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Semelhante a Vector Calculus Concepts
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Vector Calculus Concepts
- 1. 1 Chapter 4 Calculus of Vector fields 1. Introduction 2. Vector Calculus 3. Divergence and gradient of a vector 4. Line Integral 5. Volume Integral 6. Surface Integral 7. Green’s Theorem
- 2. 2 Continued 9. Divergence Theorem (Gauss’ Theorem) 10. Stokes’ Theorem
- 3. 3 1. Introduction Definition 2.1 (Scalar and vector) Vector is a directed quantity, one with both magnitude and direction. For instance acceleration, velocity, force Scalar is a quantity that has magnitude but not direction. For instance mass, volume, distance
- 4. 4 We represent a vector as an arrow from the origin O to a point A. The length of the arrow is the magnitude of the vector written as or . O A or O A OA a a OA
- 5. 5 Basic Vector System Unit vectors , , • Perpendicular to each other • In the positive directions of the axes • have magnitude (length) 1
- 6. 6 Magnitude of vectors Let P = (x, y, z). Vector is defined by with magnitude (length) OP = = + + p x i y j z k = [ ] x, y, z OP = p OP = = + + p x y z 2 2 2
- 7. 7 Calculation of Vectors 1. Vector Equation Two vectors are equal if and only if the corresponding components are equals 1 2 3 1 2 3 1 1 2 2 3 3 Let and . Then , , a a i a j a k b b i b j b k a b a b a b a b = + + = + + = = = =
- 8. 8 2. Addition and Subtraction of Vectors 3. Multiplication of Vectors by Scalars k b a j b a i b a b a ) ( ) ( ) ( 3 3 2 2 1 1 + + = k b j b i b b ) ( ) ( ) ( then scalar, a is If 3 2 1 + + =
- 9. 9 Example Given 5 3 and 4 3 2 . Find p i j k q i j k = + - = - + a p q ) + ) b p q - ) 2 10 d q p - c p ) Magnitude of vector
- 10. 10 Vector Products 1 2 3 1 2 3 ~ ~ ~ ~ ~ ~ ~ ~ If and , a a i a j a k b b i b j b k = + + = + + 1 1 2 2 3 3 ~ ~ a b a b a b a b = + + 1) Scalar Product (Dot product) 2) Vector Product (Cross product) ~ ~ ~ 1 2 3 ~ ~ 1 2 3 2 3 3 2 1 3 3 1 1 2 2 1 ~ ~ ~ i j k a b a a a b b b a b a b i a b a b j a b a b k = = - - - + - ~ ~ ~ ~ and between angle the is , cos | || | . or b a b a b a =
- 11. 11 3) Application of Multiplication of Vectors a) Given 2 vectors and , projection onto is defined by b) The area of triangle ~ ~ 1 . 2 A a b = a b a b a b b a compb a ~ ~ ~ ~ ~ ~ . comp | | | . | length ( ) | | b a b a b a b l b = =
- 12. 12 c) The area of parallelogram d) The volume of tetrahedrone e) The volume of parallelepiped a b a b x A = a b c 3 2 1 3 2 1 3 2 1 6 1 c c c b b b a a a = 6 1 = V a . b c x a b c 3 2 1 3 2 1 3 2 1 c c c b b b a a a = = V a . b c x
- 14. 2. Vector Calculus Limits and Continuity of Vectors A vector function v(t) of a real variable t is said to have the limit l as t approaches t0, if v(t) is defined in some neighborhood of t0 (possibly except at t0) and Then we write Here, a neighborhood of t0 is an interval (segment) on the t-axis containing t0 as an interior point (not as an endpoint). 14 0 lim ( ) 0. t t t - = v l 0 lim ( ) . t t t = v l
- 15. Continuity • A vector function v(t) is said to be continuous at t = t0 if it is defined in some neighborhood of t0 (including at t0 itself!) and • If we introduce a Cartesian coordinate system, we may write 15 1 2 3 1 2 3 ( ) ( ), ( ), ( ) ( ) ( ) ( ) . t v t v t v t v t v t v t = = + + v i j k
- 16. Continued Then v(t) is continuous at t0 if and only if its three components are continuous at t0. Derivative of a Vector Function • A vector function v(t) is said to be differentiable at a point t if the following limit exists: 16 0 ( ) ( ) ( ) lim . t t t t t t + - = v v v
- 17. Continuid • This vector v’(t) is called the derivative of v(t) • In components with respect to a given Cartesian coordinate system, 17
- 18. Continuid • Hence the derivative v’(t) is obtained by differentiating each component separately. For instance, if v = [t, t2, 0], then v’ = [1, 2t, 0]. 18 1 2 3 ( ) ( ), ( ), ( ) . t v t v t v t = v
- 19. Continuid Properties and in particular Partial Derivatives of a Vector Function Suppose that the components of a vector function 19 ( ) ( ) c c = + = + v v u v u v ( ) = + u v u v u v ( ) = + u × v u × v u × v ( ) ( ) ( ) ( ). = + + u v w u v w u v w u v w
- 20. Continuid • are differentiable functions of n variables t1, … , tn. Then the partial derivative of v with respect to tm is denoted by ∂v/∂tm and is defined as the vector function • Similarly, second partial derivatives are; 20 = = + + 1 2 3 1 2 3 , , v v v v v v v i j k 3 1 2 . m m m m v v v t t t t = + + v i j k
- 21. Continuid 21 2 2 2 2 3 1 2 . l m l m l m l m v v v t t t t t t t t = + + v i j k
- 23. 23 Example 2 The position of a moving particle at time t is given by x = 4t + 3, y = t2 + 3t, z = t3 + 5t2. Obtain • The velocity and acceleration of the particle. • The magnitude of both velocity and acceleration at t = 1.
- 24. 24 Solution • The parameter is t, and the position vector is • The velocity is given by • The acceleration is . ) 5 ( ) 3 ( ) 3 4 ( ) ( ~ 2 3 ~ 2 ~ ~ k t t j t t i t t r + + + + + = . ) 10 3 ( ) 3 2 ( 4 ~ 2 ~ ~ ~ k t t j t i dt r d + + + + = . ) 10 6 ( 2 ~ ~ 2 ~ 2 k t j dt r d + + =
- 25. 25 • At t = 1, the velocity of the particle is and the magnitude of the velocity is . 13 5 4 )) 1 ( 10 ) 1 ( 3 ( ) 3 ) 1 ( 2 ( 4 ) 1 ( ~ ~ ~ ~ 2 ~ ~ ~ k j i k j i dt r d + + = + + + + = . 210 13 5 4 ) 1 ( 2 2 2 ~ = + + = dt r d
- 26. 26 • At t = 1, the acceleration of the particle is and the magnitude of the acceleration is . 16 2 ) 10 ) 1 ( 6 ( 2 ) 1 ( ~ ~ ~ ~ 2 ~ 2 k j k j dt r d + = + + = . 65 2 16 2 ) 1 ( 2 2 2 ~ 2 = + = dt r d
- 29. 29 Vector Integral Calculus • The concept of vector integral is the same as the integral of real-valued functions except that the result of vector integral is a vector. ~ ~ ~ ~ ~ ~ If F( ) ( ) ( ) ( ) then ( ) ( ) ( ) ( ) . b b a a b b a a u f u i g u j h u k F u du f u du i g u du j h u du k = + + = + +
- 30. 30 Example . 80 2 42 ] [ ] 5 [ ] 2 [ 4 ) 5 2 ( ) 4 3 ( . calculate , 4 ) 5 2 ( ) 4 3 ( If ~ ~ ~ ~ 3 1 4 ~ 3 1 2 ~ 3 1 2 3 3 1 ~ 3 3 1 ~ 3 1 ~ 2 3 1 ~ 3 1 ~ ~ 3 ~ ~ 2 ~ k j i k t j t t i t t k dt t j dt t i dt t t dt F dt F k t j t i t t F + - = + - + + = + - + + = + - + + = Answer
- 31. 31 Exercise . 2 7 3 2 4 7 ) 4 ( 2 ) 3 ( . calculate , ) 4 ( 2 ) 3 ( If ~ ~ ~ 1 0 ~ 1 0 ~ 2 1 0 ~ 3 1 0 ~ 1 0 ~ ~ ~ 2 ~ 3 ~ k j i k dt t j dt t i dt t t dt F dt F k t j t i t t F - + = = = - + + + = - + + + = Answer
- 32. Curves, Arc Length and Tangent In our day to day life, a curve is defined as a line which is not straight. But, in math, a curve can also be straight or it is also called continuous line. Curves are either open or closed. Open curve is defined as a curve whose ends do not meet. Example is parabola, hyperbola. Closed curves are curves whose ends are joined. Closed curves do not have end points. Examples of closed curves are ellipse and circle. 32
- 33. • The application of vector calculus to geometry is a field known as differential geometry. • Bodies that move in space form paths that may be represented by curves C. This and other applications show the need for parametric representations of C with parameter t, which may denote time or something else 33
- 34. A typical parametric representation is given by (1) r(t) = [x(t), y(t), z(t)] = x(t)i + y(t)j + z(t)k 34
- 35. here t is the parameter and x, y, z are Cartesian coordinates, that is, the usual rectangular coordinates. The use of parametric representations has key advantages over other representations that involve projections into the xy-plane and xz-plane or involve a pair of equations with y or with z as independent variable. The projections look like this: (2) y = f(x), z = g(x). The advantages of using (1) instead of (2) are that, in (1), the coordinates x, y, z all play an equal role, that is, all three coordinates are dependent variables. Moreover, the parametric representation (1) induces an orientation on C. 35
- 36. This means that as we increase t, we travel along the curve C in a certain direction. The sense of increasing t is called the positive sense on C. The sense of decreasing t is then called the negative sense on C, given by (1). • Example: Straight Line A straight line L through a point A with position vector a in the direction of a constant vector b (see in the figure below) can be represented parametrically in the form (3) r(t) = a + tb = [a1 + tb1, a2 + tb2, a3 + tb3]. 36
- 37. Parametric representation of a straight line A plane curve is a curve that lies in a plane in space. A curve that is not plane is called a twisted curve. A simple curve is a curve without multiple points, that is, without points at which the curve intersects or touches itself. 37
- 38. Circle and helix are simple curves. Figures below shows curves that are not simple. An arc of a curve is the portion between any two points of the curve. For simplicity, we say “curve” for curves as well as for arcs. Curves with multiple points 38
- 39. Tangent to a Curve The next idea is the approximation of a curve by straight lines, leading to tangents and to a definition of length. Tangents are straight lines touching a curve. The tangent to a simple curve C at a point P of C is the limiting position of a straight line L through P and a point Q of C as Q approaches P along C. Let us formalize this concept. If C is given by r(t), and P and Q correspond to t and t + Δt, then a vector in the direction of L is (4) 39
- 40. In the limit this vector becomes the derivative (5) provided r(t) is differentiable, as we shall assume from now on. If r′(t) ≠ 0 we call r′(t) a tangent vector of C at P because it has the direction of the tangent. The corresponding unit vector is the unit tangent vector (see figure below) (6) Note that both r′ and u point in the direction of increasing t. Hence their sense depends on the orientation of C. It is reversed if we reverse the orientation. 40 0 1 ( ) lim ( ) ( ) , t t t t t t = + - r r r 1 . = u r r
- 41. continued It is now easy to see that the tangent to C at P is given by q(w) = r + wr' (Fig below). This is the sum of the position vector r of P and a multiple of the tangent vector r′ of C at P. Both vectors depend on P. The variable w is the parameter in (7). 41
- 42. Length of a Curve If r(t) has a continuous derivative r′, it can be shown that the sequence l1, l2, … has a limit, which is independent of the particular choice of the representation of C and of the choice of subdivisions. This limit is given by the integral l is called the length of C, and C is called rectifiable 42
- 43. Arc Length s of a Curve The length of a curve C is a constant, a positive number. But if we replace the fixed b in the previous integral with a variable t, the integral becomes a function of t, denoted by s(t) and called the arc length function or simply the arc length of C. Thus Here the variable of integration is denoted by because t is now used in the upper limit. 43 ( ) . t a d s t dt dt = = r r r r l l t
- 44. Arc Length(Continued) Linear Element ds. It is customary to write dr = [dx, dy, dz] = dx i + dy j + dz k and ds2 = dr • dr = dx2 + dy2 + dz2. ds is called the linear element of C. Arc Length as Parameter. The use of s in instead of an arbitrary t simplifies various formulas. For the unit tangent vector we simply obtain 44
- 45. Arc Length(Continued) u(s) = r′(s) Indeed, |r′(s)| = (ds/ds) = 1 shows that r′(s) is a unit vector. 45
- 46. 46 Del Operator Or Nabla (Symbol ) • Operator is called vector differential operator, defined as . ~ ~ ~ + + = k z j y i x
- 47. 47 3. Gradient of Scalar Functions • If f x,y,z is a scalar function of three variables and f is differentiable, the gradient of f is defined as . grad ~ ~ ~ k z j y i x + + = = f f f f f function vector a is * function scalar a is * f f
- 52. 52 Properties of gradient If A and B are two scalars, then ) ( ) ( ) ( ) 2 ) ( ) 1 A B B A AB B A B A + = + = +
- 53. 53 Directional Derivative . of direction in the r unit vecto a is which , where . is of direction in the of derivative l Directiona ~ ~ ~ ~ ~ ~ r d r d r d a grad a ds d a = = f f f
- 54. 54 Example 4.9 . 4 3 2 vector the of direction in the ) 1 , 2 , 1 ( point at the 2 of derivative l directiona the Compute ~ ~ ~ ~ 2 2 2 k j i A yz xy z x - + = - + + = f
- 55. 55 Solution Directional derivative of f in the direction of . ) 2 ( ) 4 ( ) 2 2 ( hence , 2 Given ~ 2 ~ 2 ~ 2 2 2 2 k yz x j z xy i y xz yz xy z x + + + + + = + + = f f . and where . ~ ~ ~ ~ ~ ~ ~ A A a k z j y i x grad grad a ds d = + + = = = f f f f f f f ~ a
- 58. 58 Unit Normal Vector Equation f (x, y, z) = constant is a surface equation. Since f (x, y, z) = constant, the derivative of f is zero; i.e. . 90 0 cos 0 cos grad 0 grad . ~ ~ = = = = = f f f r d r d d
- 59. 59 • This shows that when f (x, y, z) = constant, • Vector grad f = f is called normal vector to the surface f (x, y, z) = constant . ~ r d grad f y ds grad f z x
- 60. 60 Unit normal vector is denoted by Example Calculate the unit normal vector at (-1,1,1) for 2yz + xz + xy = 0. . ~ f f = n
- 61. 61 Solution Given 2yz + xz + xy = 0. Thus . 6 1 1 4 and 2 ) 1 2 ( ) 1 2 ( ) 1 1 ( (-1,1,1), At . ) 2 ( ) 2 ( ) ( ~ ~ ~ ~ ~ ~ ~ ~ ~ = + + = + + = - + - + + = + + + + + = f f f k j i k j i k x y j x z i y z ) 2 ( 6 1 6 2 is vector normal unit The ~ ~ ~ ~ ~ ~ k j i k j i n ~ + + = + + = = f f
- 62. 62 4. Divergence and gradient of a Vector . . ) .( . as defined is of divergence the , If ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ z a y a x a A A div k a j a i a k z j y i x A A div A k a j a i a A z y x z y x z y x + + = = + + + + = = + + =
- 66. 66 Curl of a Vector ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ If ( , , ) ( , , ) ( , , ) , the curl of is defined by ( ) . A f x y z i g x y z j h x y z k A curl A A i j k f i g j hk x y z i j k curl A A x y z f g h = + + = = + + + + = =
- 71. 71 5. Line Integral Ordinary integral f (x) dx, we integrate along the x-axis. But for line integral, the integration is along a curve. f (s) ds = f (x, y, z) ds A O B ~ ~ r d r+ ~ r
- 72. 72 Scalar Field, V Integral If there exists a scalar field V along a curve C, then the line integral of V along C is defined by . where ~ ~ ~ ~ ~ k dz j dy i dx r d r d V c + + =
- 75. 75 . 11 36 5 24 4 11 36 5 24 4 36 48 36 ) 3 4 3 )( 12 ( ~ ~ ~ ~ 1 0 11 ~ 1 0 10 ~ 1 0 9 1 0 ~ 10 1 0 ~ 9 1 0 ~ 8 1 0 ~ 2 ~ ~ 8 ~ k j i k u j u i u k du u j du u i du u k du u j udu i du u r d V u u B A + + = + + = + + = + + = = =
- 77. 77 Vector Field, Integral Let a vector field and The scalar product is written as . ) ).( ( . ~ ~ ~ ~ ~ ~ ~ ~ dz F dy F dx F k dz j dy i dx k F j F i F r d F z y x z y x + + = + + + + = ~ ~ ~ ~ k F j F i F F z y x + + = . ~ ~ ~ ~ k dz j dy i dx r d + + = ~ ~ . r d F
- 78. 78 . . by given is B point another A to point a from curve the along of integral line then the , curve the along is field vector a If ~ ~ ~ ~ + + = c z c y c x c dz F dy F dx F r d F C F C F
- 84. 84 * Double Integral (Revisions)* 2 2 Given ( , ) 4 in region bounded by a straight line 0, and 2. Find ( , ) in both order integrals. ( , ) 4 unit . R R f x y y R x y x y f x y dA f x y dA = - = = = = Example Answer
- 85. 85 2 2 Using double integral, find the area of a region bounded by 5 and 3. 1 The area of the region 4 unit . 2 y x y x = - = + = Example Answer
- 86. 86 2 2 Evaluate a solid which is bounded by 16 and 2. Stated ( , , ) as integral in order . z x y z f x y z dV dzdydx = - - = Example
- 87. 87 2 Describe ( , , ) as integral in order if is a solid which is bounded by 0, , and 4 2 . f x y z dV dzdydx S z z x y x = = = - Example
- 88. 88 6. Volume Integral Scalar Field, F Integral If V is a closed region and F is a scalar field in region V, volume integral F of V is = V V Fdxdydz FdV
- 89. 89 Example Scalar function F = 2 x defeated in one cubic that has been built by planes x = 0, x = 1, y = 0, y = 3, z = 0 and z = 2. Evaluate volume integral F of the cubic. z x y 3 O 2 1
- 90. 90 = = = = 2 0 3 0 1 0 2 z y x V xdxdydz FdV Solution 6 ] [ 3 3 ] [ 2 1 . 2 2 1 2 2 2 2 0 2 0 3 0 2 0 2 0 3 0 1 0 2 0 3 0 2 = = = = = = = = = = = = z dz dz y dydz dydz x z z z y z y
- 91. 91 Vector Field, Integral If V is a closed region and , vector field in region V, Volume integral of V is ~ F ~ F = V x x y y z z dzdydx F dV F 2 1 2 1 2 1 ~ ~
- 92. 92 Evaluate , where V is a region bounded by x = 0, y = 0, z = 0 and 2x + y + z = 2, and also given V dV F ~ ~ ~ ~ 2 k y i z F + = Example
- 93. 93 If x = y = 0, plane 2x + y + z = 2 intersects z-axis at z = 2. (0,0,2) If x = z = 0, plane 2x + y + z = 2 intersects y-axis at y = 2. (0,2,0) If y = z = 0, plane 2x + y + z = 2 intersects x-axis at x = 1. (1,0,0) Solution
- 94. 94 We can generate this integral in 3 steps : 1. Line Integral from x = 0 to x = 1. 2. Surface Integral from line y = 0 to line y = 2(1-x). 3. Volume Integral from surface z = 0 to surface 2x + y + z = 2 that is z = 2 (1-x) - y z x y 2 O 2 1 2x + y + z = 2 y = 2 (1 - x)
- 95. 95 Therefore, = - = - - = = V x x y y x z dzdydx F dV F 1 0 ) 1 ( 2 0 ) 1 ( 2 0 ~ ~ - = - - = = + = ) 1 ( 2 0 ) 1 ( 2 0 ~ ~ 1 0 ) 2 ( x y y x z x dzdydx k y i z ~ ~ 3 1 3 2 k i+ =
- 96. 96 Example Evaluate where and V is region bounded by z = 0, z = 4 and x2 + y2 = 9 V dV F ~ ~ ~ ~ ~ 2 2 k y j z i F + + = x z f y 4 - 3 3
- 97. 97 f cos = x f sin = y z z = z d ρdρd dV f = ; ; ; where , 3 0 , 2 0 f 4 0 z Using polar coordinate of cylinder,
- 98. 98 + + = V V dxdydz k y j z i dV F ) 2 2 ( ~ ~ ~ ~ Therefore, = = = + + = 4 0 2 0 3 0 ~ ~ ~ ) sin 2 2 ( z k j z i f f dz d d f ~ ~ 144 72 j i + =
- 99. 99 Exercise
- 100. 100 7. Surface Integral Scalar Field, V Integral If scalar field V exists on surface S, surface integral V of S is defined by = S S dS n V S Vd ~ ~ where S S n = ~
- 101. 101 Example Scalar field V = x y z defeated on the surface S : x2 + y2 = 4 between z = 0 and z = 3 in the first octant. Evaluate S S Vd ~ Solution Given S : x2 + y2 = 4 , so grad S is ~ ~ ~ ~ ~ 2 2 j y i x k z S j y S i x S S + = + + =
- 102. 102 Also, 4 4 2 2 ) 2 ( ) 2 ( 2 2 2 2 = = + = + = y x y x S Therefore, ) ( 2 1 4 2 2 ~ ~ ~ ~ ~ j y i x j y i x S S n + = + = = Then, + = S S dS j y i x xyz dS n V ) ( 2 1 ~ ~ ~ + = dS j z xy i yz x ) ( 2 1 ~ 2 ~ 2
- 103. 103 Surface S : x2 + y2 = 4 is bounded by z = 0 and z = 3 that is a cylinder with z-axis as a cylinder axes and radius, So, we will use polar coordinate of cylinder to find the surface integral. . 2 4 = = x z f y 2 2 3 O
- 104. 104 Polar Coordinate for Cylinder cos 2cos sin 2sin ρ x y z z dS d dz f f f f f = = = = = = where 2 0 f 3 0 z (1st octant) and
- 105. 105 Using polar coordinate of cylinder, f f f f f f f f cos sin 8 ) ( ) sin 2 )( cos 2 ( sin cos 8 ) sin 2 ( ) cos 2 ( 2 2 2 2 2 2 z z z xy z z yz x = = = = From = + = S S S S Vd dS j z xy i yz x dS n V ~ ~ 2 ~ 2 ~ ) ( 2 1
- 106. 106 = = + = S z dzd j z i z S Vd 2 0 3 0 ~ 2 ~ 2 ~ ) 2 )( cos sin 8 sin cos 8 ( 2 1 f f f f f f 3 2 2 2 2 2 0 ~ ~ 0 2 2 2 0 ~ ~ 1 1 8 cos sin sin cos 2 2 9 9 8 cos sin sin cos 2 2 z i z j d i j d f f f f f f f f f f = + = + 2 2 2 0 ~ ~ 3 3 2 ~ ~ 0 ~ ~ 9 8 cos sin sin cos 2 cos sin sin cos 36 3( sin ) 3(cos ) 12( ) i j d i j i j f f f f f f f f f f f = + = + - = + Therefore,
- 107. 107 2 2 2 ~ , 9 0 2 S If V is a scalar field whereV xyz evaluate V d S for surface S that region bounded by x y between z and z in the first octant. = + = = = Exercise ~ ~ : 24( ) Answer i j +
- 108. 108 Vector Field, Integral If vector field defeated on surface S, surface integral of S is defined as = S S dS n F S d F . . ~ ~ ~ ~ ~ F ~ F ~ where S n S =
- 109. 109 Example ~ ~ ~ ~ 2 2 2 ~ ~ Vector field 2 defeated on surface : 9 and bounded by 0, 0, 0 in the first octant. Evaluate . . S F y i j k S x y z x y z F d S = + + + + = = = =
- 110. 110 Solution 2 2 2 Given : 9 is bounded by 0, 0, 0 in the1st octant. This refer to sphere with center at (0,0,0) and radius, 3, in the1st octant. S x y z x y z r + + = = = = = x z y 3 3 3 O
- 111. 111 ~ ~ ~ ~ ~ ~ 2 2 2 2 2 2 So, grad is 2 2 2 , and (2 ) (2 ) (2 ) 2 2 9 6. S S S S S i j k x y z x i y j z k S x y z x y z = + + = + + = + + = + + = =
- 112. 112 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Therefore, . . 1 ( 2 ) ( ) 3 1 ( 2 ) . 3 S S S S F d S F ndS y i j k x i y j z k dS xy y z dS = = + + + + = + + ). ( 3 1 6 2 2 2 ~ ~ ~ ~ ~ ~ ~ k z j y i x k z j y i x S S n + + = + + = =
- 113. 113 Using polar coordinate of sphere, 2 sin cos 3sin cos sin sin 3sin sin cos 3cos sin 9sin where 0 , . 2 x r y r z r dS r d d d d f f f f f f f = = = = = = = =
- 114. 114 f f f f f f f f f f d d d d S d F S ] cos sin sin sin 2 cos sin sin 3 [ 9 ] sin 9 [ ] cos 3 ) sin sin 3 ( 2 ) sin sin 3 )( cos sin 3 [( 3 1 . 2 0 0 3 0 0 ~ ~ 2 2 2 2 + + = + + = = = = = + = 4 3 1 9
- 116. 116 8. Green’s Theorem If C is a closed curve in counter-clockwise on plane-xy, and given two functions P(x, y) and Q(x, y), where S is the area of C. + = - c S dy Q dx P dy dx y P x Q ) (
- 117. 117 Example 2 2 2 2 Prove Green's Theorem for [( ) ( 2 ) ] which has been evaluated by boundary that defined as 0, 0 4 in the first quarter. c x y dx x y dy x y and x y + + + = = + = y 2 x 2 C3 C2 C1 O x2 + y2 = 22 Solution
- 118. 118 1 1 2 2 2 2 1 2 3 1 2 2 2 2 0 2 3 0 Given [( ) ( 2 ) ] where and 2 . We defined curve as , . i) For : 0, 0 0 2 ( ) ( ) ( 2 ) 1 8 . 3 3 c c c x y dx x y dy P x y Q x y c c c and c c y dy and x Pdx Qdy x y dx x y dy x dx x + + + = + = + = = + = + + + = = =
- 119. 119 2 2 2 ii) For : 4 ,in the first quarter from (2,0) to (0,2). This curve actually a part of a circle. Therefore, it's more easier if we integrate by using polar coordinate of plane, 2cos , 2sin , 0 c x y x y + = = = 2 2sin , 2cos . dx d dy d = - =
- 121. 121 3 3 3 2 2 0 2 0 2 2 iii) : 0, 0, 0 2 ( ) ( ) ( 2 ) 2 4. 8 16 ( ) ( 4) 4 . 3 3 c c c For c x dx y Pdx Qdy x y dx x y dy y dy y Pdx Qdy = = + = + + + = = = - + = + - - = -
- 122. 122 b) Now, we evaluate where 1 2 . Again,because this is a part of the circle, we shall integrate by using polar coordinate of plane, cos , sin where S Q P dxdy x y Q P and y x y x r y r - = = = = 0 r 2, 0 . 2 and dxdy dS r dr d = =
- 124. 124 Therefore, ( ) 16 . 3 Green's Theorem has been proved. C S Q P Pdx Qdy dxdy x y LHS RHS + = - = - =
- 125. 125 9. Divergence Theorem (Gauss’ Theorem) If S is a closed surface including region V in vector field . . ~ ~ ~ = S V S d F dV F div ~ F ~ y x z f f f div F x y z = + +
- 126. 126 Example 2 ~ ~ ~ ~ 2 2 Prove Gauss' Theorem for vector field, 2 in the region bounded by planes 0, 4, 0, 0 4 in the first octant. F x i j z k z z x y and x y = + + = = = = + =
- 128. 128 1 2 3 4 2 2 5 ~ ~ ~ For this problem, the region of integration is bounded by 5 planes : : 0 : 4 : 0 : 0 : 4 To prove Gauss' Theorem, we evaluate both . , The answer should be the same. V S S z S z S y S x S x y div F dV and F d S = = = = + =
- 129. 129 2 ~ ~ ~ ~ ~ 2 ~ ~ 1) We evaluate . Given 2 . So, ( ) (2) ( ) 1 2 . Also, (1 2 ) . The region is a part of the cylinder. So, we integrate by using polar c V V V div F dV F x i j z k div F x z x y z z div F dV z dV = + + = + + = + = + oordinate of cylinder , ; sin ; where 0 2, 0 , 0 4. 2 x = cos y z z dV d d dz z f f f f = = =
- 130. 130 2 2 2 2 2 2 2 4 0 0 0 2 2 4 0 0 0 2 0 0 2 2 0 0 0 0 ~ Therefore, (1 2 ) (1 2 ) [ ] (20 ) [10 ] (40) 40 20 . 20 . V z V z dV z dzd d z z d d d d d d div F dV f f f f f f f f f f f = = = = = = = = = + = + = + = = = = = =
- 131. 131 1 ~ ~ ~ ~ 1 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ 2) Now, we evaluate . . . i) : 0, , 2 0 . ( 2 ).( ) 0 . 0. S S S F d S F ndS S z n k dS rdrd F x i j k F n x i j k F ndS = = = - = = + + = + - = =
- 132. 132 2 2 2 ~ ~ 2 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ 2 2 0 0 ~ ~ ii) : 4, , 2 (4) 2 16 . ( 2 16 ).( ) 16. Therefore for , 0 r 2, 0 2 . 16 16 . S r S z n k dS rdrd F x i j k x i j k F n x i j k k S F ndS rdrd = = = = = = + + = + + = + + = = = =
- 133. 133 3 3 ~ ~ 2 ~ ~ ~ ~ 2 ~ ~ ~ ~ ~ ~ 3 2 4 0 0 ~ ~ iii) : 0, , 2 . ( 2 ).( ) 2. Therefore for S , 0 2, 0 4 . ( 2) 16. S x z S y n j dS dxdz F x i j z k F n x i j z k j x z F ndS dzdx = = = = - = = + + = + + - = - = - = = -
- 134. 134 4 4 ~ ~ 2 2 ~ ~ ~ ~ ~ ~ 2 ~ ~ ~ ~ ~ ~ ~ iv) : 0, , 0 2 2 . (2 ).( ) 0. . 0. S S x n i dS dydz F i j z k j z k F n j z k i F ndS = = - = = + + = + = + - = =
- 135. 135 2 2 5 5 5 ~ ~ ~ 5 ~ ~ 5 ~ ~ 5 v) : 4, 2 2 4 2 2 4 1 ( ). 2 By using polar coordinate of cylinder : cos , sin , where for : 2, 0 , 0 4, 2 2 S x y dS d dz S x i y j and S x i y j S n S x i y j x y z z S z dS d dz f f f f f + = = = + = + = = = + = = = = =
- 136. 136 . 4 16 ) 2 )( sin )(cos 2 ( . ). sin (cos 2 2. kerana ; sin 2 cos 2 ) sin ( ) cos ( 2 1 2 1 2 1 2 1 ). 2 ( . 5 2 0 4 0 2 ~ ~ 2 2 2 2 ~ ~ ~ 2 ~ ~ ~ ~ f f f f f f f f f f + = = + = + = = + = + = + = + + + = = = S z dz d dS n F y x j y i x k z j i x n F
- 137. 137 1 2 3 4 5 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ Finally, . . . . . . 0 16 16 0 16 4 20 . . 20 . Gauss' Theorem has been proved. S S S S S S S F d S F d S F d S F d S F d S F d S F d S LHS RHS = + + + + = + - + + + = = =
- 138. 138 10. Stokes’ Theorem If is a vector field on an open surface S and boundary of surface S is a closed curve C, therefore = S c r d F S d F curl ~ ~ ~ ~ ~ F ~ ~ ~ ~ ~ x y z i j k curl F F x y z f f f = =
- 139. 139 Example Surface S is the combination of 2 2 ~ ~ ~ ~ i) part of the cylinder 9 0 and 4 0. ii) half of the circle with radius 3 at 4, and iii) 0 , prove Stokes' Theorem for this case. a x y between z z for y a z plane y If F z i xy j xz k + = = = = = = + +
- 140. 140 Solution 2 2 1 2 3 We can divide surface S as S : x y 9 0 z 4 y 0 S : z 4, half of the circle with radius 3 S : y 0 for and + = = = z y x 3 4 O S3 C2 S2 C1 S1 3
- 141. 141 We can also mark the pieces of curve C as C1 : Perimeter of a half circle with radius 3. C2 : Straight line from (-3,0,0) to (3,0,0). Let say, we choose to evaluate first. Given ~ ~ S curl F d S ~ ~ ~ ~ k xz j xy i z F + + =
- 143. 143 By integrating each part of the surface, 2 2 1 1 ~ ~ 2 2 1 2 2 ( ) : 9, 2 2 (2 ) (2 ) 2 6 i For surface S x y S x i y j and S x y x y + = = + = + = + =
- 145. 145 By using polar coordinate of cylinder ( because is a part of the cylinder), 9 : 2 2 1 = + y x S cos , sin , 3, 0 0 4. x y z z dS d dz where dan z f f f f = = = = =
- 146. 146 Therefore, ~ ~ 1 (1 ) 3 1 sin 1 3 sin (1 ) ; 3 curl F n y z z z because f f = - = - = - = Also, dz d dS f 3 =
- 147. 147 1 1 ~ ~ ~ ~ 4 0 0 4 0 0 4 0 3 sin (1 ) 3 (1 ) cos 3 (1 )(1 ( 1)) 24 S S z curl F d S curl F n dS z d dz z dz z dz f f f f = = = = - = - - = - - - = -
- 148. 148 (ii) For surface , normal vector unit to the surface is By using polar coordinate of plane , 4 : 2 = z S . ~ ~ k n = d dr r dS dan z r y = = = 4 , sin 0 r 3 and 0 . where
- 149. 149 2 2 ~ ~ ~ ~ ~ ~ ~ ~ ~ 3 0 0 3 2 0 0 (1 ) sin ( sin )( ) sin 18 S S r r curl F n z j y k k y r curl F d S curl F n dS r rdrd r d dr = = = = = - + = = = = = =
- 150. 150 (iii) For surface S3 : y = 0, normal vector unit to the surface is dS = dxdz The integration limits : . ~ ~ j n - = 3 3 0 4 x and z - So, 1 ) ( ) ) 1 (( ~ ~ ~ ~ ~ - = - + - = z j k y j z n F curl
- 151. 151 3 3 1 2 3 ~ ~ ~ ~ 3 4 3 0 ~ ~ ~ ~ ~ ~ ~ ~ Then, . . ( 1) 24. . . . . 24 18 24 18. S S x z S S S S curl F d S curl F n dS z dzdx curl F d S curl F d S curl F d S curl F d S =- = = = - = = = + + = - + + =
- 152. 152 ~ ~ 1 1 Now, we evaluate . for each pieces of the curve C. i) is a half of the circle. Therefore, integration for will be more easier if we use polar coordinate for plane with radius C F d r C C 3, that is 3cos , 3sin dan z 0 where 0 . r x y = = = =
- 153. 153 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ (3cos )(3sin ) 9sin cos and 3sin 3cos . F z i xy j xzk j j dr dx i dy j dzk d i d j = + + = = = + + = - +
- 154. 154 1 2 ~ ~ 2 0 ~ ~ 3 0 From here, . 27sin cos . . 27sin cos 9cos 18. C F d r d F d r d = = = - =
- 155. 155 2 2 ~ ~ ~ ~ ~ ~ ~ ii) Curve is a straight line defined as , 0 z 0, where 3 3. Therefore, 0. . 0. C C x t y and t F z i xy j xz k F d r = = = - = + + = =
- 156. 156 1 2 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ . . . 18 0 18. We already show that . . Stokes' Theorem has been proved. C C C S C F d r F d r F d r curl F d S F d r = + = + = =
- 157. Line Integral Independent of Path • Suppose that C is a smooth curve given by Also suppose that f is a function whose gradient vector is continuous on C. Then, • Notice the strong resemblance of this theorem to the usual FTC. Some sort • of definite integral is equal to the value of a function at one point subtracted from the 157 ( ), . r t a t b f . C f dr f r b f r a = -
- 158. Continued value of a function at another point. The role of an anti derivative for F is played by the scalar function f, and the role of the endpoints of the interval [a, b] are played by the endpoints of C, r(a), r(b). • One consequence of this theorem is that if F is conservative on R 2 , then the value of a line integral of F along C does not actually depend on the path taken by C, but only on the endpoints of C. That is, a phenomenon like that which we observed in the previous set of examples can never 158
- 159. Continued take place for a conservative vector field. We say that a line integral of F over C is independent of path if its values only depend on the endpoints of C. In this terminology, the line integral of a conservative vector field is independent of path for all curves C. • A special case of the above case occurs when we calculate the line integral of F along a curve C whose starting and endpoints are the same. If C is such a curve, we call C a closed curve, and the line integral of a conservative vector field along a closed curve C is equal to 159
- 160. Continued 160 0 C Fdr f r b f r a = - =
- 161. Continued since r(b) = r(a). Therefore, the line integral of a conservative vector field along a closed path is always equal to 0, regardless of the shape of the closed path. Example. Let and let C, be the parameterization of C,0 ≤ t ≤ 1. Calculate the line integral of F along C. 161 , x x F ye e = 17 3 ,cos r t t t =