Example. Check that the Jacobian of the transformation to spherical coordinates is

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1 lss, given on Feb 3, 2, for Mth 3, Winter 2 Recll tht the fctor which ppers in chnge of vrible formul when integrting is the Jcobin, which is the determinnt of mtrix of first order prtil derivtives. Exmple. heck tht the Jcobin of the trnsformtion to sphericl coordintes is ρ 2 sin φ. The formuls relting rectngulr to sphericl coordintes re x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ. Therefore, the Jcobin is given by x ρ x θ x φ y ρ y θ y φ z ρ z θ z φ = sin φ cos θ ρ sin φ sin θ ρ cos φ cos θ sin φ sin θ ρ sin φ cos θ ρ cos φ sin θ cos φ ρ sin φ. If one expnds this determinnt in the usul wy, fter lot of gthering terms nd using the identity cos 2 + sin 2 =, the bove expression eventully equls ρ 2 sin φ. Here re few remrks bout the Jcobin: The Jcobin nd chnge of vrible formul is generliztion of u-substitution. If we mke u-substitution u = u(x), then we hve formul f(u) du = f(u(x))u (x) dx. The Jcobin of the trnsformtion u = u(x) is u (x), which is exctly the fctor ppering on the right side of this eqution. The missing bsolute vlue sign is ccounted for by the fct tht if u (x) is negtive, the bounds of integrtion re interchnged. If you hve been crefully pying ttention to the definition of Jcobin, you might be somewht bothered by the fct tht the Jcobin seems to depend on the ordering of the vribles in ech of the two vrible systems, since the ordering determines the order of the rows nd columns of the mtrix in the Jcobin. It turns out tht interchnging rows nd columns of mtrix my chnge the sign of determinnt, but never the bsolute vlue of the determinnt, so the ordering of the vribles does not prticulrly mtter when clculting Jcobin. You my end up with n nswer which differs by minus sign from someone else with different ordering, but when using the chnge of vrible formul, you tke the bsolute vlue of the Jcobin so the sign mbiguity will dispper. Here is typicl exmple of using chnge of vribles which is not polr, cylindricl, or sphericl. You need to first decide which chnge of coordintes you should use to bring the problem to mngeble form, nd then mke the pproprite coordinte chnge. Deciding which chnge to use is more of n rt thn science, nd requires lot of prctice to get consistently right. Nevertheless there re some clues which might help you mke the correct coordinte chnge. Exmple. Evlute the double integrl e (y x)/(y+x) da. D where D is the tringle with vertices (, ), (2, ), (, 2). The presence of y x, y + x in frction suggests we should mke chnge of vrible like u = y x, v = y + x. (It is still not obvious t this point tht this is the correct chnge to mke.) We strt by determining wht the corresponding domin of integrtion, sy R,

2 2 in the uv plne is. The tringle D lies in the xy plne, so to determine R we should look t the imge of D under the mp u = y x, v = y + x. In prticulr, we cn determine the boundry of R by looking t how this chnge of vribles to uv coordintes cts on the boundry of the tringle D. For exmple, the side with endpoints (, ), (2, ) is given by x 2, y =. The corresponding uv coordintes re then u = x, v = x, x 2, so this side mps to the side v = u, 2 u. In prticulr, in the uv plne this is line segment with endpoints (, ), ( 2, 2). We lso find tht the other two line segments mp to line segments, nd tht R is ctully tringle with vertices (, ), ( 2, 2), (2, 2). Therefore we cn describe R using inequlities v 2, v u v. We now need to clculte the Jcobin of this trnsformtion. We need to solve for x, y in terms of u, v. Fortuntely, in this exmple this is esy, nd we see tht x = (v u)/2, y = (v + u)/2. The Jcobin is given by the following determinnt: x u x v = /2 /2 /2 /2 = /2. Therefore, the originl integrl is equl to the integrl y u y v R e u/v (/2) da = 2 2 v v e u/v du dv = 2 2 ve u/v u=v u= v dv = 2 2 ve ve = e /e.. Line integrls We hve spent few weeks tlking bout higher-dimensionl generliztions of definite integrls, nd discussed how to clculte them, s well s vrious pplictions of double nd triple integrls in rel life. We now try to generlize the notion of n integrl in different direction. Insted of focusing on defining integrls over high dimensionl objects, we will develop theory of how to integrte not over just n intervl on the rel xis, but over curves. Let s motivte this ide with whimsicl exmple. The Americn rtist Richrd Serr is known for his minimlist sculptures, mny of which re curving sheets of metl. If you were working for Richrd Serr question you might encounter is just how much sheet metl you need to mke one of his instlltions. In more mthemticlly precise lnguge, suppose n instlltion of sheet metl is going to be plced over curve in the xy plne. For exmple, you might be given using prmetric equtions x = x(t), y = y(t), t b. Furthermore, Serr hs told you wht the height of the instlltion is t ny point of. This might be given to you by function f(x, y), defined on. How much sheet metl (tht is, wht is the surfce re) is needed for the sculpture? One the one hnd, if were line segment; sy line segment in the x-xis, then we could just integrte f over tht line segment in the x-xis, since tht integrl is the re under the height function f, which gives the surfce re of the instlltion. On the other hnd, if f(x, y) were constnt, then the surfce re of the metl would be the length of times the height f(x, y). The length of is the rc length of, which we know how to clculte: the rc length is equl to the integrl b x (t) 2 + y (t) 2 dt. The sitution we re deling with is mix of these two simpler cses: we re llowing the function f(x, y) to vry, while letting be curve in the xy plne. In ny cse, this seems to be the type of problem integrtion is suited for.

3 The key ide behind integrtion is tht it is the limit of pproximtions mde by ssuming f is constnt over vrious pieces of the region we re integrting over. Suppose we mke n pproximtion to the surfce re of the metl s follows. We select vrious points of, sy P, P,..., P n, where P, P n re the endpoints of. We lso require tht s i increses, P i keeps moving in the sme direction. We will pproximte the re of sheet metl between points P i nd P i s follows: we pretend f(x, y) is constnt on this piece of, sy, equl to f(x i, y i ), for some (x i, y i ) between P i, P i, nd we lso pretend tht stright line from P i to P i. With both of these pproximtions in mind, the surfce re of the metl from P i to P i is given by f(x i, yi ) x 2 i + y2 i where x i is the chnge in the x coordinte from P i to P i. Now, ech of these points P i is equl to (x(t i ), y(t i ), where we ssume x(t), y(t), re given in such wy to ensure tht t i re n incresing sequence of numbers, with t =, t n = b. Therefore, if we sum ll these pproximtions, we end up with Riemnn sum ( x ) 2 f(x(t i ), y(t i )) + t i ( ) y 2 t. t The limit of this Riemnn sum is defined to be the line integrl of f on, nd is written f(x, y) ds. On the other hnd, this Riemnn sum looks like Riemnn sum for the vrible t, nd the limit s t is equl to b f(x(t), y(t)) x (t) 2 + y (t) 2 dt. If we write using vector nottion, nd let r(t) = x(t), y(t), then this eqution cn be rewritten s b f(x(t), y(t)) r (t) dt. Strictly speking, to mke sure tht the line integrl of f(x, y) over is equl to the expression we ve given bove, we wnt to be smooth, which mens tht r (t) exists nd is everywhere nonzero, nd f(x, y) to be continuous. Exmples. Line integrls re genuine generliztions of definite integrls of single vrible. Suppose the curve is n intervl [, b] on the x-xis, nd we hve function f(x, ) = f(x) defined on tht intervl. Then cn be prmeterized by x(t) = t, y(t) =, where t b. Then the line integrl of f over is equl to b f(x(t), y(t)) dt = b f(t, ) dt = b f(t) dt. Arc length integrls re specil cses of line integrls. Suppose we wnt to evlute the line integrl of the constnt function f(x, y) = on, prmeterized by r(t) = x(t), y(t). Then we hve 3

4 b x (t) 2 + y (t) 2 dt, which is just the formul for the rc length of. One consequence of the presence of the fctor of r (t) is tht line integrls re independent of the prmeteriztion of, t lest if we restrict ourselves to prmeteriztions which trverse ech point of exctly once. This is good to know, since our nottion for line integrls depends only on f(x, y) nd, nd not the choice of prmeteriztion for. However, when ctully clculting line integrl, you will need to determine some prmeteriztion for. Exmple. Richrd Serr hs given you your first ssignment! He wnts to instll piece of sheet metl over the curve given by the prt of y = x 2 between x =, x = 2, nd the height of this metl t (x, x 2 ) is given by the function f(x, y) = f(x, x 2 ) = 2x. Wht is the surfce re of the metl you must cut out for this instlltion? When strting out with line integrl problem, the very first plce to strt is to determine prmeteriztion for, if one is not given to you. In this cse, we need to prmeterize the prt of y = x 2 which lies between x = nd x = 2. Perhps the most obvious prmeteriztion for this curve is x(t) = t, y(t) = t 2, t 2. Then the corresponding line integrl is 2 2t 2 + (2t) 2 dt. This cn be evluted using u-substitution; let u = t 2 +, du = 8t dt, so teh bove integrl is equl to 7 u du = 2u 3/2 7 3 = 73/2. 6 We ve seen one interprettion of the vlue of line integrl: s the re of the surfce over the curve with height f(x, y). Another interprettion involves mss or chrge. Suppose we hve thin wire (thin enough to be thought of s one-dimensionl object) in the shpe of curve, with density ρ(x, y) t point (x, y). Then the mss of the wire is given by the line integrl m = ρ(x, y) ds. Of course, the coordintes of the center of mss re given by x = xρ(x, y) ds, y = m m with nlogous formuls for moments of inerti, etc. yρ(x, y) ds, Exmple. Suppose wire is in the shpe of the semicircle given by x 2 + y 2 =, y, nd hs uniform density. Wht is the center of mss of the wire? The x-coordinte is clerly by symmetry. The y coordinte is something we hve to clculte, though. Assume ρ(x, y) =. Then the mss of the wire is given by the rc length of, which is evidently equl to π. We lso need to clculte

5 y ds. We cn prmeterize by using x(t) = cos t, y(t) = sin t, t π. Then this line integrl is equl to π y(t) ( sin t) 2 + (cos t) 2 dt = π sin t dt = 2. Therefore, the y coordinte of the center of mss is equl to 2/π. 5