F = F x x + F y. y + F z

Transcription

1 ECTION 6: etor Calulus MATH20411 You met vetors in the first year. etor alulus is essentially alulus on vetors. We will need to differentiate vetors and perform integrals involving vetors. In partiular, we will look at two fundamental results alled The ivergene Theorem and tokes Theorem. Now would be an ideal time to revise any notes you have on vetors. ome of the basi fats that you should already know about vetors, and operations on vetors that involve derivatives divergene, gradient, url ) are summarised on the handout on div, grad and url. Exerise heet 9 also ontains questions that are intended as revision Introdution A vetor field F in three dimensions is a rule whih tells us how to assoiate a vetor with eah point x, y, z). ee the handout on div, grad and url. For example, the veloity of a fluid is a vetor field. In general, F = F x i + F y j + F z k, where F x, F y and F z are funtions of x, y, z. The handout shows the following two-dimensional examples with F z = 0). yi + xj F 1 = xi + yj, F 2 = x 2 + y. 2 The vetor field r = xi + yj + zk is alled the radial diretion vetor. We will meet it several times. Note that F 1 is the two-dimensional version. ivergene and url are two important mathematial operations on vetor fields. Reall, F = F x x + F y + F z z. The divergene gives a measure of net mass flow If F = 0 then no mass is reated or destroyed. A simple alulation reveals that F 2 = 0 hek!). If we fous on a small region in the x-y plane then we an see that mass is onserved by observing that the arrows pointing into that region are mathed by arrows of the same length pointing out of that region. The url F gives a measure of twisting or urling of a vetor field. A simple alulation reveals that F 2 = 0 hek!). The url essentially tells us how a partile released into the flow field rotates. Reall, the url is alulated via F = x i, j, ) z k F x i + F y j + F z k) i j k = = x z F x F y F z Fz F y z ) i Fz x F ) x Fy j + z x F ) x k. 1

2 Example. Consider F = x 2 yi + xzj + xyzk then F = xz x)i yzj + z x 2 )k. There are two fundamental identities involving the operators div, grad and url. funtions f = fx, y, z), f) = 0. The url of a gradient is always zero. In addition, for all vetor fields F, For all salar F) = 0. The divergene of the url of a vetor is always zero. You are asked to prove these identities on Exerise heet olume Integrals of alar Funtions) Reall, the double integral fx, y)da, where is a two-dimensional region in the x-y plane and da = dxdy the area element in Cartesian oordinates) represents the volume between the surfae z = fx, y) and the region. If is a retangle, then we have onstant limits of integration. We an then easily swap the order of integration without worrying about the limits. For example, if = [0, 1] [0, 2] and fx, y) = x 2 +y 2 then fx, y) da = x 2 + y 2 dydx = x 2 + y 2 dxdy. If is a region with a more ompliated shape, then we usually don t have onstant limits of integration. Limits for the inner integral must be expressed as funtions of the outer variables. Example. Compute the integral 1 x y da, where is the right-angled triangle with verties 0, 0), 1, 0) and 0, 1). If we hoose to perform the y integral first, then 1 1 x 1 x y da = 1 x y dydx. Alternatively, if we perform the x integral first, then 1 x y da = 1 1 y 1 x y dxdy. In both ases, you should find that the answer is 1/6. The important point is that we annot just swap the order of integration and keep the same limits. Now, a volume integral of a salar funtion f) over a three-dimensional volume is denoted f d, 2

3 where d is the so-alled volume element. We use three integral signs here to emphasis that the integral is over a three-dimensional volume, but it is also ok to only write one integral sign. In Cartesian oordinates, you already know that the volume element is d = dxdydz. Remember that an integral is really defined as a limit. We replae the integral with a sum over piees of. On eah piee we evaluate the funtion f and multiply by the volume of the piee. We then take the limit of the sum as the number of piees tends to infinity. In Cartesian oordinates, a natural way to break up a volume is into small briks. If the brik has lengths dx, dy and dz in eah of the x, y and z oordinate diretions then the volume of the piee is dxdydz. This is the volume element. If the volume is a simple brik, then we have onstant limits of integration. o, if = [a, b] [, d] [e, f] then f d b fx, y, z) d = fx, y, z) dxdydz. z=e In this ase, we an swap the order of integration easily and we don t need to worry about hanging the limits. As with double integrals, if is more ompliated then we need to pay attention to the limits of integration. rawing a piture of usually helps determine the orret limits! Example. Evaluate y= x=a 1 + xy d where is the tetrahedron with verties 0, 0, 0), 0, 0, 1), 0, 1, 0) and 1, 0, 0). Lets perform the z integral first, followed by the y integral and finally the x integral. If we perform the z integral first, then we need to provide limits for z as a funtion of the outer variables y and x. For a fixed x and a fixed y, the z oordinate is bounded by the fae of the tetrahedron in the x-y plane where z = 0) and the fae of the tetrahedron that oinides with the plane z = 1 x y. Hene, 1 x y ) 1 + xy d = 1 + xy dz dy dx. The limits for y should then be expressed as funtions of the outer variable x. We have 1 x 1 x y 1 + xy d = 1 + xy dz dy dx. Finally, the variable x varies from 0 to 1 so 1 + xy d = 1 x x y z=0 z=0 1 x 1 x y z=0 1 + xy dz dy dx = 7/40. If is not a brik but a more ompliated shape that is not easy to desribe in the Cartesian o-ordinate system, then it may be easier to work in an alternative oordinate system. If we do this, however, are must be taken to onvert the volume element d in the proper way. For instane, in ylindrial oordinates, d does not mean drdθdz. You already learned how to do double integrals in polar oordinates revise your first year notes if you have forgotten). Reall, f da = fx, y) dx dy = fxr, θ), yr, θ)) r dr dθ. A x y r θ

4 The area element here in polar oordinates is da = rdrdθ. In Cartesian oordinates, we have da = dxdy. Where does the extra fator of r ome from? Reall that r is the determinant of the Jaobian matrix x x r θ. r It aounts for the hange in area when we map a small retangle with area dxdy into polar o-ordinates. The mapped retangle is not a retangle. In three dimensions, we know that the volume element for Cartesian oordinates is θ d = dxdydz. In ylindrial oordinates we have and in spherial oordinates we have d = rdrdθdz d = ρ 2 sin φ dρdφ dθ. You should learn these formulae. To work out the last two expliitly, you an write down the Jaobian matrix and find its determinant, exatly as you did for polar o-ordinates. Let s test out the laim that d = ρ 2 sin φ dρdφ dθ for spherial o-ordinates. Note that 1d gives the volume of. If we get the volume element d and the limits of integration right, then integrating one over a sphere should give the volume of the sphere whih you already know how to ompute). Example. Find the volume of a sphere of radius two. Integrating one over where is the sphere entred at the origin with radius two gives 1 d = 1 ρ 2 sin φ dρ dθ dφ = = = π = 2π 2 φ=0 θ=0 π 2π φ=0 π φ=0 θ=0 ρ=0 16π sin φ 16π osπ) 8 sin φ + ρ 2 sin φ dρ dθ dφ dφ = 16π + 16π = 2π. ) dθ dφ 16π os0) Now, of ourse, we also know that the standard formula for the volume of a sphere is 4 πρ where ρ is the radius. ine the radius is two, the volume is 4 π 8 = 2 π, whih mathes the above alulation. There is an important theorem that onnets: the divergene F of a vetor field, a losed volume, and its surfae. The theorem is quite tehnial. We present it first, and then investigate it. 4

5 Theorem: The ivergene Theorem. Let be a bounded, losed region in spae with pieewise smooth boundary. Let ˆn be the unit normal vetor to, pointing outward. Then, if F is a differentiable vetor field, F d = F ˆn d. In essene, this result says that the total divergene of a vetor field in a bounded region in spae is equal to the net flow or flux ) aross the boundary of the surfae in the normal diretion. Note that F is a salar funtion so the integral on the left-hand side is a standard volume integral whih you know how to evaluate). Note that F ˆn is the dot produt of two vetors and this is also a salar funtion. o, to evaluate the right-hand side of the equation, we first need to know how to find the normal vetor to the surfae, and then how to evaluate surfae integrals of salar funtions. Unit Normal etors to urfaes For some surfaes, it is easy to determine the unit normal vetor. Example. Consider the unit ube. That is, the ube whose edges all have length one, and one of the verties is the origin 0, 0, 0). The ube has six faes, whih are portions of the surfaes z = 0, z = 1 the top and bottom faes), x = 0, x = 1, y = 0 and y = 1 the side faes). The ube is a lose volume. The unit vetor that points out of the ube at eah of the six faes is aligned with one of the x, y and z oordinate axes. On the top surfae, the vetor must point straight up and have length one), so ˆn = k. On the bottom surfae, the vetor must point straight down and have length one), so ˆn = k. imilarly, on the four side faes, we have ˆn = i, ˆn = i, ˆn = j, and ˆn = j. For urved surfaes it is more ompliated to determine ˆn. For surfaes of the form z = fx, y), we have the following general formula. Let be a portion of a surfae of the form z = fx, y). A unit normal vetor to is ˆn = f x i f j + k ) 2 ) f x + f Note that this vetor is simply a normal vetor to the surfae. If we reverse the sign, then it is still pointing in a diretion normal to the surfae. For losed surfaes, if we want the outward pointing normal, we must pay attention to the sign. Example. Let be the surfae of the unit sphere. On, we have x 2 + y 2 + z 2 = 1. o, the surfae an be expressed as z = ± 1 x 2 y 2 = fx, y). The positive square root orresponds to points on the upper hemisphere and the negative square root orresponds to the lower hemisphere. ifferentiating gives for both signs) and so 1 + ) f 2 + x f x = x z, f = y z, ) f 2 = 1 + x2 z 2 + y2 z 2 = 1 z z 2 + x 2 + y 2. 5

6 Using the above formula, we have ˆn = xi + yj + zk z 2 + x 2 + y. 2 We know that on, x 2 + y 2 + z 2 = 1. Hene, a unit normal vetor to is ˆn = xi + yj + zk. This is, of ourse, the radial position vetor. It points in the outward normal diretion at all points on the surfae of the sphere. 6.. urfae Integrals Let be a portion of a surfae z = fx, y). Note that here ould be open or losed. The integral 1 d is the surfae area of and Gx, y, z) d is the surfae integral of G. If the surfae is of the form z = fx, y) then we an onvert the surfae integral into a standard double integral over a flat region in the x-y plane by performing a hange of variable. To do this, we first need to relate a small area δ on the possibly urved) surfae to its projetion δ or shadow ) onto the x-y plane. It an be shown that the relationship between δ and δ is δ = δ ˆn k where ˆn is the unit normal vetor to on δ and obviously k is the unit normal vetor to δ, pointing up). o, 1 Gx, y, z) d = Gx, y, zx, y)) ˆn k dxdy. Using the formula for ˆn, we have and so, Gx, y, z) d = ˆn k = 1 + f x Gx, y, zx, y)) 1 ) 2 + f 1 + ) 2 ) f 2 + x ) f 2 dxdy. Basially, when onverting to a standard integral in the x-y plane, we just need to remember the square root fator to aount for the hange in urvature. Note that if is itself flat, and parallel to the x-y plane, then ˆn = k and ˆn k = k k = 1, so there is no extra fator. Example. Evaluate the surfae integral z2 d, where is the open surfae orresponding the portion of the surfae of the unit sphere that lies in the first otant x 0, y 0, z 0). [Note: it might help to draw a diagram of here.] 6

7 is a portion of the surfae We have already seen that and so on. Hene, 1 + ) f 2 + x z 2 d = z = + 1 x 2 y 2 = fx, y). f x = x z, f = y z ) f 2 = 1 + x2 z 2 + y2 z 2 = 1 z z 2 + x 2 + y 2 = 1 z z 2 1 z dx dy = z dx dy = 1 x 2 y 2 dx dy. This is now just a standard double integral but we need to work out the limits of integration. is the projetion of onto the x-y plane. This is a quarter irle bounded by the lines x = 0, y = 0 and x 2 + y 2 = 1. It would be easier to evaluate this integral in polar o-ordinates. o, hanging variables one again, π/2 1 z 2 d = 1 r 2 rdrdθ. θ=0 on t forget the extra fator of r in the area element when we hange variables). The integral with respet to r an be done by substitution or by just identifying a funtion whose derivative is the given integrand z 2 d = π/2 θ=0 [ 1 r 2 ) /2 r=0 ] 1 0 dθ = π/2 θ=0 1 dθ = 1 π 2 = π 6. You will find more surfae integral examples on Exerise heet 10. Note that we an perform integrals over surfaes that are open as in the above example) or losed for example, the surfae of a whole sphere). In the ivergene Theorem, however, we have a surfae integral of a funtion of the form Gx, y, z) = F ˆn, where the surfae is always losed The ivergene Theorem We now have all the ingredients we need to evaluate both integrals in the ivergene Theorem. Example. how that the ivergene Theorem F d = F ˆn d is satisfied for the vetor field F = xi + yj + zk when is the unit sphere. First, we evaluate the left-hand side. Clearly, F =. Hene, Fd = 1 d, 7

8 whih is three times the volume of the sphere. This is 4π/) = 4π. For the integral on the right, we know that ˆn = xi + yj + zk the radial diretion vetor). o, on. Hene, F ˆn = x 2 + y 2 + z 2 = 1 whih is just the surfae area of the sphere. 4π1) 2 = 4π. Both integrals math. F ˆn d = Example. erify that the ivergene Theorem holds when and is the unit ube. 1 d, Using the standard formula for surfae area, this is F = y x)i + y z)j + x y)k, First we note that F = = 0. Hene F d = 0. It is not easy to desribe the surfae of the unit ube, or the normal vetor to, with a single equation. In this ase, it is best to split into 6 distint parts the 6 faes of the ube). That is, F ˆn d = 6 i=1 i F ˆn i d i, where i is the ith fae, and ˆn i is the unit normal vetor to that fae pointing out of the ube). We evaluate these surfae integrals one at a time. For example, suppose that 1 is the top fae. Then, z = 1 on 1 and ˆn 1 = k so F ˆn 1 = x y. Now, sine 1 is flat, 1 F ˆn 1 d 1 = 1 imilarly for the other faes... ee Exerise heet Line Integrals and tokes Theorem 1 x y dxdy = 0. You studied line integrals for lines in the x-y plane) in first year alulus. If you have forgotten, please revise your notes now. ee also the handout: Revision on line integrals in the plane. Consider the integral of a salar funtion f along a line segment in three spae dimensions, where starts at point a and finishes at point b. We write fx, y, z) ds. Here, s is the ar length parameter. We an use this oordinate to parametrise any line. At point a, s = 0 and at point b, the value of s is the length of the line segment joining a and b. If the oordinates x, y and z of any point on the line an be expressed easily in terms of s, then we just have a standard one dimensional integral length of line fxs), ys), zs)) ds. 8

9 Now, let F be a vetor field and onsider F ˆt ds, where ˆt is the unit tangent vetor to the urve. This integral gives the work done by the vetor field to move an objet along. If we know how to find ˆt then, given, this is just a standard line integral, as above. raw any line segment part of a urve) and mark two points, with ar length parameter s and s + δs. Let x, y, z) be the standard Cartesian oordinates of the first point, and x + δx, y + δy, z + δz) be the oordinates of the seond. Now draw a straight line joining these two points. This straight line has the vetor equation δxi + δyj + δzk = xs + δs) xs)) i + ys + δs) ys)) j + zs + δs) zs)) k. If we divide by δs and take the limit as δs 0 then we obtain the vetor dx ds i + dy ds j + dz ds k. This is, by definition, the unit tangent vetor to the urve at x, y, z). Hene ˆt = dx ds i + dy ds j + dz ds k. Using this expression for ˆt, we an also write dx F ˆt ds = F ds i + dy ds j + dz ) ds k ds = F dr, where dr = dxi + dyj + dzk. ome textbooks use the notation F dr but we will stik to F ˆt ds. Reall also that when is a losed loop we write and not. Example. Let F = yi xj and let denote the losed urve shown in Figure 1. Note that this urve lies in the x-y plane). Evaluate F ˆt ds. Using the expression for ˆt with z = 0) gives dx F ˆt ds = yi xj) ds i + dy ) ds j ds = y dx ) ds xdy ds. ds We break the urve up into three parts, and on eah straight line segment, express x and y as funtions of s. That is, y dx ) ds xdy ds = ys) dxs) xs) dys) ) ds. ds ds ds i=1 i 9

10 0,1) 1 0,0) 2 1,0) Figure 1: A losed loop orresponding to the boundary of a triangle. The arrows give the orientation of the path taken. On 1 the line joining 0, 1) to 0, 0)), we have x = 0 and y = 1 s. [Chek: the length of the line is 1. When y = 1, s = 0 and when y = 0, s = 1.] o, 1 ys) dxs) xs) dys) ) 1 ds = 1 s)0 0) ds = 0. ds ds On 2, we have y = 0 and x = s, so ys) dxs) xs) dys) ) 1 ds = 0 s0) ds = 0. ds ds 2 Finally, on, we have x = 1 s/ 2, and y = s/ 2 so ys) dxs) xs) dys) ) 1 ds = s/ ) 2 1/ ) 2 1 s/ ) 2 1/ ) 2 ds = 1. ds ds Now, we have an important theorem that onnets: the url of a vetor field, an open surfae, and the losed urve that spans the surfae. For a flat surfae in two dimensions, the losed urve is just the boundary of as in the triangle example above). In three dimensions, an open surfae may have urvature. Consider, for example, the upper half of the surfae of the unit sphere. This surfae is spanned by the losed loop orresponding to the set of points x 2 + y 2 = 1. Theorem: tokes Theorem. Let be a losed urve and let be an open surfae spanned by. Let ˆn be the unit normal vetor to oriented with respet to the right-hand rule). Then, if F is a differentiable vetor field, F ˆt ds = F) ˆn d. Note that it is important here that the vetors ˆn and ˆt are oriented orretly with respet to one another. In words, the theorem says that the integral of F ˆt around a losed loop is equal to the integral of the normal omponent of the url of F aross the surfae spanned by. 10

11 Example. Let us return to the above example with F = yi xj, where is a right-angled triangle and is the boundary. We have already omputed F ˆt ds, where the tangent vetor points in the diretion shown in Figure 1. Let ˆn be the unit normal vetor to pointing up in the positive z diretion). That is, ˆn = k. Then, sine F = 2k, we have F) ˆn d = 2 1 d, whih is minus two times the area of the triangle. Hene, F) ˆn d = = 1. This integral mathes the line integral omputed above. We see that tokes Theorem is indeed satisfied. When is a flat surfae, lying in the x-y plane, as in the above example, the omputation is relatively straight forward. Exerise heet 10 has more hallenging examples with urved surfaes. In the next example is still a triangle but is not lying in the x-y plane. Example. erify tokes Theorem when F = y + y 2) k and is the boundary of the triangle with verties 0, 0, 1), 1, 0, 0), 0, 1, 0). First we ompute F = i j k x z 0 0 y + y 2 = 1 + 2y)i. The surfae here is a portion of the plane x + y + z = 1 or equivalently, a portion of the surfae z = fx, y) where fx, y) = 1 x y. Using the general formula for finding a unit vetor that is normal to this surfae gives 1)i 1)j + k ˆn = = 1 i + j + k). The integral on the left-hand side of the theorem is F) ˆn d = y d. We an onvert this into a standard double integral in the x-y plane as follows y d = y) dxdy. We need the orret limits for x and y. The projetion of onto the x-y plane is a triangle bounded by the lines x = 0, y = 0 and x + y = 1 so y 1 + 2y d = 1 + 2y dxdy = y x

12 Next, we ompute the line integral on the right-hand side. Note that the normal vetor hosen points in the upward diretion. The tangent vetor ˆt to the boundary of the triangle should be oriented aordingly. We must travel in an anti-lokwise diretion.) First, note that and so F ˆt = y + y 2) k dx ds i + dy ds j + dz ) ds k F ˆt ds = y + y 2 ) dz ds ds. = y + y 2) dz ds We break the boundary up into three parts, making sure to travel in the right diretion. Let 1 denote the line segment from 0, 0, 1) to 1, 0, 0), let 2 denote the line segment from and 1, 0, 0) to 0, 1, 0), and let denote the line segment from and 0, 1, 0) to 0, 0, 1). On 1, s = 0 at 0, 0, 1) and s = 2 at 1, 0, 0). Along this line, y = 0 so 1 y + y 2 ) dz ds ds = 2 0 dzs) ds ds = 0. On 2, s = 0 at 1, 0, 0) and s = 2 at 0, 1, 0). Along this line, z = 0 so 2 y + y 2 ) dz ds ds = 2 y + y 2 ) 0 ds = 0. Finally, on, s = 0 at 0, 1, 0) and s = 2 at 0, 0, 1). Along this line, y = 1 s/ 2 and z = s/ 2 so y + y 2 ) dz ds ds = We see that both integrals math. 2 = [ 1 s/ 2) + 1 s/ ] 2 2) 1/ ) 2 ds 2 s 2 + s2 2 ds =