Section 7.: Line Integrls Objectives. ompute line integrls in IR nd IR 3. Assignments. Red Section 7.. Problems:,5,9,,3,7,,4 3. hllenge: 6,3,37 4. Red Section 7.3 3 Mple ommnds Mple cn ctully evlute line integrls through the PthInt commnd. Suppose we wnt to evlute ( + x y) ds, where is the upper hlf of the unit circle x + y. We cn describe prmetriclly s x cos(t), y sin(t), where t π. Then +x y becomes +cos (t) sin(t), which is seen in the line_integrl commnd below. The rclength formul tht represents ds is given s x (t) + y (t), which is sp below. (rp contins the derivtive informtion, nd sp contins the length.) Thus, we cn evlute the line integrl by mking the right substitutions nd just using the Mple int commnd. > r : <cos(t), sin(t)>; > rp : diff(r,t); > sp : simplify(vectornorm(rp, )); > line_integrl : int(( + r[]^*r[])*sp, t..pi); We cn evlute the sme integrl using the PthInt commnd. We specify the function in non-prmetric form, nd we identify the prmetric representtion through the Pth option. The commnd for evluting the bove integrl vi PthInt is > line_integrl : PthInt( +x^*y, [x,y]pth(<cos(t),sin(t)>,t..pi));
4 Lecture Line integrls re useful when you wnt to integrte function f(x, y) or f(x, y, z) long curve tht is described by the prmetric equtions x x(t), y y(t), t b. In order for the integrl to be defined, must belong to the domin of f. 4. Forming the line integrl To integrte long, we need to divide into subrcs. We cn ccomplish this by dividing [, b] into subintervls [t i, t i ]. If t i [t i, t i ], we let P i (x i, yi ) (x(t i ), y(t i )). This defines point on the curve, nd we define s i to be the rclength of the intervl on bound by P i nd P i. We form the Riemnn sum using the function evlutions over ech of the subintervls s f(x i, yi ) s i, which, fter tking the limit s n, gives the line integrl of f long s f(x, y) ds lim f(x i, yi ) s i. n Now we hve to find s i, or ds. This is ssocited with the rclength, nd we know from erlier work tht the rclength s is given s b s + dy. Thus, we know tht ds + dy integrl of f long cn be rewritten s b f(x, y) ds f(x(t), y(t)) 4.. Exmple : Problem 7.., so tht ds + dy. + dy. Thus the line In this problem, we re sked to evlute (y/x) ds, where is defined by x(t) t4, y(t) t 3, t. Thus we hve y x ds t 3 6t6 + 9t t 4 4 t 6t + 9 (6t + 9) 3/ 3 3/ 49 4.
4.. Exmple : Problem 7..4 This problem sks us to evlute yex ds, where is the line segment joining (, ) to (4, 7). We need to find the prmeteriztion of the line segment, which is pretty esy. The prmeteriztion of line segment is defined by r(t) ( t)r() + tr(), where r() is the strting point of the line segment nd r() is the terminl point of the line segment. For this problem, we get the prmeteriztion r(t) < 3t +, + 5t >, so tht x(t) 3t +, y(t) + 5t, t. The line integrl becomes ye x ds (5t + )e 3t+ 34. I hve to integrte by prts to evlute this integrl, nd s finl nswer I get 34 9 (6e4 e). 4. hnging orienttion Note tht specifying t b defines n orienttion long the curve. We cn esily chnge the orienttion by chnging the strting nd terminl points on the curve, nd we need to know if this ffects the evlution of the line integrl. (Think of the line segment exmple bove; how does chnging the orienttion ffect the prmeteriztion of the line?) The curve hs s strting point the terminl point of, nd hs s terminl point the strting point of. However, the rclenth of is the sme s the rclength of, so f(x, y) ds f(x, y) ds. However, we cn lso define line integrls with respect to x nd y. (The line integrl defined bove is sometimes clled the line integrl of f over with respect to rclength.) The line integrl with respect to x is defined s f(x, y) b f(x(t), y(t)) x (t), nd the line integrl with respect to y is defined s f(x, y) dy b f(x(t), y(t)) y (t). Ech of these integrls is ffected by the orienttion of. To see this, reprmeterize the line segment problem bove, where the strting point is (4, 7) nd the terminl point is (, ). Evlute the rclength x (t) + y (t), x (t), nd y (t). The 3
rclength function remins the sme regrdless of the orienttion; but x (t) nd y (t) do not. Thus we hve f(x, y) f(x, y), f(x, y) dy f(x, y) dy. 4.3 Extensions to IR 3 Ech of these ides extends to IR 3 in exctly the mnner you would expect. You should see your text for the exct definitions. 4.4 Line integrls over vector fields One of the most interesting pplictions of the line integrl is in defining the work done by vector field. Let F be continuous vector field defined on curve described by the prmeteriztion r(t), t b. The line integrl of F long is defined s F dr b F(r(t)) r (t) F T ds. How do we get this definition? Remember tht the work done by constnt force F in moving n object long stright-line distnce described by vector D is W F D. If F is vector field, we no longer hve constnt force, since F P (x, y)i + Q(x, y)j, tht is, F cn chnge vlue for ech (x, y). Also, we re no longer moving the prticle long stright-pth; we wnt the prticle to move long some generl pth. However, if we consider point on, cll it (x i, yi ), then the force is constnt t tht point, nd we cn consider the tngent line to the curve T s pth for the prticle t tht point. Thus, we cn pproximte the work done by evluting F t severl points, nd using s the stright-line distnce D the tngent vector T multiplied by the rclength of the subintervl contining (x i, yi ). Of course, s we increse the number of subintervls (nd hence use more points for the evlution of F), our pproximtion for the work done becomes much better. Thus, we hve W F(x i, yi ) ( s i T(t i ) s i [F(x i, yi ) T(t i )] 4
nd W lim s i [F(x i, yi ) T(t i )] n F T ds. Note tht F(x(t), y(t)) F(r(t)), nd tht T cn be written s F T ds b b b [F(r(t)) T(t)] r (t) [ F(r(t)) r (t) r (t) F(r(t)) r (t). r (t). Thus, the line integrl r (t) ] r (t) Note: Since the line integrl over vector field is dependent on the tngent vector T, we hve tht F T ds F T ds, s the direction of the tngent vector chnges if we chnge the orienttion of our curve. 4.4. Exmple 3: Problem 7..4 We re given tht the vector field describing the force is F Kr, where r(t) is r 3 the prmeteriztion of the line from (,, ) to (,, 5). Thus, r(t) <, t, 5t >, t. Hence the work done by the vector field cn be clculted s K <, t, 5t > <,, 5 > (4 + 6t ) 3/ K(6t)(4 + 6t ) 3/ K(4 + 6t ) / ( K ) 3 ( K + ). 3 5