder from which we wish to cut the tin is grphed below: 2

3 The top of the piece of tin should be shped like f(x, y) 1 + cos(πx)/4: We wish to cut long the intersection of the two surfces: So the cut we mke yields the following shpe: 3

4 The line integrl of f over the region x 2 + y 2 9 will yield the surfce re of this shpe. We must first prmterize the curve x 2 + y 2 9; we cn think of it s x 3 cos t nd y 3 sin t (or r(t) 3 cos t i + 3 sin t j) on the intervl t 2π. Then r (t) 3 sin t i + 3 cos t j, nd r (t) ( 3 sin t) cos 2 t 9 3. Using the prmeteriztion, we rewrite f s f(x(t), y(t)) 1 + cos(π(3 cos t))/4. So the line integrl is f(x, y) ds 3 2π 2π f(x(t), y(t)) r (t) dt (1 + cos(π(3 cos t))/4)(3) dt 1 + cos(3π cos t) 4 dt. This is not n elementry integrl, but Mthemtic cn pproximte the vlue: 2π cos(3π cos t) dt So the surfce re of the piece of tin is pproximtely squre inches. Evlute the line integrl of f(x, y) 2 + x 2 y over the upper hlf of the unit circle x 2 + y 2 1. This is ctully the exmple from the grph t the beginning of the section: 4

5 The line integrl of f over this region will yield the re of the region below f tht lies bove the upper hlf of the circle. We need to hve prmeteriztion of the curve; we cn think of it s x cos t nd y sin t (or r(t) cos t i + sin t j) on the intervl t π. Then r (t) sin t i + cos t j, nd r (t) ( sin t) 2 + cos 2 t 1 1. Using the prmeteriztion, we rewrite f s f(x(t), y(t)) 2 + cos 2 t sin t. So the line integrl is f(x, y) ds π π f(x(t), y(t)) r (t) dt (2 + cos 2 t sin t)(1) dt 2 + cos 2 t sin t dt 2t 1 3 cos3 t π using the substitution u cos θ 2π π The number 2π is the re of the region below f tht lies bove the upper hlf of the circle. Suppose tht the curve is mde up of severl different components, s in the following grph: We cn still evlute the line integrl of function f over ; if is mde up of different curves 1, 2,..., n, then the line integrl of f over is just the sum of the line integrls over the 5

6 components; we write f(x, y) ds f(x, y) ds + 1 f(x, y) ds f(x, y) ds. n Exmple: Evlute 8xyz ds, where consists of the line segments 1 from (,, 3) to (24, 1, 3) followed by the verticl line segment 2 from (24, 1, 3) to (24, 1, ). Since we wnt to evlute line integrl over two different curves, we will write seprte integrl for ech of 1 nd 2. We need to strt by prmeterizing the line segments bove. To do so, recll tht we cn find the eqution of line in spce by finding vector prllel to the line nd point on the line. For the first line segment 1, the vector 24, 1, is prllel to the line, so the vector eqution for the line is r 1 (t) 24t i + 1t j + 3 k. Similrly,,, 3 is prllel to 2, whose vector eqution is r 2 (t) 24 i + 1 j + (3 3t) k. We will need to know r (t) for ech eqution: r 1 (t) 24 i + 1 j, nd r 2 (t) 3 k, so tht r 1 (t) nd r 2 (t) 9 3. The prmeteriztion for 1 is r 1 (t) 24t i + 1t j + 3 k, t 1, nd the prmeteriztion for 2 is r 2 (t) 24 i + 1 j + (3 3t) k, t 1. We will need this prmeteriztions to rewrite f. In the first cse, with x 24t, y 1t, nd z 3, f(x, y, z) xyz becomes 24t 1t 3 72t 2. In the second cse, with x 24, y 1, nd z 3 3t, f becomes 24 1 (3 3t) 72 72t. So the first integrl is 1 xyz ds 72t 2 26 dt t t 2 dt The second integrl is 2 xyz ds (72 72t) 3 dt ( t t t dt ) 1 6

7 So the line integrl of f over the two curves is Work As we hve seen, force cting on points in spce cn be represented by vector field. If we consider in prticulr the wy tht the force F P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k cts long curve r(t) g(t) i + h(t) k + j(t) k, it mkes sense to think bout the work done by the force in moving prticle long the curve. We cn clculte the work quite esily using the following theorem: Definition 13. The work done by force F P (x, y, z) i + Q(x, y, z) j + R(x, y, z) k in moving n object over smooth curve r(t) g(t) i + h(t) k + j(t) k from t to t b is Recll tht W F ( r(t)) r (t)ds T r (t) r (t). F T ds. The theorem sys tht work cn be thought of s the line integrl of the tngentil component of the force. When clculting the integrl, we must be creful bout the order of integrtion; switching the direction of trvel long the curve r will chnge the sign of the nswer. lculting the work integrl in the given form my be difficult (since we prefer not to work in terms of the rc length prmeter s), so using one of the following equivlent forms my be helpful: W F T ds F ( r(t)) r (t)dt F d r ( P dg dt + Qdh dt + Rdj ) dt dt P dx + Q dy + R dz. Exmple Find the work done by the force field F (x, y) x 2 i xy j in moving prticle long the curve r(t) cos t i + sin t j on t π 2. To use the formul W F ( r(t)) r (t)dt, we will need to clculte F ( r(t)) nd r (t), both 7

8 of which re quite simple to do. The quntity F ( r(t)) is given by Next, we find r (t): Finlly, to evlute So the vlue for work is F ( r(t)) F (cos t, sin t) cos 2 t i cos t sin t j. r (t) sin t i + cos t j. F ( r(t)) r (t)dt, we will need to tke the dot product: F ( r(t)) r (t) (cos 2 t i cos t sin t j) ( sin t i + cos t j) W π 2 sin t cos 2 t sin t cos 2 t 2 sin t cos 2 t. F ( r(t)) r (t)dt 2 sin t cos 2 t dt 2 3 cos3 t π 2 using the substitution u cos t