Iterated, Line, and Surface Integrals

Transcription

1 hapter Iterated, Line, and Surface Integrals. Iterated Integrals Iterated Double Integrals Double Integrals over General egions hanging the Order of Integration Triple Integrals Triple Integrals over General egions Iterated Double Integrals Let f be a real-valued function of two variables, defined on a rectangular region {(, ) :a b, c d} where a, b, c, d are real numbers. We partition [a, b] and [c, d] as follows. Let m, n be positive integers, let a,,..., m b be a partition of [a, b], and let c,,..., n d be a partition of [c, d]. For i m and j n, consider the rectangular subregions i,j {(, ) : i i, j j }. The collection { i,j : i,...,m and j,...,n} is called a partition of into sub-rectangles. Let i i i, j j j, and let be the norm of which is defined as the maimum of all the i s and j s. We denote the area of rectangle i,j b A i,j i j. Let L be a real number. We smbolicall write n m L lim f ( i,j, i,j ) A i,j i j if for ever ε>thereeistsδ> such that n m f ( i,j, i,j ) A i,j L <ε i j for all ( i,j, i,j )in i,j whenever <δ. If such a limit L eists, we sa f is iemann integrable or integrable on, and we write n m f(, )da lim f ( i,j, i,j ) A i,j. i j We must confess in all humilit that, while number is a product of our mind alone, space has a realit beond the mind whose rules we cannot completel prescribe. -arl Gauss 59

2 6 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS We quote a theorem from advanced calculus about iemann integrable functions. We omit the proof of the theorem since it is beond the scope of this book. Theorem. iemann Integrable Functions Let z f(, ) be a bounded real-valued function on a rectangular region {(, ) :a b, c d}. If the points of discontinuities of f in lie on a finite union of graphs of continuous functions of one independent variable, then f is iemann integrable on. In particular, ever continuous real-valued function on a rectangular region is integrable on. However, in order to evaluate f(, )da, weintroducethe concept of an iterated integral. If we fi a value of, thenf(, ) is a function of onl, and we have definite integral b f(, )d. To illustrate, we assume is constant in the integration below: a 4 sin d 4 sin (4) sin 6sin. () sin Likewise, d f(, )d is a definite integral if is held constant. The net theorem c implies that a double integral f(, )da ma be evaluated as an iterated integral if f is continuous on. Also, we omit the proof of the theorem that is usuall discussed in advanced calculus tets. Theorem. Fubini s Theorem and Iterated Integrals If z f(, ) is the function in Theorem., then the identit below holds: f(, )da b d f(, )dd d b a c c a In particular, if f is continuous on, the above identit applies. f(, )dd Aboundedreal-valuedfunctiononarectangularregion whose set of discontinuities is of measure zero is iemann integrable. Also, if f(), a b, isacontinuousfunction, then {(, f()) : a b} is a set of measure zero in.

3 .. ITEATED INTEGALS 6 Eample Evaluating an Iterated Integral Evaluate the integral Solution 3 π cos()dd. For the inner integral, we have cos()d sin +. onsequentl, we evaluate the iterated integral as follows: 3 π 3 cos()dd sin() 3 Then we integrate b parts using the substitutions: π sin(π)d. d u dv sin(π)d du d v π cos(π) Since udv uv vdu, we obtain 3 sin(π)d π cos(π) 3 + π 3 cos(π)d π + π sin(π) 3 π. Thus, we find 3 π cos()dd π. Tr This Evaluate 4 9 dd. Double Integrals over General egions Let φ,φ,ψ,ψ be continuous real-valued functions satisfing φ () φ () and ψ () ψ () for all in [a, b], and in [c, d]. We evaluate double integrals of a real-valued continuous function z f(, ) over elementar regions in the -plane, see Figures -. We classif these regions as either of tpe or, or possibl of both tpes, depending on whether the boundar of the region are graphs of functions of or, respectivel: A) {(, ) :a b, φ () φ ()} B) {(, ) :c d, ψ () ψ ()}

4 6 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Φ Ψ d Φ a b c Ψ Figure egion of tpe Figure egion of tpe To evaluate f(, )da, let be a rectangular region that contains.we etend f to a function F on : f(, ) if (, ) F (, ) if (, ) / Since f is bounded and continuous on, the function F is bounded and possibl discontinuous onl on points in the graph of φ or φ.then F (, )da eists b Theorem.. Suppose we choose such that is consists of points (, ) satisfing a b and p q. Notice, for an in [a, b], we find F (, ) if does not belong to the interval [φ (),φ ()]. Thus, q p F (, )d φ() φ () F (, )d φ() φ () f(, )d. and the latter integral eists b the continuit of f on. B Fubini s Theorem, we obtain b q b φ() F (, )da F (, )dd f(, )dd. a p a φ () We have a similar identit for double integrals over a tpe region. Theorem.3 Integrating over egions of Tpe or If z f(, ) is a continuous real-valued function on a region of tpe or, then b φ() a) f(, )da f(, )dd a φ () d ψ() b) f(, )da c ψ () f(, )dd In Theorem.3, iff(, ) for all,, then da represents the area of region where or.

5 .. ITEATED INTEGALS 63 Also, we ma appl double integrals to define the volume of a solid. Definition Volume of a Solid and Double Integrals Let be a region in the -plane of either tpe or. Let z f(, ) be a nonnegative real-valued continuous function defined on. Let S be the solid consisting of the points (,, f(, )) in 3-space where (, ). The volume of S is defined as Volume f(, )da. Eample Evaluating the Volume of a Solid Find the volume of the solid that lies below the plane f(, ) (6 ) and 3 above the region {(, ) :, /} in the -plane, see Figure 3. Solution The region is both tpe and, see Figure 4. We choose to evaluate the integral for the volume as a tpe region. Volume f(, )da 3 3 / Volume 8 9 units3 (6 )dd 3 / (6 ) d 3 4 d Figure 3 Asolidboundedaboveba plane, and with a triangular base in the -plane Figure 4 The base of the above solid in the -plane. Tr This A solid S is bounded above b the plane z and bounded below b the region {(, ) :, }. Find the volume of S. Figure 5 The solid for Tr This.

6 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS hanging the Order of Integration To be able to integrate f(, )da, there ma be an advantage to evaluating one iterated integral over another. The reason is the order of integration, i.e., integrating with respect to then integrating with respect to, ma be easier than integrating first with respect to, thenwithrespectto secondl. Eample 3 Switching the Order of Integration Sketch the region of integration of 4 /4 sin(π )dd. Then switch the order of integration, and evaluate the resulting integral. Figure 6 egion of integration Solution In Figure 6, we see the triangular region. To switch the order of integration to dd, we partition the interval [, ] in the -ais into smaller subintervals. If the order of integration is dd, partition [, 4] in the -ais. Then draw a tpical rectangle such that its base is a subinterval in the partition, and its length etends from the lower boundar to the upper boundar of the region. Since the order of integration is dd, each point (, ) in a tpical rectangle satisfies and 4. Then we integrate as follows: 4 /4 sin(π )dd 4 sin(π )dd sin(π ) 4 d 4 4 sin(π )d π cos(π ) /4 sin(π )dd 4 π Figure 7 egion of integration for Tr This 3. Tr This 3 Sketch the region of integration of 4 / 3 + dd. Then switch the order of integration, and evaluate the resulting integral.

7 .. ITEATED INTEGALS 65 Triple Integrals Let w g(,, z) be a real-valued function that is defined on a artesian product S [a, b] [c, d] [e, f] {(,, z) :a b, c d, e z f} We partition each of [a, b], [c, d], and [e, f] into finitel man subintervals. Let [ i, i+ ], [ j, j+ ], and [z k,z k+ ] be subintervals in the partitions, and we denote their lengths b i, j, z k, respectivel. We consider a bo S ijk [ i, i+ ] [ j, j+ ] [z k,z k+ ] whose volume is denoted b V ijk i j z k. Let p ijk be a point in S ijk, and let be the maimum of the norms of the partitions of [a, b], [c, d], and [e, f]. We define g to be iemann integrable or integrable on the bo S if there is a number L such that for each ε> there eists a positive number δ> satisfing g p ijk Vijk L <ε i j k for all p ijk in S ijk whenever <δ. In the above sums, we evaluate over all the values of i, j, k. In such a case, we write S g(,, z)dv lim g p ijk Vijk L. We state an integrabilit condition for g similar to Theorem.. Also, we state a Fubini s theorem for triple integrals g(,, z)dv. We omit the proofs since S the are usuall discussed in advanced calculus tets. i j k Theorem.4 iemann Integrable Functions Let w g(,, z) be a bounded real-valued function on a bo S [a, b] [c, d] [e, f]. If the points of discontinuities of g in S lie on a finite union of graphs of continuous functions of two independent variables, then g is integrable on S. Theorem.5 Fubini s Theorem for Triple Integrals If w g(,, z) is the function in Theorem.4, then g(,, z)dv b d f S a c e Also, the si iterated triple integrals eist and are equal. g(,, z)dzdd

8 66 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Eample 4 Evaluating a Triple Integral Evaluate the integral Solution 3 3 We integrate as follows: + z dddz + z dddz. 3 + z ddz 3 + z ddz z dddz. 3 +z ddz +z z 9 dz z dz 3 dz Tr This 4 Evaluate the triple integral ze dddz. Triple Integrals over General egions We wish to etend and evaluate triple integrals g(,, z)dv over elementar S regions S 3 that we are about to describe. We consider an elementar region in the -plane: {(, ) :a b, φ () φ ()} where φ,φ are continuous functions of such that φ () φ (). Then we associate an elementar region in 3-space such as S {(,, z) :(, ), Φ (, ) z Φ (, )}. where Φ, Φ are continuous functions on satisfing Φ (, ) Φ (, ). Using similar ideas leading to Theorem.3, wehave b φ() Φ(,) g(,, z)dv g(,, z)dzdd S provided g is continuous on S. a φ () Φ (,)

9 .. ITEATED INTEGALS 67 Similarl, if an elementar region in the -plane is given b {(, ) :c d, ψ () ψ ()} where ψ,ψ are continuous functions of such that ψ () ψ (), we associate an elementar region S {(,, z) :(, ), Ψ (, ) z Ψ (, )}. where Ψ, Ψ are continuous functions on satisfing Ψ (, ) Ψ (, ). Likewise, we obtain d ψ() Ψ(,) g(,, z)dv g(,, z)dzdd S c ψ () Ψ (,) whenever g is continuous on S. There are si possible iterated triple integrals, and another one of these is f γ(z) Γ(,z) g(,, z)dv g(,, z)dddz S e γ (z) Γ (,z) where γ,γ are continuous functions of z satisfing γ (z) γ (z), and Γ, Γ are continuous functions of (, z) in some elementar region in the z-plane such that Γ (, z) Γ (, z). We summarize and define an elementar region S 3 in 3-space. For an point in S, two of its coordinates lie in an elementar region in a plane such as the -, z-, or z-plane, and the third coordinate lies between two continuous functions of the first two variables. Specificall, let,, 3 be a permutation of,, z, and let p<qbe real constants. Let f,f,f,f be continuous functions satisfing f ( 3 ) f ( 3 )wheneverp 3 q, and further F (, 3 ) F (, 3 ) if f ( 3 ) f ( 3 ). A point lies in S if its coordinates satisf p 3 q, f ( 3 ) f ( 3 ), and F (, 3 ) F (, 3 ). Following the idea of Theorem.3, we have the following theorem: Theorem.6 Triple Integrals over Elementar egions If g is a real-valued continuous function on an elementar region S 3,then q f( 3) F(, 3) g dv g d d d 3 S p f ( 3) F (, 3) In Theorem.6, ifg(,, z) for all,, then dv is the volume of S. Eample 5 Triple Integral Let S 3 be a region in the first octant that is bounded above b the plane + + z, and bounded below b the -plane. Then evaluate 6dV. Solution In Figure 8, the base of solid S in the -plane ma be epressed as {(, ) :, }. Then region S is given b S {(,, z) :(, ), z }. S S Figure 8 Solid of integration for Eample 5.

10 68 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS We appl Theorem.6, and evaluate the inner-most integral. 6dV S 6 dzdd 6( )dd Then we evaluate the inner-integral with respect to : (( ) 6) dd 6dV S ( ) 3 d 4( ) ( ) d 6dV. S ( ) d Tr This 5 A solid S 3 lies in the first octant, bounded b the surfaces z and +, and the coordinate planes. Then evaluate S dv. Figure 9 The solid of integration for Tr This 5.. heck-it Out Evaluate the iterated integral.. ( + ) 3 dd. 3 dd 3. ( )dzdd ewrite the integral b switching the order of integration. 4. f(, )dd 5. / f(, )dd

11 .. ITEATED INTEGALS 69 True or False. If false, revise the statement to make it true or eplain... b d f(, )dd b d a c a c ( + )dd d + f(, )dd d 3. z dzdd 8 4. If {(, ) : + r },then da πr. 5. The volume of the solid bounded b the hemisphere z 4 and the plane z is given b 4 dd. Eercises for Section. In Eercises -8, evaluate the iterated integrals π/ π/6 3 6 sin()dd. (4 +z)dzdd 4. 6e + dd 6. /3 /4 z e π sec (π)dd dddz dd + 7. z / 4 e dddz 8. 4 z z dzdd In Eercises 9-4, we see a region of integration that is bounded b the graph of the indicated equation. Epress the double integral of f(, ) over using dd. Then switch the order of integration to dd / 3 4 For No. 9 For No. For No.

12 7 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS ( ) For No. For No. 3 For No. 4 In Eercises 5-6, switch the order of integration, and evaluate the integral with the new limits. 5. dd dd 7. / 8dd 8. 4 dd 9. 4 dd. dd dd dd 5. 9 dd 3. 4e dd 6. π/ sin cos()dd π cos 4 dd In Eercises 7-34, find the volume of the solid that is bounded b the coordinate planes, and the graph of the equations in 3-space. 7. z,, z 3,, For No. 7 For No z 3, z For No. 9 For No. 3

13 .. ITEATED INTEGALS 7 3. z, 4, z For No. 3 For No z, + z 34. z, z For No. 33 For No. 34 iscellaneous Problems 35. Find the volume of the solid that is bounded b the plane z 8, and the coordinate planes. 36. Find the volume of the solid bounded b the plane + + z, and the coordinate planes. 37. Find the volume of the solid in the first octant bounded b the plane +4 + z 4, and the coordinate planes. 38. Find the volume of the solid in the first octant bounded b the plane + + z 4, and the coordinate planes. 39. Find the volume of a solid in the first octant that is bounded b the surfaces z and. 4. Find the volume of a solid in the first octant that is bounded b the graphs of z,,, and z. 4. Evaluate da where (, ) :, arcsin π. 6

14 4. Evaluate 7 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS sin da where {(, ) :, }. 43. Evaluate ( z)dv where S is the solid in 3-space that is bounded S b the graphs of, 4, z, and the coordinate planes. 44. Evaluate S zdv where S is the solid in 3-space that is bounded b the graphs of,, z, and the coordinate planes. 45. Find the volume of a solid that is bounded b the circular clinder +, the plane z +, and the -plane. 46. Evaluate 6dV where S is the solid in the first octant bounded b the paraboloid z +, S the plane z, and the coordinate planes. 47. A solid S consists of the points (,, z) satisfing,, z and z, see Figure for No. 47. Then find the volume of S. In terms of probabilit, if,, z are independent uniform random variables on [, ], then the volume of S is the probabilit that the solutions t of the quadratic equation t + zt + are real numbers.. Figure for No. 47

15 .. HANGE OF VAIABLES IN INTEGATION 73. hange of Variables in Integration Jacobian Determinant lindrical oordinates Spherical oordinates Jacobian Determinant The substitution method is an important technique in the integration of a function of one variable, namel, (b) (a) f()d b a f((u)) (u)du. We etend the above method to the integration of multivariable functions. Let D {(u, v) : a u b, c u d} be a rectangular region in the uv-plane, see Figure. Let (u, v) and (u, v) be real-valued functions on D with continuous first partial derivatives. Let T be a transformation from D into the -plane defined b T (, ) ((u, v),(u, v)). Further, suppose T is a one-to-one function on D, i.e.,t maps distinct points in D to distinct points in the -plane. Let be the image of D under the transformation T,i.e, T (D) {(, ) : (u, v), (u, v), (u, v) D}. If the sides of rectangle D are small, we approimate the area of. The idea is to use the linear approimation of T near (a, c) as described in (36), page 39. oreover, we use a result from linear algebra claiming that a one-to-one linear transformation maps a rectangle to a parallelogram. In Figure, let v be the vector from point T (a, c) tot (b, c), and let w be the vector from T (a, c) tot (a, d). We ma approimate v and w b the tangent vectors along the boundar of, respectivel,i.e., v u T and w v T u v (a,c) (a,c) d c v a D Figure egion S in the uv-plane Ta,d Ta,c Tb,c Figure egion T () inthe-plane b u if u b a and v d c are small. The area of D is u v, and the area of is approimatel the area of the parallelogram defined b v and w. ecall, the area of a parallelogram is the magnitude of the cross product of the vectors defining the sides of the parallelogram. Then Area of v w T u (a,c) T v (a,c) (Area of D) Det i j k u v u v (Area of D) The linear approimation of T near (a, c) isdefinedbt () A( )+T ( ) where A is a b matri, and, are realized as column vectors.

16 74 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS where the partial derivatives are evaluated at (a, c). onsequentl, we find Area of (, ) (u, v) (Area of D) where The factor (, ) (u, v) Det u v u (, ) is called the Jacobian determinant of the transformation T. (u, v) Let z f(, ) be a real-valued continuous function on an elementar region in the -plane. We sketch a proof of the change of variables theorem. For a detailed proof, consult an advanced calculus tetbook. f(, )dd n i j n i j v m f( i, j ) i j We summarize the discussion into a theorem.. m f((u i,v j ),(u i,v j )) (, ) (u, v) u v S f((u, v),(u, v)) (, ) (u, v) dudv Theorem.7 hange of Variables Let z f(, ) be a continuous real-valued function defined on an elementar region in the -plane. Let (, ) T (u, v) be a one-to-one function defined on an elementar region S in the uv-plane. Suppose the components of T, namel, (u, v) and (u, v) have continuous partial derivatives and T (S). Then f(, )dd f((u, v),(u, v)) (, ) (u, v) dudv. S.5.5 Eample Appling a hange of Variables Evaluate da where is the rectangular region with vertices Figure 3 egion in -plane at (, ), (, ), ( 3, ), and (, ). Let (u + v) and (u v).

17 .. HANGE OF VAIABLES IN INTEGATION 75 Solution Solving for u and v, wefindu + and v. The four vertices of correspond to four points in the uv-plane: (, ) (u, v) (, ) (, ) (, ) (, ) ( 3, ) (, ) (, ) (, ) The region is mapped to region S in the uv-plane, see Figure. Since (u+v)/ and (u v)/. The Jacobian determinant is given b (, ) (u, v) Det u v Det u v Notice, uv. Appling Theorem.7, we obtain da (, ) uv (u, v) dvdu udu vdv v S Figure 4 egion in uv-plane u 4 9 Tr This Evaluate the same integral in Eample, butlet u + v and u v. Eample Appling Polar oordinates Evaluate e ( + ) da where is the region between the circles + 9 and + 4, where. See Figure 5. Solution We appl the change of variables r cos θ and r sin θ, where r. These equations arise from trigonometr, as shown below. 3 r Figure 5 A region S between two concentric circles. Θ Figure 6 ight triangle trigonometr

18 76 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Π Θ S 3 Figure 7 ectangular region S in the rθ-plane. r Solving for r and θ, we have r + and tan θ / if. In particular, r is the distance between point (, ) and the origin. In Figure 5, we deduce that r satisfies r 3. Also, θ is the standard angle that the line through the origin and point (, ) makes with the positive -ais. From Figure 5, we see θ satisfies θ π. The region S in the rθ-plane corresponding to region is shown in Figure 7. Since r cos θ and r sin θ, the Jacobian determinant is given b (, ) (r, θ) Det r r θ θ Det r cos θ +sin θ r. cos θ sin θ r sin θ rcos θ Appling r + and the change of variables, we obtain e ( + ) (, ) da e r (u, v) drdθ S π 3 re r drdθ π r3 e r dθ r e ( +) da π e 4 e 9. Tr This Evaluate + da where is the circular region bounded b +. Appl the transformation defined b r cos θ and r sin θ. Net, we discuss the change of variables for triple integrals. Let T be a one-toone transformation from a rectangular bo B {(u, v, w) :a u b,a v b,a 3 w b 3 } into 3 where a i <b i are real constants, i,, 3. We investigate the effect of T on the volume of B. Let (u, v, w), (u, v, w), z z(u, v, w) be real-valued functions on B with continuous partial derivatives such that (,, z) T (u, v, w). The derivative T of T at (a,a,a 3 ) provides the best linear approimation 3 of T near (a,a,a 3 ). 3 The linear approimation of T near (a,a,a 3 )isdefinedbt () A( )+T ( ) where A is a 3 b 3 matri, and, 3 are realized as column vectors. See (36), page 39.

19 .. HANGE OF VAIABLES IN INTEGATION 77 We appl a result from linear algebra stating that an injective linear transformation in 3-space should map B onto a parallelepiped. If all the sides of B are small, then the volume of T (B) is approimatel the volume of the parallelepiped T (B) whose three defining edges are vectors v u T u, T v v v, T v3 w w. where the partial derivatives are evaluated at (a, b, c), u b a, v b a, and w b 3 a 3. B Theorem., the volume of T (B) is the absolute value of the triple scalar product of v, v, and v 3. Appling identit () in page 5, the triple scalar product is the determinant of the 3 b 3 matri whose rows are the vectors. That is, (v v ) v 3 (,, w) (u, v, w) Vol(B) where the volume of B is Vol(B) u v w, and the Jacobian determinant is (,, z) (u, v, w) Det and the partial derivatives are evaluated at (a, b, c). Then an effect of transformation T on the volume of B is given b Vol(T (B)) Vol(T (B)) (,, w) (u, v, w) Vol(B). Furthermore, let A F (,, z) be a real-valued function. We sketch a proof of the change of variables for triple integrals. n m p F (,, z)dv F ( i, j,z k ) i j z k i j k n m i j k N u u z u v v z v w w z w p (F T )(u i,v j,w k ) (,, z) (u, v, w) u v w (F T )(u, v, w) (,, z) (u, v, w) dv A detailed proof of the change of variables is found in advanced calculus books. Theorem.8 hange of Variables II Let A F (,, z) be a continuous real-valued function defined on an elementar region in the z-space. Let (,, z) T (u, v, w) be a one-to-one function defined on an elementar region N in the uvw-space. Suppose the components of T, namel, (u, v, w), (u, v, w) and z (u, v, w) have continuous partial derivatives and T (N). Then F (,, z)dv (F T )(u, v, w) (,, z) (u, v, w) dv. N

20 78 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Figure 8 lindrical solid of radius Eample 3 Triple Integral and hange of Variables Evaluate + dv where is the solid bounded b the circular clinder + 4, and the planes z and z, see Figure 8. Appl the transformation r cos θ, r sin θ, and z z. Solution The base of the solid lies in the -plane. Notice, + r.then each point (,, ) in the base satisfies + 4, or equivalentl r and θ π. In addition, the z-component of an point in satisfies z. Then the solid defined b N {(r, θ, z) : r, θ π, z } Figure 9 ectangular solid N is mapped into b the transformation, see Figure 9. The Jacobian determinant of the transformation is given b r θ z cos θ r sin θ (,, z) (r, θ, z) Det r θ z sin θ rcos θ r z r z θ z z Appling the change of variables theorem, we obtain + dv r (,, z) (r, θ, z) dv N π r drdθdz 6π 3. Figure Washer between two clindrical clinders Tr This 3 Evaluate e z dv where is the solid between two circular clinders + and + 4, and the planes z and z. See Figure, and appl the same transformations used in Eample 3.

21 .. HANGE OF VAIABLES IN INTEGATION 79 Eample 4 hanging the Variables in a Triple Integral dv Evaluate + + where the solid S is the unit ball bounded b + z S + + z. Appl the transformation ρ sin φ cos θ, ρ sin φ sin θ, and z ρ cos φ where ρ, φ π, and θ π. Solution We find + + z (ρ sin φ cos θ) +(ρ sin φ sin θ) +(ρ cos φ) (ρ sin φ) (cos θ +sin θ)+(ρ cos φ) (ρ sin φ) +(ρ cos φ) Figure Solid ball S of radius. ρ. Observe, ρ is the distance between (,, z) and the origin. Also, θ is the angle between the -ais and the vector from the origin to (,, ) for tan θ /. The dot product of (,, z) withk satisfies (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) k ρ cos φ. In particular, φ is the angle between vector (,, z) and k. The solid ball S is mapped into a rectangular solid T, see Figure. The Jacobian determinant of the transformation is given below but we postpone the proof to page 83 of the section: (,, z) (ρ, φ, θ) Det ρ ρ z ρ φ φ z φ θ θ z θ ρ sin φ. Figure ectangular solid T Appling the change of variables, we obtain dv z S T +ρ (,, z) (ρ, φ, θ) dρdφdθ π π ρ sin φ +ρ dρdφdθ S ρ 4π +ρ dρ 4π +ρ dρ 4π(ρ arctan ρ dv z π(4 π) ρ ρ

22 8 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Tr This 4 Evaluate dv where S is the unit ball that is bounded b the sphere S + + z. Appl the transformation ρ sin φ cos θ, ρ sin φ sin θ, and z ρ cos φ where ρ, φ π, and θ π. lindrical oordinates The clindrical coordinates of a point (,, z) 3 in artesian coordinates are (r, θ, z) where r +, tan θ if. Figure 3 lindrical coordinates (r, θ, z) ofpointp. In particular, θ is the angle between the positive -ais and the line segment joining the origin to (,, ). Usuall, we require θ<π. Also, (r, θ) represent the polar coordinates of (, ). Note, r is a real number representing the directed distance from the origin to (, ). The clindrical coordinates of a point are not unique. For instance, the clindrical coordinates (, π/6, 3), (, 3π/6, 3), and (, 7π/6, 3) represent the same point in 3-space. The identities below r cos θ, r sin θ, z z. are helpful when converting to artesian coordinates from clindrical coordinates. Eample 5 Switching between artesian and lindrical oordinates Find the clindrical coordinates of the point P (,, z) (, 3, 4) in artesian coordinates. Then find the artesian coordinates of the point Q(r, θ, z) 4, 5π 6, given in clindrical coordinates. Solution Since P (,, z) (, 3, 4), we find r Figure 4 lindrical point (r, θ, z) 4, π 3, 4. From the identit tan θ / 3, we ma choose θ π 3, see Figure 4. Then the clindrical coordinates of point P are (r, θ, z) 4, π 3 +kπ, 4, 4, π3 +kπ, 4 where k is an integer. Using the clindrical coordinates Q(r, θ, z) 4, 5π 6,,wefind r cos θ 4 cos 5π 6 3 r sin θ 4sin 5π 6. Thus, the artesian coordinates of Q are Q(,, z) ( 3,, ) since z.

23 .. HANGE OF VAIABLES IN INTEGATION 8 Tr This 5 Find the clindrical coordinates of point P (,, z) (,, ). Also, find the artesian coordinates of point 3, 3π, given in clindrical coordinates. Eample 6 Finding the Volume of a Solid Find the volume of the solid S bounded b the cone + z and the sphere + + z wherez, see Figure 5. Solution The volume of S is dv. In clindrical coordinates, we have the S identit + r. Then an equation of the cone + z in clindrical coordinates is z r. The clindrical coordinates for the sphere is r + z. To find where the surfaces intersect, substitute the former equation into the later one. Then z and z for z. Thus, ever point (,, z) ins satisfies + and + z. Since +, we have r and θ π, see the base of the solid in Figure 6. The solid T corresponds to solid S under the change in coordinates from rectangular to clindrical coordinates. In Eample 3, the Jacobian determinant associated to the change of variables is (,, z) (r, θ, z) r. From + z, we obtain r z r. Appling the change of variables theorem for integration, we find dv (,, z) (r, θ, z) dv S T π r π r rdzdrdθ r r r drdθ Figure 5 Solid S bounded between a cone and a sphere. Figure 6 Solid T between z r, r + z,andθ π in the first octant. Volume 4π 3 ( ). Tr This 6 Find the volume of the solid bounded b the ellipsoid + +4z 4. Epress the volume as a triple integral that uses clindrical coordinates. See Figure 7. Figure 7 The ellipsoid + +4z 4.

24 8 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Spherical oordinates The spherical coordinates of a point (,, z) 3 in artesian coordinates are (ρ, θ, φ) where ρ + + z, tan θ if, and ρ cos φ (,, z) k. Figure 8 Spherical coordinates (ρ, θ, φ) ofpointp. The coordinate θ<π is the angle between the positive -ais and a vector from the origin to point (,, ), see Figure 8. The coordinate ρ is the distance between (,, z) and the origin. Further, we require φ π and φ is the angle between the positive z-ais and the line segment joining the origin to (,, z). The identities below ρ sin φ cos θ, ρ sin φ sin θ, z ρ cos φ are useful when converting to artesian coordinates from spherical coordinates. Eample 7 Switching between artesian and Spherical oordinates 3 Find the spherical coordinates of point A(,, z), 3., Then find the artesian coordinates of the point B(ρ, θ, φ) 8, 7π 4, π 6. Solution Since A(,, z) 3, 3,,wefind ρ + + z Using the identit tan θ 3, we choose θ π 6, see Figure 9. Also, we find ρ cos φ (,, z) k 3 3 cos φ,, k Figure 9 Spherical coordinates of point A, π 6, π 3. φ π 3. Then the spherical coordinates of point A are A(ρ, θ, φ), π 6, π. 3 Net, we determine the artesian coordinates of B(ρ, θ, φ) 8, 7π 4, π 6. ρ sin φ cos θ 8sin π 6 cos 7π 4 ρ sin φ sin θ 8sin π 6 sin 7π 4 z ρ cos φ 8 cos π Thus, the artesian coordinates are B(,, z) (,, 4 3).

25 .. HANGE OF VAIABLES IN INTEGATION 83 Tr This 7 Find the spherical coordinates of point (,, z) (3, 3 3, 6). Also, find the rectangular (or artesian) coordinates of D (ρ, θ, φ), 3π 4, π. 3 For the purpose of evaluating triple integrals, we evaluate the Jacobian determinant of the transformation to artesian from spherical coordinates. The spherical coordinates (ρ, φ, θ) of a point (,, z) satisf ρ sin φ cos θ, ρ sin φ sin θ, z ρ cos φ. We compute the determinant below b epanding the minors in the third row, i.e., ρ φ θ sin φ cos θ ρcos φ cos θ ρ sin φ sin θ (,, z) Det (ρ, φ, θ) ρ φ θ sin φ sin θ ρcos φ sin θ ρsin φ cos θ cos φ ρ sin φ z ρ z φ z θ Det[A 3, ] cos φ Det [A 3, ]( ρ sin φ) where the b matrices A 3,j are obtained from the above 3 b 3 matri b crossing out the third row and jth column, j,. Namel, the matrices A 3,j are ρ cos φ cos θ A 3, ρ cos φ sin θ ρ sin φ sin θ sin φ cos θ ρsin φ cos θ, A 3, sin φ sin θ ρ sin φ sin θ ρsin φ cos θ Then Det[A 3, ] ρ sin φ cos φ cos θ + ρ sin φ cos φ sin θ Using a similar calculation, we find ρ sin φ cos φ(cos θ +sin θ) ρ sin φ cos φ. Det[A 3, ] ρ sin φ. onsequentl, the Jacobian determinant for the change of variables to artesian coordinates from spherical coordinates is given b (,, z) (ρ, φ, θ) Det[A 3, ] cos φ Det [A 3, ]( ρ sin φ) (ρ sin φ cos φ) cos φ (ρ sin φ)( ρ sin φ) ρ sin φ. ()

26 84 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Eample 8 Triple Integral and Spherical oordinates Evaluate zdv where S is the solid between the two S spheres of radii and, and centered at the origin. See Figure. Solution In spherical coordinates, we have z ρ cos φ. Also, the Jacobian determinant for spherical coordinates is ρ sin φ, see(). Figure AsolidS bounded b two hemispheres. Analzing Figure, the spherical coordinate ρ satisfies ρ sincethe distance between a point in S and the origin is a value between and, inclusive. The coordinate φ satisfies φ π/ for the angle between vector k and a vector from the origin to a point in S is a value from to π/ radians. Also, the polar angle of an point is S satisfies θ<π. In Figure, we see that the spherical coordinates of S describe a rectangular bo. Then b the change of variables theorem, we obtain S zdv π π/ π π/ ρ 4 4 ρ 3 sin φ cos φdρdφdθ ρ ρ sin φ cos φdφdθ Figure The solid S in spherical coordinates π π/ π sin (φ) sin φ cos φdφdθ φπ/ φ dθ S 5 8 π zdv 5π 4. dθ Tr This 8 Set up a triple integral in spherical coordinates for the volume of the part of solid S in Figure that lies in the first octant. Then evaluate the integral.

27 .. HANGE OF VAIABLES IN INTEGATION 85. heck-it Out Evaluate the integral b appling the indicated transformation... e + da where is the circular region of radius and centered at the origin. Appl the transformations r cos θ and r sin θ. S + + z dv where S is the unit ball, i.e., the solid bounded b the unit sphere + + z. Appl a change of variables b using spherical coordinates, see Eample 8. True or False. If false, revise the statement to make it true or eplain.. If u v and u + v, the Jacobian determinant satisfies. If r cos θ and r sin θ, the Jacobian determinant satisfies 3. If {(, ) : + 5}, then e + da π 5 4. The Jacobian determinant for spherical coordinates satisfies 5. If u and 3v, then b the change of variables we have 3 f(, )dd Eercises for Section. f(u, 3v)dudv. (, ) (u, v). (, ) (r, θ) r. re r drdθ. (,, z) ρ sin φ. (ρ, φ, θ) In Eercises -8, evaluate the integral b appling the indicated change of variables. The region of integration in artesian coordinates is shown. The region S to where is mapped b the change of variables is shown, too. In addition, state the Jacobian determinant of the change of variables.. da where is the quadrilateral region with vertices at (, ), (4, ), (, ), and (5, ). Let v u and v. v S u Figure for

28 . 86 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS ( )da where is the quadrilateral region bounded b the lines +, +,, and +. Let u v and u v. v 5 3 u 3. Figure for + da where is the quadrilateral region with vertices at (, ), (, ), (, ), and (, ). Let u + v and u v. v S u 4. Figure for 3 + da where is the circular unit circle centered at the origin. Let r cos θ and r sin θ. Θ Π S r Figure for 4

29 5... HANGE OF VAIABLES IN INTEGATION 87 zdv where is the solid in the first octant bounded b the surfaces z +, +, and the coordinate planes. Let r cos θ, r sin θ, and z z. 6. Figure for 5 e ( + +z ) 3/ dv where is the solid bounded b the spheres + + z and + + z 4. hange the variables to spherical coordinates. 7. Figure for 6 zdv where is the solid bounded b the surfaces z 4, z,,, and. hange the variables to clindrical coordinates. Figure for 7

30 8. 88 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS zdv where is the solid bounded b the surfaces z +, +, and z. hange the variables to spherical coordinates. Figure for 8 In Eercises 9-4, if a point is defined in artesian coordinates, epress the point in clindrical coordinates. If the point is given in clindrical coordinates, change to artesian coordinates. 9. (,, z) ( 3,, ). (r, θ, z), 5π3, 3. (r, θ, z), 3π, 4 3. (r, θ, z), π 3,. (,, z) ( 6, 3, ) 4. (,, z) 3, 3, 3 In Eercises 5-, if a point is given in artesian coordinates, epress the point in spherical coordinates. If the point is epressed in spherical coordinates, convert to artesian coordinates. 5. (,, z) (, 3, 3) 6. (ρ, θ, φ) 8. (ρ, θ, φ), 7π 6, π 3 9. (ρ, θ, φ) 4, π 3, 5π 6,π, 3π 4 7. (,, z) 3 3,,. (,, z) 6 3, 6, 4 3 In Eercises -4, evaluate the integral b appling the indicated change of variables. Sketch the indicated region. Also, sketch the image S of the transformation defined b the change of variables.. ( )da where is a quadrilateral region with vertices (, ), (, ), ( 5, ), and ( 6, ). Appl the change of variables u + v and v.. ( )da where is a quadrilateral region with vertices (, ), (, ), (3, ), and (, ). 3. Appl the change of variables u + v and v. da where is the region bounded b the lines, +,, and +. Let u + v and u + v.

31 4... HANGE OF VAIABLES IN INTEGATION 89 +dawhere is the region bounded b the lines 4 +, 4 + 5, + 3, and +. Let u + v 6 and u +v. 3 In Eercises 5-8, solve for and in the given substitution. Then evaluate the integral b appling a change of variables. Sketch the region in the artesian plane, and the image S of the transformation defined b the change of variables dawhere is a trapezoid with vertices (, ), (, ), (, ) and (, ). Substitute u + and v. da where is the region in the first quadrant that is bounded b the graphs of, 3, , and 3. Substitute u and v. ( + ) da where is the triangular region with vertices (, ), (, ), and (, ). Substitute u + and v +. ( )( + + )da where is the triangular region with vertices (, ), (, ), and (, ). Substitute u + and v. ( )da where is a quadrilateral region bounded b the lines,,, and + 3. Substitute u + v and u + v. 3. Verif the identit da π e n where n {(, ) : + n }, n. n e 3. We adopt the notation of Eercise 3. Appling the onotone onvergence Theorem 4,wehave da lim da. e oreover, b Fubini s theorem, we have da n e Then verif the identit e d π n e e dd. 4 The onotone onvergence Theorem and a version of Fubini s Theorem used in Eercise 3 are proved in a course such as integration theor or real analsis. The proof is beond the scope of this tet.

32 9 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS 3. A solid is bounded b the surfaces +, z, and z. Sketch the solid. Then find the volume of the solid b evaluating a triple integral in clindrical coordinates. In Eercises 33-36, evaluate the integral b changing the variables to spherical coordinates. Include a graph of solid z dv where is the unit ball of radius and centered at the origin. zdv where is a solid in the first octant, and is one-eighth of a solid ball of radius dv which is the volume of the solid inside the sphere ρ and above the cone φ π 3. dv which is the volume of the solid inside the sphere ρ, above the plane z, and below the cone φ π 4. In Eercises 37-38, evaluate the integral b changing the variables to clindrical coordinates. Also, sketch the solid dv where is the solid in the first octant bounded b + 4, z, and z. z + + dv where is the solid bounded b z + and z.

33 .3. LINE INTEGALS AND A FUNDAENTAL THEOE 9.3 Line Integrals and a Fundamental Theorem The Arc Length Function Line Integral e-parametrizing a urve Fundamental Theorem for Line Integrals Path Integrals The Arc Length Function Geometricall, a curve is an arc of a graph such as a parabola, circle, line, and others. In this section we stud integrals of functions that are defined on curves. We begin with a definition of the parametrization of a curve. z r t r' t Definition Parametrization of a urve Let r be a continuous function from a closed interval [a, b] to n where n, 3. Suppose r satisfies the two conditions below. a) r is one-to-one on [a, b], or r is one-to-one on [a, b) and r(a) r(b). b) r (t) is continuous on (a, b) such that r (t) and r (t) In such a case, r parametrizes the smooth curve defined b {r(t) a t b}. are bounded. Figure Apositionvectorr(t) and atangentvectorr (t) For simplicit, we ma write that a smooth curve is a curve. If r is one-to-one on [a, b], we sa is a non-intersecting curve. If r is one-to-one on [a, b) and r(a) r(b), we sa is a simple closed curve. Suppose n 3 and r(t) ((t),(t),z(t)) is a vector in standard position, i.e., the initial point of r(t) is the origin, and the terminal point is ((t),(t),z(t)). If a<t<band t > is a small, the difference quotient r (t + t) r(t) t is a vector whose initial point is the terminal point of r(t). As t, the limit r (t) is a vector that is tangent to the curve at point r(t), see Figure. To define the arc length of, partition [a, b] intom subintervals. Let t a, t (b a)/m, and t k t k + t for k,...,m. Using the partition, subdivide into m subarcs, and find the sum of the distances between the initial and terminal points of each subarc. The distance between the endpoints of r(t k ) and r(t k ) satisfies r (t k ) r(t k ) (p k ) + (q k ) + z (v k ) t because of the ean Value Theorem, where t k <p k,q k,v k <t k. We recall r (t) ( (t)) +( (t)) +(z (t)). Since,,z are continuous, we estimate each of p k,q k,v k b a single value t k in (t k,t k ). onsequentl, r (t k ) r(t k ) r (t k) t.

34 9 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Then a reasonable approimation for the arc length of is the iemann sum m k r (t k ) r(t k ) m k r (t k) t. B definition, r is bounded and continuous, and consequentl, integrable. We let m, and define the limit of the iemann sums as the arc length of. Length() b a r (t) dt. The arc length of is independent of the parametrization; this follows from a modification of Theorem.9 to be discussed later in the section. oreover, the arc length function is defined b s(t) t a r (w) dw. () Eample Arc Length Function z Figure A curve emanating from the origin. Find the arc length function for the curve parametrized b r(t) t 3/, t 3/, 3t, t 3 5. Then find the arc length from r() to r(3/5), see Figure. Solution The derivative is r (t) 3 3 t, 3 t,. Then the magnitude of derivative is 9 r (t) 4 t +9t t +. Thus, the arc length function satisfies 45 4 t s(t) t r (w) dw t 3 5w +dw 3 (5w + )3/ 3 5 wt w (5t + ) 3/ Hence, the arc length from r() to r(3/5) is given b s / v 7 5.

35 .3. LINE INTEGALS AND A FUNDAENTAL THEOE 93 Tr This Find the arc length function for r(θ) (cos θ, sin θ, θ), θ π. The graph of r is a heli, see Figure 3. The Line Integral A line integral generalizes the concept of an integral b f()d of a real-valued a function f defined on [a, b]. In a line integral, we integrate a vector field F that is defined on a curve. Definition 3 Line Integral Let n be a curve parametrized b a function r defined on I [a, b] where n, 3. Let F be a vector field defined on such that F r is continuous from I to n.thelineintegral of F along is denoted b F dr, and defined b F dr b a F (r(t)) r (t)dt Figure 3 The heli r lies on a circular clinder of radius. Later in the section, we show the line integral of a vector field along a curve is independent of the parametrization provided the parameterization is orientation preserving, see Theorem.. We discuss an application of line integrals. Suppose a vector field F represents a force that causes a particle to move along a curve parametrized b r. That is, the particle is at position r(t) at time t. We show the line integral F dr represents the work done b the force on the particle. The unit tangent vector to r(t) isdefinedb T (t) r (t) r (t). (3) Since the differential of the arc length function is ds r (t) dt, wewrite F dr b a F (r(t)) r (t) r (t) r (t)dt b a F (r(t)) T (t) ds. (4)

36 94 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS ecall, the projection of vector v F (r(t)) onto unit vector w T (t) satisfies Proj w (v) (F (r(t)) T (t)) T (t) see identit (5) in page 5. The scalar F (r(t)) T (t) ds in the integrand of (4) represents the work done b the force F along the tangential component T (t), as the particle moves through an arc length of ds. Then the line integral (4) is the work done b a force F acting on a particle that is moving along curve. z Eample Evaluating a Line Integral Let F (,, z) (z,, ) be a vector field. Let be the curve parametrized b r(t) t, t,t 3, t, see Figure 4. Then evaluate F dr. Solution Let r(t) ((t),(t),z(t)) where (t) t, (t) t, and z(t) t 3. Evaluating the vector field, we obtain F (r(t)) F ((t),(t),z(t)) Figure 4 The curve r(t) t, t,t 3, t ((t)z(t), (t), (t)) t 5,t,t. Also, the derivative of the curve is r (t), t, 3t. Appling Definition 3, the line integral is given b F dr F (r(t)) r (t)dt t 5,t,t, t, 3t dt z t 6 F dr 7 3. t 5 +t +6t 3 dt 6 + t t4 t t Tr This Figure 5 For Tr This Let F (,, z) (, z, ) be a vector field. Let r(t) t, t, t, t, be a parametrization of curve. Then evaluate the line integral F dr.

37 .3. LINE INTEGALS AND A FUNDAENTAL THEOE 95 e-parametrizing a urve We stud the effect of re-parametrizing a curve has on a line integral. Let be a curve that is parametrized b a function r defined on [a, b]. The orientation of with respect to r is the direction along that the points r(t) trace as t increases. For instance, the orientation ma be counter clockwise on a simple closed curve. Let α be a one-to-one continuous function from [c, d] onto [a, b] such that the derivative α and α are bounded and continuous on (c, d). We sa the composite function r α is a re-parametrization of r. Notice, r α parametrizes. Wesar α is orientation preserving if the orientations of with respect to r and to r α are the same. Otherwise, we sa r α is orientation reversing. The line integral of vector field F using the parametrization r r α satisfies d c F (r (t)) r (t) dt d c F (r(α(t)) r (α(t)) α (t) dt hain ule α(d) α(c) F (r(w)) r (w) dw where w α(t),dw α (t)dt ± b a F (r(w)) r (w) dw (5) for b a f()d a f()d. The sign in (5) depends on whether α(c) a or b α(c) b. We choose the plus sign if α(c) a, i.e.,r is orientation preserving. Otherwise, we choose the negative sign if r is orientation reversing. oreover, an two parametrizations of a curve with the same initial points are re-parametrizations of each other, see theorem below. Theorem.9 Parametrization of urves Let r and r be functions defined on I [a, b] and I [c, d], respectivel, that parametrize a curve. Ifr (a) r (c), then r is a re-parametrization of r. Proof Let t (a, b), and choose s (c, d) such that r (t) r (s). Then the function given b s f(t), f(a) c, and f(b) d is a bijection from I onto I. B definition, r r f. Weshowf is differentiable. If t is small enough, choose s such that r (t + t) r (s + s). Then r (t + t) r (t) r (s + s) r (s). B definition, we have f (t + t) s + s.

38 96 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Then f f (t + t) f (t) (t) lim t t s lim t t r c r d r w r v r a r b Figure 6 Parametrizations r and r with different initial points. lim t r (t + t) r (t) t r (s + s) r (s) s B the continuit of the curves, t if and onl if s. onsequentl, f (t). r (t) r (s). (6) B definition, r i and r i are bounded for i,. Then f and f are bounded on (a, b). Hence, r r f, i.e.,r is a re-parametrization of r. Acurve has eactl two orientations. If we fi an orientation for, the other orientation is called the opposite orientation and which we denote b. Let be a non-intersecting oriented curve, and let F be a continuous vector field on. Appling identit (5) and Theorem.9, the line integral of F along is independent of an orientation preserving parametrization r of. Net, we analze the line integral on a simple closed curve. Note, the initial points of two functions r and r that parametrize ma not be the same. We assume the orientations of with respect to r and r are the same. Suppose r is defined on [a, b], and r is defined on [c, d]. hoose w, v such that r (w) r (c) r (d), and r (v) r (a) r (b), as in Figure 6. Then b a F (r (t)) r (t)dt w a d v d c F (r (t)) r (t)dt + F (r (t)) r (t)dt + F (r (t)) r (t)dt b w v c F (r (t)) r (t)dt F (r (t)) r (t)dt where in the middle equation we applied identit (5) with the plus sign. Thus, the line integral of a vector field along an oriented simple closed curve is independent of the orientation preserving parametrization of the curve. We summarize below. Theorem. Line Integrals and e-parametrization of urves Let be an oriented curve, and let F be a continuous vector field defined on. Then the line integral F dr is independent of the orientation preserving parametrization of. Also, F dr F dr. oreover, the lengths of curves and are the same.

39 .3. LINE INTEGALS AND A FUNDAENTAL THEOE 97 Proof We onl have to prove the last statement. Let r be a parametrization of, and let r r α be a re-parametrization where α is a one-to-one function from [c, d] onto [a, b] such that α and /α are continuous and bounded on (a, b). Since α is one-to-one, α is entirel nonpositive or entirel nonnegative. Then d r (t) dt d c c r (α(t)) α (t) dt d c d c r (α(t)) α (t) dt if α is increasing r (α(t)) α (t) dt if α is decreasing b a a b r (w) dw if w α(t), and α is increasing r (w) dw if α is decreasing d c r (t) dt b a r (t) dt Hence, the arc lengths of and are equal. Given a parametrization r of a curve defined on [a, b], there is a natural orientationreversing parametrization r of. Namel, r is defined on [a, b] and r (t) r (a + b t). (7) Eample 3 Evaluating a Line Integral z Let F (,, z) (sinπ, cos πz, ) be a vector field. Let be the oriented line segment from A(,, ) to B(,, ), see Figure 7. Then evaluate F dr. Solution The vector from A to B is (,, ). Then a parametric equation of the line segment from A to B is r(t) (,, )t +(,, ) ( t, t, t), t. Since r (t) (,, ), we obtain F dr (sin π( t), cos πt, t) (,, )dt Figure 7 Line segment from A(,, ) to B(,, ). ( sin π( t) + cos πt +t) dt F dr π. π cos(π( t)) + π sin(πt)+t

40 98 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Tr This 3 Evaluate the line integral in Eample 3 where is the oriented line segment from A(,, ) to B(, 4, ). Eample 4 Evaluating a Line Integral Let F (, ) (, ) be a vector field. Let be a circle of radius, centered at the origin, and oriented counter clockwise. Then evaluate the line integral F dr. Solution The standard equation of circle is + 4, see Figure 8. Using the identit cos θ +sin θ, the parametrization r(θ) ( cos θ, sinθ), θ π Figure 8 A counter-clockwise path around a circle of radius traces the circle in a counter clockwise motion. The derivative is r (θ) ( sinθ, cos θ). According to Theorem., the line integral is independent of the orientation preserving parametrization. Thus, π F dr F ( cos θ, sinθ) ( sinθ, cos θ) dθ π ( sinθ, cos θ) ( sinθ, cos θ) dθ π F dr 8π. 4sin θ + 4 cos θ dθ π 4 dθ Tr This 4 Let F (, ) (, ) be a vector field. Let be the unit circle oriented counter clockwise. Then evaluate the line integral F dr. We etend the definition of a line integral of a vector field. Let be a finite union of smooth curves i, i,...,n. n We denote the line integral of F along b F ds, and define it b F dr F dr + + F dr. (8) n

41 .3. LINE INTEGALS AND A FUNDAENTAL THEOE 99 Using Theorem., it can be shown that the left side is independent of the finite union of smooth curves i.wesa is a piecewise-smooth curve. Eample 5 Line Integral Along a Triangular Image Let F (,, z) (z,, z) be a vector field. Let be a simple closed triangular path that bounds the plane + + z 4, and the coordinate planes. The orientation is counter clockwise as shown in Figure 9. Then evaluate F dr. Solution We parametrize the segment from (,, ) to (,, ). Since is increasing from to, let t where t. Notice, the plane + contains.then t, and we obtain a parametrization of : r (t) ( t, t, ), t. Appling Definition 3 and F (,, z) (z,, z), we find F dr (, t, ) (,, ) dt 4. Secondl, consider the segment from (,, ) to (,, 4). Then z is increasing z toz 4. Let z t for t 4. Note, the plane + z 4 contains. Then (4 t)/, and we have a parametrization for : r (t), 4 t,t, t 4. The line integral along satisfies 4 F dr, 4 t, 4t t,, dt r3 4 z r r Figure 9 A triangular path in the plane + + z 4. 4 t +5t 4 dt 4 3. Similarl, a parametrization of the segment 3 from (,, 4) to (,, ) is given b r 3 (t) (t,, 4 t), t. Then the line integral along 3 satisfies F dr F (t,, 4 t) (,, ) dt 3 Finall, we obtain F dr 4t t,, (,, ) dt 8 3. F dr + F dr + F dr

42 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Tr This 5 Let F (, ) (, ) be a vector field. Let be a counter clockwise closed curve bounded and, see Figure. Then evaluate the line integral F dr. Figure An oriented closed path along and. Fundamental Theorem for Line Integrals If a vector field F is conservative, the line integral F dr depends onl on F, and the initial and terminal points of, as the net theorem shows. Theorem. Fundamental Theorem for Line Integrals Let F be a continuous vector field satisfing F f for some real-valued function f. Let be an oriented curve that is parametrized b a function r with initial point r(a) and terminal point r(b). Then F dr f (r(b)) f (r(a)). Proof For simplicit, let F be a function of two variables. Let F B the chain rule and the Fundamental Theorem of alculus, we find b f F dr, f r (t)dt r(t) a r(t) f, f. b a (f r) (t) dt f (r(b)) f (r(a)). We recall a characterization of conservative vector fields, see Theorem.. Let G be a vector field defined on 3 with continuous first partial derivatives ecept for finitel man points. Then curl(g) if and onl if G g for some real-valued function g. If G (,N,P) where,n,p are real-valued functions of,, z, then the proof of Theorem. shows curl G P N z P i N j + z k.

43 .3. LINE INTEGALS AND A FUNDAENTAL THEOE Thus, curl(g) if and onl if P N z, P z, N (9) Often, we ma denote the line integral of G (,N,P) along b G dr (d+ Nd + Pdz). For vector fields F (,N) of two variables and, we ma denote the line integral b F dr (d+ Nd). Eample 6 Fundamental Theorem for Line Integrals Let be a curve that is parametrized b r(t) (sint, cos t), t π.then evaluate (d +( )d) b the Fundamental Theorem for Line Integrals. Solution We appl (9) to show that F is the gradient of a real-valued function. We write F (,N,P) where, N, and P. Since,N are functions of and onl, we find curl(f ) eactl when N. learl, the above statement is true. Then curl(f ). Now, we ma eeball and directl find a function f whose gradient is F. In fact, ( ) (, ) F and we let f(, ) + where is a constant. We present a different method for finding f. Notice, f f and. Integrating with respect to, wefind f(, ) + φ() where φ is a function of onl. Since f + φ (), we obtain In an case, we have + φ () φ () φ() +. f(, ) + φ() +. Hence, b the Fundamental Theorem for Line Integrals we find F dr f (r(π/)) f (r()) f (, ) f (, ).

44 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Tr This 6 Let be an oriented curve with initial point (, ) and terminal point (, ). Then evaluate d +d b the Fundamental Theorem for Line Integrals. Eample 7 Fundamental Theorem for Line Integrals Let be an oriented path from point A(3,, ) to B(,, ) where. If z F (,, z), z, is a vector field, evaluate F dr b the Fundamental Theorem for Line Integrals. Solution We find a function f satisfing f F, but we omit the details in verifing (9). We ma eeball f and choose f(,, z) z/ z. Alternativel, since f/ z/, we integrate with respect to. Then f(,, z) z + φ(, z) where φ is a function of and z onl. Note, f z + φ f z. Also, z for f F.Thus, + φ z φ z. Then φ(, z) z + ψ() whereψ is a function of onl. Thus, f(,, z) z Observe, f z + ψ () and f z z + ψ () z z + ψ(). () for f F. ombining, we obtain ψ() where is a constant. Then we rewrite () as follows: f(,, z) z z +. Hence, b the Fundamental Theorem for Line Integrals we find F dr f (,, ) f (3,, ).

45 .3. LINE INTEGALS AND A FUNDAENTAL THEOE 3 Tr This 7 Appl the Fundamental Theorem for Line Integrals in evaluating d + z d z dz where is an oriented smooth path from A(,, ) to B(4, 4, ) where z. Path Integrals Let be a curve parametrized b a function r. ecall, the differential of the arc length function in () satisfies ds r (t) dt. Net, we define the path integral of real-valued function f along. Definition 4 Path Integral Let be a smooth curve parametrized b a function r defined on [a, b]. Let f be a real-valued function such that f r is continuous on [a, b]. The path integral of f along is denoted b f()ds, and defined b f()ds b a f(r(t)) r (t) dt. The line integral of f along is independent of the parametrization r of. The proof of which is similar to that of Theorem.. That is, if r and r are two functions that parametrize, and defined on [a,b ], [a,b ], respectivel, then b a f(r (t)) r (t) dt b a f(r (t)) r (t) dt. The above identit holds even if r is orientation reversing of r. Geometricall, if f(r(t)) represents the height of a fence at r(t), the path integral f()ds is the area of the fence, see Figure. The product f()ds is the area of a rectangle with base ds and height f(). A parameter s for a parametrizing function r is called arc length parameter if r (s) for all s ecept possibl at the endpoints. onsequentl, the arc length of r in subinterval [c, d] isd c. To re-parametrize r(t) b using the arc length parameter s, we suggest the following guidelines. z Figure If f(), the path integral f()ds is the area of a fence built on a curve.

46 4 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Parametrizing b the Arc Length Parameter Let r be a smooth curve on [a, b].. Evaluate the arc length function, s φ(t) t a r (t) dt.. Since s φ(t) is an increasing function, solve for the inverse function t φ (s). 3. The arc length parametrization is r (s) r φ (s). Let be a curve parametrized b r(s) wheres is the arc length parameter. From definition (3), the unit tangent vector to the curve at r(s) satisfies T (s) r (s) r (s) r (s) see (3). The curvature κ of is a function that is defined on where κ() T (s) () and r(s). The curvature κ is independent of the parametrization of. In particular, let r (s ) and r (s ) be two functions that parametrize where s and s are the arc length parameters. If r and r have the same orientations, then r r f where f (s ), see identit (6). B the chain rule, we find dr ds dr ds f (s ) dr ds. If r and r have opposite orientations, r and r have the same orientations where r (s )r (a + b s ), see (7). Similarl, r r f and dr dr f (s ) dr dr ds ds ds ds. a+b s Thus, we find sf(s ) dr ds ± dr ds. sf(s ) where the sign depends on whether r and r are orientation preserving or reversing. In an case, the magnitudes of the second derivatives are equal, i.e., d r ds d r ds. Thus, the curvature κ is independent of the parametrization.

47 .3. LINE INTEGALS AND A FUNDAENTAL THEOE 5 Eample 8 Evaluating a Path Integral Find the curvature function κ of the curve parametrized b r(s) (cos s, sin s, s) where s π. Then evaluate the path integral κds. Solution and Indeed, s is the arc length parameter for r (s) ( sin s, cos s, ) r (s) ( sin s, cos s, ). Since T (s) r (s), the curvature κ at r(s) satisfies κ() T (s) r (s) ( cos s, sin s, ). Appling Definition 4 and since r (s), the path integral of κ along is given b κ()ds π π. κ(r(t)) r (t) dt π dt Tr This 8 Find the curvature function κ for the circle defined b s s r(s) cos, sin where s π, and >. Then evaluate the path integral κds..3 heck-it Out. Find the arc length function for the curve r(t) (t, mt + b).. Let F (, ) (, 3) be a vector field. Let be the curve parametrized b r(t) (cos t, sin t), t π. Then evaluate F dr.

48 6 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS 3. Let F (, ) (, + ) be a vector field. Let be the curve parametrized b r(t) (3 cos t, 3sint), t π 3. Then evaluate F dr b the Fundamental Theorem for Line Integrals. 4. Let G(,, z) ( +,, z) be a vector field, and let be the line segment that joins point A(,, ) to B(,, ). Then evaluate F dr. True or False. If false, revise the statement to make it true or eplain.. A parametrization of the unit circle that is oriented in the counter clockwise direction is given b r(θ) (sinθ, cos θ), θ π.. The line segment from (,, ) to (,, ) is parametrized b r(t) ( t, t, t) where t. 3. The unit tangent vector to a curve parametrized b r(t), a t b, is given b r (t). 4. If is a curve parametrized b r(t) (t,, ) where t, then the line integral of a vector field F on satisfies F dr F (t,, ) i dt 5. A definite integral b a f()d is a line integral. 6. If is a curve parametrized b r(t) (t,, ) where t, then the line integral of a vector field F on satisfies F dr F (t,, ) i dt 7. If F (, ) (, ), the Fundamental Theorem for Line Integrals applies to F dr. 8. Let r(t) be a parametrization of a curve where a t b.thentisthearc length parameter if there is a constant c> satisfing r (t) c for all a<t<b. 9. Let r(t) be a parametrization of a curve where t is the arc length parameter. Then the curvature of the curve is r (t).. Let be an oriented curve with initial point (a, b) and terminal point (c, d). If F (, ) (, ), then F dr cd ab. Eercises for Section.3 In Eercises -4, evaluate the line integral of the vector field along the curve parametrized b r... F dr where F (, ) (, ), and r(t) (sint, cos t), t π 6. (d + d) wherer(t) (sint, cos t), t π 6.

49 LINE INTEGALS AND A FUNDAENTAL THEOE 7 G dr where G(,, z) (, + z,), and r(t) ( t, t +,t), t. (sin d+ cos zd sin dz)wherer(t) (t, t, t), t π 3. In Eercises 5-, parametrize the path b a curve r that is one-to-one ecept possibl at the endpoints. Then evaluate the line integral of the indicated vector field F or G along the path F dr where F (, ) (, ), and is the directed line segment from (, ) to (3, 4). F dr where F (, ) (, ), and is the directed line segment from (, ) to (, 3). (d + d) where is the oriented line segment from the origin to (, 4). (3d +( + )d) where is the line segment from the point (, ) to (, 4). F dr where F (, ) ( 4, 4), is a path along the unit circle that joins (, ) to in the counter clockwise direction. F dr where F (, ) ( +, ), and is a path along a circle of radius, centered, at the origin, and joining (, ) to (, 3) in the counter clockwise direction. G dr where G(,, z) ( +, z, + z), and is the line segment from (,, 3) to (3, 6, 5). (d + z d + dz) where is the line segment from (,, ) to (,, ). In Eercises 3-6, evaluate the line integral b the Fundamental Theorem for Line Integrals. State the function f such that the gradient f is the conservative vector field in the integral ( d +d) where is a path from the origin to (, 3). ( d d) where is an oriented path from the (, ) to (3, 6). (cos cos d sin sin d)where is the path defined b r(t) t, t, t π 3. F dr where is a path from point A(, ) to B(, ), and F (, ) (e +,e ). F dr where is the image of the curve r(t) (arccos t, arcsin t), t, F (, ) (e cos, e sin ).

50 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS 6e d +3e d where is the path defined b r(t) t, t +, t. 3 cos()d 6 sin()d where is the path defined b r(t) (t, (t )π), t. F dr where F (, ) ( e sin(), e cos ), and is defined b r(t) (t, 6t/π), t π 3. F dr where F (, ) cos()e sin, sin()e sin, and is defined b r(t) (t, t), t π 3. arcsin d + d where is a path joining point (, ) to ( 3, ) with ±. 4 z d + z d +zdz where is a path from point P (,, ) to Q (3,, ). (e cos d+ e sin d+ dz) where is a path from point A π,, to B π 3,,. F dr where is a path from point P (,, ) to Q(,, ), and F (,, z) (e z,e z. e z ). F dr where is a path from point P (,, ) to Q( z 3,, ), and F (,, z),, arctan. + In Eercises 7-34, evaluate the path integral of the given real-valued function along the smooth curve parametrized b r. See Definition f(, )ds where f(, ), r(t) (3t, 4t), t 3. f(, )ds where f(, ) 4 3 ( ), r(t) (t, 5t), t. ( + )ds where r(t) (4sint, 4 cos t), t π 8. f(,, z)ds where f(,, z) + z, r(t) (cos t, sin t, 3 t), t π. f(,, z)ds where f(,, z), r(t) ( cos t, sint, 5 t), t π 6. f(,, z)ds where f(,, z) cos 6 +, r(t) 8t, t,t 3 /3, t π 6. g(,, z)ds where g(,, z) z, r(t) (t, t, t), t. g(,, z)ds where g(,, z) z, r(t) t, t, t, t 3.

51 .3. LINE INTEGALS AND A FUNDAENTAL THEOE 9 In Eercises 35-4, find the unit tangent vector T (t) to the smooth curve parametrized b r. See identit (3) in page r(t) t, t 36. r(t) (8t, 4t, 8t) 37. r(t) (a cos t, asin t), a > 38. r(t) (b cos t, bsin t, t), b > 39. r(t) 3 t, t,t 3 4. r(t) t, e t,e t In Eercises 4-44, find the curvature of the curve parametrized b r at the indicated point P. B the chain rule, the curvature κ satisfies κ T (s) dt/dt ds/dt T (t) r (t). 4. r(t) t, t, P (, ) 4. r(t) (t, mt), P (t, mt) 43. r(t) (3 cos t, 4sint), P (, 4) 44. r(t) t, t, P (, ) In Eercises 45-46, a fence is built on a curve parametrized b r. The height of the fence at r(t) is given b f(r(t)) where f is the indicated real-valued function. Find the area of the fence. 45. r(t) (cos t, sin t), t π, f(, ) r(t) (t, 3t), t, f(, ) + z Figure for 45 Figure for 46 In Eercises 47-48, a curve is parametrized b a function r. Findthework done b the given force F on a particle that is moving along. 47. r(t) t,t ln, t, F (, ) (, ) 48. r(t) (e t,t), t, F (, ),+ 49. Let r be a smooth curve defined on [a, b], and let T (t) be the unit tangent vector at r(t), see definition (3). Let s be the arc length parameter where ds/dt r (t). a) Verif the identities r (t) ds T (t), and dt r (t) d s dt T (t)+ds dt T (t).

52 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS b) Using the fact that T (t) T (t), prove r (t) r (t) ds T (t) T (t). dt c) Using the fact that T (t) T (t), prove T (t) is perpendicular to T (t). d) Show r (t) r (t) ds T (t). dt e) Using the fact that curvature satisfies κ(t) T (t) r (t),provetheidentit κ(t) r (t) r (t) r (t) 3. () 5. Let be a parabolic curve parametrized b r,t,t. a) Appling the curvature identit (), verif κ(t) b) Evaluate κ(t)dt. c) Evaluate the path integral κds. ( + 4t ) 3/. 5. Let be a hperboloid curve parametrized b r (sinht, cosh t, ). a) Appling the curvature identit (), verif κ(t) (secht) 3/. b) Evaluate the path integral κds. 5. Let r be a smooth curve defined on [a, b], and let T (t) be the unit tangent vector at r(t). a) The normal vector to the curve at r(t) isdefinedb N(t) T (t) T (t). (3) b) Using the fact that T (t) T (t), prove N(t) is perpendicular to T (t). c) The binormal vector to the curve at r(t) isdefinedb B(t) T (t) N(t). (4) Using the fact that B(t) B(t), show B (t) B(t). d) Using the fact that B(t) T (t) B(t) T (t), prove B (t) T (t). e) Appling c) and d), prove B (t) is multiple of N(t). 53. If r(t) (t, t,t ), find the unit tangent vector T (t), the normal vector N(t), and the binormal vector B(t) at r(t).

53 .4. SUFAE INTEGALS.4 Surface Integrals Area of a Surface Surface Integrals of Vector Fields Surface Integrals of eal-valued Functions Area of a Surface In Section.3, we discussed line integrals along smooth curves. In this section, we stud surface integrals, or integrals of vector fields defined on surfaces. When the surface is a subset of the -plane, the surface integral is a double integral as we will see. We begin with the definition of a parametrized surface. Definition 5 Parametrization of a Surface Let be an elementar region 5. Let r be a continuous function from to 3 that satisfies the two conditions below ecept possibl on the boundar of. Figure Surface parametrized b two independent variables. a) r is one-to-one with continuous partial derivatives, and b) the functions and φ(u, v) are nonzero and bounded. φ(u, v) r u r v (5) In such a case, we sa r parametrizes the surface defined b {r(u, v) (u, v) }. The function φ in (5) is integrable on, and we omit its proof since it is similar to the application of Theorem. to Theorem.3 in pages 6 and 6. Net, we define the surface area of. For the moment, let be a rectangular region. Partition into sub-rectangles ij with uniform width u and uniform height v. Let (u i,v j ) be the verte of ij nearest the origin, as in Figure. We approimate the image r( ij ) b a rectangular region A ij whose sides are the vectors u r and v r. The area of rectangle A (ui,v ij j) is the magnitude u (ui,v j) v of the cross product of the vectors that define the rectangle. That is, Area (A ij ) u r u (ui,v j) v r v (ui,v j) r u (ui,v j) r v (ui,v j) u v 5 Elementar regions are discussed in page 6

54 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS The iemann sum of the areas of rectangles A ij approaches a limit as the norm of the partition of approaches zero, due to the boundedness and continuit in (5). The limit which is an integral is a reasonable definition for the surface area of. Notice, the integral of r u r v over eists even if is an elementar region. In general, let r be a function satisfing Definition 5. Let be the corresponding surface parametrized b r. We define the surface area of as follows: (Surface Area of ) r u r v da (6) where da dudv or da dvdu. The surface area of is independent of the parametrization r in Definition 5; the proof is left as an eercise and depends on the change of variables theorem. We consider a special case where a surface is the graph of a real-valued function z f(, ) with continuous partial derivatives. We parametrize b the one-to-one function r(, ) (,, f(, )). The cross product of the partial derivatives of r is given b i j k r r f f f, f,. Appling (6), the area of the surface defined b z f(, ), (, ), satisfies (Surface Area of ) r r da + f + f da. (7) z Eample Evaluating a Surface Area Find the surface area of the portion of the plane + + z in the first octant, Figure The plane + + z in the first octant. see Figure. Solution Solving for z, we find and write z f(, ). Let be the triangular region in the first octant defined b z f(, ). The subset of the -plane below the triangular region satisfies (, ),.

55 .4. SUFAE INTEGALS 3 We appl identit (7) to find the surface area of. Notice, f f + + +( ) +( ) 6. Then the surface area of is given b (Surface Area of ) + f + f da 6 ( )/ dd 6 (Area of ) 6 sq. units since the area of triangle is square unit. Tr This Find the surface area of the portion of the plane + + z 4 in the first octant above the triangular region in the -plane with vertices (,, ), (,, ), and (,, ). Eample Surface Area Find the surface area of the part of the sphere + + z 5 where + 6 and z, see Figure 3. Solution We solve for z and write Figure 3 A dome above a circle of radius 4. z f(, ) 5. Evaluating the partial derivatives, we find a) f b) f 5 5 Then we obtain + f + f 5 5.

56 4 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Let be the upper part of the sphere that lies above circular region (, ) + 6. We appl identit (7) to find the surface area of. (Surface Area of ) + f + 5 da 5. We appl a change of variables to polar coordinates. Let f da a) r cos θ b) r sin θ, c) da rdrdθ where r 4 and θ π. Since + r, we obtain (Surface Area of ) 4 π 5r dθdr 5 r 4 rdr π 5 r π 5 r r4 (Surface Area of ) π sq. units r Tr This Figure 4 For Tr This Find the area of the surface z See Figure 4. + that lies below the plane z. Net, we evaluate the area of a surface of revolution. Let be a curve in the z-plane that is parametrized b r (t) (,(t),z(t)), a t b (8) Figure 5 The curve z f() is revolved about the z-ais. where (t), as in Figure 5. A surface of revolution is generated when is revolved about the z-ais. Then surface is parametrized b r(θ, t) ((t) cos θ, (t) sin θ, z(t)) (9) where θ<π, t [a, b]. We leave the verification of (9) as an eercise.

57 .4. SUFAE INTEGALS 5 oreover, the cross product of the partial derivatives of r satisfies r θ r t i j k (t) sin θ (t) cos θ (t) cos θ (t)sinθ z (t) ((t)z (t) cos θ, (t)z (t)sinθ, (t) (t)). Since (t), we find r θ r t (t) [ (t)] +[z (t)] (t) r (t) () Appling (6), the area of the surface of revolution defined b (8) and (9) is (Area of Surface of evolution) b π r θ r t dθdt a π b a (t) r (t) dt. () Eample 3 Surface Area of a Sphere Parametrize the sphere S of radius that is centered at the origin. Then find the surface area of S. Solution Let be the semicircle of radius in the z-plane given b r (ψ) (, cos ψ, sinψ), π ψ π. The sphere S is generated when is revolved about the z-ais, see Figure 6. Appling (9) we parametrize S as follows Figure 6 Semicircle generates a sphere b a rotation. r(θ, ψ) ( cos ψ cos θ, cos ψ sin θ, sinψ) where θ<π. From the definition of r (ψ), we find r (ψ) and let (ψ) cos ψ be the -component of r (ψ). Using (), we obtain (Surface Area of S) π π b a π/ π/ 8π sin ψ (ψ) r (ψ) dψ ( cos ψ)() dψ π/ π/ 6π sq. units

58 6 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Tr This 3 Parametrize a right circular cone S whose base is a circle of radius and the height is units. Assume the base lies on the -plane and is centered at the origin, see Figure 7. Then find the surface area of the cone S. Figure 7 ight circular cone with radius and height units. Surface Integrals of Vector Fields Let r be a function from an elementar region into 3 as in Definition 5. Let 3 be the surface that is parametrized b r. Let F be a vector-valued function defined on with values in 3 such that F r is continuous. In such a case, we sa F is a continuous vector field defined on. Definition 6 Surface Integral of a Vector Field Let 3 be a surface parametrized b a function r satisfing Definition 5. Let F be a continuous vector field defined on, where the values of F lie in 3.Thesurface integral of F on denoted b F ds is defined b r F ds F (r(u, v)) u r da v where da dudv or da dvdu, and is the domain of definition of r. An orientation on a surface is a continuous function that assigns to each interior point p of a unit vector N(p) that is perpendicular to the tangent plane at p. We claim the surface integral F ds is independent of the orientation preserving parametrization r of, as we briefl eplain below. B Definition 5, r is defined on an elementar region. Givenp r(u, v) where (u, v) lies in the interior of, the unit vector below Nr(p) r u r v r u r v is normal to the tangent plane at p, where the partial derivatives are evaluated at (u, v). For the details, see Eercise 5. We sa r is orientation preserving if Nr(p) N(p) for all interior points p. Whiler is orientation reversing if for all interior points p. Nr(p) N(p)

59 .4. SUFAE INTEGALS 7 Eample 4 Evaluating a Surface Integral Let F (,, z) (,,) be a vector field. Let be a triangular region in the -plane with vertices at (, ), (, ), and the origin, see Figure 8. Let be the surface that is oriented, and parametrized b r(, ),, + +, (, ). Then evaluate the surface integral F ds. Solution The surface is a portion of the larger surface z + + that lies above the region in the -plane, see Figure 9. The cross product of the partial derivatives of r satisfies i j k r r + +, ,. We appl Definition 6 to evaluate the surface integral. Notice, F (r(, )) (,,). r F ds F (r(, )) r da Figure 8 Triangular region Figure 9 Surface above region in Figure 8 (,,) + +, + +, dd dd d d F ds 4 3. Tr This 4 Let F (,, z) (,, ) be a vector field. Let be a triangular region in the -plane with vertices at (, ), (, ), and the origin. Let be the surface parametrized b r(, ),,, (, ). Then evaluate the surface integral F ds.

60 8 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Eample 5 Evaluating a Surface Integral Let F (,, z) (,, z) be a vector field on the unit sphere centered at (,, ). Assume the sphere is oriented b unit vectors that point awa from the origin. Then evaluate the surface integral F ds. Solution In Eample 3, we parametrized a sphere of radius centered at the origin. Likewise, we obtain a parametrization r of the unit sphere, namel, r(θ, ψ) (cos ψ cos θ, cos ψ sin θ, sin ψ) where θ<π and π ψ π. For the cross product, we simplif and find r θ r i j k ψ cos ψ sin θ cos ψ cos θ sin ψ cos θ sin ψ sin θ cos ψ cos ψ cos θ, cos ψ sin θ, cos ψ sin ψ. Analzing the third component of r θ r ψ, we observe the cross product vector points awa from the origin. In particular, r is orientation preserving. oreover, we find r F (r(θ, ψ)) θ r ψ (cos ψ cos θ, cos ψ sin θ, sin ψ) cos ψ cos θ, cos ψ sin θ, cos ψ sin ψ cos 3 ψ(cos θ +sin θ) + cos ψ sin ψ cos 3 ψ + cos ψ sin ψ cos ψ(cos θ +sin θ) cos ψ. Appling Definition 6, we evaluate the surface integral as follows: r F ds F (r(θ, ψ)) θ r da ψ π/ π π/ cos ψdθdψ π/ π cos ψdψ π/ F ds 4π.

61 .4. SUFAE INTEGALS 9 Tr This 5 The orientation on the unit sphere are unit vectors that point awa from the origin. If F (,, z) (,,z), evaluate the surface integral F ds. In the short sequel, we give a geometric interpretation of a surface integral F ds. Suppose a fluid is flowing through a surface such that the fluid s direction and velocit at point p in S is given b vector F (p) ft/sec. The component of F (p) in the vector r r normal to S at p is given b the dot product r F (p) r r r ft/sec We partition into rectangular-like subregions where r r dd ft is the area of a subregion. When we multipl the above dot product to the area of a subregion, we obtain r F (p) r dd ft 3 /sec that describes the volume of fluid flowing through the subregion per second. Summing up and taking the limit as the norm of the partition approaches zero, we obtain the surface integral F ds that is the volume of fluid flowing through surface per second. Surface Integrals of eal-valued Functions Definition 7 The Integral of a eal-valued Function Defined on a Surface Let f be a real-valued function on a surface. Suppose is parametrized b a function r, as in Definition 5, such that f r is continuous. We denote the integral of f on b fds, and define it b fds f(r(, )) r r da () where is an elementar region in the -plane on which r is defined. The surface integral of a real-valued function f on is independent of the parametrization r of. That is, whether r is orientation preserving or reversing, the integral () is invariant. See Eercise 5 for a proof. In particular, if f is a nonnegative function, the surface integral of f describes the volume of a solid. oreover, the base of the solid is the surface and the height of the solid at point r(, ) isf(r(, )). The integrand in () represents the volume of a rectangular bo whose base has area r r da and the height of the bo is f(r(, )).

62 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Eample 6 Evaluating a Surface Integral Let be the rectangular region in the plane z where,. Let f(,, z) be a real-valued function that is defined on. Then evaluate the surface integral fds. Solution To parametrize where z, welet r(, ) (,, ) where (, ) lies in a rectangular region in the -plane defined b,. For the cross product of the partial derivatives of r, wefind r r i j k Using the given function f, we obtain (,, ). f(r(, )) f (,, ). Appling Definition 7, the surface integral satisfies fds f(r(, )) r r da Figure The base of a solid is the surface in the plane z.. dd The surface integral fds represents the volume of a solid in Figure. The base of the solid is the surface, and the height at (,, z) of the solid is f(,, z). Tr This 6 Let be the rectangular region in the plane z where,, i.e., is the base of the solid in Figure. If f(,, z) + z is defined on, evaluate the surface integral fds.

63 .4. SUFAE INTEGALS.4 heck-it Out In Eercises -3, let be the triangular part of the plane + + z in the first octant. Suppose is oriented b unit vectors normal to that point awa from the origin.. Find the surface area of.. If F (,, z) (,,+ z), evaluate the surface integral F ds 3. If f(,, z) + z, then evaluate the surface integral fds True or False. If false, revise the statement to make it true or eplain.. The plane + + z 4 in the first octant is parametrized b r(, ) (,, 4 ) where, 4.. Let be a surface parametrized b r(s, t) (s, t, ) where s, t. If F is vector field on, then F ds F (s, t, ) k dsdt 3. The surface area of z f(, ) is given b for some region in the -plane. 4. The surface area of is given b + f + f da r(s, t) da where r is a function that parametrizes, and r is defined on some region in the st-plane. 5. If F (,, z) (,, z), and is parametrized b r(, ) (,, ), then F ds ( + + )da for some region in the -plane. 6. If f(,, z) + + z, and is parametrized b r(s, t) (s, t, s + t ), then fds (s + t ) +4s +4t da for some region in the st-plane. 7. If F is a vector field on a surface defined b z f(, ), then F ds F (,, f(, )) f, f, da for some region in the -plane.

64 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Eercises for Section.4 In Eercises -6, find a parametrization of the indicated surface. Then find the area of the surface.. The quadrilateral region defined b + + z 4where and. Figure for Figure for. The surface in the parabolic clinder z 3 where and. 3. The elliptical-shaped surface + + z 3 such that +. Figure for 3 Figure for 4 4. The dome-shaped surface + + z 4 such that + and z>. 5. The surface consisting of points (,, z) satisfing + and z. Figure for 5 Figure for 6 6. The cone defined b z + where + 6. The top is not included.

65 .4. SUFAE INTEGALS 3 In Eercises 7-6, evaluate the surface integral F ds for the indicated vector field F defined on the oriented surface. Assume is oriented b unit vectors pointing upward. 7. Let F (,, z) (,,z), and let be parametrized b r(t, θ) (t cos θ, t sin θ, ) where θ π and t. 8. Let F (,, z) (,+ z,), and let be parametrized b r(t, θ) (tcos θ, t sin θ, t sin θ) where θ π and t. 9. Let F (,, z) (,, z), and let be parametrized b r(ψ, θ) (sinψ cos θ, sin ψ sin θ, cos ψ) where ψ π and θ π. Note, is the unit sphere.. Let F (,, z) /z, /z, /z, and let be parametrized b r(, ),, + + where a, b.. Let F (,, z) (,, ), and let be the triangular planar region defined b + +z inthe first octant.. Let F (,, z) (,, ), and let be upper unit hemisphere z. 3. Let F (,, z) ( z,, ), and let be the triangular planar region defined b + z in the octant where, z > and <. 4. Let F (,z,), and let be the portion of the plane + + z 4 in the first octant. 5. Let F curl(,,), and let be the portion of the sphere + + z 5 in the first octant. 6. Let F curl(,, ), and let be the hemisphere z 4. In Eercises 7-4, evaluate the surface integral fds for the indicated real-valued function f defined on the given surface. 7. Let f(,, z) 3z, and let be the surface parametrized b r(θ, t) (t cos θ, t sin θ, t) where t and θ π. 8. Let f(,, z), and let be the surface parametrized b r(θ, t) (tcos θ, t sin θ, t + ) where t and θ π. 9. Let f(,, z) +, and let be the surface parametrized b r(, ) (,, + ) such that +.. Let f(,, z) z +, and let be the disk of radius defined b + 4 and z 3.

66 4 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS. Let f(,, z) + + z, and let be the plane + + z 4 in the first octant.. Let f(,, z), and let be the plane + + z 3 in the first octant. 3. Let f(,, z) 6, and let be the triangular region with vertices (,, ), (,, ), and (,, ). 4. Let f(,, z) 6z, and let be the triangular region with vertices (,, ), (,, 4), and (,, ). The Effects of e-parametrizations on Surface Integrals 5. Let r, r be two functions that parametrize a surface 3, as in Definition 5. Suppose r, r are defined on elementar regions,, respectivel. To each interior point (s, t) in, choose an interior point (u, v) in such that r (u, v) r (s, t). a) Prove there eists a function h that is one-to-one and differentiable in the interior of such that (u, v) h(s, t). We sketch a proof below. Let r (s,t )r (u,v )(,,z ). Since the cross product r s s t r (s,t ) t is (s,t ) nonzero, without loss of generalit we assume the third component of the cross product is nonzero, s t Det s t t s where r (s, t) (,,z ). We appl the Inverse Function Theorem to the sstem of equations: (s, t) (s, t) Since the above determinant is nonzero, it is possible to epress (s, t) as a differentiable function of (, ) near (, ). That is, (s, t) f(, ) for some invertible function f defined on an open disk containing (, ) with continuous partial derivatives. Similarl, since r u r (u,v ) v is nonzero, we write (u, v) g(, ), (u, v) g(, z), or (u,v ) (u, v) g(, z) for some function g with continuous partial derivatives in an open disk containing (, ), (,z ), or (,z ), respectivel. In an case, we obtain (s, t) f g (u, v). b) Appl the hain ule to verif r s r t s t t (, ) r (s, t) r r s r where (,) (s,t) is the Jacobian determinant of the transformation from (s, t) to (, ). Thus, the tangent plane to the surface is well-defined since the cross-products r s r t and r r are parallel.

67 .4. SUFAE INTEGALS 5 c) Let F be a continuous vector field defined on and with values in 3. Appling part a) and the change of variables theorem, we find r F (r (s, t)) s r (, ) r dsdt F (r (, )) t (s, t) r (s, t) (, ) dd F (r (, )) r r (, ) (s, t) (s, t) (, ) dd r ± F (r (, )) r dd where the sign is positive if the Jacobian determinant (,) (s,t) is positive, and the sign is negative if the Jacobian determinant is negative. The Jacobian determinant is positive when r and r are orientation preserving. Then the integral of F on surface is independent of the orientationpreserving parametrization r of. d) Appling the change of variable theorem, we find f (r (s, t)) r s r t dsdt f (r (, )) (, ) (s, t) r r f (r (, )) r r dd. In particular, the integral of the scalar-valued function f over, namel, fds, is independent of the parametrization r of. oreover, the path integral is independent of the orientation of r, i.e., the orientation ma be preserving or reversing. (s, t) (, ) dd

68 6 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS.5 Integral Theorems of Green, Stokes, and Gauss Stokes Theorem Stokes Theorem Applied to Surfaces that are Graphs of Functions Green s Theorem Gauss Divergence Theorem Proofs of the Integral Theorems Stokes Theorem Figure An oriented surface, and its boundar is oriented positivel. Stokes Theorem relates a surface integral to a line integral along the boundar of the surface. In the short sequel, we develop the relationship. Let be an oriented surface parametrized b a function r satisfing Definition 5 in page. onsequentl, there are unit vectors N(p) that are normal to at p such that N(p) varies continuousl as a function of p. Furthermore, suppose the boundar of is a simple closed curve. We define a positive orientation for the curve using the orientation of, see Figure. That is, using our right hand, let our thumb point to the direction of the unit vectors N(p), and the direction along the curve as ou curl our fingers is the positive orientation for. In words, Stokes Theorem states that the line integral of a differentiable vector field F in 3-space along a positivel oriented boundar is the equal to the surface integral of the curlf on the surface. Theorem. Stokes Theorem Let be an oriented surface such that its boundar is a simple closed curve. Suppose has the positive orientation. Let F be a vector field defined on with values in 3.IfF has continuous partial derivatives, then F dr curlf ds. z N Figure The boundar is oriented positivel. Eample Illustrating Stokes Theorem Verif Stokes Theorem if F (,, z) (, z, ), is the plane + + z in the first octant, and is oriented b unit vectors that point awa from the origin. Notice, the boundar of is oriented counter-clockwise, see Figure. Solution Let r parametrize the straight path from (,, ) to (,, ). r (t) ( t, t, ), t. Likewise, let r parametrize the straight path from (,, ) to (,, ). r (t) (, t, t), t. Similarl, let r 3 parametrize the straight path from (,, ) to (,, ). r 3 (t) (t,, t), t.

69 .5. INTEGAL THEOES OF GEEN, STOKES, AND GAUSS 7 The boundar of is the union of the oriented lines, i.e., 3. Then the line integral of F along satisfies F dr F dr + F dr + F dr (3) 3 Since F (,, z) (, z, ), the line integral along r is given b F dr F (r (t)) r (t)dt (t,,t ) (,, )dt tdt. Similarl, the line integral along r satisfies F dr F (r (t)) r (t)dt ( t, t, ) (,, )dt tdt. Likewise, we find 3 F dr. Appling (4), the sum of the line integrals equals F dr. (4) Since the plane is z, we parametrize b r(, ) (,, ) where (, ) liesinthe-plane directl below, see Figure a. Then r r i j k (,, ). The curl of F satisfies curlf F i j k z (,, ). z Using the region of integration in Figure a, the surface integral is given b curlf ds (,, ) (,, )dd dd (Area of ). Hence, the surface integral agrees with the line integral (4), thereb, verifing Stokes Theorem. Figure a The region in the -plane below.

70 8 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS z z N Tr This Figure b A unit disk centered at (,, ), and oriented b vector N k. Let be a circular region of radius in the plane z, and centered at (,, ). Assume is oriented b unit vectors N that point upward, as seen in Figure b. Verif Stokes Theorem for the vector field F (,, z) ( /,/, ). Stokes Theorem Applied to Surfaces that are Graphs of Functions Let be a surface that is the graph of a function, i.e., is parametrized b r(, ) (,, f(, )) (5) where f is a function on an elementar region in the -plane with continuous partial derivatives. Suppose the boundar of is a simple closed curve. Notice, is oriented b unit vectors pointing awa from the origin. Then the positive orientation on is the counter clockwise direction in the -plane. In Section.4, wehaveseen i j k r r Det f f f, f, Let F (,, z) be a vector field defined on and with values in 3. Suppose curlf (r(, )) (L, N, P )wherel, N, P are functions of (, ). Then the surface integral of curlf on satisfies curlf ds curlf (r(, )) r r da L f N f + P da (6) where da dd or da dd. Thus,if is a surface parametrized b (5), is a simple closed curve oriented in the counter clockwise motion, and curlf (L, N, P ), then we ma rewrite Stokes Theorem as follows: F ds curlf ds L f N f + P da (7)

71 .5. INTEGAL THEOES OF GEEN, STOKES, AND GAUSS 9 Eample Surfaces Defined b Functions Let be the intersection of the surfaces + and z 3, as in Figure 3. Assume curve is oriented b the counterclockwise direction on the -plane. Appl Stokes Theorem in evaluating the line integral 3 d + 3 d +( 3 + z)dz. Solution The vector field in the line integral is F (,, z) 3, 3, 3 + z. We evaluate the curl of F : i j k curlf F z 3,, z Notice, the surface that is enclosed b is the graph of f(, ) 3 where +. If (L, N, P )curlf,then L f N f + P 3 +(3 3 ) 3. In appling Stokes Theorem, we use the special case (7). 3 d + 3 d +( 3 + z)dz curlf ds L f N f + P da 3 da. The region of integration is the unit disk of radius centered at the origin. Using a change of variables to polar coordinates with r cos θ and da rdrdθ, we obtain 3 da π 3r 3 cos θdrdθ z N Figure 3 The region in the -plane below π 4. π cos θdθ Hence, b Stokes Theorem we obtain 3 d + 3 d +( 3 + z)dz 3π 4.

72 3 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS z N Tr This Figure 4 A b rectangular region and a normal vector N. Let be an oriented rectangular region with vertices (,, ), (,, ), (,, ), and (,, ), see Figure 4. Let the boundar of be oriented positivel. Then appl Stokes Theorem in evaluating the line integral (zd + dz). Green s Theorem We analze a special case of Stokes Theorem. In the parametrization (5), let f(, ). In particular, each point in has the form (,, ) where the set of of (, ) s lie in an elementar region in the -plane, see Figure 5. We assume is oriented b the unit vector k, and the boundar is a simple closed curve oriented with the counter clockwise direction. oreover, we consider vector fields of the form F (,, ) (f (, ),f (, ), ) where f,f are real-valued functions defined on. Then the curl is given b i j k curlf F z f (, ) f (, ) z,, f f. k Figure 5 A surface in the -plane oriented b the unit vector k. Appling a special case of Stokes Theorem, i.e., (7), we obtain f (f (, ),f (, ), ) ds f da. Since z, the vector field F ma be realized as having values in. Then Stokes Theorem reduces to the following: Theorem.3 Green s Theorem Let be an elementar region in the -plane whose boundar is a simple closed curve, oriented b the counter clockwise motion. If f,f are real-valued functions defined on with continuous partial derivatives, then f (f (, )d + f (, )d) f da. Notice, the line integral (f (, )d + f (, )d) reduces to the area of if f f for all (, ).

73 .5. INTEGAL THEOES OF GEEN, STOKES, AND GAUSS 3 Eample 3 Appling Green s Theorem Appl Green s Theorem in evaluating the line integral d +3d where is a triangular path from the origin to point (, ) to point (, ) and to the origin. Solution Let be the triangular region enclosed b, see Figure 6. Notice,, i.e, the boundar of, is oriented in the counter clockwise direction. Appling Green s Theorem, we obtain d +3d (3) (3 ) da ( ) da Figure 6 Triangular region with hpotenuse. dd d d d +3d 3. Tr This 3 Appl Green s Theorem in evaluating the line integral sin d+3 d where is a triangular path from (, ) to (, ) to (, ) to (, ). See Figure 7. Figure 7 Triangular path with hpotenuse /. Eample 4 Illustrating Green s Theorem Verif Green s Theorem if F (, ) 3 3, 3 3 and is a circular region of radius and centered at the origin. Suppose the unit circle, i.e., boundar of, is oriented in the counter clockwise direction. See Figure 8. Solution The unit circle is parametrized in the counter clockwise motion b r(t) (cos t, sin t) where t π. We evaluate the line integral of F along as follows: Figure 8 ounter clockwise circular path

74 3 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS F dr π F (r(t)) r (t) dt F dr π. π π π π π 3t + sin 3 t, cos 3 t ( sin t, cos t) dt cos 4 t +sin 4 t dt + cos t cos t + dt + cos t 4 (3 + cos 4t) dt sin 4t 4 tπ t dt To verif Green s Theorem, we evaluate a double integral as in Theorem.3. We appl a change of variables b changing from artesian to polar coordinates. Notice, is the circular region centered at the origin and radius da + da π π r 3 drdθ 4 dθ π. since + r and da rdrdθ. Then the double integral and line integral have the same values. This completes the verification of Green s Theorem. Tr This 4 Verif Green s Theorem if F (, ), and is the circular region of radius and centered at (, ). Assume circle, i.e., the boundar of is oriented in the counter clockwise direction.

75 .5. INTEGAL THEOES OF GEEN, STOKES, AND GAUSS 33 Gauss Divergence Theorem We consider a solid in 3 whose boundar consists of finitel man surfaces that are oriented b unit vectors pointing outward from the solid. An eample of such a solid is a cube, and its boundar consists of si planar faces. Another eample of a solid is the ball + + z r, and its boundar is the sphere + + z r where r>. See Figures 9 and. z N Theorem.4 Gauss Divergence Theorem Let be a solid region in 3 such that its boundar consists of finitel man surfaces oriented b unit vectors pointing outward from. Let F be a vector field defined on and with values in 3.IfF has continuous partial derivatives, then F ds div(f ) dv. N Figure 9 A cube oriented b unit vectors pointing outward. Eample 5 Illustrating Gauss Divergence Theorem Let be the ball + + z r of radius r>. Verif the Divergence Theorem for F (,, z) (,,z). Assume the boundar of is oriented b unit vectors pointing outward from the ball, see Figure. Solution We parametrize using spherical coordinates, i.e., r(φ, θ) (r sin φ cos θ, r sin φ sin θ, r cos φ) where φ π and θ<π, see Section.. Furthermore, we find r φ r θ i j k r cos φ cos θ rcos φ sin θ r sin φ r sin φ sin θ rsin φ cos θ r sin φ cos θ, r sin φ sin θ, r sin φ cos φ Notice, the above cross product is a vector that points awa from the ball. Appling Definition 6 in Section.4, the surface integral of F on satisfies π π r F ds F (r(φ, θ)) φ r dθdφ θ π π π π r (,,rcos φ) φ r dθdφ θ r 3 cos φ sin φdθdφ π πr 3 cos φ sin φdφ F ds 4πr3 3. Figure Sphere of radius r and outward pointing vectors. Φ Π r Ρ Π Figure ectangular solid T Θ

76 34 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Net, we evaluate the triple integral as in Theorem.4. Sincediv(F ), we find div(f ) dv dv. We appl a change of variables and use the spherical coordinates ρ sin φ cos θ, ρ sin φ sin θ, and z ρ cos φ where ρ r, φ π, and θ<π. The ball is mapped into a rectangular solid T, see Figure, b the transformation that sends (,, z) into (φ, θ, ρ) T. In page 83, the Jacobian determinant of the transformation is given b (,, z) (ρ, φ, θ) Det ρ ρ z ρ φ φ z φ T θ θ z θ ρ sin φ. Appling the change of variables, we obtain dv (,, z) (ρ, φ, θ) dρdφdθ π π r ρ sin φ dρdφdθ πr3 3 π sin φdφ 4πr3 3 Hence, in the case of Eample 5, wehaveverified F ds div(f ) dv. Tr This 5 Let be the cube in the first octant bounded b s, s, z s where s> as in Figure 9. Let be oriented b unit vectors that point awa from. If F (,, z) (,, ), evaluate F ds b Gauss Divergence Theorem. Eample 6 Appling Gauss Divergence Theorem Let be the solid circular clinder bounded b + 4, z and z. Let the boundar of be oriented b unit vectors that point awa from, see Figure. If F (,, z) 3,, z, appl Gauss Divergence Theorem in evaluating the surface integral F ds.

77 .5. INTEGAL THEOES OF GEEN, STOKES, AND GAUSS 35 Solution The divergence of F is given b div 3,, z (3 )+ ( )+ z ( z) Appling the Gauss Divergence Theorem, we rewrite the surface integral. F ds div(f ) dv 6 dv. To evaluate the above triple integral, we use clindrical coordinates Figure A disk oriented b unit vectors pointing outward. r cos θ, r sin θ, z z as discussed in Section.. Notice, r, θ π, and z. The Jacobian determinant of the transformation from cartesian to clindrical coordinates is given b (,, z) (r, θ, z) Det r r z r θ θ z θ z z z z cos θ r sin θ sin θ rcos θ r onsequentl, we obtain 6 dv π π π 3r 4 6(r cos θ) (,, z) (r, θ, z) drdθdz 6r 3 cos θdrdθdz r r cos θdθdz 4 π π cos θdθdz ( + cos θ) dθdz θ + sin θ θπ θ dz 6 dv 4π. π dz Hence, we have F ds 4π.

78 36 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS Tr This 6 Let be the solid bounded b +, z and z. Suppose the boundar is oriented b unit vectors pointing awa from. If F (,, z) z, 3,, appl Gauss Divergence Theorem in evaluating the surface integral F ds. a Figure 3a g() f() for a b d c 3 b 4 Figure 3b m() n() for c d Proof of Green s Theorem We prove Green s Theorem for the case when is an elementar region that is both of tpe and, see page 6. That is, we ma epress as {(, ) :a b, g() f()} Tpe {(c, d) :c d, m() n()} Tpe where f,g and m, n are continuous real-valued functions on [a, b] and [c, d], respectivel. Assume the boundar is a simple closed curve oriented b the counter clockwise motion. In Figure 3a, let where and are parametrized b r (, ) (, g()) r (, ) (, f()) and a b. Iff is a real-valued function defined on, the line integral of the vector field (f, ) along satisfies f d f d f d f d b a f (, g())d b f() a g() b f dd f da. a f (, f())d Similarl, if f is defined on, the line integral of (,f ) along satisfies f f d da. Adding the last two line integrals, we obtain f d + f d f f da. This proves Green s Theorem when is an elementar region that is of both tpe and.

79 .5. INTEGAL THEOES OF GEEN, STOKES, AND GAUSS 37 Proof of Stokes Theorem We prove Stokes Theorem for surfaces that are graphs of functions. Following (5), we parametrize b r(, ) (,, f(, )) where f is a function with continuous partial derivatives defined on an elementar region in the -plane. In Section.4, weproved r r f, f,. (8) Notice, is oriented b unit vectors pointing awa from the origin. Assume the boundar of is a simple closed curve. Then the positive orientation on is the counter clockwise direction in the -plane. Let F be a vector field defined on such that the values of F lie in 3.Wewrite F (P, Q, ) where P, Q, are real-valued functions with continuous partial derivatives defined on. Then the curl is given b curlf Q i z P z see (3) in Section.3. j + On the other hand, the boundar of is parametrized b c(t) ((t),(t),f((t),(t)) Q P k (9) where (t),(t) are certain real-valued function defined on [a, b]. Then the line integral of F along the boundar satisfies b F dr (F c)(t) c (t)dt a b a ((P, Q, ) c)(t) (t), (t), f (t)+ f (t) dt c(t) c(t) where the chain rule is used in the last line. Evaluating the dot product, we find b F dr P (c(t)) (t)+(c(t)) f (t) dt + c(t) a b a Q(c(t)) (t)+(c(t)) f (t) dt c(t) P + f d + Q + f d where we used the definition of the line integral in the previous line.

80 38 HAPTE. ITEATED, LINE, AND SUFAE INTEGALS B Green s Theorem, chain rule, and the product rule, we obtain F dr Q + f P + f da Q + Q f z + f + f f z + f da P + P f z + f + z Q + Q f z + f da P + P f z + f da. f f + f da Grouping and factoring f and f,wefind F dr Q f P + z z f + Q P da Q z, P z, Q P f, f, da r (curlf ) r da where in the last line we used the curl of F in (9), and the cross product in (8). Note, the right side of the last line is the definition of the surface integral of curlf on surface. Finall, we obtain F dr curlf ds. This completes the proof of Stokes Theorem for the case when the surface is the graph of a function z f(, ). Proof of Gauss Divergence Theorem Let be an elementar solid in 3-space, see page 66 in Section.. Assume is oriented b unit vectors pointing awa from. Let F (P, Q, ) be a vector field where P, Q, are real-valued functions on. We epress the surface integral of F on the boundar of as a sum: F ds (P i) ds + (Qj) ds + (k) ds.

81 .5. INTEGAL THEOES OF GEEN, STOKES, AND GAUSS 39 Using the definition of the divergence, we obtain P divf dv dv + Q dv + z dv. The Gauss Divergence Theorem will follow if the three statements below are true. (P i) ds P dv (3) (Qj) ds Q dv (3) (k) ds dv z (3) We onl prove (3), and in a special case. Suppose consists of three faces F i for i,, 3 such that F 3 is a lateral surface, and k is perpendicular to the normal vectors to F 3, see Figure 4. If r 3 is an orientation preserving parametrization of F 3 defined on D 3,then r3 k r 3. Figure 4 Solid with three faces F,F,F 3, and outward pointing vectors. onsequentl, b the definition of the surface integral we obtain r3 k ds k F 3 D 3 r 3 da. (33) Also, we know the surface integral of k on is the sum of the surface integrals on F,F,F 3. ombined with (33), we obtain k ds k ds + k ds. (34) F F Let D be an elementar region in the -plane, and let Φ, Φ be continuous realvalued functions on D satisfing {(,, z) :(, ) D, Φ (, ) z Φ (, )}. For i,, we parametrize F i as follows: r i (, ) (,, Φ i (, )), (, ) D. Since F is oriented b upward pointing unit vectors, we have r k ds k F D r da k Φ, Φ, da D D (,, Φ (, ))da.