SECTIONS and 14.12

Transcript

1 HAPTE4 ETION 4. and Flu and irculation FIGUE 4.5 ector v represents direction of gas flow at point P. A is a unit area at P perpendicular to v A P FIGUE 4.52 Flu for a surface is the mass of gas passing through per unit time d n FIGUE 4.53 Flu for a closed surface is the mass of gas flowing out of the surface per unit time v n v v In man branches of engineering and phsics we encounter the concepts of flu and circulation. In this section we discuss the relationships between flu and divergence and between circulation and curl. We find a phsical setting mentioned in ection 4. of the tetbook most useful in developing these ideas. Fluid Flow onsider a gas flowing through a region D of space. At time t and point P,, in D,gas flows through P with velocit v,,,t.ifais a unit area around P perpendicular to v Figure 4.5 and ρ,,,t is densit of the gas at P, then the amount of gas crossing A per unit time is ρv. At ever point P in D, then, the vector ρv is such that its direction v gives velocit of gas flow, and its magnitude ρ v describes the mass of gas flowing in that direction per unit time. onsider some surface in D Figure If ˆn is a unit normal to, then ρv ˆn is the component of ρv normal to the surface. Ifd is an element of area on, then ρv ˆn d describes the mass of gas flowing through d per unit time. onsequentl, ρv ˆn d is the mass of gas flowing through per unit time. This quantit is called flu for surface. If is a closed surface Figure 4.53 and ˆn is the unit outer normal to, then ρv ˆn d is the mass of gas flowing out of surface per unit time. If this flu for closed is positive, then there is a net outward flow of gas through i.e., more gas is leaving the volume bounded b than is entering; if the flu is negative, the net flow is inward. If we appl the divergence theorem to the flu integral over the closed surface,wehave ρv ˆn d = ρvd Now the flu on the left side of this equation is the mass of gas per unit time leaving. In order for the right side to represent the same quantit, ρv must be interpreted as the mass of gas leaving unit volume per unit time, because then ρvd represents the mass per unit time leaving d, and the triple integral is the mass per unit time leaving. We have obtained, therefore, an interpretation of the divergence of ρv. The divergence of ρv is the flu per unit volume per unit time at a point: the mass of gas leaving unit volume in opright 2008 Pearson Education anada 5

2 6 hapter 4 ETION 4. and 4.2 FIGUE 4.54 irculation for radial gas flow is ero i j, FIGUE 4.55 irculation for circular gas flow is not ero i j, unit time. We can use this idea of flu to derive the equation of continuit for fluid flow. The triple integral ρ t d = ρd t measures the time rate of change of mass in a volume. If this triple integral is positive, then there is a net inward flow of mass; if it is negative, the net flow is outward. We conclude, therefore, that this triple integral must be the negative of the flu for the volume ; that is, ρ d = ρvd t or [ ρv ρ ] d = 0. t If ρv and ρ/t are continuous functions, then this equation can hold for arbitrar volume onl if ρv ρ = t This equation, called the equation of continuit, epresses conservation of mass; it is basic to all fluid flow. If is a closed curve in the flow region D, then the circulation of the flow for the curve is defined b Ɣ = v dr To obtain an intuitive feeling for Ɣ, we consider two ver simple two-dimensional flows. First, suppose v = î ĵ, so that all particles of gas flow along radial lines directed awa from the origin Figure In this case, the line integral defining Ɣ is independent of path and Ɣ = 0 for an curve whatsoever. econd, suppose v = î ĵ, so that all particles of the gas flow counterclockwise around circles centred at the origin Figure In this case Ɣ does not generall vanish. In particular, if is the circle 2 2 = r 2, then v and dr are parallel, and Ɣ = v dr = v ds = 2 2 ds = rds = 2πr 2. FIGUE 4.56 v is an indication of the circulator nature of a flow d These two flow patterns indicate perhaps that circulation is a measure of the tendenc for the flow to be circulator. If we appl tokes s theorem to the circulation integral for the closed curve,wehave v dr = v ˆn d, 4.63 where is an surface in the flow with boundar. If the right side of this equation is to represent the circulation for also, then v ˆn d must be interpreted as the circulation for the curve bounding d or simpl for d itself. Then the addition process of the surface integral Figure 4.56 gives the circulation around, the circulation around all internal boundaries cancelling. But if v ˆn d is the circulation for d, then it follows that v ˆn must be the circulation for unit area perpendicular to ˆn. Thus v describes the circulator nature of the flow v; its component v ˆn in an direction describes the circulation for unit area perpendicular to ˆn. Electromagnetic Theor The concepts of flu and circulation also pla a prominent role in electromagnetic theor. For eample, suppose a dielectric contains a charge distribution of densit ρ charge per unit opright 2008 Pearson Education anada

3 4. Flu and irculation 7 volume. This charge produces an electric field represented b the electric displacement vector D.If is a surface in the dielectric, then the flu of D through is defined as D ˆn d, and in the particular case in which is closed, as D ˆn d. Gauss s law states that this flu integral must be equal to the total charge enclosed b. If is the region enclosed b, we can write D ˆn d = ρd On the other hand, if we appl the divergence theorem to the flu integral, we have D ˆn d = Dd. onsequentl, Dd = ρd D ρd = 0. If D and ρ are continuous functions, then the onl wa this equation can hold for arbitrar volume in the dielectric is if D = ρ This is the first of Mawell s equations for electromagnetic fields. Another of Mawell s equations can be obtained using tokes s theorem. The flu through a surface of a magnetic field B is defined b B ˆn d. If B is a changing field, then an induced electric field intensit E is created. Farada s induction law states that the time rate of change of the flu of B through must be equal to the negative of the line integral of E around the boundar of : E dr = B ˆn d t But tokes s theorem applied to the line integral also gives E dr = E ˆn d. It follows, therefore, that if is stationar, E ˆn d = B ˆn d t E B ˆn d = 0. t Once again, this equation can hold for arbitrar surfaces,if E and B/t are continuous, onl if E = B t opright 2008 Pearson Education anada

4 8 hapter 4 ETION 4. and ector Analsis in Orthogonal oordinates We used polar, clindrical, and spherical coordinates in hapter 3 to evaluate multiple integrals; the are useful in man other contets. In this section we discuss the gradient, divergence, and curl in these coordinate sstems. We do so in detail for spherical coordinates, and develop similar results for clindrical coordinates and other orthogonal coordinate sstems in the eercises. pherical coordinates,,are related to artesian coordinates b the equations =sin cos, =sin sin, =cos With the restrictions, 0, 0 π, π < π, ever point in space ecept those on the -ais has eactl one set of spherical coordinates. Inverse to transformation 4.68 are the equations = 2 2 2, = os 2 2 2, = Tan ± π, 4.69 where in the formula for, it must be decided, depending on,, and, whether π should be added, π should be subtracted, or neither should be done. To epress a scalar function f,, in terms of spherical coordinates, we use equations 4.68 to write f sin cos, sin sin, cos. Epressing vectors in spherical coordinates is more complicated. ectors î, ĵ, and ˆk form a basis for all vectors in space; that is, ever vector v can be epressed as a linear combination v = v î v ĵ v ˆk of î, ĵ, and ˆk, where v, v, and v are the artesian components of v. To find the corresponding basis for spherical coordinates, we begin b calling a line with equations = constant, = constant a coordinate curve in artesian coordinates; it is a straight line parallel to the -ais. Ever point on the line has the same - and -coordinates; onl the -coordinate varies. A unit tangent vector to this line in the direction of increasing is î. imilarl, ĵ is tangent to the coordinate curve line = constant, = constant, and ˆk is tangent to = constant, = constant, and this is true at ever point in space. FIGUE 4.57a which onl varies oordinate curve along FIGUE 4.57b which onl varies oordinate curve along ˆ : oordinate curve along which onl varies = constant come = constant plane opright 2008 Pearson Education anada

5 4.2 ector Analsis in Orthogonal oordinates 9 FIGUE 4.57c which onl varies oordinate curve along FIGUE 4.58 curves are mutuall perpendicular Unit tangent vectors to coordinate ˆ γ The equations = constant, = constant define a half straight line from the origin; it is called a coordinate curve in spherical coordinates along which onl varies Figure 4.57a. Parametric equations for are equations 4.68 with and fied. If r = î ĵ ˆk is the position vector of points on, then a unit tangent vector to, pointing in the direction in which increases, at an point on is ˆ = r/ r/ = sin cos,sin sin,cos sin 2 cos 2 sin 2 sin 2 cos 2 = sin cos,sin sin,cos. 4.70a A unit tangent vector to the coordinate curve in Figure 4.57b along which varies, but and are constant, and points in the direction in which increases, is ˆ = r/ r/ = cos cos, cos sin, sin 2 cos 2 cos 2 2 cos 2 sin 2 2 sin 2 = cos cos,cos sin, sin. 4.70b Finall, a unit tangent vector to the coordinate curve in Figure 4.57c is ˆ = r/ r/ = sin sin, sin cos,0 2 sin 2 sin 2 2 sin 2 cos 2 = sin,cos, c All three unit vectors ˆ, ˆ, and ˆ are drawn at the same point in Figure The are mutuall perpendicular, as can also be seen algebraicall b noting that ˆ ˆ = ˆ ˆ = ˆ ˆ = Because this is true at ever point in space ecept on the -ais, spherical coordinates are said to constitute a set of orthogonal curvilinear coordinates. It is also a right-handed coordinate sstem in that ˆ ˆ = ˆ. Unlike artesian coordinates where î, ĵ, and ˆk alwas have the same direction, directions of ˆ, ˆ, and ˆ var from point to point. ectors ˆ, ˆ, and ˆ form a basis for all vectors in space; that is, ever vector v can be epressed in the form v = v ˆ v ˆ v ˆ, where v, v, and v are called the spherical opright 2008 Pearson Education anada

6 20 hapter 4 ETION 4. and 4.2 components of v. Given the artesian components of v, we can find its spherical components at an point,,b taking scalar products with ˆ, ˆ, and ˆ: v = v ˆ = v,v,v sin cos,sin sin,cos = v sin cos v sin sin v cos, 4.72a v = v ˆ = v,v,v cos cos,cos sin, sin = v cos cos v cos sin v sin, 4.72b v = v ˆ = v,v,v sin,cos,0 = v sin v cos. 4.72c. These equations define the spherical components of a vector in terms of its artesian components. It is understood that v, v, and v are to be epressed in terms of,, and. We can invert these equations and epress v, v, and v in terms of v, v, and v. calar products of v = v ˆ v ˆ v ˆ with î, ĵ, and ˆk give v = v ˆ î v ˆ î v ˆ î = v sin cos v cos cos v sin, 4.73a v = v ˆ ĵ v ˆ ĵ v ˆ ĵ = v sin sin v cos sin v cos, 4.73b v = v ˆ ˆk v ˆ ˆk v ˆ ˆk = v cos v sin. 4.73c There is an important difference between artesian and spherical components of a vector. No matter where the tail of a vector is placed, its artesian components are alwas the same; spherical components 4.72 of a vector depend on where the tail is placed. This is a direct result of the fact that î, ĵ, and ˆk alwas have the same direction, whereas ˆ, ˆ, and ˆ do not. This is illustrated in the following eample. EXAMPLE 4.29 What are the spherical components of the vector î? Evaluate these components at the points with artesian coordinates, 0, 0, 0,, 0, and,,. In each case draw î, ˆ, ˆ, and ˆ at the point. OLUTION ince artesian components of î are, 0, 0, equations 4.72 give i = sin cos, i = cos cos, i = sin. Thus, î = sin cos ˆ cos cos ˆ sin ˆ. pherical coordinates of the point with artesian coordinates, 0, 0 are,π/2, 0 so that at this point, î = ˆ Figure 4.59a. pherical coordinates of the point with artesian coordinates 0,, 0 are,π/2,π/2, and at this point î = ˆ Figure 4.59b. Finall, the point with artesian coordinates,, has spherical coordinates 3, os / 3, π/4, and at this point i = sin [os / 3] cos π/4 = / 3, i = / 3/ 2 = / 6, i = / 2; that is, î = / 6 2 ˆ ˆ 3ˆ Figure 4.59c. opright 2008 Pearson Education anada

7 4.2 ector Analsis in Orthogonal oordinates 2 FIGUE 4.59a at, 0, 0 î = ˆ FIGUE 4.59b at 0,, 0 î = ˆ FIGUE 4.59c î = 6 2 ˆ ˆ 3ˆ at,, i = i i Because spherical coordinates are orthogonal equation 4.7, calculating the scalar product of two vectors with spherical components u = u,u,u and v = v,v,v is much like that with artesian components: multipl corresponding components and add, u v = u ˆ u ˆ u ˆ v ˆ v ˆ v ˆ = u v u v u v Furthermore, using the facts that ˆ ˆ = ˆ, ˆ ˆ = ˆ, and ˆ ˆ = ˆ, we obtain u v = u ˆ u ˆ u ˆ v ˆ v ˆ v ˆ = u v u v ˆ u v u v ˆ u v u v ˆ. This is more easil remembered in determinant form, ˆ ˆ ˆ u v = u u u v v v Once again, the simplicit here is due to equations 4.7: spherical coordinates are orthogonal. EXAMPLE 4.30 Find the scalar and vector products of the vectors u = î ĵ 2 ˆk and v = î ĵ 2 ˆk in spherical coordinates b a calculating the products in artesian coordinates and transforming to spherical coordinates and b transforming u and v to spherical components and using 4.74 and OLUTION a In artesian coordinates u v = 2 2 4, and when we transform to spherical coordinates, u v = 2 sin 2 4 cos 4. Furthermore, î ĵ ˆk u v = 2 2 = 22 î 2 2 ĵ. Using 4.72 we can epress this in terms of ˆ, ˆ, and ˆ: u v = 2 2 sin cos 2 2 sin sin ˆ 2 2 cos cos 2 2 cos sin ˆ 2 2 sin 2 2 cos ˆ opright 2008 Pearson Education anada

8 22 hapter 4 ETION 4. and 4.2 = 2 2 cos 2 sin sin sin cos sin cos sin ˆ 2 2 cos 2 cos sin sin cos sin cos sin ˆ 2 2 cos 2 sin sin 2 sin cos 2 ˆ = 2 3 sin cos 2 ˆ. b pherical components of u and v are u = sin cos sin sin 2 cos ˆ cos cos cos sin 2 sin ˆ sin cos ˆ = sin 2 cos 2 sin 2 sin 2 2 cos 3 ˆ sin cos cos 2 sin cos sin 2 2 sin cos 2 ˆ sin sin cos sin sin cos ˆ = sin 2 2 cos 3 ˆ sin cos 2 sin cos 2 ˆ, and similarl, v = sin 2 2 cos 3 ˆ sin cos 2 sin cos 2 ˆ. onsequentl, u v = sin 2 2 cos 3 sin 2 2 cos 3 2 sin cos 2 sin cos 2 sin cos 2 sin cos = 2 sin 4 4 cos 6 4 sin 2 cos 4 2 sin 2 cos 2 = 2 sin 2 cos 2 4 cos 4 sin 2 4 sin 2 cos 4 2 sin 2 cos 2 = 2 sin 2 4 cos 2 and ˆ ˆ ˆ u v = sin 2 2 cos 3 2 sin cos 2 sin cos 0 sin 2 2 cos 3 2 sin cos 2 sin cos 0 = [ sin 2 2 cos 3 2 sin cos 2 sin cos sin 2 2 cos 3 2 sin cos 2 sin cos ]ˆ = 2 3 sin cos 2 ˆ. We now discuss the gradient, divergence, and curl in spherical coordinates. According to equations 4.72, the spherical components of the gradient vector f = /î / ĵ /ˆk of a function f,, are f f = sin cos sin sin cos, 4.76a f = cos cos cos sin sin, 4.76b f = sin cos. 4.76c opright 2008 Pearson Education anada

9 4.2 ector Analsis in Orthogonal oordinates 23 But with the schematic to the left = = sin cos sin sin cos, and similarl, Therefore, = cos cos cos sin sin, = sin sin sin cos. grad f = ˆ ˆ sin ˆ Hence to find spherical components of the gradient of a function f,,, we can transform its artesian components according to 4.76, or we can transform f,, to f cos sin, sin sin, cos and use EXAMPLE 4.3 Use equations 4.76 and 4.77 to find the spherical components of f when f,, = OLUTION With f = 2î 2ĵ 3 2 ˆk, equations 4.76 give f = 2 sin cos 2 sin sin 3 2 cos = 2 sin 2 cos 2 2 sin 2 sin cos 2 cos = 2 sin cos 2 cos, f = 2 cos cos 2 cos sin 3 2 sin = 2 sin cos cos 2 2 sin cos sin sin cos 2 = 2 sin cos 3 2 sin cos 2, f = 2 sin 2 cos = 2 sin sin cos 2 sin sin cos = 0. Thus, grad f = 2 sin cos 3 ˆ 2 sin cos 3 2 sin cos 2 ˆ. Alternativel, with f cos sin, sin sin, cos = 2 sin 2 3 cos 3, equations 4.77 give grad f = 2 sin cos 3 ˆ 22 sin cos 3 3 cos 2 sin ˆ. The divergence of a vector function v = v î v ĵ v ˆk is opright 2008 Pearson Education anada v = v v v. 4.78

10 24 hapter 4 ETION 4. and 4.2 This can be changed to spherical coordinates b replacing,, and according to Alternativel, using the schematic to the left ields With 4.69, v = v v v. v n, v, v = = = 2 / sin cos = = sin cos, [ ] = /2 = 2 2 sin sin = 2 sin 2 cos cos, = sin sin, and therefore v = sin cos v cos cos v sin v sin. imilarl, v = sin sin v cos sin v cos v sin, v = cos v sin v, When we substitute for v, v, and v from equations 4.73, v = sin cos v sin cos v sin v cos cos cos cos sin sin = sin cos v sin cos v sin v cos cos v sin cos v sin v cos cos sin cos v sin v cos cos v cos cos sin cos v v cos cos sin v cos cos v v sin cos sin sin cos v sin v sin sin sin v v cos cos cos v v cos sin, opright 2008 Pearson Education anada

11 4.2 ector Analsis in Orthogonal oordinates 25 v = sin sin v sin sin v cos v cos sin cos sin cos sin = sin sin v sin sin v cos v cos sin v sin sin v cos v cos sin sin sin v cos v cos sin v cos sin sin sin v v cos sin cos v cos sin v v sin sin cos sin sin v sin v sin cos cos v v sin cos sin v v cos cos, v = cos v cos v sin sin = cos cos v sin v sin When these are added together and simplified, the result is v cos v sin cos v v sin sin v v cos div v = v 2 v v cot v v sin a This can be epressed in a form resembling that for divergence in artesian coordinates, [ div v = 2 sin 2 sin v sin v ] v. 4.79b Thus, to find the divergence of a vector function in spherical coordinates, we use 4.78 when its artesian components v, v, and v are known, and then substitute for,, and in terms of,, and, or we use 4.79 when its spherical components v, v, and v are known. EXAMPLE 4.32 Find div v in spherical coordinates using 4.78 and 4.79 when v = 2 2 î ˆk. OLUTION Using 4.78, div v = 2 = 2 sin cos. Alternativel, since v = 2 2 sin cos cos = 2 sin 3 cos cos 2, v = 2 2 cos cos sin = 2 sin 2 cos cos sin cos, v = 2 2 sin = 2 sin 2 sin, opright 2008 Pearson Education anada

12 26 hapter 4 ETION 4. and 4.2 equation 4.79a gives div v = 2 sin 3 cos cos sin 3 cos cos 2 22 sin cos 2 cos 2 sin 3 cos cos 2 sin 2 sin 2 sin 2 cos = 2 sin cos. The curl of a vector function v = v î v ĵ v ˆk is î ĵ ˆk curl F = = v v v v v v î v v ĵ v ˆk It can be changed to spherical coordinates b replacing,, and with,, and, and transforming components according to equations Alternativel, according to 4.72, the spherical components of vare v v v = v sin cos v v = v cos cos v v = v sin Using chain rules gives v v = v v = sin sin v v v v v v cos v sin v sin v v cos v cos sin sin sin cos sin cos. v cos sin v v v v v cos v v sin cos v sin v cos v cos sin sin v cos sin v sin sin sin v v cos sin cos sin v v sin sin cos v opright 2008 Pearson Education anada ; cos, sin,

13 v v = v = cos v v 4.2 ector Analsis in Orthogonal oordinates 27 v v sin cos v cos cos v sin v v sin sin cos v v cos cos cos cos v v sin cos sin v sin cos cos v sin v cos cos sin sin v v = v v = sin cos These can be simplified to cos v v sin sin v v cos cos v sin v v ; v v sin sin v cos sin v cos v v cos cos sin sin v v cos sin cos sin v v sin sin cos v sin sin sin v sin v sin cos cos sin v v cos cos cos v v sin sin sin sin cos v cos cos v sin v cos sin sin cos v v cos cos cos cos v v sin cos sin v cos sin cos v sin v sin sin cos cos v v cos sin sin v v cos. v = v v sin cot v, v = opright 2008 Pearson Education anada v sin v v,

14 28 hapter 4 ETION 4. and 4.2 Thus, curl v = v = v v v. v sin v v v v v cot ˆ ˆ. sin v v v ˆ 4.8a This can be epressed in determinant form as is the case in artesian coordinates. Epanding the following determinant gives ˆ ˆ sin ˆ [ 2 sin = sin v 2 sin cos v v ˆ v v sin v v sin v ] sin v ˆ v v v sin ˆ Hence, we ma write that curl v = = v sin sin 2 sin v cot v v v v v v v ˆ. ˆ ˆ ˆ ˆ sin ˆ. 4.8b v v sin v EXAMPLE 4.33 Find the spherical components of the curl of the vector function in Eample 4.32 using 4.80 and 4.8b. OLUTION Using 4.80, î ĵ ˆk curl v = = 2ˆk Equations 4.72 give the spherical components of this vector, v = 2 cos = 2 sin cos sin, v = 2 sin = 2 sin 2 sin, v = 0. opright 2008 Pearson Education anada

15 4.2 ector Analsis in Orthogonal oordinates 29 Hence, curl v = 2 sin cos sin ˆ 2 sin 2 sin ˆ = 2 sin sin cos ˆ sin ˆ. Alternativel, using 4.8b and the spherical components of v calculated in Eample 4.32, ˆ ˆ sin ˆ curl v = 2 sin 2 sin 3 cos 3 sin 2 cos cos 3 sin 3 sin cos 2 2 sin cos = 2 sin [ 33 sin 2 cos sin 3 sin 2 cos sin ˆ 3 2 sin 3 sin 2 sin 3 sin ˆ 3 2 sin 2 cos cos 2 sin cos 3 2 sin 2 cos cos 2 cos sin sin ˆ] = 2 sin sin cos ˆ sin ˆ. In Eample 2.9 of ection 2.6 we used chain rules to find the Laplacian in polar coordinates. hain rules could also be used to find the Laplacian in spherical coordinates. It is much easier, however, to use 4.77 and 4.79b: [ 2 = = 2 sin sin ] 2 sin sin = sin sin 2 2 sin EXEIE 4.2 In Eercises 7 we develop results for clindrical coordinates parallel to those for spherical coordinates.. how that unit tangent vectors to coordinate curves in clindrical coordinates have artesian components ˆr = cos î sin ĵ, ˆ = sin î cos ĵ, ˆk = ˆk. 2. If v = v r ˆr v ˆ v ˆk are clindical components of a vector with artesian components v = v î v ĵ v ˆk, show that v r = v cos v sin, v = v r cos v sin, 3. how that the scalar and vector products of vectors u = u r ˆr u ˆ u ˆk and v = v r ˆr v ˆ v ˆk are ˆr ˆ ˆk u v = u r v r u v u v, u v = u r u u. v r v v 4. how that clindrical components of the gradient of a function f,, are v = v sin v cos, v = v, v = v r sin v cos, v = v. grad f = r ˆr r ˆ ˆk. opright 2008 Pearson Education anada

16 30 hapter 4 ETION 4. and how that the divergence of a vector function with clindrical components v = v r ˆr v ˆ v ˆk is div v = [ r r rv r v r v ]. 6. how that the curl of a vector function with clindrical components v = v r ˆr v ˆ v ˆk is curl v = ˆr r ˆ ˆk. r r v r rv v 7. Use Eercises 4 and 5 to develop the Laplacian in clindrical coordinates, 2 = 2 r 2 r 8. epeat Eample 4.29 for ĵ and ˆk. r r epeat Eample 4.29 for î and ĵ, and clindrical coordinates. 0. Find the scalar and vector products of the vectors u = î ĵ ˆk and v = 2 2 ˆk in clindrical coordinates b a calculating the products in artesian coordinates and transforming to clindrical coordinates and b transforming u and v to clindrical coordinates and using Eercise 3.. epeat Eample 4.30 in clindrical coordinates. 2. epeat Eercise 0 for spherical coordinates. In Eercises 3 6 find clindrical and spherical components of the vector see Eercise F = 2î 3ĵ ˆk 4. F = 3î 2ĵ 2 ˆk 5. F = î ˆk 6. F = î 2 2 ĵ ˆk In Eercises 7 20 find components of f in clindrical and spherical coordinates see Eercise 4 for clindrical coordinates. 7. f,, = f,, = 2 9. f,, = f,, = 2 2 e In Eercises 2 24 find the divergence of the vector function in clindrical and spherical coordinates. Do so b a using 4.78 and then transforming coordinates and b finding clindrical and spherical components of v, and then using Eercise 5 and equation 4.79b. 2. v = 2 2 î ĵ 22. v = î ĵ ˆk 23. v = 2 î 2 ĵ 2 ˆk 24. v = î ĵ ˆk In Eercises find the curl of the vector function in clindrical and spherical coordinates. Do so b a using 4.80 and then transforming coordinates and b finding clindrical and spherical components of v, and then using Eercise 6 and equation 4.8b. 25. v = î ĵ ˆk 26. v = î ĵ ˆk 27. When v = v î v ĵ is the fluid velocit in two-dimensional, stead-state, incompressible flow, the equation of continuit demands that v v = 0 see equation 4.6 with ρ constant. how that if v = v r ˆr v ˆ are polar components of v, then the equation of continuit takes the form v r r v r v r = 0. r 28. The velocit and acceleration of a particle with displacement r = tî tĵ tˆk t being time are v = ṙ = ẋî ẏĵ ż ˆk, a = r = v = ẍî ÿĵ ˆk, where indicates differentiation with respect to t. how that velocit and acceleration in clindrical coordinates are v = ṙ ˆr r ˆ ż ˆk, a = r r 2 ˆr r 2ṙ ˆ ˆk. 29. The velocit and acceleration of a particle with displacement r = tî tĵ tˆk t being time are v = ṙ = ẋî ẏĵ ż ˆk, a = r = v = ẍî ÿĵ ˆk, where indicates differentiation with respect to t. how that velocit and acceleration in spherical coordinates are v = ˆ ˆ sin ˆ, a = 2 sin 2 2 ˆ 2 sin cos 2 ˆ 2 sin 2 cos sin ˆ. opright 2008 Pearson Education anada

17 4.2 ector Analsis in Orthogonal oordinates 3 In Eercises we develop results for an arbitrar set of orthogonal curvilinear coordinates analogous to those for spherical and clindrical coordinates. uppose that u, v, w are curvilinear coordinates in some region of space related to artesian coordinates b equations = u, v, w, = u, v, w, = u, v, w, If r = î ĵ ˆk, let û = ˆv = ŵ = r/u r/u = r/v r/v = u = u,,, v = v,,, w = w,,. h u r u, h v r v, r/w r/w = h w r w, be unit tangent vectors to the coordinate curves. These vectors are mutuall perpendicular when the curvilinear coordinates are orthogonal. 30. how that curvilinear components q = q u û q v ˆv q w ŵ of a vector function with artesian components q = q î q ĵ q ˆk are related b q u = q h u u q q v = q h v q w = q h w u q u v q v q v,, w q w q w q = q u h u u q v h v v q w h w w,, q = q u h u u q v h v v q w h w w, q = q u h u u q v h v v q w h w w. 3. how that curvilinear components of the gradient of a function f,, are grad f = h u uû h v v ˆv h w wŵ. 32. The divergence of a vector function with curvilinear components q = q u û q v ˆv q w ŵ is div q = h u h v h w [ u h vh w q u v h uh w q v ] w h uh v q w. how that this reduces to equation 4.79b in spherical coordinates. 33. The curl of a vector function with curvilinear components q = q u û q v ˆv q w ŵ is h u û h v ˆv h w ŵ curl q =. h u h v h w u v w h u q u h v q v h w q w how that this reduces to equation 4.8b in spherical coordinates. 34. Use Eercises 3 and 32 to show that the Laplacian in u, v, w coordinates is 2 = [ hv h w h u h v h w u h u u ] hu h v. w w h w hu h w v h v v opright 2008 Pearson Education anada