The formulas in Table 16.1 then give. M = d ds = s2 - zd ds = s2 - sin td dt = 2p - 2. = s2 sin t - sin 2 td dt = 8 - p. = 8 - p 2.

Transcription

1 6. ine Integrals 93 The formulas in Table 6. then give p M = d ds = s - d ds = s - sin td dt = p - p M = d ds = s - d ds = ssin tds - sin td dt p = s sin t - sin td dt = 8 - p = M M = 8 - p # p - = 8 - p 4p With to the nearest hundredth, the center of mass is (,,.57). t b t a (, ) s k height f (, ) Plane curve IGURE 6.5 The line integral ƒ ds gives the area of the portion of the clindrical surface or wall beneath = ƒs, d Ú. ine Integrals in the Plane There is an interesting geometric interpretation for line integrals in the plane. If is a smooth curve in the -plane parametried b rstd = stdi + stdj, a t b, we generate a clindrical surface b moving a straight line along orthogonal to the plane, holding the line parallel to the -ais, as in Section.6. If = ƒs, d is a nonnegative continuous function over a region in the plane containing the curve, then the graph of ƒ is a surface that lies above the plane. The clinder cuts through this surface, forming a curve on it that lies above the curve and follows its winding nature. The part of the clindrical surface that lies beneath the surface curve and above the -plane is like a winding wall or fence standing on the curve and orthogonal to the plane. At an point (, ) along the curve, the height of the wall is ƒs, d. We show the wall in igure 6.5, where the top of the wall is the curve ling on the surface = ƒs, d. (We do not displa the surface formed b the graph of ƒ in the figure, onl the curve on it that is cut out b the clinder.) rom the definition n ƒ ds = lim n: q a ƒs k, k d s k, k = where s k : as n : q, we see that the line integral ƒ ds is the area of the wall shown in the figure. Eercises 6. Graphs of Vector Equations Match the vector equations in Eercises 8 with the graphs (a) (h) given here. a. b. c. d. (,, )

2 94 hapter 6: Integration in Vector ields e. f. (,, ) (,, ) (,, ) (,, ) g. h.. rstd = ti + s - tdj, t. rstd = i + j + tk, - t 3. rstd = s cos tdi + s sin tdj, t p 4. rstd = ti, - t 5. rstd = ti + tj + tk, t 6. rstd = tj + s - tdk, t 7. rstd = st - dj + tk, - t 8. rstd = s cos tdi + s sin tdk, t p Evaluating ine Integrals over Space urves 9. Evaluate s + d ds where is the straight-line segment = t, = s - td, =, from (,, ) to (,, ).. Evaluate s d ds where is the straight-line segment = t, = s - td, =, from (,, ) to (,, ).. Evaluate s + + d ds along the curve rstd = ti + tj + s - tdk, t.. Evaluate + ds along the curve rstd = s4 cos tdi + s4 sin tdj + 3tk, -p t p. 3. ind the line integral of ƒs,, d = + + over the straightline segment from (,, 3) to s, -, d. 4. ind the line integral of ƒs,, d = 3>s + + d over the curve rstd = ti + tj + tk, t q. 5. Integrate ƒs,, d = + - over the path from (,, ) to (,, ) (see accompaning figure) given b : rstd = ti + t j, t : rstd = i + j + tk, t (,, ) (,, ) (,, ) 3 (,, ) (,, ) (a) (b) The paths of integration for Eercises 5 and Integrate ƒs,, d = + - over the path from (,, ) to (,, ) (see accompaning figure) given b : rstd = tk, t : rstd = tj + k, t 3 : rstd = ti + j + k, t 7. Integrate ƒs,, d = s + + d>s + + d over the path rstd = ti + tj + tk, 6 a t b. 8. Integrate ƒs,, d = - + over the circle rstd = sa cos tdj + sa sin tdk, t p. ine Integrals over Plane urves 9. Evaluate ds, where is a. the straight-line segment = t, = t>, from (, ) to (4, ). b. the parabolic curve = t, = t, from (, ) to (, 4).. Evaluate + ds, where is a. the straight-line segment = t, = 4t, from (, ) to (, 4). b. ; is the line segment from (, ) to (, ) and is the line segment from (, ) to (, ).. ind the line integral of ƒs, d = e along the curve rstd = 4ti - 3tj, - t.. ind the line integral of ƒs, d = along the curve rstd = (cos t)i + (sin t)j, t p. 3. Evaluate ds, where is the curve = t, = t 3, for 4>3 t. 4. ind the line integral of ƒs, d = > along the curve rstd = t 3 i + t 4 j, > t. 5. Evaluate A + B ds where is given in the accompaning figure. (, ) 5 5 (, )

3 6. Vector ields and ine Integrals: Work, irculation, and lu Evaluate where is given in the accompaning + + ds figure. (, ) (, ) (, ) (, ) In Eercises 7 3, integrate ƒ over the given curve. 7. ƒs, d = 3 >, : = >, 8. ƒs, d = s + d> +, : = > from (, > ) to (, ) 9. ƒs, d = +, : + = 4 in the first quadrant from (, ) to (, ) 3. ƒs, d = -, : + = 4 in the first quadrant from (, ) to s, d 3. ind the area of one side of the winding wall standing orthogonall on the curve =,, and beneath the curve on the surface ƒs, d = ind the area of one side of the wall standing orthogonall on the curve + 3 = 6, 6, and beneath the curve on the surface ƒs, d = Masses and Moments 33. Mass of a wire ind the mass of a wire that lies along the curve rstd = st - dj + tk, t, if the densit is d = s3>dt. 34. enter of mass of a curved wire A wire of densit ds,, d = 5 + lies along the curve rstd = st - dj + tk, - t. ind its center of mass. Then sketch the curve and center of mass together. 35. Mass of wire with variable densit ind the mass of a thin wire ling along the curve rstd = ti + tj + s4 - t dk, t, if the densit is (a) d = 3t and (b) d =. 36. enter of mass of wire with variable densit ind the center of mass of a thin wire ling along the curve rstd = ti + tj + s>3dt 3> k, t, if the densit is d = 35 + t. 37. Moment of inertia of wire hoop A circular wire hoop of constant densit d lies along the circle + = a in the -plane. ind the hoop s moment of inertia about the -ais. 38. Inertia of a slender rod A slender rod of constant densit lies along the line segment rstd = tj + s - tdk, t, in the -plane. ind the moments of inertia of the rod about the three coordinate aes. 39. Two springs of constant densit A spring of constant densit lies along the heli a. ind. b. Suppose that ou have another spring of constant densit d that is twice as long as the spring in part (a) and lies along the heli for t 4p. Do ou epect I for the longer spring to be the same as that for the shorter one, or should it be different? heck our prediction b calculating I for the longer spring. 4. Wire of constant densit A wire of constant densit d = lies along the curve ind and I. 4. The arch in Eample 3 ind for the arch in Eample enter of mass and moments of inertia for wire with variable densit ind the center of mass and the moments of inertia about the coordinate aes of a thin wire ling along the curve if the densit is d = >st + d. OMPUTER EXPORATIONS In Eercises 43 46, use a AS to perform the following steps to evaluate the line integrals rstd = st cos tdi + st sin tdj + A>3Bt 3> k, t. a. ind ds = ƒ vstd ƒ dt for the path rstd = gstdi + hstdj + kstdk. b. Epress the integrand ƒsgstd, hstd, kstdd ƒ vstd ƒ as a function of the parameter t. c. Evaluate ƒ ds using Equation () in the tet. ƒs,, d = ; rstd = ti + t j + 3t k, t ƒs,, d = ; rstd = ti + 3 t j + tk, t ƒs,, d = - 3 ; rstd = scos tdi + ssin tdj + 5tk, t p ƒs,, d = a + 9 >4 4 >3 b ; rstd = scos tdi + ssin tdj + t 5> k, rstd = scos tdi + ssin tdj + tk, t p. I rstd = ti + 3 t p I t 3> j + t k, t, d 6. Vector ields and ine Integrals: Work, irculation, and lu Gravitational and electric forces have both a direction and a magnitude. The are represented b a vector at each point in their domain, producing a vector field. In this section we show how to compute the work done in moving an object through such a field b using a line integral involving the vector field. We also discuss velocit fields, such as the vector

4 6. Vector ields and ine Integrals: Work, irculation, and lu 935 alculating lu Across a Smooth losed Plane urve slu of = Mi + Nj across d = M d - N d (7) The integral can be evaluated from an smooth parametriation = gstd, = hstd, a t b, that traces counterclockwise eactl once. EXAMPE 8 ind the flu of = s - di + j across the circle + = in the -plane. (The vector field and curve were shown previousl in igure 6.9.) Solution The parametriation rstd = scos tdi + ssin tdj, t p, traces the circle counterclockwise eactl once. We can therefore use this parametriation in Equation (7). With M = - = cos t - sin t, d = dssin td = cos t dt N = = cos t, d = dscos td = -sin t dt, we find p lu = M d - N d = scos t - sin t cos t + cos t sin td dt p = cos p t dt = + cos t dt = c t + sin t 4 p d = p. Eq. (7) The flu of across the circle is p. Since the answer is positive, the net flow across the curve is outward. A net inward flow would have given a negative flu. Eercises 6. Vector ields ind the gradient fields of the functions in Eercises ƒs,, d = s + + d -> ƒs,, d = ln + + gs,, d = e - ln s + d gs,, d = Give a formula = Ms, di + Ns, dj for the vector field in the plane that has the propert that points toward the origin with magnitude inversel proportional to the square of the distance from (, ) to the origin. (The field is not defined at (, ).) 6. Give a formula = Ms, di + Ns, dj for the vector field in the plane that has the properties that = at (, ) and that at an other point (a, b), is tangent to the circle + = a + b and points in the clockwise direction with magnitude ƒ ƒ = a + b. ine Integrals of Vector ields In Eercises 7, find the line integrals of from (,, ) to (,, ) over each of the following paths in the accompaning figure. a. The straight-line path : rstd = ti + tj + tk, t b. The curved path rstd = ti + t j + t 4 : k, t c. The path 3 4 consisting of the line segment from (,, ) to (,, ) followed b the segment from (,, ) to (,, ) 7. = 3i + j + 4k 8. = [>s + d]j 9. = i - j + k. = i + j + k.. = s3-3di + 3j + k = s + di + s + dj + s + dk (,, ) (,, ) 4 3 (,, )

5 936 hapter 6: Integration in Vector ields ine Integrals with Respect to,, and In Eercises 3 6, find the line integrals along the given path. 3. ( - ) d, where : = t, = t +, for t 3 4. d, where : = t, = t, for t 5. ( + ) d, where is given in the accompaning figure d, where is given in the accompaning figure. 7. Along the curve r(t) = ti - j + t k, t, evaluate each of the following integrals. a. b. c. 8. Along the curve r(t) = (cos t)i + (sin t)j - (cos t)k, t p, evaluate each of the following integrals. a. d b. d c. Work In Eercises 9, find the work done b over the curve in the direction of increasing t ( + - ) d ( + - ) d ( + - ) d (, ) = i + j - k rstd = ti + t j + tk, t = i + 3j + s + dk rstd = scos tdi + ssin tdj + st>6dk, = i + j + k rstd = ssin tdi + scos tdj + tk, = 6i + j + k rstd = ssin tdi + scos tdj + st>6dk, (, ) (3, ) (, 3) (, 3) 5 3 (3, 3) d t p t p t p ine Integrals in the Plane 3. Evaluate d + s + d d along the curve = from s -, d to (, 4). 4. Evaluate s - d d + s + d d counterclockwise around the triangle with vertices (, ), (, ), and (, ). 5. Evaluate # T ds for the vector field = i - j along the curve = from (4, ) to s, -d. 6. Evaluate # dr for the vector field = i - j counterclockwise along the unit circle + = from (, ) to (, ). Work, irculation, and lu in the Plane 7. Work ind the work done b the force over the straight line from (, ) to (, 3). 8. Work ind the work done b the gradient of ƒs, d = s + d counterclockwise around the circle + = 4 from (, ) to itself. 9. irculation and flu ind the circulation and flu of the fields around and across each of the following curves. a. The circle rstd = scos tdi + ssin tdj, b. The ellipse rstd = scos tdi + s4 sin tdj, 3. lu across a circle ind the flu of the fields = i - 3j and = i + s - dj across the circle rstd = sa cos tdi + sa sin tdj, t p. In Eercises 3 34, find the circulation and flu of the field around and across the closed semicircular path that consists of the semicircular arch r std = sa cos tdi + sa sin tdj, t p, followed b the line segment r std = ti, -a t a. 3. = i + j 3. = i + j 33. = -i + j 34. = - i + j 35. low integrals ind the flow of the velocit field = s + di - s + dj along each of the following paths from (, ) to s -, d in the -plane. a. The upper half of the circle + = b. The line segment from (, ) to s -, d c. The line segment from (, ) to s, -d followed b the line segment from s, -d to s -, d 36. lu across a triangle ind the flu of the field in Eercise 35 outward across the triangle with vertices (, ), (, ), s -, d. 37. ind the flow of the velocit field = i + j along each of the following paths from (, ) to (, 4). a. b. (, ) = i + j and = -i + j (, 4) 5 (, ) c. Use an path from (, ) to (, 4) different from parts (a) and (b). t p (, 4) 5 = i + s - dj t p

6 6. Vector ields and ine Integrals: Work, irculation, and lu ind the circulation of the field = i + ( + )j around each of the following closed paths. a. (, ) (, ) (, ) (, ) 43. Unit vectors pointing toward the origin ind a field = Ms, di + Ns, dj in the -plane with the propert that at each point s, d Z s, d, is a unit vector pointing toward the origin. (The field is undefined at (, ).) 44. Two central fields ind a field = Ms, di + Ns, dj in the -plane with the propert that at each point s, d Z s, d, points toward the origin and ƒ ƒ is (a) the distance from (, ) to the origin, (b) inversel proportional to the distance from (, ) to the origin. (The field is undefined at (, ).) 45. Work and area Suppose that ƒ(t) is differentiable and positive for a t b. et be the path rstd = ti + ƒstdj, a t b, and = i. Is there an relation between the value of the work integral b. 5 4 # dr and the area of the region bounded b the t-ais, the graph of ƒ, and the lines t = a and t = b? Give reasons for our answer. 46. Work done b a radial force with constant magnitude A particle moves along the smooth curve = ƒsd from (a, ƒ(a)) to (b, ƒ(b)). The force moving the particle has constant magnitude k and alwas points awa from the origin. Show that the work done b the force is c. Use an closed path different from parts (a) and (b). Vector ields in the Plane 39. Spin field Draw the spin field =- (see igure 6.) along with its horiontal and vertical components at a representative assortment of points on the circle + = Radial field Draw the radial field (see igure 6.) along with its horiontal and vertical components at a representative assortment of points on the circle + =. 4. A field of tangent vectors a. ind a field G = Ps, di + Qs, dj in the -plane with the propert that at an point sa, bd Z s, d, G is a vector of magnitude a + b tangent to the circle + = a + b and pointing in the counterclockwise direction. (The field is undefined at (, ).) b. How is G related to the spin field in igure 6.? 4. A field of tangent vectors + i + = i + j + j a. ind a field G = Ps, di + Qs, dj in the -plane with the propert that at an point sa, bd Z s, d, G is a unit vector tangent to the circle + = a + b and pointing in the clockwise direction. b. How is G related to the spin field in igure 6.? low Integrals in Space In Eercises 47 5, is the velocit field of a fluid flowing through a region in space. ind the flow along the given curve in the direction of increasing t # T ds = ksb + sƒsbdd d > - sa + sƒsadd d > D. = -4i + 8j + k rstd = ti + t j + k, t = i + j + k rstd = 3tj + 4tk, t = s - di + k rstd = scos tdi + ssin tdk, = -i + j + k rstd = s - cos tdi + s sin tdj + tk, 5. irculation ind the circulation of = i + j + k around the closed path consisting of the following three curves traversed in the direction of increasing t. : rstd = scos tdi + ssin tdj + tk, t p> : rstd = j + sp>ds - tdk, t 3 : rstd = ti + s - tdj, t t p,, (,, ) (,, ) 3 t p

7 938 hapter 6: Integration in Vector ields 5. Zero circulation et be the ellipse in which the plane = meets the clinder + =. Show, without evaluating either line integral directl, that the circulation of the field = i + j + k around in either direction is ero. 53. low along a curve The field = i + j - k is the velocit field of a flow in space. ind the flow from (,, ) to (,, ) along the curve of intersection of the clinder = and the plane =. (Hint: Use t = as the parameter.) 54. low of a gradient field ind the flow of the field = s 3 d: a. Once around the curve in Eercise 5, clockwise as viewed from above b. Along the line segment from (,, ) to s,, -d. (,, ) OMPUTER EXPORATIONS In Eercises 55 6, use a AS to perform the following steps for finding the work done b force over the given path: a. ind dr for the path rstd = gstdi + hstdj + kstdk. b. Evaluate the force along the path. c. Evaluate = 6 i + 3s 5 + dj; rstd = s cos tdi + ssin tdj, t p 3 = + i + j; rstd = scos tdi + ssin tdj, + t p = s + cos di + s + cos dj + s + cos dk; rstd = ( cos t)i + (3 sin t)j + k, t p = i - j + e k; rstd = -ti + tj + 3tk, t 4 = s + sin di + s + s>3dcos dj + 4 k; rstd = ssin tdi + scos tdj + ssin tdk, -p> t p> = s di j + k; rstd = scos tdi + ssin tdj + s sin t - dk, # dr. t p 6.3 Path Independence, onservative ields, and Potential unctions A gravitational field G is a vector field that represents the effect of gravit at a point in space due to the presence of a massive object. The gravitational force on a bod of mass m placed in the field is given b = mg. Similarl, an electric field E is a vector field in space that represents the effect of electric forces on a charged particle placed within it. The force on a bod of charge q placed in the field is given b = qe. In gravitational and electric fields, the amount of work it takes to move a mass or charge from one point to another depends on the initial and final positions of the object not on which path is taken between these positions. In this section we stud vector fields with this propert and the calculation of work integrals associated with them. Path Independence If A and B are two points in an open region D in space, the line integral of along from A to B for a field defined on D usuall depends on the path taken, as we saw in Section 6.. or some special fields, however, the integral s value is the same for all paths from A to B. DEINITIONS et be a vector field defined on an open region D in space, and suppose that for an two points A and B in D the line integral # dr along a path from A to B in D is the same over all paths from A to B. Then the integral # dr is path independent in D and the field is conservative on D. The word conservative comes from phsics, where it refers to fields in which the principle of conservation of energ holds. When a line integral is independent of the path from

8 6.3 Path Independence, onservative ields, and Potential unctions 947 EXAMPE 6 Show that d + d + 4 d is eact and evaluate the integral s,3, -d s,,d d + d + 4 d over an path from (,, ) to s, 3, -d. Solution We let M =, N =, P = 4 and appl the Test for Eactness: P = = N, M = = P, N = = M. These equalities tell us that d + d + 4 d is eact, so for some function ƒ, and the integral s value is ƒs, 3, -d - ƒs,, d. We find ƒ up to a constant b integrating the equations rom the first equation we get ƒ =, The second equation tells us that d + d + 4 d = dƒ ƒ =, ƒs,, d = + gs, d. ƒ = 4. ƒ = + g =, or g =. (4) Hence, g is a function of alone, and The third of Equations (4) tells us that ƒ = + dh = 4, or hsd = 4 +. d Therefore, ƒs,, d = + hsd. ƒs,, d = The value of the line integral is independent of the path taken from (,, ) to (, 3, -), and equals ƒs, 3, -d - ƒs,, d = + - s5 + d = -3. Eercises 6.3 Testing for onservative ields Which fields in Eercises 6 are conservative, and which are not? = i + j + k = s sin di + s sin dj + s cos dk = i + s + dj - k = -i + j = s + di + j + s + dk = se cos di - se sin dj + k inding Potential unctions In Eercises 7, find a potential function ƒ for the field. 7. = i + 3j + 4k = s + di + s + dj + s + dk = e + si + j + kd = s sin di + s sin dj + s cos dk = sln + sec s + ddi + = asec s + d + + bj + + i + a bj + + k a - + bk

9 948 hapter 6: Integration in Vector ields Eact Differential orms In Eercises 3 7, show that the differential forms in the integrals are eact. Then evaluate the integrals inding Potential unctions to Evaluate ine Integrals Although the are not defined on all of space R 3, the fields associated with Eercises 8 are simpl connected and the omponent Test can be used to show the are conservative. ind a potential function for each field and evaluate the integrals as in Eample Applications and Eamples 3. Revisiting Eample 6 Evaluate the integral from Eample 6 b finding parametric equations for the line segment from (,, ) to s, 3, -d and evaluating the line integral of = i + j + 4k along the segment. Since is conservative, the integral is independent of the path. 4. Evaluate along the line segment joining (,, ) to (, 3, 4). Independence of path Show that the values of the integrals in Eercises 5 and 6 do not depend on the path taken from A to B In Eercises 7 and 8, find a potential function for. 7. s,,d s3,5,d s,,d s,,3d s,,d s3,3,d s,,d s,,d s,,d s,,3d s,,d s,,d s,,d s,,d s,,d s,,d A B A s,3, -6d s,,d s,p>,d s-, -, -d B d + d + d d + d + d d + s - d d - d d - d - sin cos d + cos sin d + d 3 d + d + ln d s ln - d d + a d + a - b d - d s,3, -d s,,d d + d + d d + d + d + + = - i + a bj, {(, ): 7 } 8. = se ln di + a e cos d + a - sin b d + d d + d + d d - b d - d d + d + 4 d d + d + s >d d + sin bj + s cos dk 9. Work along different paths ind the work done b = s + di + s + dj + e k over the following paths from (,, ) to (,, ). a. The line segment =, =, b. The heli rstd = scos tdi + ssin tdj + st>pdk, t p c. The -ais from (,, ) to (,, ) followed b the parabola =, = from (,, ) to (,, ) 3. Work along different paths ind the work done b = e i + se + cos dj + se + sin dk over the following paths from (,, ) to s, p>, d. a. The line segment =, = pt>, = - t, t (,, ) b. The line segment from (,, ) to the origin followed b the line segment from the origin to s, p>, d (,, ) c. The line segment from (,, ) to (,, ), followed b the -ais from (,, ) to the origin, followed b the parabola = p >, = from there to s, p>, d (,, ) (,, ) (,, ) 5 (,, ) (,, ) p,, (,, ) (,, ) 5 p p,, p,, p p

10 6.4 Green s Theorem in the Plane Evaluating a work integral two was et = s 3 d and let be the path in the -plane from s -, d to (, ) that consists of the line segment from s -, d to (, ) followed b the line segment from (, ) to (, ). Evaluate # dr in two was. a. ind parametriations for the segments that make up and evaluate the integral. b. Use ƒs, d = 3 as a potential function for. 3. Integral along different paths Evaluate the line integral cos d - sin d along the following paths in the -plane. a. The parabola = s - d from (, ) to (, ) b. The line segment from s -, pd to (, ) c. The -ais from s -, d to (, ) d. The astroid rstd = scos 3 tdi + ssin 3 tdj, t p, counterclockwise from (, ) back to (, ) (, ) 33. a. Eact differential form How are the constants a, b, and c related if the following differential form is eact? sa + cd d + sb + cd d + sa + c d d b. Gradient field or what values of b and c will = s + cdi + sb + cdj + s + c dk be a gradient field? (, ) (, ) (, ) 34. Gradient of a line integral Suppose that = ƒis a conservative vector field and Show that g =. 35. Path of least work You have been asked to f ind the path along which a force field will perform the least work in moving a particle between two locations. A quick calculation on our part shows to be conservative. How should ou respond? Give reasons for our answer. 36. A revealing eperiment B eperiment, ou find that a force field performs onl half as much work in moving an object along path from A to B as it does in moving the object along path from A to B. What can ou conclude about? Give reasons for our answer. 37. Work b a constant force Show that the work done b a constant force field = ai + bj + ck in moving a particle along an path from A to B is W = # AB. 38. Gravitational field s,,d gs,, d = # dr. a. ind a potential function for the gravitational field i + j + k = -GmM s + + d 3> (G, m, and M are constantsd. b. et P and P be points at distance s and s from the origin. Show that the work done b the gravitational field in part (a) in moving a particle from to is P s,,d P GmM a s - s b. 6.4 Green s Theorem in the Plane If is a conservative field, then we know = ƒfor a differentiable function ƒ, and we can calculate the line integral of over an path joining point A to B as # dr = ƒ(b) - ƒ(a). In this section we derive a method for computing a work or flu integral over a closed curve in the plane when the field is not conservative. This method, known as Green s Theorem, allows us to convert the line integral into a double integral over the region enclosed b. The discussion is given in terms of velocit fields of fluid flows (a fluid is a liquid or a gas) because the are eas to visualie. However, Green s Theorem applies to an vector field, independent of an particular interpretation of the field, provided the assumptions of the theorem are satisfied. We introduce two new ideas for Green s Theorem: divergence and circulation densit around an ais perpendicular to the plane. Divergence Suppose that s, d = Ms, di + Ns, dj is the velocit field of a fluid flowing in the plane and that the first partial derivatives of M and N are continuous at each point of a region R. et (, ) be a point in R and let A be a small rectangle with one corner at (, ) that, along with its interior, lies entirel in R. The sides of the rectangle, parallel to the coordinate aes, have lengths of and. Assume that the components M and N do not

11 958 hapter 6: Integration in Vector ields R (a) b a R a b (b) IGURE 6.35 Other regions to which Green s Theorem applies. This shows the curve of igure 6.33 decomposed into the two directed parts : œ = g and œ s d, d Ú Ú c : = g s d, c d. The result of this double integration is N d = 6R N d d. Summing Equations (6) and (7) gives Equation (5). This concludes the proof. Green s Theorem also holds for more general regions, such as those shown in igures 6.35 and 6.36, but we will not prove this result here. Notice that the region in igure 6.36 is not simpl connected. The curves and h on its boundar are oriented so that the region R is alwas on the left-hand side as the curves are traversed in the directions shown. With this convention, Green s Theorem is valid for regions that are not simpl connected. While we stated the theorem in the -plane, Green s Theorem applies to an region R contained in a plane bounded b a curve in space. We will see how to epress the double integral over R for this more general form of Green s Theorem in Section 6.7. h R h (7) IGURE 6.36 Green s Theorem ma be applied to the annular region R b summing the line integrals along the boundaries and h in the directions shown. Eercises 6.4 Verifing Green s Theorem In Eercises 4, verif the conclusion of Green s Theorem b evaluating both sides of Equations (3) and (4) for the field = Mi + Nj. Take the domains of integration in each case to be the disk R: + a and its bounding circle : r = sa cos tdi + sa sin tdj, t p.. = -i + j. = i 3. = i - 3j 4. = - i + j irculation and lu In Eercises 5 4, use Green s Theorem to find the counterclockwise circulation and outward flu for the field and curve = s - di + s - dj : The square bounded b =, =, =, = = s + 4di + s + dj : The square bounded b =, =, =, = 7. = s - di + s + dj : The triangle bounded b =, = 3, and = 8. = s + di - s + dj : The triangle bounded b =, =, and = 9. = ( + )i + ( - )j. = ( + 3)i + ( - )j (, ). = 3 i +. = 4 j (, ) (, ) (, ) 5 + i + Atan- Bj 5 5

12 6.4 Green s Theorem in the Plane = s + e sin di + s + e cos dj 4. : The right-hand loop of the lemniscate r = cos u : The boundar of the region defined b the polar coordinate inequalities r, u p 5. ind the counterclockwise circulation and outward flu of the field = i + j around and over the boundar of the region enclosed b the curves = and = in the first quadrant. 6. ind the counterclockwise circulation and the outward flu of the field = s -sin di + s cos dj around and over the square cut from the first quadrant b the lines = p> and = p>. 7. ind the outward flu of the field across the cardioid r = as + cos ud, a ind the counterclockwise circulation of = s + e ln di + se >dj around the boundar of the region that is bounded above b the curve = 3 - and below b the curve = 4 +. Work In Eercises 9 and, find the work done b in moving a particle once counterclockwise around the given curve. 9.. : The boundar of the triangular region in the first quadrant enclosed b the -ais, the line =, and the curve = 3 : The circle s - d + s - d = 4 Using Green s Theorem Appl Green s Theorem to evaluate the integrals in Eercises = atan - bi + ln s + dj : The triangle bounded b =, + =, = : The boundar of p, sin : The circle s - d + s - 3d = 4 = a3 - = 3 i + 4 j = s4 - di + s - 4dj s d + dd s3 d + dd s6 + d d + s + d d s + d d + s + 3d d + bi + se + tan - dj : An simple closed curve in the plane for which Green s Theorem holds alculating Area with Green s Theorem If a simple closed curve in the plane and the region R it encloses satisf the hpotheses of Green s Theorem, the area of R is given b Green s Theorem Area ormula The reason is that b Equation (3), run backward, Use the Green s Theorem area formula given above to find the areas of the regions enclosed b the curves in Eercises The circle rstd = sa cos tdi + sa sin tdj, 6. The ellipse rstd = sa cos tdi + sb sin tdj, 7. The astroid rstd = scos 3 tdi + ssin 3 tdj, 8. One arch of the ccloid = t - sin t, = - cos t 9. et be the boundar of a region on which Green s Theorem holds. Use Green s Theorem to calculate a. b. Area of R = d d = a 6R 6R + b d d ƒsd d + gsd d k d + h d sk and h constantsd. 3. Integral dependent onl on area Show that the value of = around an square depends onl on the area of the square and not on its location in the plane. 3. What is special about the integral Give reasons for our answer. 3. What is special about the integral Give reasons for our answer. Area of R = d - d 33. Area as a line integral Show that if R is a region in the plane bounded b a piecewise smooth, simple closed curve, then Area of R = d + s + d d 34. Definite integral as a line integral Suppose that a nonnegative function = ƒsd has a continuous first derivative on [a, b]. et be the boundar of the region in the -plane that is bounded below b the -ais, above b the graph of ƒ, and on the sides b the lines = a and = b. Show that b d - d. 4 3 d + 4 d? - 3 d + 3 d? d = - d. ƒsd d = - d. a t p t p t p

13 96 hapter 6: Integration in Vector ields 35. Area and the centroid et A be the area and the -coordinate a. et ƒs, d = ln s + d and let be the circle of the centroid of a region R that is bounded b a piecewise + = a. Evaluate the flu integral smooth, simple closed curve in the -plane. Show that d = - d = 3 d - d = A. 36. Moment of inertia et I be the moment of inertia about the -ais of the region in Eercise 35. Show that 3 3 d = - d = 4 3 d - d = I. 37. Green s Theorem and aplace s equation Assuming that all the necessar derivatives eist and are continuous, show that if ƒ(, ) satisfies the aplace equation then ƒ + ƒ =, ƒ ƒ d - d = for all closed curves to which Green s Theorem applies. (The converse is also true: If the line integral is alwas ero, then ƒ satisfies the aplace equation.) 38. Maimiing work Among all smooth, simple closed curves in the plane, oriented counterclockwise, find the one along which the work done b = a bi + j is greatest. (Hint: Where is scurl d # k positive?) 39. Regions with man holes Green s Theorem holds for a region R with an finite number of holes as long as the bounding curves are smooth, simple, and closed and we integrate over each component of the boundar in the direction that keeps R on our immediate left as we go along (see accompaning figure). b. et K be an arbitrar smooth, simple closed curve in the plane that does not pass through (, ). Use Green s Theorem to show that has two possible values, depending on whether (, ) lies inside K or outside K. 4. Bendison s criterion The streamlines of a planar fluid flow are the smooth curves traced b the fluid s individual particles. The vectors = Ms, di + Ns, dj of the flow s velocit field are the tangent vectors of the streamlines. Show that if the flow takes place over a simpl connected region R (no holes or missing points) and that if M + N Z throughout R, then none of the streamlines in R is closed. In other words, no particle of fluid ever has a closed trajector in R. The criterion M + N Z is called Bendison s criterion for the noneistence of closed trajectories. 4. Establish Equation (7) to finish the proof of the special case of Green s Theorem. 4. url component of conservative fields an anthing be said about the curl component of a conservative two-dimensional vector field? Give reasons for our answer. OMPUTER EXPORATIONS In Eercises 43 46, use a AS and Green s Theorem to find the counterclockwise circulation of the field around the simple closed curve. Perform the following AS steps. a. Plot in the -plane. b. Determine the integrand sn>d - sm>d for the curl form of Green s Theorem. c. Determine the (double integral) limits of integration from our plot in part (a) and evaluate the curl integral for the circulation. 43. = s - di + s + 3dj, : The ellipse + 4 = = s3-3 di + s dj, : The ellipse = - e i + se ln + dj, : The boundar of the region defined b = + 4 (below) and = (above) = e i + (4 ln )j, ƒ # n ds. ƒ # n ds K : The triangle with vertices (, ), (, ), and (, 4) =

14 6.5 Surfaces and Area 969 Eercises 6.5 inding Parametriations In Eercises 6, find a parametriation of the surface. (There are man correct was to do these, so our answers ma not be the same as those in the back of the book.). The paraboloid = +, 4. The paraboloid = 9 - -, Ú 3. one frustum The first-octant portion of the cone + > between the planes = and = 3 4. one frustum The portion of the cone between the planes = and = 4 5. Spherical cap The cap cut from the sphere + + = 9 b the cone = + 6. Spherical cap The portion of the sphere + + = 4 in the first octant between the -plane and the cone = + 7. Spherical band The portion of the sphere + + = 3 between the planes = 3> and = -3> 8. Spherical cap The upper portion cut from the sphere + + = 8 b the plane = - 9. Parabolic clinder between planes The surface cut from the parabolic clinder = 4 - b the planes =, =, and =. Parabolic clinder between planes The surface cut from the parabolic clinder = b the planes =, = 3, and =. ircular clinder band The portion of the clinder + = 9 between the planes = and = 3. ircular clinder band The portion of the clinder + = 4 above the -plane between the planes = - and = 3. Tilted plane inside clinder The portion of the plane + + = a. Inside the clinder + = 9 b. Inside the clinder + = 9 4. Tilted plane inside clinder The portion of the plane - + = a. Inside the clinder + = 3 b. Inside the clinder + = = + 5. ircular clinder band The portion of the clinder s - d + = 4 between the planes = and = 3 6. ircular clinder band The portion of the clinder + s - 5d = 5 between the planes = and = Surface Area of Parametried Surfaces In Eercises 7 6, use a parametriation to epress the area of the surface as a double integral. Then evaluate the integral. (There are man correct was to set up the integrals, so our integrals ma not be the same as those in the back of the book. The should have the same values, however.) 7. Tilted plane inside clinder The portion of the plane + = inside the clinder + = = 8. Plane inside clinder The portion of the plane = - inside the clinder + = 4 9. one frustum The portion of the cone = + between the planes = and = 6. one frustum The portion of the cone = + >3 between the planes = and = 4>3. ircular clinder band The portion of the clinder + = between the planes = and = 4. ircular clinder band The portion of the clinder + = between the planes = - and = 3. Parabolic cap The cap cut from the paraboloid = - - b the cone = + 4. Parabolic band The portion of the paraboloid = + between the planes = and = 4 5. Sawed-off sphere The lower portion cut from the sphere + + = b the cone = + 6. Spherical band The portion of the sphere + + = 4 between the planes = - and = 3 Planes Tangent to Parametried Surfaces The tangent plane at a point P sƒsu, d, gsu, d, hsu, dd on a parametried surface rsu, d = ƒsu, di + gsu, dj + hsu, dk is the plane through P normal to the vector r u su, d * r su, d, the cross product of the tangent vectors r u su, d and r su, d at P. In Eercises 7 3, find an equation for the plane tangent to the surface at P. Then find a artesian equation for the surface and sketch the surface and tangent plane together. 7. one The cone rsr, ud = sr cos udi + sr sin udj + rk, r Ú, u p at the point P A,, B corresponding to sr, ud = s, p>4d 8. Hemisphere The hemisphere surface rsf, ud = s4 sin f cos udi + s4 sin f sin udj + s4 cos fdk, f p>, u p, at the point P A,, 3B corresponding to sf, ud = sp>6, p>4d 9. ircular clinder The circular clinder rsu, d = s3 sin udi + s6 sin udj + k, u p, at the point P A33>, 9>, B corresponding to su, d = sp>3, d (See Eample 3.) 3. Parabolic clinder The parabolic clinder surface rs, d = i + j - k, - q 6 6 q, - q 6 6 q, at the point P s,, -d corresponding to s, d = s, d More Parametriations of Surfaces 3. a. A torus of revolution (doughnut) is obtained b rotating a circle in the -plane about the -ais in space. (See the accompaning figure.) If has radius r 7 and center (R,, ), show that a parametriation of the torus is rsu, d = ssr + r cos udcos di + ssr + r cos udsin dj + sr sin udk, where u p and p are the angles in the figure.

15 97 hapter 6: Integration in Vector ields b. Show that the surface area of the torus is A = 4p Rr. 3. Parametriation of a surface of revolution Suppose that the parametried curve : (ƒ(u), g(u)) is revolved about the -ais, where gsud 7 for a u b. a. Show that rsu, d = ƒsudi + sgsudcos dj + sgsudsin dk is a parametriation of the resulting surface of revolution, where p is the angle from the -plane to the point r(u, ) on the surface. (See the accompaning figure.) Notice that ƒ(u) measures distance along the ais of revolution and g(u) measures distance from the ais of revolution. f (u) R r(u, ) ( f (u), g(u), ) g(u) b. ind a parametriation for the surface obtained b revolving the curve =, Ú, about the -ais. 33. a. Parametriation of an ellipsoid The parametriation = a cos u, = b sin u, u p gives the ellipse s >a d + s >b d =. Using the angles u and f in spherical coordinates, show that rsu, fd = sa cos u cos fdi + sb sin u cos fdj + sc sin fdk is a parametriation of the ellipsoid s >a d + s >b d + s >c d =. b. Write an integral for the surface area of the ellipsoid, but do not evaluate the integral. r r(u, ) u u 34. Hperboloid of one sheet a. ind a parametriation for the hperboloid of one sheet + - = in terms of the angle u associated with the circle + = r and the hperbolic parameter u associated with the hperbolic function r - =. (Hint: cos h u - sin h u =.) b. Generalie the result in part (a) to the hperboloid s >a d + s >b d - s >c d =. 35. (ontinuation of Eercise 34.) ind a artesian equation for the plane tangent to the hperboloid + - = 5 at the point s where +,, d, = Hperboloid of two sheets ind a parametriation of the hperboloid of two sheets s >c d - s >a d - s >b d =. Surface Area for Implicit and Eplicit orms 37. ind the area of the surface cut from the paraboloid + - = b the plane =. 38. ind the area of the band cut from the paraboloid + - = b the planes = and = ind the area of the region cut from the plane + + = 5 b the clinder whose walls are = and = ind the area of the portion of the surface - = that lies above the triangle bounded b the lines = 3, =, and = in the -plane. 4. ind the area of the surface - - = that lies above the triangle bounded b the lines =, =, and = 3 in the plane. 4. ind the area of the cap cut from the sphere + + = b the cone = ind the area of the ellipse cut from the plane = c (c a constant) b the clinder + =. 44. ind the area of the upper portion of the clinder + = that lies between the planes = ;> and = ;>. 45. ind the area of the portion of the paraboloid = that lies above the ring + 4 in the -plane. 46. ind the area of the surface cut from the paraboloid + + = b the plane =. 47. ind the area of the surface - ln = above the square R:,, in the -plane. 48. ind the area of the surface 3> + 3> - 3 = above the square R:,, in the -plane. ind the area of the surfaces in Eercises The surface cut from the bottom of the paraboloid = + b the plane = 3 5. The surface cut from the nose of the paraboloid = - - b the -plane 5. The portion of the cone = + that lies over the region between the circle + = and the ellipse = 36 in the -plane. (Hint: Use formulas from geometr to find the area of the region.) 5. The triangle cut from the plane = 6 b the bounding planes of the first octant. alculate the area three was, using different eplicit forms. 53. The surface in the first octant cut from the clinder = s>3d 3> b the planes = and = 6>3

16 6.6 Surface Integrals The portion of the plane + = 4 that lies above the region cut from the first quadrant of the -plane b the parabola = Use the parametriation r(, ) = i + ƒ(, )j + k and Equation (5) to derive a formula for ds associated with the eplicit form = ƒ(, ). 56. et S be the surface obtained b rotating the smooth curve = ƒ(), a b, about the -ais, where ƒ() Ú. a. Show that the vector function r(, u) = i + ƒ() cos uj + ƒ() sin uk is a parametriation of S, where u is the angle of rotation around the -ais (see the accompaning figure). (,, ) b. Use Equation (4) to show that the surface area of this surface of revolution is given b b A = pƒ() + [ƒ ()] d. a u f() 6.6 Surface Integrals To compute quantities such as the flow of liquid across a curved membrane or the upward force on a falling parachute, we need to integrate a function over a curved surface in space. This concept of a surface integral is an etension of the idea of a line integral for integrating over a curve. ur u P k s k 5 s u r ( k, k, k ) IGURE 6.47 The area of the patch s k is the area of the tangent parallelogram determined b the vectors u r u and r. The point ( k, k, k ) lies on the surface patch, beneath the parallelogram shown here. Surface Integrals Suppose that we have an electrical charge distributed over a surface S, and that the function G(,, ) gives the charge densit (charge per unit area) at each point on S. Then we can calculate the total charge on S as an integral in the following wa. Assume, as in Section 6.5, that the surface S is defined parametricall on a region R in the u-plane, r(u, ) = ƒ(u, )i + g(u, )j + h(u, )k, (u, ) H R. In igure 6.47, we see how a subdivision of R (considered as a rectangle for simplicit) divides the surface S into corresponding curved surface elements, or patches, of area s u ƒ r u * r ƒ du d. As we did for the subdivisions when defining double integrals in Section 5., we number the surface element patches in some order with their areas given b s, s, Á, s n. To form a Riemann sum over S, we choose a point ( k, k, k ) in the kth patch, multipl the value of the function G at that point b the area s k, and add together the products: n a G( k, k, k ) s k. k = Depending on how we pick ( k, k, k ) in the kth patch, we ma get different values for this Riemann sum. Then we take the limit as the number of surface patches increases, their areas shrink to ero, and both u : and :. This limit, whenever it eists independent of all choices made, defines the surface integral of G over the surface S as n G(,, ) ds = lim 6 n: q a G( k, k, k ) s k. k = S ()

17 978 hapter 6: Integration in Vector ields M = 6S d ds = p rr dr du r = p p = du = p, = M M = p p ln = ln. The shell s center of mass is the point (,, > ln ). dr du Eercises 6.6 Surface Integrals In Eercises 8, integrate the given function over the given surface.. Parabolic clinder Gs,, d =, over the parabolic clinder =,, 3. ircular clinder Gs,, d =, over the clindrical surface + = 4, Ú, 4 3. Sphere Gs,, d =, over the unit sphere + + = 4. Hemisphere Gs,, d =, over the hemisphere + + = a, Ú 5. Portion of plane s,, d =, over the portion of the plane + + = 4 that lies above the square,, in the -plane 6. one s,, d = -, over the cone = +, 7. Parabolic dome Hs,, d = 5-4, over the parabolic dome = - -, Ú 8. Spherical cap Hs,, d =, over the part of the sphere + + = 4 that lies above the cone = + 9. Integrate Gs,, d = + + over the surface of the cube cut from the first octant b the planes = a, = a, = a.. Integrate Gs,, d = + over the surface of the wedge in the first octant bounded b the coordinate planes and the planes = and + =.. Integrate Gs,, d = over the surface of the rectangular solid cut from the first octant b the planes = a, = b, and = c.. Integrate Gs,, d = over the surface of the rectangular solid bounded b the planes = ;a, = ;b, and = ;c. 3. Integrate Gs,, d = + + over the portion of the plane + + = that lies in the first octant. 4. Integrate Gs,, d = + 4 over the surface cut from the parabolic clinder + 4 = 6 b the planes =, =, and =. 5. Integrate G(,, ) = - over the portion of the graph of = + above the triangle in the -plane having vertices (,, ), (,, ), and (,, ). (See accompaning figure.) (,, ) (,, ) (,, ) (,, ) (,, ) 6. Integrate G(,, ) = over the surface given b = + for, Integrate G(,, ) = over the triangular surface with vertices (,, ), (,, ), and (,, ). (,, ) (,, ) (,, ) 8. Integrate G(,, ) = - - over the portion of the plane + = in the first octant between = and = (see the accompaning figure).

18 6.6 Surface Integrals 979 inding lu Across a Surface In Eercises 9 8, use a parametriation to find the flu across the surface in the given direction. 9. Parabolic clinder = i + j - 3k outward (normal awa from the -ais) through the surface cut from the parabolic clinder = 4 - b the planes =, =, and =. Parabolic clinder = j - k outward (normal awa from the -plane) through the surface cut from the parabolic clinder =, -, b the planes = and =. Sphere = k across the portion of the sphere + + = a in the first octant in the direction awa from the origin. Sphere = i + j + k across the sphere + + = a in the direction awa from the origin 3. Plane = i + j + k upward across the portion of the plane + + = a that lies above the square a, a, in the -plane 4. linder = i + j + k outward through the portion of the clinder + = cut b the planes = and = a 5. one = i - k outward (normal awa from the -ais) through the cone = +, 6. one = i + j - k outward (normal awa from the -ais) through the cone = +, 7. one frustum = -i - j + k outward (normal awa from the -ais) through the portion of the cone = + between the planes = and = 8. Paraboloid = 4i + 4j + k outward (normal awa from the -ais) through the surface cut from the bottom of the paraboloid = + b the plane = In Eercises 9 and 3, find the flu of the field across the portion of the given surface in the specified direction S: rectangular surface direction k In Eercises 3 36, find the flu of the field across the portion of the sphere + + = a in the first octant in the direction awa from the origin. 3. s,, d = -i + j + 3k s,, d = i - j + k S: rectangular surface =, -, 7, direction -j s,, d = k (,, ) 3. s,, d = -i + j (,, ) 4S # n ds =,, 3, 33. s,, d = i - j + k s,, d = i + j + k s,, d = i + j + k s,, d = 37. ind the flu of the field s,, d = i + j - 3k outward through the surface cut from the parabolic clinder = 4 - b the planes =, =, and =. 38. ind the flu of the field s,, d = 4i + 4j + k outward (awa from the -ais) through the surface cut from the bottom of the paraboloid = + b the plane =. 39. et S be the portion of the clinder = e in the first octant that projects parallel to the -ais onto the rectangle R :, in the -plane (see the accompaning figure). et n be the unit vector normal to S that points awa from the -plane. ind the flu of the field s,, d = -i + j + k across S in the direction of n. i + j + k + + e 4. et S be the portion of the clinder = ln in the first octant whose projection parallel to the -ais onto the -plane is the rectangle R : e,. et n be the unit vector normal to S that points awa from the -plane. ind the flu of = j + k through S in the direction of n. 4. ind the outward flu of the field = i + j + k across the surface of the cube cut from the first octant b the planes = a, = a, = a. 4. ind the outward flu of the field = i + j + k across the surface of the upper cap cut from the solid sphere b the plane = 3. Moments and Masses 43. entroid ind the centroid of the portion of the sphere + + = a that lies in the first octant. 44. entroid ind the centroid of the surface cut from the clinder + = 9, Ú, b the planes = and = 3 (resembles the surface in Eample 5). 45. Thin shell of constant densit ind the center of mass and the moment of inertia about the -ais of a thin shell of constant densit d cut from the cone + - = b the planes = and =. 46. onical surface of constant densit ind the moment of inertia about the -ais of a thin shell of constant densit d cut from the cone =, Ú, b the circular clinder + = (see the accompaning figure). S R

19 98 hapter 6: Integration in Vector ields 4 4 or r cos 47. Spherical shells a. ind the moment of inertia about a diameter of a thin spherical shell of radius a and constant densit d. (Work with a hemispherical shell and double the result.) b. Use the Parallel Ais Theorem (Eercises 5.6) and the result in part (a) to find the moment of inertia about a line tangent to the shell. 48. onical Surface ind the centroid of the lateral surface of a solid cone of base radius a and height h (cone surface minus the base). 6.7 Stokes Theorem As we saw in Section 6.4, the circulation densit or curl component of a two-dimensional (,, ) url IGURE 6.55 The circulation vector at a point (,, )in a plane in a threedimensional fluid flow. Notice its right-hand relation to the rotating particles in the fluid. field = Mi + Nj at a point (, ) is described b the scalar quantit sn> -M>d. In three dimensions, circulation is described with a vector. Suppose that is the velocit field of a fluid flowing in space. Particles near the point (,, ) in the fluid tend to rotate around an ais through (,, ) that is parallel to a certain vector we are about to define. This vector points in the direction for which the rotation is counterclockwise when viewed looking down onto the plane of the circulation from the tip of the arrow representing the vector. This is the direction our right-hand thumb points when our fingers curl around the ais of rotation in the wa consistent with the rotating motion of the particles in the fluid (see igure 6.55). The length of the vector measures the rate of rotation. The vector is called the curl vector and for the vector field = Mi + Nj + Pk it is defined to be curl = a P - N bi + am - P bj + an - M bk. () This information is a consequence of Stokes Theorem, the generaliation to space of the circulation-curl form of Green s Theorem and the subject of this section. Notice that scurl d # k = sn> -M>d is consistent with our definition in Section 6.4 when = Ms, di + Ns, dj. The formula for curl in Equation () is often written using the smbolic operator =i () + j + k. (The smbol is pronounced del. ) The curl of is *: * = 4 i j k = a P - N bi + am - P bj + an - M bk = curl. M N P 4

20 988 hapter 6: Integration in Vector ields S onservative ields and Stokes Theorem In Section 6.3, we found that a field being conservative in an open region D in space is equivalent to the integral of around ever closed loop in D being ero. This, in turn, is equivalent in simpl connected open regions to saing that * = (which gives a test for determining if is conservative for such regions). (a) THEOREM 7 url = Related to the losed-oop Propert If * = at ever point of a simpl connected open region D in space, then on an piecewisesmooth closed path in D, # dr =. (b) IGURE 6.66 (a) In a simpl connected open region in space, a simple closed curve is the boundar of a smooth surface S. (b) Smooth curves that cross themselves can be divided into loops to which Stokes Theorem applies. Sketch of a Proof Theorem 7 can be proved in two steps. The first step is for simple closed curves (loops that do not cross themselves), like the one in igure 6.66a. A theorem from topolog, a branch of advanced mathematics, states that ever smooth simple closed curve in a simpl connected open region D is the boundar of a smooth two-sided surface S that also lies in D. Hence, b Stokes Theorem, # dr = 6S * # n ds =. The second step is for curves that cross themselves, like the one in igure 6.66b. The idea is to break these into simple loops spanned b orientable surfaces, appl Stokes Theorem one loop at a time, and add the results. The following diagram summaries the results for conservative fields defined on connected, simpl connected open regions. conservative on D Theorem 3, Section 6.3 Theorem, Section 6.3 f on D Vector identit (Eq. 8) (continuous second partial derivatives) dr E over an closed path in D Theorem 7 Domain's simple connectivit and Stokes' Theorem throughout D Eercises 6.7 Using Stokes Theorem to ind ine Integrals In Eercises 6, use the surface integral in Stokes Theorem to calculate the circulation of the field around the curve in the indicated direction.. = i + j + k : The ellipse 4 + = 4 in the -plane, counterclockwise when viewed from above. = i + 3j - k : The circle + = 9 in the -plane, counterclockwise when viewed from above 3. = i + j + k :The boundar of the triangle cut from the plane + + = b the first octant, counterclockwise when viewed from above

21 6.7 Stokes Theorem = s + di + s + dj + s + dk :The boundar of the triangle cut from the plane + + = b the first octant, counterclockwise when viewed from above 5. = s + di + s + dj + s + dk : The square bounded b the lines = ; and = ; in the -plane, counterclockwise when viewed from above 6. = 3 i + j + k : The intersection of the clinder + = 4 and the hemisphere + + = 6, Ú, counterclockwise when viewed from above lu of the url 7. et n be the outer unit normal of the elliptical shell and let ind the value of (Hint: One parametriation of the ellipse at the base of the shell is = 3 cos t, = sin t, t p. ) 8. et n be the outer unit normal (normal awa from the origin) of the parabolic shell and let = a- + ind the value of S: = 36, Ú, = i + j + s + 4 d 3> sin e k. S: = 4, Ú, 9. et S be the clinder + = a, h, together with its top, + a, = h. et = -i + j + k. Use Stokes Theorem to find the flu of *outward through S.. Evaluate where S is the hemisphere + + =, Ú.. lu of curl Show that has the same value for all oriented surfaces S that span and that induce the same positive direction on.. et be a differentiable vector field defined on a region containing a smooth closed oriented surface S and its interior. et n be the unit normal vector field on S. Suppose that S is the union of two surfaces S and S joined along a smooth simple closed curve. an anthing be said about Give reasons for our answer. 6 S * # n ds. + bi + stan- dj + a + 6 S * # n ds. 6 S *s id # n ds, 6 S * # n ds 6 S * # n ds? 4 + bk. Stokes Theorem for Parametried Surfaces In Eercises 3 8, use the surface integral in Stokes Theorem to calculate the flu of the curl of the field across the surface S in the direction of the outward unit normal n = i + 3j + 5k S: rsr, ud = sr cos udi + sr sin udj + s4 - r dk, r, u p = s - di + s - dj + s + dk S: rsr, ud = sr cos udi + sr sin udj + s9 - r dk, r 3, u p = i + 3 j + 3k S: rsr, ud = sr cos udi + sr sin udj + rk, r, u p = s - di + s - dj + s - dk S: rsr, ud = sr cos udi + sr sin udj + s5 - rdk, r 5, u p = 3i + s5 - dj + s - dk S: rsf, ud = A 3 sin f cos ubi + A 3 sin f sin ubj + A 3 cos fbk, Theor and Eamples 9. Zero circulation Use the identit * ƒ = (Equation (8) in the tet) and Stokes Theorem to show that the circulations of the following fields around the boundar of an smooth orientable surface in space are ero. a. = i + j + k b. = s 3 d c. = *si + j + kd d. = ƒ. Zero circulation et ƒs,, d = s + + d ->. Show that the clockwise circulation of the field = ƒaround the circle + = a in the -plane is ero a. b taking r = sa cos tdi + sa sin tdj, t p, and integrating # dr over the circle. b. b appling Stokes Theorem.. et be a simple closed smooth curve in the plane + + =, oriented as shown here. Show that f p>, u p = i + j + k S: rsf, ud = s sin f cos udi + s sin f sin udj + s cos fdk, f p>, u p d + 3 d - d O a depends onl on the area of the region enclosed b and not on the position or shape of.. Show that if = i + j + k, then * =. 5

22 99 hapter 6: Integration in Vector ields 3. ind a vector field with twice-differentiable components whose curl is i + j + k or prove that no such field eists. 4. Does Stokes Theorem sa anthing special about circulation in a field whose curl is ero? Give reasons for our answer. 5. et R be a region in the -plane that is bounded b a piecewise smooth simple closed curve and suppose that the moments of inertia of R about the - and -aes are known to be I and I. Evaluate the integral sr 4 d # n ds, where r = +, in terms of I and I. 6. Zero curl, et field not conservative Show that the curl of is ero but that = - + i + + j + k # dr is not ero if is the circle + = in the -plane. (Theorem 7 does not appl here because the domain of is not simpl connected. The field is not defined along the -ais so there is no wa to contract to a point without leaving the domain of.) 6.8 The Divergence Theorem and a Unified Theor The divergence form of Green s Theorem in the plane states that the net outward flu of a vector field across a simple closed curve can be calculated b integrating the divergence of the field over the region enclosed b the curve. The corresponding theorem in three dimensions, called the Divergence Theorem, states that the net outward flu of a vector field across a closed surface in space can be calculated b integrating the divergence of the field over the region enclosed b the surface. In this section we prove the Divergence Theorem and show how it simplifies the calculation of flu. We also derive Gauss s law for flu in an electric field and the continuit equation of hdrodnamics. inall, we unif the chapter s vector integral theorems into a single fundamental theorem. Divergence in Three Dimensions The divergence of a vector field = Ms,, di + Ns,, dj + Ps,, dk is the scalar function div = # = M + N + P. () The smbol div is read as divergence of or div. The notation # is read del dot. Div has the same phsical interpretation in three dimensions that it does in two. If is the velocit field of a flowing gas, the value of div at a point (,, ) is the rate at which the gas is compressing or epanding at (,, ). The divergence is the flu per unit volume or flu densit at the point. EXAMPE The following vector fields represent the velocit of a gas flowing in space. ind the divergence of each vector field and interpret its phsical meaning. igure 6.67 displas the vector fields. (a) Epansion: (,, ) = i + j + k (b) ompression: (,, ) = -i - j - k

23 6.8 The Divergence Theorem and a Unified Theor 999 The undamental Theorem now sas that sbd # n + sad # n = 3 [a,b] # d. The undamental Theorem of alculus, the normal form of Green s Theorem, and the Divergence Theorem all sa that the integral of the differential operator # operating on a field over a region equals the sum of the normal field components over the boundar of the region. (Here we are interpreting the line integral in Green s Theorem and the surface integral in the Divergence Theorem as sums over the boundar.) Stokes Theorem and the tangential form of Green s Theorem sa that, when things are properl oriented, the integral of the normal component of the curl operating on a field equals the sum of the tangential field components on the boundar of the surface. The beaut of these interpretations is the observance of a single unifing principle, which we might state as follows. A Unifing undamental Theorem The integral of a differential operator acting on a field over a region equals the sum of the field components appropriate to the operator over the boundar of the region. Eercises 6.8 alculating Divergence In Eercises 4, find the divergence of the field.. The spin field in igure 6.. The radial field in igure The gravitational field in igure 6.8 and Eercise 38a in Section The velocit field in igure 6.3 alculating lu Using the Divergence Theorem In Eercises 5 6, use the Divergence Theorem to find the outward flu of across the boundar of the region D. 5. ube 6. D: The cube bounded b the planes = ;, = ;, and = ; = i + j + k a. ube D: The cube cut from the first octant b the planes =, =, and = b. ube D: The cube bounded b the planes = ;, = ;, and = ; c. lindrical can D: The region cut from the solid clinder + 4 b the planes = and = 7. linder and paraboloid D: The region inside the solid clinder + 4 between the plane = and the paraboloid = + 8. Sphere = s - di + s - dj + s - dk = i + j + 3k = i + j - k D: The solid sphere Portion of sphere D: The region cut from the first octant b the sphere + + = 4. lindrical can D: The region cut from the first octant b the clinder + = 4 and the plane = 3. Wedge D: The wedge cut from the first octant b the plane + = 4 and the elliptical clinder 4 + = 6. Sphere D: The solid sphere + + a 3. Thick sphere D: The region Thick sphere D: The region Thick sphere = s5 3 + di + s 3 + e sin dj + s5 3 + e cos dk D: The solid region between the spheres + + = and + + = 6. Thick clinder + k = i - j + 3k = s6 + di + s + dj k = i - j - k = 3 i + 3 j + 3 k = + + si + j + kd = si + j + kd> + + = ln s + di - a tan- bj + D: The thick-walled clinder +, -

24 hapter 6: Integration in Vector ields Properties of url and Divergence 7. div (curl G) is ero a. Show that if the necessar partial derivatives of the components of the field G = Mi + Nj + Pk are continuous, then # *G =. b. What, if anthing, can ou conclude about the flu of the field *Gacross a closed surface? Give reasons for our answer. 8. et and be differentiable vector fields and let a and b be arbitrar real constants. Verif the following identities. a. b. c. 9. et be a differentiable vector field and let g(,, ) be a differentiable scalar function. Verif the following identities. a. b.. If = Mi + Nj + Pk is a differentiable vector field, we define the notation # to mean M + N + P. or differentiable vector fields and, verif the following identities. a. b. # sa + b d = a # + b # *sa + b d = a * + b * # s * d = # * - # * # sgd = g # + g # *sgd = g * + g * *s * d = s # d - s # d + s # d - s # d s # d = s # d + s # d + * s * d + * s * d Theor and Eamples. et be a field whose components have continuous first partial derivatives throughout a portion of space containing a region D bounded b a smooth closed surface S. If ƒ ƒ, can an bound be placed on the sie of Give reasons for our answer.. The base of the closed cubelike surface shown here is the unit square in the -plane. The four sides lie in the planes =, =, =, and =. The top is an arbitrar smooth surface whose identit is unknown. et = i - j + s + 3dk and suppose the outward flu of through Side A is and through Side B is -3. an ou conclude anthing about the outward flu through the top? Give reasons for our answer. Side A 9 D # dv? Top (,, ) Side B 3. a. Show that the outward flu of the position vector field = i + j + k through a smooth closed surface S is three times the volume of the region enclosed b the surface. b. et n be the outward unit normal vector field on S. Show that it is not possible for to be orthogonal to n at ever point of S. 4. Maimum flu Among all rectangular solids defined b the inequalities a, b,, find the one for which the total flu of = s - - 4di - 6j + k outward through the si sides is greatest. What is the greatest flu? 5. Volume of a solid region et = i + j + k and suppose that the surface S and region D satisf the hpotheses of the Divergence Theorem. Show that the volume of D is given b the formula 6. Outward flu of a constant field Show that the outward flu of a constant vector field = across an closed surface to which the Divergence Theorem applies is ero. 7. Harmonic functions A function ƒ(,, ) is said to be harmonic in a region D in space if it satisfies the aplace equation throughout D. Volume of D = 3 6S a. Suppose that ƒ is harmonic throughout a bounded region D enclosed b a smooth surface S and that n is the chosen unit normal vector on S. Show that the integral over S of ƒ # n, the derivative of ƒ in the direction of n, is ero. b. Show that if ƒ is harmonic on D, then 8. Outward flu of a gradient field et S be the surface of the portion of the solid sphere + + a that lies in the first octant and let ƒs,, d = ln + +. alculate ƒ # n ds. # n ds. ƒ = # ƒ = ƒ + ƒ + ƒ = ƒ ƒ # n ds = ƒ ƒ ƒ dv. 6 9D S 6 S ( ƒ # n is the derivative of ƒ in the direction of outward normal n.) 9. Green s first formula Suppose that ƒ and g are scalar functions with continuous first- and second-order partial derivatives throughout a region D that is bounded b a closed piecewise smooth surface S. Show that 6 S ƒ g # n ds = 9D sƒ g + ƒ # gd dv. Equation (9) is Green s first formula. (Hint: Appl the Divergence Theorem to the field = ƒ g.) 3. Green s second formula (ontinuation of Eercise 9.) Interchange ƒ and g in Equation (9) to obtain a similar formula. Then subtract this formula from Equation (9) to show that 6 S sƒ g - g ƒd # n ds = 9D sƒ g - g ƒd dv. This equation is Green s second formula. (9) ()