lculus III prt 4): Vector lculus by Evn Dummit, 8, v. 3.) ontents 4 Vector lculus 4. Line Integrls................................................. 4. Surfces nd Surfce Integrls........................................ 5 4.. Prmetric Surfces......................................... 5 4.. Surfce Integrls........................................... 7 4.3 Vector Fields, Work, ircultion, Flux................................... 4.3. ircultion nd Work Integrls................................... 4.3. Flux Across urve......................................... 3 4.3.3 Flux Across Surfce........................................ 4 4.4 onservtive Vector Fields, Pth-Independence, nd Potentil Functions................ 6 4.5 Green's Theorem............................................... 9 4.6 Stokes's Theorem nd Guss's Divergence Theorem............................ 4.6. Stokes's Theorem........................................... 4.6. Guss's Divergence Theorem.................................... 3 4 Vector lculus Our motivting problem for multivrible integrtion ws to generlize the ide of integrtion to more complicted regions in spce, or more succinctly) to integrte function over region. We might lso sk whether there is simple wy to integrte function over n rbitrry curve in the plne or in spce, nd whether there is wy to integrte function over n rbitrry surfce in spce. The nswer s it lwys hs been to this point) is yes: the generliztion of single-vrible integrtion to rbitrry curves is clled line integrl, nd the generliztion of double integrtion to rbitrry surfces is clled surfce integrl. After introducing line nd surfce integrls, we will then discuss vector elds which re vector-vlued functions in -spce nd 3-spce) which provide useful model for the ow of uid through spce. The principl pplictions of line nd surfce integrls re to the clcultion of the work done by vector eld on prticle trveling through spce, the ux of vector eld cross curve or through surfce, nd the circultion of vector eld long curve. Finlly, we discuss severl generliztions of the Fundmentl Theorem of lculus: the Fundmentl Theorem of lculus for line integrls, Green's Theorem, Guss's Divergence Theorem, nd Stokes's Theorem. ollectively, these theorems unify ll of the dierent notions of integrtion, s they ech relte the integrl of function on region to the integrl of n ntiderivtive of the function on the region's boundry. 4. Line Integrls The motivting problem for our discussion of line integrls is: given prmetric curve rt) = xt), yt) nd function fx, y), if we build surfce long the curve with height given by the function z = fx, y), how cn we clculte the re of this surfce? This is nturl generliztion of our typicl single-vrible integrtion problem, in which we build the surfce inside plne, thus mking it the re under curve.)
Here is n exmple for visuliztion), with rt) = t, t cosπt), fx, y) = t +, for t 3 : Another closely relted question is: given prmetric curve rt) = xt), yt), zt) nd function fx, y, z), how cn we clculte the verge vlue of fx, y, z) on the curve? A third question: given thin wire shped long some curve rt) = xt), yt) with vrible density δx, y), wht is the wire's mss, nd wht re its moments bout the coordinte xes? As with ll other types of integrls we hve exmined so fr, we use Riemnn sums to give the forml denition of the line integrl of function fx, y) on plne curve. Also s before, we will use the forml denition s infrequently s possible!) The ide is to pproximte the curve with stright line segments, sum over ll the segments) the function vlue times the length of the segment, nd then tke the limit s the segment lengths pproch zero. Denition: For curve, prtition of into n pieces is list of points x, y ),..., x n, y n ) on, with the nth segment hving length s i = x i ) + y i ). The norm of the prtition P is the lrgest number mong ll of the segment lengths in P. Denition: For fx, y) continuous function nd P prtition of the curve, we dene the Riemnn sum n of fx, y) on D corresponding to P to be RS P f) = fx k, y k ) s k. ˆ Denition: For function fx, y), we dene the line integrl of f on the curve, denoted k= fx, y) ds, to be the vlue of L such tht, for every ɛ >, there exists δ > depending on ɛ) such tht for every prtition P with normp ) < δ, we hve RS P f) L < ɛ. Remrk: It cn be proven with signicnt eort) tht, if fx, y) is continuous nd the curve is smooth, then vlue of L stisfying the hypotheses ctully does exist. Remrk: The dierentil ds in the denition of the line integrl is the dierentil of rclength, which we discussed erlier in our study of vector-vlued functions. In exctly the sme wy, we cn use Riemnn sums to give forml denition of the line integrl long curve in 3-spce. We simply dd the pproprite z-terms to ll the denitions.) Like with the other types of integrls, line integrls hve number of forml properties which cn be deduced from the Riemnn sum denition. Speciclly, for D n rbitrry constnt nd fx, y) nd gx, y) continuous functions, the following properties hold: Integrl of constnt: D ds = D Arclength). onstnt multiple of function: D fx, y) ds = D fx, y) ds. Addition of functions: fx, y) ds + gx, y) ds = [fx, y) + gx, y)] ds. Subtrction of functions: fx, y) ds gx, y) ds = [fx, y) gx, y)] ds. Nonnegtivity: if fx, y), then fx, y) ds. Union: If nd re curves such tht strts where ends, nd is the curve obtined by gluing the curves end-to-end, then fx, y) ds + fx, y) ds = fx, y) ds. Remrk: These sme properties lso ll hold for line integrls of function fx, y, z) in 3-spce.
The key observtion is tht we cn reduce clcultions of line integrls to trditionl single integrls: Proposition Line Integrls in the Plne): If the curve cn be prmetrized s x = xt), y = yt) for ˆ ˆ b t b, then fx, y) ds = fxt), yt)) ds ds, where = x t) + y t) is the derivtive of rclength. Proposition Line Integrls in 3-Spce): If the curve cn be prmetrized s x = xt), y = yt), z = zt) ˆ ˆ b for t b, then fx, y, z) ds = fxt), yt), zt)) ds ds, where = x t) + y t) + z t) is the derivtive of rclength. The proof of both of these results is simply to observe tht the Riemnn sum integrl fx, y) ds is lso Riemnn sum n n fx k, y k ) s k for the line k= fx k, y k ) s k t k for the integrl b t k k= fxt), yt)) ds. Equivlently: we hve mde substitution in the integrl by chnging from s-coordintes to t-coordintes, where the dierentil chnges using the rule ds = ds. Thus, to evlute the line integrl of f on the curve ; nmely the line fx, y, z) ds, follow these integrl steps: Step : Prmetrize the curve s function of t i.e., in the form rt) = xt), yt), zt), for t b. Step : Write f s function of t: fx, y, z) = fxt), yt), zt)). Step 3: Write ds = ds = vt) = x t) + y t) + z t). Step 4: Evlute the resulting integrl b fxt), yt), zt)) x t) + y t) + z t). Exmple: Integrte the function fx, y, z) = yz 6x long the curve rt) = t 3, 6t, 3t from t = to t =. We hve fx, y, z) = yz 6x = 6t)3t ) 6t 3 = t 3, nd we lso hve ds = 3t ) + 6) + 6t) = 9t4 + 36t + 36 = 3t + 6. The integrl is therefore t3 )3t + 6) = 36t5 + 7t 3 ) = 4. Exmple: Integrte the function fx, y) = x + y long the top hlf of the unit circle x + y =, strting t, ) nd ending t, ). The unit circle is prmetrized by rt) = cos t, sin t : the rnge we wnt is t π. We hve fx, y) = x + y = cos t + sin t, nd we lso hve ds = sin t) + cos t) =. The integrl is therefore π [ cos t + sin t ] = [ ] π + cos t + sin t = π +. To nd the verge vlue of function on curve, we simply integrte the function over the curve, nd then divide by the curve's rclength. Exmple: Find the verge vlue of the function fx, y, z) = x +y +z long the line segment from,, ) to,, ). The direction vector for the line is v =,,,, =, 3,. Thus, we cn prmetrize the line segment s x, y, z =,, + t, 3, for t. So the line segment is prmetrized explicitly by x = + t, y = + 3t, z = t for t. Now we set up the integrl: the function is fx, y, z) = x + y + z = + t) + + 3t) + t) = t 4t +. Since x t) =, y t) = 3, nd z t) =, we lso hve ds = + 3 + =. 3
The integrl of f is therefore [ t 4t + ] = [ ] 3 t3 t + t =. t= 3 To compute the verge vlue, we divide by the rclength, which is ds = =. Thus, the verge vlue is 3. We lso hve formuls for the mss nd moments of wire of vrible density: enter of Mss nd Moment Formuls Thin Wire): Given -dimensionl wire of vrible density δx, y, z) long prmetric curve in 3-spce: The totl mss M is given by M = δx, y, z) ds. The x-moment M yz is given by M yz = x δx, y, z) ds. The y-moment M xz is given by M xz = y δx, y, z) ds. The z-moment M xy is given by M xy = z δx, y, z) ds. Myz The center of mss x, ȳ, z) hs coordintes M, M xz M, M ) xy. M Note: For wire in -spce, the formuls re essentilly the sme except without the z-coordinte), though the x-moment is denoted M y nd the y-moment is denoted M x. Exmple: Find the totl mss, nd the center of mss, of thin wire hving the shpe of the unit circle nd vrible density δx, y) = + x. We cn prmetrize the unit circle with x = cos t, y = sin t, so ds = sin t) + cos t) =. The totl mss M is M = δx, y) ds = π + cos t) = π. The x-moment M y is M y = The y-moment M x is M x = Therefore, the center of mss is x δx, y) ds = π cos t + cos t) = y δx, y) ds = π sin t + cos t) = Mx M, M ) y = M ),. [ sin t + t + 4 sint) ] π [ cos t 4 cost) ] π We will lso be interested in computing ˆ line integrls involving the dierentils dx, dy, nd dz rther thn ds: nmely, expressions of the form f dx + g dy + h dz. We evlute such line integrls by mking the pproprite ˆ substitutions: if is prmetrized by x = xt), y = yt), z = zt) for t b, then the line integrl f dx + g dy + h dz is given by the single-vrible ˆ b [ integrl f dx + g dy + h dz ]. Exmple: Find y dx + z dy + x dz, where is the curve x, y, z) = t, t, t 3 ) rnging from t = to t =. We hve x = t, y = t, nd z = t 3, so tht dx =, dy = t, nd dz = 3t. The integrl is [ t + 3t t + t 3t ] = [ t + 6t 3 + 3t 4] = 73 3. Exmple: Find x dy y dx, where is the upper hlf of the ellipse x /9 + y /6 =, strting t 3, ) nd ending t 3, ). This ellipse is prmetrized by rt) = 3 cos t, 4 sin t : the rnge we wnt is t π. We hve x = 3 cos t nd y = 4 sin t, so tht dx = 3 sin t nd dy = 4 cos t. The desired integrl is π [3 cos t 4 cos t ) 4 sin t 3 sin t )] = π [ cos t + sin t ] = π. t= t= =. = π. 4
4. Surfces nd Surfce Integrls We would now like to consider the problem of computing the integrl of function on surfce in 3-dimensionl spce. In similr wy to how we computed line integrls using single) integrls, we will be ble to compute surfce integrls s double integrls. There re essentilly two wys to describe surfce in 3-spce: either s n implicit surfce of the form fx, y, z) = c, or s prmetric surfce rs, t) = xs, t), ys, t), zs, t) for two prmeters s nd t. Note tht the explicit surfce z = gx, y) is simply specil cse of the generl implicit surfce, since gx, y) z = hs the form fx, y, z) = c with fx, y, z) = gx, y) z nd c =. In cses where the functions x, y, nd z re suciently simple or nice, it cn be possible to eliminte the vribles s nd t from the system x = xs, t), y = ys, t), z = zs, t), nd obtin n eqution for the surfce s n implicit surfce fx, y, z) = c. However, if we cn nd prmetric description of surfce, it is often esier to work with it thn with n implicit description. For exmple, grphing prmetric surfce requires only plugging in vlues nd plotting points, wheres grphing n implicit surfce requires nding solutions to the implicit eqution. We will describe how to nd prmetriztions of some common surfces, nd then give the denition of surfce integrl nd show how to compute them on both prmetric nd implicit surfces. 4.. Prmetric Surfces If we grph vector-vlued function of two vribles rs, t) = xs, t), ys, t), zs, t) s s nd t vry, we will obtin surfce in spce brring something strnge hppening). Exmple: The surfce rs, t) = x, y, z +t v, v, v 3 +s w, w, w 3 is the plne pssing through the point x, y, z ) tht contins the two vectors v = v, v, v 3 nd w = w, w, w 3, provided tht v nd w re not prllel. We could lso describe the plne s n implicit surfce of the form x+by+cz = d, where, b, c = v w is the norml vector to the plne nd d = x + by + cz. There re mny wys to describe given plne s prmetric surfce. For exmple, both of the prmetriztions rs, t) = s, t, s t nd rs, t) = 3 + s t, + t + s, + t 3s describe the sme plne x + y + z =. Exmple: For two positive rdius prmeters r nd R with r < R, the surfce dened prmetriclly by rs, t) = cost) [R + r sins)], sint) [R + r coss)], r sins), for t π nd s π is donutshped surfce clled torus. It is the surfce obtined by tking verticl circle of rdius r nd moving its center long the circle x + y = R in the xy-plne. Four tori, with respective prmeters r, R) equl to, 5),, 5), 3, 5), nd 4, 5), re plotted below: Exmple: The surfce dened prmetriclly by rs, t) = coss) + cost), s + t, sins) + sint), for t 4π nd s 4π is helicl ribbon: 5
In generl, it cn be somewht involved problem to convert geometric or verbl description of surfce into prmetriztion. It is relly more of n rt form thn generl procedure.) To prmetrize prts of cylinders, cones, nd spheres, it is lmost lwys very good ide to consider whether cylindricl or sphericl coordintes cn be of ssistnce. Using trnsltions nd resclings, we cn lso prmetrize surfces like ellipsoids. There re mny dierent wys to prmetrize the sme surfce, nd which description is best will depend on wht the prmetriztion will be used for. For exmple, x = s, y = t, z = s + t prmetrizes the cone z = x + y, but so does the prmetriztion x = s cos t, y = s sin t, z = s. If we wnt to describe the points lying over rectngulr region in the xy-plne, the rst prmetriztion is more useful, but if we wnt to describe the points on the cone up to specic height, the second prmetriztion is more useful. Exmple: Prmetrize the portion of the cylinder x + y = 4 lying between the plnes z = nd z =. In cylindricl coordintes, we know tht x = r cos θ, y = r sin θ, nd z = z. Since the given cylinder hs eqution r = in cylindricl coordintes, we see tht prmetriztion of the full cylinder is x = cos t, y = sin t, z = s, where t π but with no restrictions on s. Here we think of t s θ nd s s z.) To obtin just the portion with z we just restrict the rnge for s. Thus the prmetriztion of the desired portion of the cylinder is x = cos t, y = sin t, z = s, where t π nd s. Exmple: Prmetrize the portion of the cylinder x +y = 4 lying between the plnes z = y nd z = x+4. Like in the previous exmple, we tke the prmetriztion of the full cylinder s x = cos t, y = sin t, z = s, nd then restrict the rnges for s nd t ppropritely. In this cse, we wnt the portion of the surfce where y z x + 4. It is strightforwrd to check tht the two plnes do not intersect inside the cylinder since y inside the cylinder, while x + 4 ). So in this cse, we tke t π nd sin t s cos t + 4. Exmple: Prmetrize the sphere x + y + z = 9. In sphericl coordintes, we know tht x = ρ cosθ) sinϕ), y = ρ sinθ) sinϕ), z = ρ cosϕ). The sphere hs eqution ρ = 3, so we cn immeditely see tht x = 3 cost) sins), y = 3 sint) sins), z = 3 coss), with t π nd s π, will prmetrize the sphere. Here, we re thinking of t s θ nd s s ϕ.) 6
Exmple: Prmetrize the sphere x ) + y + ) + z 6) = 4. It is not so esy to describe this sphere using sphericl coordintes directly. However, if we shift the coordintes to center the sphere t the origin, we cn esily write down the prmetriztion. By trnslting bck, we cn see tht x = + cost) sins), y = + sint) sins), z = 6 + coss), with t π nd s π, will prmetrize the sphere. Exmple: Prmetrize the ellipsoid x 4 + y 9 + z 6 =. It is gin not so esy to write down the prmetriztion using ny of our coordinte systems directly. However, if we rescle the coordintes by setting x = x/, y = y/3, nd z = z/4, then we see x ) + y ) + z ) =, nd we cn use sphericl coordintes to prmetrize the coordintes x, y, z. By rescling bck, we cn see tht x = cost) sins), y = 3 sint) sins), z = 4 coss), with t π nd s π, will prmetrize this ellipsoid. Exmple: Prmetrize the portion of the cone z = 3 x + y tht lies below the plne z = + x + y. In cylindricl, the equtions re z = 3r nd z = + r cos θ + r sin θ. They re equl when 3r = + r cos θ + r sin θ, or r = 3 cos θ sin θ. Note tht sin θ + cos θ, so the denomintor is never zero.) The full surfce is prmetrized by x = s cost), y = s sint), z = 3s. The portion under the plne corresponds to s, with t π. 3 cos t sin t 4.. Surfce Integrls The motivting problem for our discussion of surfce integrls is: given prmetric surfce rs, t) = xs, t), ys, t), zs, t) nd function fx, y, z), we would like to integrte the function on tht surfce. Like with line integrls, we hve two nturl pplictions: computing the verge vlue of function on the surfce, nd nlzying the physicl properties of thin surfce with vrible density. As with ll the other types of integrls, the ide is to pproximte the surfce with smll ptches, sum over ll the ptches) the function vlue times the re of the ptch, nd then tke the limit s the ptch sizes pproch zero. Denition: For prmetric surfce S, prtition of S into n pieces is list of disjoint subregions inside S, where the kth subregion corresponds to s k s s k, t k t t k, nd hs re σ k. The norm of the prtition P is the lrgest number mong the res of the rectngles in P. Denition: For fx, y, z) continuous function nd P prtition of the surfce S, we dene the n Riemnn sum of fx, y, z) on R corresponding to P to be RS P f) = frs k, t k )) σ k. Denition: For function fx, y, z), we dene the surfce integrl of f on S, denoted fx, y, z) dσ, to be the vlue of L such tht, for every ɛ >, there exists δ > depending on ɛ) such tht for every prtition P with normp ) < δ, we hve RS P f) L < ɛ. Remrk: It cn be proven with signicnt eort) tht, if fx, y, z) is continuous, then vlue of L stisfying the hypotheses ctully does exist. As with ll of the other types of integrls, surfce integrls possess some forml properties: Integrl of constnt: dσ = AreS). onstnt multiple of function: fx, y) dσ = fx, y) dσ. S Addition of functions: fx, y) dσ + gx, y) dσ = [fx, y) + gx, y)] dσ. S S k= S 7
Subtrction of functions: fx, y) dσ gx, y) dσ = [fx, y) gx, y)] dσ. S S Nonnegtivity: if fx, y), then fx, y) dσ. Union: If S nd S don't overlp nd hve union S, then S fx, y) dσ+ S fx, y) dσ = fx, y) dσ. Like with line integrls, we cn reduce clcultions of surfce integrls to trditionl double integrls: Proposition Prmetric Surfce Integrls): If fx, y, z) is continuous on the surfce S which is prmetrized s rs, t) = xs, t), ys, t), zs, t), where S is described by region R in st-coordintes, then the surfce integrl of f on S is fx, y, z) dσ = fxs, t), ys, t), zs, t)) r s r t ds. S R The key step is to recognize the Riemnn sum for the surfce integrl s the Riemnn sum for prticulr double integrl. Ultimtely, the dierentil of surfce re dσ = r s r t ds rises from computing the re of smll ptch in st-coordintes: when s chnges slightly, the chnge in r is given by r, nd when t s chnges slightly, the chnge in r is given by r t. These two vectors form smll prllelogrm tht closely pproximtes the surfce S, so the dierentil of surfce re dσ is roughly equl to the re of this prllelogrm, which is r s r t, times the dierentil ds. There is lso formul for clculting the surfce integrl for n implicit surfce of the form gx, y, z) = c: Proposition Implicit Surfce Integrls): If fx, y, z) is continuous on the implicit surfce S dened by gx, y, z) = c, R is the projection of S into the xy-plne, nd g/ z on R, then the surfce integrl of f on S is fx, y, z) dσ = fx, y, z) g dy dx g k S where g is the grdient of g nd k =,,. Thus, g k = g/ z.) R The sttement tht g/ z on R is equivlent to sying tht the tngent plne to gx, y, z) = c is never verticl bove R. In prticulr this implies tht the surfce never doubles bck on itself over the region R. Thus for exmple, we could not use the method directly to compute surfce integrl on the entire unit sphere, becuse it hs verticl tngent plne bove its projection x + y in the xy-plne. This formul cn be derived from the prmetric surfce integrl formul: fter some simpliction, it is wht one obtins by using the prmetriztion rs, t) = s, t, zs, t), where zs, t) is dened implicitly vi the reltion fs, t, zs, t)) = c. Using these two results, we cn reduce clcultions of surfce integrls to trditionl double integrls: given description of the surfce S, we cn convert it to double integrl using one of two methods: For prmetric surfce given in the form rs, t) = xs, t), ys, t), zs, t) : Step : Find the bounds on s nd t which prmetrize the portion of the surfce tht's being integrted over. Step : Express the function fx, y, z) to be integrted in terms of s, t). Step 3: Find the dierentil of surfce re dσ = r s r t ds. Step 4: Write down the integrl fxs, t), ys, t), zs, t)) r s r t ds nd evlute. S 8
For n implicit surfce given in the form gx, y, z) = c: Step : Sketch the surfce, determine the shpe of its projection R into the xy-plne, nd mke sure tht the surfce does not cover ny prt of the projection more thn once. Step : Evlute the integrl fx, y, z) g dy dx, where g is the grdient of g nd k = R g k,,. It is very importnt to note tht the only vribles llowed in the integrl re x nd y, so if the function being integrted hs ny z terms we must use the implicit eqution gx, y, z) = c to get rid of them. Note tht, by swpping z with x or with y, the implicit surfce procedure cn lso be used with projection into the xz-plne or the yz-plne. Also note tht for surfce of the form z = fx, y), either method will pply. Exmple: Integrte the function gx, y, z) = z over the surfce with prmetriztion rs, t) = sint), cost), s + t for t π nd s π. We hve n explicit prmetriztion of the surfce, so we use the prmetric formul. On the surfce, we hve z = s + t so gx, y, z) = z = s + t. We hve r r r =,, nd = cost), sint),, so s t s r t = Then r s r t =. The integrl is therefore given by i j k cost) sint) = sint), cost),. ˆ π ˆ π s + t) ds = = = ˆ π [ ] s + st π s= [ ] π + πt [ π t + π ] π t = 3π 3. t= ˆ π Exmple: Integrte the function fx, y, z) = 8xy over the portion of the plne x + y + z = with x, y. We use the implicit surfce formul, with gx, y, z) = x + y + z. We hve g =,, so g = + + = 3 nd g k =. The desired integrl is therefore 8xy 3/) dy dx = 6x dx = 3. Exmple: Integrte the function fx, y, z) = xz over the portion of the plne 4x + y + z = with x, y. We use the implicit surfce formul, with gx, y, z) = 4x + y + z. We hve g = 4,, so g = 4 + + = nd g k =. Since the function involves z, we must use the implicit reltion to eliminte it. In this cse, z = 4x y, so fx, y, z) = xz = x 4x xy. The desired integrl is therefore x 4x xy) dy dx = 4x ) dx = 4. 3 To compute surfce re, we cn simply integrte the function on the surfce, in exctly the sme wy tht integrting on plne region gives its re or integrting on solid region gives its volume. Exmple: Find the re of the portion of the surfce z = x y tht lies bove the xy-plne. 9
We cn rewrite the eqution of the surfce implicitly s x + y + z =, so we use the implicit surfce formul. The projection of the surfce into the xy-plne is the region R on which x y, which is the sme s x + y, nd this describes the disc of rdius centered t the origin. Since this surfce is explicit we do not need to worry bout hving verticl tngent plne. We hve g = x, y, so g = 4x + 4y + nd g k =. The desired integrl is therefore 4x + 4y R + dy dx, since to clculte surfce re we simply integrte the function. To evlute this integrl, we chnge to polr coordintes, since both the region nd the function to be integrted will become simpler: the region is r, θ π, nd the function is 4r +. Since the re dierentil in polr is r dr dθ, we obtin the polr integrl π To evlute this new integrl, we mke nother) substitution u = 4r +, with du = 8r dr: ˆ π ˆ 4r + r dr dθ = = = ˆ π ˆ 9 ˆ π ˆ π 8 u du dθ 8 ) 9 3 u3/ 6 3π dθ = 3 dθ u=. 4r + r dr dθ. Remrk: Alterntively, we could hve prmetrized this surfce using cylindricl coordintes, s x = s cost), y = s sint), z = s for s, t π. This would hve led us directly to the integrl tht showed up t the end though with s nd t in plce of r nd θ). To nd the verge vlue of function, we integrte the function on the surfce nd then divide by the surfce re. Exmple: Find the verge vlue of fx, y, z) = z on the surfce S given by the portion of the cone z = x + y tht lies inside the cylinder x + y = 4. By using cylindricl coordintes we see tht we cn prmetrize this portion of the cone s x = s cost), y = s sint), z = s, for s nd t π. We then hve rs, t) = s cost), s sint), s, so dr dr = cost), sint), nd = s sint), s cost),. ds Then r s r t = i j k cost) sint) = s cost), s sint), s, so the mgnitude is given by s sint) s cost) r s r t s = cos t) + s sin t) + s = s. We lso hve fx, y, z) = z = s. So Also, the surfce re is Thus, the verge vlue is S dσ = π Are s s ds = π S z dσ = π 8 3 s ds = π = 4π. z dσ = 6π /3 4π = 4 3. 6π =. 3 Like with double, triple, nd line integrls, we hve mss nd moment formuls for surfce integrls: enter of Mss nd Moment Formuls Thin Surfce): Given surfce S of vrible density δx, y, z) in 3- spce: The totl mss M is given by M = δx, y, z) dσ. The x-moment M yz is given by M yz = x δx, y, z) dσ. The y-moment M xz is given by M xz = y δx, y, z) dσ. The z-moment M xy is given by M xy = z δx, y, z) dσ. Myz The center of mss x, ȳ, z) hs coordintes M, M xz M, M ) xy. M
4.3 Vector Fields, Work, ircultion, Flux Denition: A vector eld in the plne is function Fx, y) = P x, y), Qx, y) tht ssocites vector to ech point in the plne. A vector eld in 3-spce is function Fx, y, z) = P x, y, z), Qx, y, z), Rx, y, z) tht ssocites vector to ech point in 3-spce. One vector eld we hve lredy encountered is the vector eld ssocited to the grdient of function fx, y) or fx, y, z): for exmple, if fx, y) = x + xy, then fx, y) = x + y, x. To represent vector eld visully, we choose some nice) collection of points generlly in grid) nd drw the vectors corresponding to those points s rrows pointing in the pproprite direction nd with the pproprite length. Exmple: The three vector elds Fx, y) = x, y, Gx, y) = y, x, nd Hx, y) = x + y, xy re plotted below on the region with x, y : We cn lso produce these plots for 3-dimensionl vector elds, but the digrms tend to be quite cluttered; here is such digrm for Fx, y, z) = x, z y, x + y : We cn think of vector eld s describing the ow of n incompressible uid through spce: the vector Fx, y) t ny point x, y) gives the direction nd velocity of the uid's ow there. In this context, if we hve prticle tht trvels long some given pth rt) through the uid, we might like to know how much work the uid does on the prticle, or essentilly equivlently) how much the uid is pushing the prticle long its pth. This is the centrl ide behind work integrls nd circultion integrls. Intuitively, we see tht the more the vector eld F ligns with the tngent vector T to the prticle's pth, the more work it does. In the picture, prticle moving counterclockwise round the circle will be pushed long its pth by the vector eld:
Alterntively, if we hve prticle trveling long pth, we could lso sk: how much is the uid pushing the prticle o of the pth? This is the centrl ide behind ux integrl. Another wy of thinking bout this is to imgine the pth s being thin membrne, nd sking how much uid is pssing cross the membrne. Here, we see tht more uid is owing cross the membrne if the vector eld F ligns with the norml vector N to the prticle's pth: We cn lso formulte these ides in 3-dimensionl spce: the ide of circultion nd work remin the sme, but the notion of ux requires surfce for the uid to ow cross. 4.3. ircultion nd Work Integrls Denition: The counterclockwise) circultion or ow) of the vector eld F long the curve is dened to ˆ be ircultion = F T ds, where T is the unit tngent to the curve. Wht this sys is: the circultion is given by integrting the dot product function ft) = Fxt), yt)) Tt) long the curve. In order to evlute the integrl s written, we would need to prmetrize, nd the unit tngent vector Tt) to the curve, nd then integrte the dot product Fxt), yt)) Tt) long the curve. We would like to see if there is simpler wy, so let us suppose tht Fx, y) = P, Q, where P nd Q re functions of x nd y, nd sy is prmetrized by rt) = xt), yt) from t = to t = b. dx Then Tt) = vt) vt) =, dy dx P, Q, dy P dx, so F T = = + Q dy. vt) vt) vt) We cn then write F T ds = P dx b + Q dy vt) = [ b P dx vt) + Q dy ]. ˆ b [ Thus, the circultion integrl cn be written more explicitly s ircultion = P dx + Q dy ], where P, Q hve been rewritten s functions of t. Note tht this expression is lso equl to P dx+q dy. Note: The denition lso holds for curve in 3-spce, provided we simply dd the corresponding z-terms: ˆ b [ if F = P, Q, R, then ircultion = P dx + Q dy + R dz ]. Terminology Note: Some uthors reserve the term circultion for closed curves, nd use ow to refer to the generl cse. This terminology cn be somewht confusing given tht there is lso ux integrl, nd the words ux nd ow in the dictionry sense) re synonyms. Exmple: Find the circultion of the vector eld Gx, y) = y, x round pth tht winds once counterclockwise round the unit circle. We cn prmetrize the pth s x = cos t, y = sin t for t π.
Thus, P = y = sin t nd Q = x = cos t, nd lso dx So ircultion = b = cos t. P dx + Q dy ) = π sin t) sin t) + cos t)cos t)) = π = π. = sin t nd dy Denition: The work performed on prticle by vector eld F s the prticle trvels long curve is ˆ ˆ Work = F dr = F T ds. Note tht the work integrl hs the sme form s the circultion integrl. Nottion: The vector dierentil dr is dened s dr = dx, dy or s dr = dx, dy, dz for eld in 3-spce). ˆ ˆ ˆ b [ Then F dr = P dx + Q dy, so the work integrl is F dr = P dx + Q dy = P dx + Q dy ] ˆ ˆ ˆ b [ in the plne, or s F dr = P dx + Q dy + R dz = P dx + Q dy + R dz ] in 3-spce. Exmple: Find the work done by the vector eld Fx, y, z) = x + z, yz, xy on prticle trveling long the pth rt) = t, t, t from t = to t =. We hve P = x + z = 3t, Q = yz = t 3, nd R = xy = t 3. Also, dx dy dz =, = t, nd =. Therefore, the work is b P dx + Q dy + R dz ) = [ 3t)) + t 3 )t) + t 3 )) ] = 3t + 4t 4 + t 3 ) = 4 5. 4.3. Flux Across urve ˆ Denition: The ux of the vector eld F cross the curve is Flux = norml to the curve. F N ds, where N is the unit As with the circultion integrl, we would like n esier wy to evlute the ux integrl. If Fx, y) = P, Q nd is prmetrized by rt) = xt), yt) from t = to t = b, fter some lgebr dy we cn clculte tht Nt) = vt), dx. At the very lest, it is esy to observe tht this is unit vector tht is orthogonl to T.) dy P, Q, dx P dy Then F N = = Q dx. vt) vt) Plugging this in gives F N ds = P dy b Q dx vt) = [ b P dy vt) Q dx ]. ˆ ˆ b [ Thus, the ux integrl cn be written more explicitly s Flux = P dy Q dx = P dy Q dx ]. Note: The ux integrl s dened here only mkes sense for curves in the plne. In 3-dimensionl spce, the corresponding notion requires surfce integrl, since membrne will be surfce, rther thn curve. Exmple: Find the ux of the vector eld Gx, y) = x, y cross pth tht winds once counterclockwise round the unit circle. 3
We cn prmetrize the pth s x = cos t, y = sin t for t π. Thus, P = x = cos t nd Q = y = sin t, nd lso dx dy = sin t nd ) So Flux = b P dy Q dx = cos t. = π cos t)cos t) sin t) sin t)) = π = π. Exmple: For the vector eld Fx, y) = x + y, y x, nd the ux cross, nd circultion long, the portion of the curve rt) = t, t between, ) nd, 4). Here is plot of the vector eld, long with the curve: 4 3..5..5. From the picture, we would expect the circultion nd ux to be roughly equl, since the vector eld mkes roughly 45-degree ngle with the pth ner the end. The prmetriztion given sys x = t nd y = t, so tht P = x + y = t + t nd Q = y x = t t. Also, the strt is t = nd the end is t =. Then the circultion is F T ds = b P dx ) = t + t ) + t t) t ) = 4t 3 t + t ) = t 4 3 t3 + t ) t= = 5 3. The ux is F N ds = b P dy Q dx ) t4 + 3 t3 + t ) = 5 t= 3. = + Q dy Indeed, we see tht the ux nd circultion re roughly nd exctly) equl. ˆ t + t ) t t t) ) ˆ = t 3 + t + t ) = 4.3.3 Flux Across Surfce In 3-dimensionl spce, the corresponding notion of circultion long curve remins the sme. However, in order to mke sense of ux, we must insted tlk bout ux through surfce rther thn through curve): Denition: The norml) ux of the vector eld F cross the surfce S is Flux cross S = n is the outwrd unit norml to the surfce. S F n dσ, where Recll tht the norml vector to surfce is orthogonl to the tngent plne it is in fct the norml vector to the tngent plne s we dened it erlier). If the surfce is n implicit surfce gx, y, z) = c, then the norml vector n is the grdient g. Nottion: When speking of unit norml to surfce we will use lowercse n, to keep the nottion dierent from the unit norml N to curve which is n uppercse N). Remrk: The integrl F n dσ computes the ux through the surfce in the direction of the outwrd norml. It is lso possible to sk bout ux in the direction of prticulr unit vector u; the integrl in tht cse is F u dσ, insted. In generl, when it is not specied wht type of ux integrl is ment, the ux in the direction of the outwrd norml is intended. 4
If S is prmetrized by rs, t) = xs, t), ys, t), zs, t), then the two vectors r r nd spn the tngent s t plne to the surfce, nd so norml vector to the surfce is given by the cross product r s r t. r The unit norml to the surfce S t rs, t) is therefore n = s r )/ r t s r t. Importnt Wrning: If we write the fctors of the cross product in the opposite order, the cross product vector is multiplied by. To remedy this mbiguity, one must specify which of these two orienttions is being sked for. One should lwys check to ensure tht the norml vector is pointing in the correct direction typiclly, we intend for it to be pointing outwrd or upwrd.) r Plugging into the surfce integrl formul for n yields Flux cross S = F S s r ) ds, provided tht r s r is the outwrd-pointing norml vector to the surfce. Note in prticulr tht the t t ugly prt of the surfce-re dierentil cncels out the normliztion in the unit norml vector.) We lso hve formul for ux through n implicitly-dened surfce: it sys Flux cross S = where the surfce S is dened implicitly by fx, y, z) = c nd R is the projection of S in the xy-plne. Exmple: Find the outwrd ux of the vector eld F = xz, yz, x 3 e y through the portion of the cylinder x + y = 4 tht lies between the plnes z = nd z =. R F g dy dx, g k From cylindricl coordintes, we cn prmetrize the cylinder s rs, t) = cos t, sin t, s, where the desired portion corresponds to s nd t π. Then r r r = sin t, cos t, nd =,,, so t s t r s = i j k sin t cos t = cos t, sin t,. This is indeed n outwrd-pointing norml vector since it is the vector pointing from,, s) to the point rs, t) = cos t, sin t, s) on the surfce. r Then F t r ) = s cos t, s sin t, cos t) 3 e sin t cos t, sin t, = 4s cos t + 4s sin t = s 4s. The ux integrl is thus π 4s ds = π 8 6π =. 3 3 Exmple: Find the outwrd ux of the vector eld F = x z, y, x + z through the portion of the sphere x + y + z = 4 tht lies bove the plne z =. We use the ux cross n implicit surfce formul. On the sphere, z = corresponds to x + y = 3, nd s z increses to, the vlue of x + y decreses to. Thus the projection of the surfce into the xy-plne is the region R : x + y 3. We hve g = x, y, z, so F g g k = x xz + y + xz + z 4 = z 4 x y. The ux integrl is therefore given by R coordintes. 4 dy dx. We will evlute this integrl using polr 4 x y In polr coordintes, the region is r 3 nd θ π, so the integrl is π 3 4 r dr dθ. 4 r Substituting u = 4 r in the inner integrl gives π 3 4 r dr dθ = π 4 r du dθ = u π 4 dθ = 8π. Alterntively, we could hve observed tht for sphere of rdius ρ centered t the origin, the outwrd unit norml vector is n = x, y, z. ρ 5
The desired integrl is therefore x, y, z x z, y, x + z dσ = x + y + z ) dσ = S dσ. This is twice the surfce re of S, which we could compute using simpler surfce integrl) to be 4π, mening tht the desired ux is gin 8π. 4.4 onservtive Vector Fields, Pth-Independence, nd Potentil Functions If we hve vector eld Fx, y) nd two dierent pths nd between the sme two points, we might wonder if there is ny reltion between the work integrls F dr nd F dr. Exmple: For the elds Fx, y) = y, x nd Gx, y) = y, x evlute the work integrls from, ) to, ) long the the three dierent pths : x, y) = t, t), : x, y) = t 3, t ), nd 3 : x, y) = t 7, t ), for t. Along we hve F = t, t, G = t, t, dx Then F dr = [t + t ] =, nd =, nd dy =. G dr = Along we hve F = t, t 3, G = t 4, t 3, dx = 3t, nd dy = t. Then F dr = [ t + t ] = 5 6. [ t 3t + t 3 t ] =, nd G dr = [ t4 3t + t 3 t ] = 9 35. Along 3 we hve F = t, t 7, G = t, t 7, dx = 7t6, nd dy = t9. Then 3 F dr = [ t 7t 6 + t 7 t 9] =, nd 3 G dr = [ t 3 7t 6 + t 7 t 9] = 389 459. Observe tht for F, ll three pths give the sme vlue, while for G, ech pth gives dierent vlue. Denition: A vector eld F is conservtive on region R if, for ny two pths nd inside R) from P to P, it is true tht F dr = F dr. In other words, F is conservtive if ny two pths yield the sme work integrl. Equivlent to the bove denition is the following: F is conservtive on region R if, for ny closed curve in R, F dr =. A closed curve is one whose strt nd end points re the sme.) Nottion: For line integrl round closed curve, we often use the nottion, the circle being suggestive exmple of closed curve. These two sttements re equivlent becuse, if nd re two pths from P to P, then we cn construct closed pth by following from P to P nd then following from P bck to P. Then for this curve, we hve F dr = F dr F dr, nd so the left-hnd side is zero if nd only if the right-hnd side is zero. Theorem Fundmentl Theorem of lculus for Line Integrls): The vector eld F is conservtive on simply-connected region R if nd only if there exists function U, clled potentil function for F, such tht F = U. If such function U exists, then b F dr = Ub) U) long ny pth from to b. Notice the similrity of the sttement b F dr = Ub) U) to the Fundmentl Theorem of lculus, which reltes the integrl of derivtive of function to its vlues t the endpoints of pth. Technicl Note: The term simply-connected is technicl requirement needed for the proof of the theorem: intuitively, simply-connected region consists of single piece tht does not hve ny holes in it. More rigorously, it mens tht the region is connected contins only one piece) nd tht if we tke ny closed loop in the region, we cn shrink it to point without leving the region. The disc x + y 4 is simply-connected, wheres the nnulus x + y 4 is not. The full proof is not especilly enlightening. We will insted show one direction of the proof. 6
U Proof Reverse Direction in 3-Spce): Suppose tht F = U = x, U, U. z By the multivrible) hin Rule, if is the pth with x = xt), y = yt), nd z = zt) for t b, then du = U x dx + U dy + U z dz. Now we cn write ˆ F dr = = = ˆ b ˆ b ˆ b U x, U, U dx z, dy, dz [ U x dx + U dy + U z dz ] [ ] du = Urb)) Ur)) where we used the Fundmentl Theorem of lculus for the lst step. Notice tht this expression does not depend on : it only involves the potentil function U nd the two endpoints rb) nd r). Hence we see tht the integrl is independent of the pth, so F is conservtive. If we cn see tht vector eld is conservtive, then it is very esy to compute work integrls: we just need to nd potentil function for the vector eld. Exmple: Find the work done by the vector eld Fx, y) = x + y, x on prticle trveling long the pth rt) = cosπe t ), tn t) from t = to t =. If we try to set up the integrl directly using the prmetriztion, it will be rther unplesnt. However, this vector eld is conservtive: it is esily veried tht for Ux, y) = x + xy, then U = x + y, x = F. By the Fundmentl Theorem of lculus for line integrls, the work done by the vector eld is then simply the vlue of Ur)) Ur)). Since r) =, π/4 nd r) =,, the work is U, π/4) U, ) = π. An immedite question is whether there n esy wy to determine whether given vector eld is conservtive. There is, but to stte the result we rst need to dene the curl of vector eld: Denition: If F = P, Q, R then the curl of F is dened to be the vector eld curl F = F = i j k / x / / z R P Q R = Q z, P z R x, Q x P = R y Q z, P z R x, Q x P y. Exmple: If F = 3x y, xyz, e xy then curl F = F = xe xy xy, ye xy, yz 3x. If F = P, Q is vector eld in the plne then we dene the curl of F to be the curl of the vector eld P, Q, : nmely,,, Q x P. Since this vector only hs one nonzero component, some uthors dene the curl of vector eld in the plne to be the sclr quntity Q x P. We will not do this: for us, the curl of vector eld will lwys be new vector eld.) Theorem Zero url Implies onservtive): A vector eld on simply-connected region in the plne or in 3-spce is conservtive if nd only if its curl is zero. More explicitly, the vector eld F = P, Q is conservtive on simply-connected region R in the plne if nd only if P y = Q x, nd the vector eld F = P, Q, R is conservtive on simply-connected region D in 3-spce if nd only if P y = Q x, P z = R x, nd Q z = R y. It is firly esy to see why we need the equlity of the derivtives of the components: if F = P, Q = U then P = U x nd Q = U y, so by the equlity of mixed prtil derivtives, we see tht P y = U xy = U yx = Q x. 7
The three necessry equlities when F = P, Q, R follow in the sme wy: if F = U then P = U x, Q = U y, nd R = U z, so P y = U xy = U yx = Q x, P z = U xz = U zx = R x, nd Q z = U yz = U zy = R y. The converse sttement tht zero curl implies the eld is conservtive) is more dicult, nd we omit the veriction. The two theorems give us n eective procedure for determining whether eld is conservtive: we rst check whether its curl is zero, nd then if it is) we cn try to nd potentil function by computing ntiderivtives. Exmple: Determine whether Fx, y) = x + y, x + y is conservtive. If it is, nd potentil function. For F, we see [ x + y ] = = [ ] x + y, so the eld is conservtive. x To nd potentil function U with U = F, we need to nd U such tht U x = x + y nd U y = x + y. Tking the ntiderivtive of U x = x + y with respect to x yields U = 3 x3 + xy + fy), for some function fy). To nd fy) we dierentite: U y = x + f y), so we get f y) = y so fy) = 3 y3. Plus n rbitrry constnt, but we cn ignore it.) Thus we see tht potentil function for F is Ux, y) = 3 x3 + xy + 3 y3. Exmple: Determine whether Gx, y) = x + y, x + y is conservtive. If it is, nd potentil function. For G, we see [ ] x + y = y x = [ x + y ], so the eld is not conservtive. x Exmple: Determine whether Hx, y, z) = y + z, x + 3z, x + 3y is conservtive. If it is, nd potentil function. For H, we hve [y + z] = = x [x + 3z], [y + z] = = z x [x + 3y], nd [x + 3z] = 3 = z [x + 3y], so the eld is conservtive. To nd potentil function U with U = H, we need to nd U such tht U x = y + z, U y = x + 3z, nd U z = x + 3y. Tking the ntiderivtive of U x = y + z with respect to x yields U = xy + xz + fy, z), for some function fy, z). To nd fy, z) we dierentite: x+f y = x+3z nd x+f z = x+3y, so f y = 3z nd f z = 3y. Repeting the process yields f = 3yz + gz), where g z) =. Thus we see tht potentil function for H is Ux, y, z) = xy + xz + 3yz. If we cn nd potentil function for conservtive vector eld, then s we sw bove) it is very esy to compute work integrls. Exmple: If F = x 3 + 4x 3 sin y sin z + y z, xyz + y + x 4 cos y sin z, z 3 + x 4 sin y cos z + xy, nd the work done by F on prticle tht trvels long the curve : rt) = sinπt), t t + 3, t 3 + for t. In theory we could compute the work integrl using the prmetriztion of the pth, but this seems quite unplesnt. Insted, we will check whether this vector eld is conservtive: then determining the nswer only requires us to nd the potentil function of the eld. We hve P y = 4x 3 cos y sin z + yz nd Q x = yz + 4x 3 cos y sin z so they re equl. We hve P z = 4x 3 sin y cos z + y nd R x = 4x 3 sin y cos z + y so they re lso equl. Finlly we hve Q z = xy + x 4 cos y cos z nd R y = 4x 3 cos y cos z + xy, nd these re lso equl. Thus, the eld is conservtive. 8
To nd potentil function U with F = U = U x, U y, U z : We know U x = x 3 + 4x 3 sin y sin z + y z so tking the ntiderivtive with respect to x yields U = 4 x4 + x 4 sin y sin z + xy z + y, z). We then see U y = x 4 cos y sin z + xyz + y y, z) must equl xyz + y + x 4 cos y sin z so we see y = y. Then tking the ntiderivtive with respect to y yields y, z) = y + Dz). We now hve U = 4 x4 + x 4 sin y sin z + xy z + y + Dz). Then U z = x 4 sin y cos z + xy + D z) must equl z 3 + x 4 sin y cos z + xy so we see D z) = z 3 so we cn tke Dz) = 4 z4. We conclude tht potentil function for F is Ux, y, z) = 4 x4 + x 4 sin y sin z + xy z + y + 4 z4. Then the desired work integrl is equl to U,, 4) U,, ) = 6. 4.5 Green's Theorem Green's Theorem is -dimensionl version of the Fundmentl Theorem of lculus tht reltes line integrl of function round closed curve to the double integrl of relted function over the region R tht is enclosed by the curve. Green's Theorem: If is simple closed rectible curve oriented counterclockwise, nd R is the region it ˆ Q encloses, then for ny dierentible functions P x, y) nd Qx, y), P dx + Q dy = x P ) dy dx. Here is n exmple of curve nd its corresponding region R: R Remrk: The hypotheses bout the curve simple closed rectible, oriented counterclockwise) re to ensure the curve is nice enough for the theorem to hold. Simple mens tht the curve does not cross itself, closed mens tht its strting point is the sme s its ending point e.g., circle), rectible mens piecewise-dierentible i.e., dierentible except t nite number of points), nd oriented counterclockwise mens tht runs round the boundry of R in the counterclockwise direction. Proof: It suces to prove Green's Theorem for rectngulr regions, s more complicted regions cn be built by gluing together simpler ones in much the mnner of Riemnn sum); overlpping boundry pieces on two rectngles shring side will hve opposite orienttions nd will therefore cncel out. For rectngulr region x b, c y d, we then hve = + + 3 + 4, where,, 3, nd 4 re the four sides of the rectngle with the proper orienttion), nd the function to be integrted on ech curve is P dx + Q dy. Setting up prmetriztions shows [P dx + Q dy] + 3 [P dx + Q dy] = b [P x, c) P x, d)] dx, nd [P dx + Q dy] + 4 [P dx + Q dy] = d [Qb, y) Q, y)] dy. c For the double integrl we hve R P dy dx = b d c P dy dx = b [P x, c) P x, d)] dx, nd Q R x dx dy = c b Q d x dx dy = d [Qb, y) Q, y)] dy. c By compring the expressions, we see tht Q [P dx + Q dy] = R x P ) dy dx, s desired. 9
Green's Theorem cn be seen s generliztion of the Fundmentl Theorem of lculus: both theorems show tht the integrl of the derivtive of function in n pproprite sense) on region cn be computed using only the vlues of the function on the boundry of the region. Green's Theorem cn be used to convert line integrls into double integrls which cn often be esier to evlute if the curve is complicted). Exmple: Evlute the integrl 3x dx + xy dy, where is the counterclockwise boundry of the tringle hving vertices, ),, ), nd, ). We will evlute the integrl both s line integrl nd using Green's Theorem. Green's Theorem sys tht P dx + Q dy = R Q x P y ) dy dx, so setting P = 3x nd Q = xy produces 3x dx + xy dy = y dy dx, where R is the interior of the tringle. R To compute the double integrl, we need to describe the region R: A quick sketch shows tht R is dened by the inequlities x nd y x. Thus, the integrl is x y dy dx = y ) x dx = y= x) dx = 4 3. To compute the line integrl, we need to prmetrize ech piece of the boundry. omponent #, joining, ) to, ): This component is prmetrized by x = t, y = for t. Then dx = nd dy =, so the integrl is 3t =. omponent #, joining, ) to, ): This component is prmetrized by x =, y = t for t. Then dx = nd dy =, so the integrl is t = 4. omponent #3, joining, ) to, ): This component is prmetrized by x = t, y = t for t. Then dx = nd dy =, so the integrl is [ 3 t) ) + t) t) ) ] = [ t + t t ] = 3. Thus, the vlue of the integrl over the entire boundry is + 4 3 = 4 3. We frequently pply Green's Theorem to simplify the clcultion of circultion nd ux integrls: we cn use the theorem to give expressions for circultion nd ux either s line integrls or s double integrls over region. Depending on the shpe of the region nd its boundry, nd the nture of the eld F, either the line integrl or the double integrl cn be esier. Green's Theorem Tngentil form): ircultion round = F T ds = curl F) k da. Recll tht if F = P, Q, then curl F = F =,, Q x P nd curl F) k = Q x P. The curl mesures how much the vector eld is rotting round given point. Thus, if we write everything out in terms of vector eld components, the tngentil form of Green's Theorem reds Q P dx + Q dy = R x P ) dy dx, which is just the sttement we gve bove. Green's Theorem Norml form): Flux cross = F N ds = R R div F) da. Here, if F = P, Q then div F = F = P x + Q. This is clled the divergence of F nd mesures how much the vector eld is pushing inwrd or outwrd t the given point. Explicitly, the norml form of Green's Theorem reds P P dy Q dx = R x + Q ) dy dx, which we cn recognize s the originl sttement of Green's Theorem except with Q in plce of P nd P in plce of Q.
There is nice interprettion of the norml form of Green's Theorem: imgine tht F is modeling popultion movement, nd tht is the border of country tking up the region R. At city long the border, the vlue F N mesures the immigrtion in or out) to tht city from cross the border. At city inside the country, the vlue div F mesures the net immigrtion into or out of) tht city. The norml form of Green's Theorem then sys: if we dd up the net immigrtion long the border, this equls the totl popultion ow inside the country. These two quntities re denitely equl, since they both tlly the net immigrtion into the country s whole.) Exmple: Find the outwrd ux through, nd the counterclockwise) circultion round, the squre with vertices, ),, ),, ), nd, ), for the vector eld Fx, y) = x xy, y 3 x. We could prmetrize the boundry of this region nd evlute the line integrls to nd the ux nd circultion. However, this would be very tedious, s it requires computing four line integrls ech time one for ech side of the squre). We cn sve lot of eort by using Green's Theorem, which pplies becuse the boundry is closed curve. Flux: Green's Theorem sys tht Flux cross = F N ds = R P x + Q ) dy dx. Here, we hve P = x xy nd Q = y 3 x, nd the region is x nd y. Therefore, since P Q = x y nd x = 3y, the ux is ˆ ˆ x y + 3y ) dy dx = xy y + y 3) dx = 4x + 4) dx = 6. y= ircultion: Green's Theorem sys tht ircultion round = Q F T ds = R x P ) dy dx. Since Q x = nd P = x, the circultion is ˆ ˆ + x) dy dx = + 4x) dx = 4. Exmple: For Fx, y) = x y, xy, nd the outwrd ux through nd the counterclockwise) circultion round the circle x + y = 4. We pply Green's theorem: in this cse, the region R is the region x + y 4. The ux is R P x + Q ) dy dx = xy + xy) da = da =. R R The circultion is R Q x P ) dy dx = R y + x ) da = π r r dr dθ = 8π, upon switching to polr coordintes. One of the mny pplictions of Green's Theorem is to give vrious wys to compute the re of plnr region using line integrl round its boundry. Speciclly, if is the counterclockwise boundry curve of the region R nd nd R stisfy the hypotheses of Green's Theorem), then Are of R = x dy = y dx = x dy y dx) becuse by Green's Theorem, ech of the line integrls is equl to R dy dx, which is the re of R. Exmple: ompute the re enclosed by the ellipse x = cos t, y = b sin t, t π. Using the third formul, we compute A = x dy y dx) = π [ cos t)b cos t) b sin t) sin t)] = π 4.6 Stokes's Theorem nd Guss's Divergence Theorem b = πb. Stokes's Theorem nd Guss's Divergence Theorem re generliztions of the two forms of Green's Theorem to the 3-dimensionl setting. As with Green's Theorem, these theorems cn be used in either direction, depending on which integrl is esier to set up nd evlute.
4.6. Stokes's Theorem Stokes's Theorem: If is simple closed rectible curve in 3-spce tht is oriented counterclockwise round the surfce S, then we hve ircultion round = F T ds = S curl F) n dσ where T is the unit tngent to the curve nd n is the unit norml to the surfce. Notice the similrity of the sttement of Stokes's Theorem to the tngentil form of Green's Theorem. Importnt Note: The curve must run counterclockwise round S in other words, when wlking long, the surfce should be on its left-hnd side. If one wishes to set up problem where curve runs clockwise round surfce, it is equivlent to trversing the curve in the opposite direction, nd so the integrl will be scled by. Remrk: The hypotheses bout the curve simple closed rectible, oriented counterclockwise) re the sme s in Green's Theorem, nd they ensure the curve is nice enough for the theorem to hold. i j k Recll curl F = F = / x / / z R P Q R = Q z, P z R x, Q x P if F = P, Q, R. Intuitively, if we think of vector eld s modeling the ow of uid, the quntity curl F) n t x, y, z) mesures how much the uid is circulting round the point x, y, z) long the surfce. Stokes's Theorem then sys: we cn mesure how much the uid circultes round the whole surfce by mesuring how much it circles round its boundry. The proof of Stokes's Theorem which we omit) is essentilly the sme s the proof of Green's Theorem: we cn reduce to the cse of showing the result for simple ptches on the surfce. Then, by prmetrizing the ptches explicitly, we cn show Stokes's Theorem is essentilly the sme s the tngentil form of Green's Theorem on ech ptch. Stokes's Theorem generlizes the tngentil form of Green's Theorem to cover 3-dimensionl closed curves nd the surfces they bound. Note tht unlike in Green's Theorem, there re mny possible surfces tht ny given curve cn bound. For exmple, the unit circle x + y =, z = in the xy-plne bounds the upper portions i.e., where z ) of the sphere x + y + z =, the prboloid z = x y ), nd the cone z = x + y, s pictured below: Typiclly, we use Stokes's theorem when the line integrl over the boundry is dicult, but there is nice surfce vilble. Exmple: Find the circultion of the eld Fx, y, z) = y z 3, xyz 3, 3xy z round the ellipse given by the intersection of the upper hlf of the ellipsoid x + y + z = with the cone x + y = z.
Here is picture of the surfces nd the ellipse: We could write down prmetriztion for this ellipse with little bit of eort: substituting the cone's eqution into the sphere's eqution gives 3z = hence z =. Then using the fct tht x + y = 4 is prmetrized by x = cost) nd y = sint) gives us prmetriztion for the curve s rt) = cost), sint),. The resulting circultion integrl does not look so wonderful, lthough it is possible to evlute it. Another wy is to try to use Stokes's Theorem. We hve two obvious surfces to choose from ellipsoid nd cone); since the curve runs counterclockwise round the ellipsoid, we will use tht. Stokes's Theorem tells us tht ircultion round = F T ds = curl F) n dσ. We hve curl F = i j k / x / / z y z 3 xyz 3 3xy z = 6xyz 6xyz, 3y z 3y z, yz 3 yz 3 =,,. So the curl of F is zero. Hence curl F) n will lso be zero, so we see tht the circultion is, without even hving to evlute the surfce integrl. Exmple: Find the ux of the curl curlf) n dσ, where F = yzi xzj + ex+y k, S is the surfce which is the prt of the sphere x + y + z = 5 below the plne z = 3, nd n is the outwrd norml. We will use Stokes's Theorem. In this cse, we wnt S to be the prt of the sphere x + y + z = 5 which is below the plne z = 3. The boundry of this surfce will be the intersection of the plne nd the sphere: we see tht the curve is the set of points { x, y, z) : x + y = 6, z = 3 }, which is circle tht we cn prmetrize s rt) = 4 cost), 4 sint), 3 for t π. However: the surfce S lies below the curve, not bove it: so, when viewed from below which is required becuse we re using the the outwrd norml), the curve runs clockwise round the surfce. In order to pply Stokes's Theorem, we need to reverse the orienttion of the curve, which we cn do by interchnging the limits of integrtion: thus we strt t t = π nd end t t =. From Stokes's Theorem, the ux of the curl is given by the line integrl F dr = P dx+q dy +R dz. We hve P = sint), Q = cost), nd R = e 4 cost)+4 sint), nd lso dx = 4 sint), dy = 4 cost), nd dz =. We get F dr = [ π sint)) 4 sint) ) + cost)) 4 cost) )) + e 4 cost)+4 sint) ] = 48 = 96π. π 4.6. Guss's Divergence Theorem Guss's Divergence Theorem: If S is closed nd bounded) piecewise-smooth surfce which fully encloses solid region D, nd F is continuously dierentible vector eld, then we hve Flux cross S = where n is the outwrd unit norml to the surfce. S F n dσ = D div F) dv Notice the similrity of the sttement of the Divergence Theorem to the norml form of Green's Theorem. 3