Calculus 2 Nguyễn Văn Hộ

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Giáo trình Giải tích dành cho các CTTT Đại học Bách khoa Hà Nội bằng Tiếng Anh được biên soạn bởi thầy Nguyễn Văn Hộ Giáo trình Giải tích dành cho các CTTT Đại học Bách khoa Hà Nội bằng Tiếng Anh được biên soạn bởi thầy Nguyễn Văn Hộ Giáo trình Giải tích dành cho các CTTT Đại học Bách khoa Hà Nội bằng Tiếng Anh được biên soạn bởi thầy Nguyễn Văn Hộ Giáo trình Giải tích dành cho các CTTT Đại học Bách khoa Hà Nội bằng Tiếng Anh được biên soạn bởi thầy Nguyễn Văn Hộ Giáo trình Giải tích dành cho các CTTT Đại học Bách khoa Hà Nội bằng Tiếng Anh được biên soạn bởi thầy Nguyễn Văn Hộ Giáo trình Giải tích dành cho các CTTT Đại học Bách khoa Hà Nội bằng Tiếng Anh được biên soạn bởi thầy Nguyễn Văn Hộ Giáo trình Giải tích dành cho các CTTT Đại học Bách khoa Hà Nội bằng Tiếng Anh được biên soạn bởi thầy Nguyễn Văn Hộ Giáo trình Giải tích dành cho các CTTT Đại học Bách khoa Hà Nội bằng Tiếng Anh được biên soạn bởi thầy Nguyễn Văn Hộ Giáo trình Giải tích dành cho các CTTT Đại học Bách khoa Hà Nội bằng Tiếng Anh được biên soạn bởi thầy Nguyễn Văn Hộ HANOI UNIVERSITY OF TECHNOLOGY NGUYỄN VĂN HỘ A COURSE IN CALCULUS 2009 Calculus – Chapter 1: Vector and Geometry of Space Nguyen Van Ho - 2009 Chapter Vector and Geometry of Space 1.1 VECTORS In this chapter we introduce vectors and coordinate system for three-dimensional space We will see that vectors provide simple descriptions of lines and planes in space We first choose in space a fixed point O (the origin) and three directed lines through O that are perpendicular to each other, called the coordinate axes, labeled the x-axis, y-axis, and zaxis The direction of z-axis is determined by the right-hand-rule: If you curl the fingers of your right hand around the z-axis in the direction of a 90o counterclockwise rotation from the positive x-axis to the positive y-axis, then your thumb points the positive direction of the zaxis The three coordinate planes divide space into octants There is a one-to-one correspondence between any point P in space and a triple ( xP , yP , z P ) of real numbers, the coordinates of P See Figure 1.1.1 Figure 1.1.1: The coordinate axes  Figure 1.1.2: Vector a = AB A VECTOR The term vector is used to indicate a quantity that has both magnitude and direction (velocity,  force, ) A vector is used to be denoted by AB or a See Figure 1.1.2 DEFINITION 1.1.1 VECTOR AND COMPONENTS A three-dimensional vector is an ordered triple a = a1 , a2 , a3 of real numbers The umbers a1 , a2 , a3 are called the components of a PROPERTY 1.1.1 a b Given A ( x A , y A , z A ) , B ( xB , yB , z B ) , then  AB = xB − x A , yB − y A , z B − z A (1.1.1) The length of the vector a = a1 , a2 , a3 is | a |= a12 + a22 + a32 (1.1.2) Calculus – Chapter 1: Vector and Geometry of Space DEFINITION 1.1.2 Nguyen Van Ho - 2009 VECTOR ADDITION If a = a1 , a2 , a3 , b = b1 , b2 , b3 , then a + b = a1 + b1 , a2 + b2 , a3 + b3 DEFINITION 1.1.3 MULTIPLICATION A VECTOR BY A SCALAR If a = a1 , a2 , a3 , and c is a scalar (number), then ca = ca1 , ca2 , ca3 PROPERTY 1.1.2 If a, b, and c are vectors in space, k, l are scalars (numbers), then a a+b =b+a b a + (b + c) = (a + b) + c c d a + = a , where is the zero vector a + ( −a ) = e k (a + b) = ka + kb f (k + l )a = ka + la g (kl )a = k (la) h 1a = a DEFINITION 1.1.4 STANDARD BASIC VECTORS The vectors i = 1, 0, , j = 0,1, , k = 0, 0,1 are called the standard basic vectors PROPERTY 1.1.3 If a = a1 , a2 , a3 , then a = a1i + a2 j + a3k EXAMPLE 1.1.1  Given A ( −1, 2, −2 ) , B ( 2, −3, ) , find a = AB, and | a |   Solution: a = AB = 3, −5, , | a |= AB = 32 + (−5) + 22 = 38 a b A 100-kG weight hangs from two wires as shown in Figure 1.1.3 Find the tensions (forces) T1 and T2 in both wires and their magnitudes Figure 1.1.3: Figure 1.1.4: Solution: Look at Figure 1.1.4 Express first the forces in terms of their components, then equate the total components to zero (balance the forces) T1 = (− | T1 | cos 600 ) i + (| T1 | cos 300 ) j = (− 12 | T1 |) i + ( 23 | T1 |) j , T2 = (| T2 | cos 300 ) i + (| T2 | cos 600 ) j = ( | T2 |) i + ( 12 | T2 |) j , Calculus – Chapter 1: Vector and Geometry of Space Nguyen Van Ho - 2009 T1 + T2 + W = ⇒ T1 = −25 i + 75 j and T2 = 25 i + 25 j B DOT PRODUCT DEFINITION 1.1.5 DOT PRODUCT If a = a1 , a2 , a3 and b = b1 , b2 , b3 , then the dot product of a and b is the number a ⋅ b given by a ⋅ b = a1b1 + a2b2 + a3b3 (1.1.3) PROPERTY 1.1.4 If a, b, and c are vectors in space and k is a constant, then a a ⋅ a = | a |2 b c a ⋅b = b ⋅a a ⋅ (b + c) = a ⋅ b + a ⋅ c d (ka) ⋅ b = k (a ⋅ b) = a ⋅ ( kb) e 0⋅a = THEOREM 1.1.1 a If  , ≤  ≤  , is the angle between a, b, then a ⋅ b = | a | | b | cos  cos  = b a1b1 + a2b2 + a3b3 a ⋅b = |a| |b| a12 + a22 + a32 b12 + b22 + b32 (1.1.4) (1.1.5) Vectors a and b are orthogonal if and only if a ⋅b = i.e a1 b1 + a 2b2 + a 3b3 = (1.1.6) Proof Apply the Law of Cosines to a triangle: | b - a |2 = | a |2 + | b |2 − | a | | b |cos  On the other hand: | b - a |2 = (b - a) ⋅ (b - a) = b ⋅ b - b ⋅ a - a ⋅ b + a ⋅ a = b ⋅ b - 2a ⋅ b + a ⋅ a =| a |2 + | b |2 - 2a ⋅ b Therefore we obtain (1.1.4) Figure 1.1.5 Calculus – Chapter 1: Vector and Geometry of Space DEFINITION 1.1.6 Nguyen Van Ho - 2009 DIRECTION ANGLES AND DIRECTION COSINES a The direction angles of a nonzero vector a are the angles , , and  in the interval [0,π] that the vector a makes with the positive x- , y- , and z- axes b The cosines of these direction angles, cos, cos, and cos are called the direction cosines of the vector a It follows from (1.1.5) that cos  = a a a a⋅i a⋅ j a ⋅k = ; cos  = = ; cos  = = ; |a| |i| |a| | a | | j| | a | |a| |k | |a| a1 = | a | cos  ; a2 = | a | cos  ; a3 = | a | cos  ; (1.1.7) (1.1.8) Therefore a = | a | cos  , cos  , cos  (1.1.9) The unit vector of a : a = cos  , cos  , cos  |a| (1.1.10) cos  + cos  + cos  = (1.1.11) DEFINITION 1.1.7 a SCALAR PROJECTION AND VECTOR PROJECTION The scalar projection of a onto b (also called the component of a along b): OP = compb a = | a | cos  = b a ⋅b |b| (1.1.12) The vector projection of a onto b:   a ⋅b  b  a ⋅b  OP = projb a =  =  b   |b| |b| |b|   OP > , OP = kb, k > (1.1.13)  OP < , OP = kb, k < Figure 1.1.6: Projection EXAMPLE 1.1.2 a Find the scalar projection of a = 1, −2, −3 onto b = −2,3, −1 b The force (in newtons) F = 3, 2,5 moves a particle from the point A(2, 1, 4) to the point B(6, 3, 5) , (the length unit is meter) Find the work done c Find the direction cosines of a = 2,3,1 Calculus – Chapter 1: Vector and Geometry of Space c Find the angle between a = −1,3, −2 and b = 1, −2, −1 d Find the unit vector of a = 2, 4, −4 e Are a = −1,3, −7 and b = 1, −2, −1 orthogonal? Nguyen Van Ho - 2009 Let a = −1, 4,3 and b = 1, 1, Find the scalar and vector projections of b onto a and of a onto b 9 18 −9 36 27 Answer: compb a = / ; projb a = , , ; compab = / 26 ; proja b = , , 6 26 26 26 f C CROSS PRODUCT DEFINITION 1.1.8 CROSS PRODUCT The cross product of a and b is a vector, denoted by a × b , that satisfies (i) a × b is orthogonal to both vectors a and b (ii) The direction of a × b is given by the right-hand rule (the fingers of the right hand curl in the direction of a rotation through an angle less than 180o from a to b, then the thumb points the direction of a × b ) (iii) | a × b | = | a || b | sin  , where  , ≤  ≤  , is the angle between a, b (1.1.14) Note that the condition (1.1.14) means that the magnitude (length) of a × b is equal to the area of the parallelogram determined by a and b See Figure 1.1.7 Figure 1.1.7 PROPERTIES 1.1.5 CROSS PRODUCT a i × j = k , j × k = i, k × i = j b a × b = if and only if a and b are parallel c b × a = − (a × b) d a × (b + c) = (a × b) + (a × c) e (ka) × b = k (a × b) f (ka) × (lb) = (kl )(a × b) , where k and l are numbers From these properties, it is easy to obtain the following component expression for the cross product of two vectors a = a1 , a2 , a3 and b = b1 , b2 , b3 Calculus – Chapter 1: Vector and Geometry of Space THEOREM 1.1.2 Nguyen Van Ho - 2009 MATRIX EXPRESSION OF THE CROSS PRODUCT If a = a1 , a2 , a3 and b = b1 , b2 , b3 , then the cross product of a and b is determined by i a × b = a1 j a2 b1 b2 k a a3 = b2 b3 a3 b3 i− a1 a3 b1 b3 j+ a1 b1 a2 k b2 = (a2b3 − a3b2 ) i + (a3b1 − a1b3 ) j + (a1b2 − a2b1 ) k = a2 b3 − a3b2 , a3b1 − a1b3 , a1b2 − a2 b1 (1.1.15) Proof a × b = a1 , a2 , a3 × b1 , b2 , b3 = (a1i + a2 j + a3k ) × (b1i + b2 j + b3k ) Apply properties 1.1.5 EXAMPLE 1.1.3 a Given a = 1,3, −2 , b = −2, 4,1 , find a × b , | a × b | Answer: 11,3,10 , 230 b Find the area of the triangle ABC, A(2,8,12), B(4,5,8), C(1,4,10) Answer: 285 / c Find the height AH of the triangle ABC, A(1,6,4), B(2,5,8), C(-1,4,0)   | BA × BC | 176 88  Answer: AH = = = 74 37 | BC | d Prove that a × (b × c) = (a ⋅ c) b − (a ⋅ b) c (1.1.16) Hint: Apply (1.1.3) and (1.1.15) D SCALAR TRIPLE PRODUCT DEFINITION 1.1.9 SCALAR TRIPLE PRODUCT The scalar triple product of three vectors a, b, and c, denoted by (a, b, c) , is a number that is defined by the scalar product of a and a × b : (a, b, c) = a ⋅ (b × c) THEOREM 1.1.3 a (1.1.17) EXPRESSION OF THE SCALAR TRIPLE PRODUCT Let a = a1 , a2 , a3 , b = b1 , b2 , b3 , and c = c1 , c2 , c3 Then the scalar triple product of three vectors a, b, and c is determined by the determinant Calculus – Chapter 1: Vector and Geometry of Space a1 (a, b, c) = a ⋅ (b × c) = b1 c1 a2 b2 c2 Nguyen Van Ho - 2009 a3 b3 c3 (1.1.18) b (b, a, c) = −(a , b, c) , i.e., b (a × c) = −a ⋅ (b × c) (1.1.19) c (a , b, c) = (b , c, a) = (c, a, b), i.e, a ⋅ (b × c) = b ⋅ (c × a) = c ⋅ (a × b) (1.1.20) Proof The conclusion a follows directly from (1.1.3), (1.1.15), and (1.1.16) The conclusions b and c follow from the determinant properties EXAMPLE 1.1.4 Let a = 2, −1, −2 , b = −1,3,1 , and c = −3, 2, Find (a , b, c) , (b, c, a) , (b, a , c) Answer: -5, -5, These results justify (1.1.18) and (1.1.19) PROPERTIES 1.1.6 SCALAR TRIPLE PRODUCT a The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude (absolute value) of their scalar triple product: V = | (a , b, c) | b (1.1.21) The vectors a, b, and c are coplanar if and only if (a, b, c) = (1.1.22) Proof a (1.1.20) follows directly from (1.1.4) and (1.1.14): (a , b, c) = a ⋅ (b × c) = | a | | (b × c) |cos  , where  , ≤  ≤  , is the angle between a and b × c The length | b × c | is equal to the area of the parallelogram determined by b and c; the height h of the parallelepiped is the magnitude of | a | cos  Figure 1.1.8 shows that when ≤  ≤  / : | a | cos  ≥ ⇒ (a,b,c) ≥ and when  / <  ≤  : | a | cos  < ⇒ (a,b,c) < c It is the consequence of a a) ≤  ≤  / : (a, b, c) = a ⋅ (b × c) ≥ Figure 1.1.8: b)  / <  ≤  : (a, b, c) = a ⋅ (b × c) < The parallelepiped determined by the vectors a, b, and c Calculus – Chapter 1: Vector and Geometry of Space Nguyen Van Ho - 2009 EXAMPLE 1.1.5 Let a = 3, −3, −4 , b = 1, −3, −1 , and c = k , −2, a Find the volume of the parallelepiped determined by a, b, and c, if k = b Determine k so that a, b, and c are coplanar c Determine whether the points A(1, 0, 1), B(2, 4, 6), C(3, -1, 2), and D(6, 2, 8) lie in the same plane Answer: 1.2 a 67; b -22/9; c Yes EQUATIONS OF LINES AND PLANES A EQUATIONS OF LINES A line L is determined by a point P0 ( x0 , y0 , z0 ) on it and its direction vector v = a, b, c   Let P ( x, y, z ) be an arbitrary point on L Let r = OP = x, y, z , and r0 = OP0 = x0 , y0 , z0 be  the position vectors of P and P0 Vector P0 P = r - r0 is parallel to v Look at Figure 1.2.1 Figure 1.2.1 The vector equations of L : r = r0 + t v (1.2.1) The parametric equations of L : x = x0 + at , (t is the parameter, t ∈ y = y0 + bt , z = z0 + ct (1.2.2) ) The symmetric equations of L : x − x0 y − y0 z − z0 = = a b c (1.2.3) If one of a, b, or c = 0, these equations become, for instance if a = 0: (The line lies in the plane x = x0 that is parallel to the xy-plane) x = x0 y − y0 z − z0 = b c (1.2.4) 10 Calculus – Chapter 1: Vector and Geometry of Space Nguyen Van Ho - 2009 If two among a, b, or c = 0, these equations become, for instance if a = and b = 0: x = x0 y = y0 (1.2.5) EXAMPLE 1.2.1 a Find the vector equation and the parametric equations for the line that passes through the point (3, 2, -1) and is parallel to the vector v = −2, − 5, At what points does the line passes through the xy-plane? Solution: The vector equation: r = − 2t , − 5t , − + t = (3 − 2t )i + ( − 5t ) j + ( − + t )k The parametric equations: x = − 2t , y = − 5t , z = − + t Put z = ⇒ t = ⇒ x = 1, y = −3, z = ⇒ the line passes through the xy-plane at the point P(1, -3, 0) b Find the vector equation, the parametric equation, and the symmetric equation for the line that passes through the points A(4, 3, 6), B(2, -5, 6) Is this line parallel to the line in question a? B EQUATIONS OF PLANES A plane P is determined by a point P0 ( x0 , y0 , z0 ) in it and its normal vector n = a, b, c   Let P ( x, y, z ) be an arbitrary point in P Let r = OP = x, y, z , and r0 = OP0 = x0 , y0 , z0 be  the position vectors of P and P0 Vector P0 P = r - r0 is orthogonal to n = a, b, c and so we have The vector equations of P : n ⋅ (r − r0 ) = The scalar equation of P : a ( x − x0 ) + b( y − y0 ) + c( z − z0 ) = (1.2.7) ax + by + c z + d = (1.2.8) or : where or n ⋅ r = n ⋅ r0 (1.2.6) d = −(ax0 + by0 + cz0 ) EXAMPLE 1.2.2 a  Find an equation of the plane that passes through A(1, 3, -4) and is orthogonal to BC , where B(-1, -3, 1), C(3, 4, -2) Solution: 11 Calculus – Chapter 6: Surface Integrals Nguyen Van Ho GRADIENT, TANGENT PLANE, AND NORMAL LINE Suppose that the smooth surface S in space is defined by the differentiable function z = z ( x, y ), or, g ( x, y, z ) = z − z ( x, y ) = 0, ( x, y ) ∈ D ⊂ Suppose S is a smooth surface with equation g ( x, y, z ) = k Let P ( x0 , y0 , z0 ) be a point on S Let C be a curve on S and passes through the point P The curve C is described by a continuous vector function r (t ) = x(t ), y (t ), z (t ) Let t0 be the parameter value corresponding to P: P = ( x(t0 ), y (t0 ), z (t0 ) ) = ( x0 , y0 , z0 ) Since C is on S , any point ( x(t ), y (t ), z (t ) ) must satisfy the equation of S : g ( x(t ), y (t ), z (t )) = k Then ∂g dx ∂g dy ∂g dz + + = ⇒ ∇g ⋅ r′(t ) = ∂x dt ∂y dt ∂z dt ⇒ at P = ( x(t0 ), y (t0 ), z (t0 ) ) = ( x0 , y0 , z0 ) : ∇g ( x0 , y0 , z0 ) ⊥ r′(t0 ) But r′(t0 ) is the tangent vector to C at P, then the gradient vector ∇g ( x0 , y0 , z0 ) is perpendicular to any curve C on S passing through P Thus the gradient vector ∇g ( x0 , y0 , z0 ) is perpendicular to the tangent plane to the surface S at P Therefore the equation of the tangent plane at the point P ( x0 , y0 , z0 ) is g x ( x0 , y0 , z0 )( x − x0 ) + g y ( x0 , y0 , z0 )( y − x0 ) + g z ( x0 , y0 , z0 )( z − x0 ) = (6.2.2) and the equation of the normal line to S at P ( x0 , y0 , z0 ) is ( x − x0 ) ( y − x0 ) ( z − x0 ) = = g x ( x0 , y0 , z0 ) g y ( x0 , y0 , z0 ) g z ( x0 , y0 , z0 ) EXPRESSION 6.2.1 (6.2.3) OF SURFACE INTEGRALS a Suppose that S is determined by z = z ( x, y ) or g ( x, y, z ) = z − z ( x, y ) = 0, ( x, y ) ∈ D ⊂ , and is upward oriented It means that the z-component of the unit normal vector n is positive (see Figure 6.2.1 below) Let F = P( x, y, z ), Q( x, y, z ), R( x, y, z ) be defined on S Then n= − z x ( x, y ), − z y ( x, y ), ∇g = 2 | ∇g | + ( z x ( x, y ) + ( z y ( x, y ) dS = + [ z x ( x, y ) ] +  z y ( x, y )  dx dy 2 (6.2.4) Therefore (6.2.1) has the expression as the double integral below ∫∫ F ⋅ n dS = ∫∫ − P ( x, y, z ( x, y) ) z ( x, y) − Q ( x, y, z ( x, y) ) z x S y + R ( x, y, z ( x, y ) )  dx dy (6.2.5) D 76 Calculus – Chapter 6: Surface Integrals Nguyen Van Ho Suppose S is oriented by the downward unit normal vector n, (see Figure 6.2.2), then b n=− z x ( x, y ), z y ( x, y ), − a×b = , 2 | a×b | + ( z x ( x, y ) + ( z y ( x, y ) dS = + [ z x ( x, y ) ] +  z y ( x, y )  dx dy , see (6.2.4) 2 ∫∫ F ⋅ n dS = ∫∫  P ( x, y, z ( x, y) ) z ( x, y) + Q ( x, y, z ( x, y) ) z x S y − R ( x, y, z ( x, y ) )  dx dy (6.2.6) D You can write similarly the expressions for the surface integrals if S is given by an equation of the form x = x( y, z ), ( y, z ) ∈ D ⊂ or y = y ( x, z ), ( x, z ) ∈ D ⊂ EXPRESSION 6.2.2 OF SURFACE INTEGRALS Let S be the oriented smooth surface in space Let F = P( x, y, z ), Q( x, y, z ), R( x, y, z ) be the vector field defined on S Let the unit normal vector of S be expressed by the direction cosines, (see 1.1, Chapter 1) n = cos  , cos  , cos  , (6.2.7) Then F ⋅ n = P( x, y, z ), Q( x, y, z ), R( x, y, z ) ⋅ cos  , cos  , cos  = P ( x, y, z ) cos  + Q ( x, y , z ) cos  + R ( x, y , z ) cos  (6.2.8) Therefore (6.2.1) can be expressed as below ∫∫ F ⋅ dS = ∫∫ F ⋅ n dS = ∫∫  P ( x, y, z ) cos  + Q ( x, y, z ) cos  S S S + R ( x, y, z ) cos   dS (6.2.9) Remark To evaluate the integral in the right hand side of (6.2.9), we need to divide the surface S into sub-surfaces that can be expressed by one of the following equations: z = z ( x, y ), ( x, y ) ∈ Dxy ⊂ or x = x( y, z ), ( y, z ) ∈ Dyz ⊂ or y = y ( x, z ), ( x, z ) ∈ Dxz ⊂ 77 Calculus • – Chapter 6: Surface Integrals Nguyen Van Ho If S has the equation x = x( y, z ), ( y , z ) ∈ Dy z then    ∫∫ P ( x( y, z ), y, z ) dydz if ≤  <  Dyz   P x , y , z cos  dS = if  = ( )  ∫∫S     − ∫∫ P ( x( y, z ), y, z ) dydz if <  ≤   Dyz • If S has the equation y = y ( x, z ), ( x, z ) ∈ Dx z then    ∫∫ Q ( x, y ( x, z ), z ) dxdz if ≤  <  Dxz   if  = ∫∫S Q ( x, y, z ) cos  dS =      − ∫∫ Q ( x, y ( x, z ), z ) dxdz if <  ≤   Dxz • (6.2.10) (6.2.11) If S has the equation z = z ( x, y ), ( x, y ) ∈ Dx y then    ∫∫ R ( x, y, z ( x, y ) ) dxdy if ≤  <  Dxy   R x , y , z cos  dS = if  = ( )  ∫∫S     − ∫∫ R ( x, y, z ( x, y ) ) dxdy if <  ≤   Dxy (6.2.12) EXAMPLE 6.2.1 a Evaluate I = ∫∫ F ⋅ dS , where F = x, y , z S and S is the upper half-sphere, upward oriented, z = 16 − x − y , x + y ≤ 16 Solution Apply (6.2.5), I = ∫∫  − x(−2 x) − y (−2 y ) + (16 − x − y )  dx dy = ∫∫ (16 + x + y ) dx dy = 384 D b Evaluate I = ∫∫ F ⋅ dS , where F = D x / z, y / z, z − S and S is the upper surface, upward oriented, of z = − x − y , x + y ≤ 78 Calculus – Chapter 6: Surface Integrals Nguyen Van Ho Solution Apply (6.2.5),       x y (−2 x) −  (−2 y ) + ( − x − y − )  dx dy = (8ln − 2) I = ∫∫  −  2  2  4− x − y   4− x − y  D    c Find the flux of the vector field F = x, z , y outward through the sphere x + y + z = a Solution For the upper half of the sphere, S1 : z = a − x − y , ( x, y ) ∈ D : x + y ≤ a , apply (6.2.5):   −x − x  = F ⋅ d S ∫∫ ∫∫ 2  S1 D    a −x −y  −   (  −y a2 − x2 − y   a − x2 − y  )    + y  dx dy      x2 = ∫∫  + y  dx dy  a2 − x2 − y  D   For the lower half of the sphere, S : z = − a − x − y , x + y ≤ a , apply (6.2.6):   x x  = F ⋅ d S ∫∫ ∫∫  a − x2 − y S2 D      + − a2 − x2 − y   (   x2 = ∫∫  − y  dx dy  a2 − x2 − y  D   d )  y   a − x2 − y     − y  dx dy    ⇒ ∫∫ F ⋅ dS = ∫∫ F ⋅ dS + ∫∫ F ⋅ dS = S S1 S2 4 a 3 A fluid with density k flows with velocity v = y, 1, z Find the rate of flow upward through the paraboloid S : z = − ( x + y ) / 4, x + y ≤ 36 Solution The rate of flow upward through the paraboloid S is, applying (6.2.5),   −x   − y   x2 + y  R = k ∫∫ v ⋅ dS = k ∫∫  − y  − + −   dxdy = 162k            S D  e Find the flux of the vector field F = x, y, z outward through the cube [−1, 2]3 Solution: Apply (6.2.10) – (6.2.12) to six faces of the cube: For the face S : z = 2, ( x, y ) ∈ D1 = [−1, 2]2 ,  =  =  / 2,  = 0, ∫∫ F ⋅ dS = ∫∫ 3z dxdy = ∫∫ × dxdy = × = 54 S1 S1 D1 For the face S : z = −1, ( x, y ) ∈ D1 = [−1, 2]2 ,  =  =  / 2,  =  , 79 Calculus – Chapter 6: Surface Integrals Nguyen Van Ho ∫∫ F ⋅ dS = − ∫∫ 3z dxdy = − ∫∫ × (−1) dxdy = × = 27 S2 S2 D1 ⇒ ∫∫ F ⋅ dS = ∑ S f ∫∫ F ⋅ dS = 162 i =1 S i Find the flux of the vector field F = x + y, y + z , z + x outward through the triangular pyramid OABC, O (0, 0, 0), A(1, 0, 0), B (0,1, 0), C (0, 0,1) 1 Answer: ∫∫ F ⋅ dS = ∑ ∫∫ F ⋅ dS = − − − + = i =1 S i S EXAMPLE 6.2.2 THE ELECTRIC FIELD Suppose an electric charge Q is located at the origin According to Coulomb’s Law, the electric force F exerted by the charge Q on a charge q located at M ( x, y, z ) is  qQ (6.2.13) F (r ) = r, where r = x, y, z | r |3 where  is a constant The force exerted by the charge Q (located at the origin) on a unit chart, q = 1, located at any point M ( x, y, z ), ( x, y, z ) ∈ , is called the electric field of Q: Q (6.2.14) E(r ) = r, where r = x, y, z |r| Find the electric flux of E given in (6.2.14) outward through the sphere S : x + y + z = a Solution Method On the sphere S : E(r ) = Q a r= Q a x, y, z , where x + y + z = a By the same way in calculating the flux in Example 6.2.1 b: • For the upper half of the sphere, S1 : z = a − x − y , ( x, y ) ∈ D : x + y ≤ a , apply (6.2.5), ∫∫ E ⋅ dS = S1   −x − x  ∫∫ a D   a − x2 − y   Q   −y − y   a − x2 − y      + a − x − y  dx dy    = 2 Q • For the lower half of the sphere, S : z = − a − x − y , x + y ≤ a , apply (6.2.6), 80 Calculus – Chapter 6: Surface Integrals ∫∫ E ⋅ dS = S2   x x  ∫∫ a D   a − x2 − y   Nguyen Van Ho   y + y   a − x2 − y   Q   − − a2 − x2 − y2   ( )   dx dy  = 2 Q Therefore the electric flux of E given in (6.2.11) through the sphere S : x + y + z = a is ∫∫ E ⋅ dS = 4 Q (6.2.15) S Method Note that, E have the same direction as the unit outward normal vector n of the sphere S x + y + z = a , see (6.2.14) Therefore E(r ) ⋅ n = Q |r| r ⋅n = Q | r |= |r| Q |r| = Q on the sphere , a2 and Q ∫∫ E ⋅ dS = ∫∫ E ⋅ n dS = ∫∫ a S S S = Q a 2 dS Area(S ) = Q a 4 a = 4 Q From (6.2.15), Q can be expressed by the electric flux as below Q= ∫∫ E ⋅ dS =  ∫∫ E ⋅ dS (6.2.16) ≈ 8.8546 ×10 −2 C / N m (6.2.17) 4 S S where 0 = 4 Formula (6.2.16) gives a result of Gauss’s Law, one of the important laws of electrostatics The Law says that the net charge enclosed by any closed surface S is expressed by (6.2.16) 6.3 STOKES’ THEOREM Looking back formula (5.7.2) in Chapter 5, for the vector field F in the plane ∫ C+ F ⋅ dr = ∫ C+ ( F ⋅ T ) ds = ∫C + P ( x, y ) dx + Q ( x, y ) dy = ∫∫ D ( curl F ) ⋅ k dxdy (6.3.1) Because k is the unit normal vector of the upper side of “surface” D on the xy-plane, the above formula can be rewrite in surface integral sense as below ∫ C+ F ⋅ dr = ∫ C+ ( F ⋅ T ) ds = ∫C + P ( x, y ) dx + Q ( x, y ) dy 81 Calculus – Chapter 6: Surface Integrals = ∫∫ D Nguyen Van Ho ( curl F ) ⋅ k dxdy = ∫∫D ( curl F ) ⋅ n dS (6.3.2) This formula says that the line integral around the boundary C  of D of the tangential component of F is equal to the surface integral of the normal component of the curl of F across the D We have also the same conclusion for the vector field F in the space as given in the Theorem by Stokes that we accept THEOREM 6.3.1 STOKES’ THEOREM Let S be an oriented piecewise-smooth surface that is bounded by a simple, closed, piecwise-smooth curve C +, where the positive direction of C coincides with the direction of n in the right hand rule sense Let F be a vector field with components having continuous partial derivatives on an open region in that contains S Then ∫ C+ F ⋅ dr = ∫ C+ = ∫ ( F ⋅ T ) ds C+ P ( x, y, z ) dx + Q ( x, y, z ) dy + R ( x, y, z ) dz = ∫∫ curl F ⋅ dS = ∫∫ curl F ⋅ n dS S (6.3.3) S COROLLARY 6.3.1 If curl F = , then F is conservative and ∫ C F ⋅ dr = for any closed curve C Proof If curl F = , then by (6.3.3), ∫ C+ F ⋅ dr = for any closed curve C Therefore, by Corollary 5.4.1, F is conservative Remark This Corollary is the converse statement of Corollary 5.5.1, from which follows Theorem 5.5.2 EXAMPLE 6.3.1 a Evaluate ∫ C+ F ⋅ dr where F = − y , x, yz and C + is the intersection of the cylinder x + y = and the plane x + y + z = The direction of C + coincides with the positive direction of the z-axis Solution 82 Calculus – Chapter 6: Surface Integrals Nguyen Van Ho Method 1: Direct calculation of the line integral The parametric equations of C + : x = cos t , y = sin t , z = − cos t − sin t , ≤ t ≤ 2 ∫ C+ 2 ∫ (− y F ⋅ dr = (t ) x′(t ) + x(t ) y ′(t ) + y (t ) z (t ) z ′(t ) ) dt = 2 ∫ sin t + cos t + sin t (1 − cos t − sin t )(1 − cos t − sin t )′ dt = 2 Method 2: Apply Stokes’Theorem’ i j k F = − y , x, xz ⇒ curl F = ∂ / ∂x ∂ / ∂y ∂ / ∂z = z , 0, (1 + y ) − y2 x yz ∫ C+ F ⋅ dr = ∫∫ curl F ⋅ dS = ∫∫ curl F ⋅ n dS , apply (6.2.5) where z = − x − y , S S = ∫∫ [ −(1 − x − y ) (−1) + 0(−1) + (1 + y ) ] dxdy D = ∫∫ (2 − x + y ) dxdy = D b Evaluate ∫∫ curl F ⋅ n dS 2 0 ∫ d ∫ (2 − r cos  + r sin  ) rdr = 2 where F = − yz , xz , xy and S is the upper part, upward S oriented, of the sphere x + y + z = 25 that lies inside the cylinder x + y = Solution Method 1: Direct calculation of the surface integral: F = − yz , xz , xy ⇒ curl F = 0, − y, z   −y 2 curl F ⋅ d S = curl F ⋅ n dS = − ( − y ) + 25 − x − y   dxdy = 72 ∫∫S ∫∫S ∫∫D  2  25 − x − y  Method 2: Apply Stokes’Theorem’ The parametric equations of the boundary C + of S : x = 3cos t , y = 3sin t , z = 4, ≤ t ≤ 2 ∫∫ curl F ⋅ dS = ∫ + F ⋅ dr = S c C 2 ∫ [− y(t ) z (t ) x′(t ) + x(t ) z (t ) y′(t ) +x(t ) y (t ) z ′(t )] dt = 72 Use Stockes’ Theorem to evaluate ∫ C+ F ⋅ dr where 83 Calculus – Chapter 6: Surface Integrals Nguyen Van Ho F = x + y , y + z , z + x and C is the triangle with vertices (1, 0, 0), (0, 1, 0), (0, 0, 1), oriented counterclockwise as viewed from above Solution: i j k curl F = ∂ / ∂x ∂ / ∂y ∂ / ∂z = −2 z , − x, − y x + y2 y + z z + x2 The equation of the plane containing the triangle: x + y + z = or z = − x − y, ( x, y ) ∈ D = {( x, y ) : ≤ y ≤ − x, ≤ x ≤ 1} Applying (6.2.4) and (6.3.3), we obtain ∫ C+ 6.4 F ⋅ dr = ∫∫ curl F ⋅ n dS = ∫∫ [ −2(1 − x − y ) − x − y ] dxdy = ∫∫ −2 dxdy = −1 S D D THE DIVERGENCE THEOREM Formula (5.7.3) in Chapter expresses the integral of the normal component of F along the closed curve C + in the xy-plane as the double integral of div F over the domain D enclosed by the curve (6.4.1) ∫ + F ⋅ n ds = ∫∫ div F( x, y) dxdy C D Note that C + is the positively oriented boundary of the domain D in the xy-plane The following theorem gives the similar result for the vector field F in space, the domain D in space, and the boundary surface S of D with positive orientation, i.e with outward orientation THEOREM 6.4.1 THE DIVERGENCE THEOREM Suppose that D is a simple solid region (or D is a finite union of simple solid regions), and S is its boundary surface positively oriented, and F = P( x, y, z ), Q( x, y, z ), R( x, y, z ) is a vector field where P ( x, y, z ), Q ( x, y, z ), and R ( x, y, z ) have continuous partial derivatives on an open region that contains D Then  ∂P ∂Q ∂R  div F dxdydz = ∫∫∫  + + (6.4.2)  dxdydz ∫∫S F ⋅ dS = ∫∫S F ⋅ n dS = ∫∫∫ ∂x ∂y ∂z  D D  We accept the Theorem It follows from the Midpoint Rule For Triple Integrals, see Property 4.1.2 of Chapter 4, and the Divergence Theorem that, there exist a point M = ( x0 , y0 , z0 ) in D so that the flux of F across outward the surface S can be express by ∫∫ F ⋅ dS = ∫∫ F ⋅ n dS = ∫∫∫ div F dxdydz = div F(M ) × V(D) S S (6.4.3) D where V(D) denotes the volume of D 84 Calculus – Chapter 6: Surface Integrals Nguyen Van Ho If D is a ball of a given center M = ( x, y, z ) and of a radius a then from (6.4.3) and the continuity of div F , by the hypotheses of the Divergence Theorem, F ⋅ dS a → V(D ) ∫∫ S div F( M ) = lim (6.4.4) This equation says that div F at a given point M = ( x, y, z ) , div F ( M ) , is the net rate of outward flux per unit volume at M If div F ( M ) > , M is called a source If div F ( M ) < , M is called a sink EXAMPLE 6.4.1 a Find the flux of the vector field F = x, z , y outward through the sphere x + y + z = a Solution We have done the direct calculation in Example 6.2.1b and obtained ∫∫ F ⋅ dS = S 4 a Let us calculate it applying (6.4.2) Since F = x, z , y then div F = and obtain easily that result: ∫∫ F ⋅ n dS = ∫∫∫ div F dxdydz = ∫∫∫ dxdydz = Volume( D) = S b D D 4 a 3 Find the flux of the vector field F = x, y, z outward through the cube [−1, 2]3 Solution The direct calculation in Example 6.2.1d gives ∫∫ F ⋅ dS = 162 S Let us calculate it applying (6.4.2) Since F = x, y, z then div F = and we obtain easily the same result: ∫∫ F ⋅ n dS = ∫∫∫ div F dxdydz = ∫∫∫ dxdydz = ×Volume( D) = × 3 S c D = 162 D Find the flux of the vector field H = x + y, y + z , z + x outward through the triangular pyramid OABC, O (0, 0, 0), A(1, 0, 0), B (0,1, 0), C (0, 0,1) Solution ∫∫ F ⋅ dS = Example 6.2.1.f gives We can obtain easily the result applying (6.4.2): S F = x + y, y + z , z + x , then div F = , and then 1 1 ∫∫ F ⋅ n dS = ∫∫∫ div F dxdydz = ∫∫∫ dxdydz = ×Volume( D) =  ×  = S D D 85 Calculus – Chapter 6: Surface Integrals Nguyen Van Ho EXERCISES 6.1 6.2 Find the area of a The ellipse cut from the plane z = cx by the cylinder x + y = b The surface x − ln x + 15 y − z = above the square D :1 ≤ x ≤ 2, ≤ y ≤ 1, in the xy-plane c The part of the paraboloid z = x + y which lies under the plane z = Evaluate the surface integral a I = ∫∫ ( x z + y z ) dS , where S is the surface (hemisphere) x + y + z = 4, z ≥ S b I = ∫∫ xyz dS , where S is the surface with parametric equations S x = uv, y = u − v, z = u + v, (u , v) ∈ D : u + v ≤ 1, u ≥ v ≥ c I = ∫∫ y dS , where S : x + y + z = a S d I = ∫∫ z dS , where S is the closed surface that contains the cylinder x + y = a and S the planes z = and z = y + a, a > e I = ∫∫ yz dS , where S is the surface with parametric equations S x = uv, y = u + v, z = u − v, u + v ≤ 6.3 Use Stockes’ Theorem to evaluate a ∫∫ curl F ⋅ n dS , where F = x e yz , y e xz , z e xy and S is the hemisphere with equation S x + y + z = 4, z ≥ , oriented upward b The work done by the force field F = x x + z , y y + x , z z + y when a particle moves under its influence around the close edge of the part of the sphere x + y + z = that lies in the first octant, in a counterclockwise direction as viewed from above 6.4 Use the Divergence Theorem to evaluate ∫∫ F ⋅ dS S a F = z x, 13 y + sin z , x z + y S is the upward oriented top half of the sphere x + y + z = b F = z y , x y + sin z , x z + y and S is the outward oriented surface of the cube [−1, 1]3 86 Calculus c – Chapter 6: Surface Integrals F = − z x, z y , z + y Nguyen Van Ho and S is the outward oriented surface of the sphere x2 + y + z = a2 d F = − xy, y , zy and S is the outward oriented surface of the tetrahedron with vertices (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1) e F = xy , x e z , z and S is the outward oriented surface of the solid bounded by the cylinder x + y = and the planes x = −1, and x = f F = x5 , 10 x y , zy and S is the outward oriented surface of the solid bounded by the paraboloid z = − x − y and the plane z = 6.5 Use the Divergence Theorem to evaluate a ∫∫ (2 x + y + z ) dS , S is the sphere x + y + z = a S b ∫∫ S c − x2 + y + 4z x2 y z + + 16 81 625 ∫∫ F ⋅ n dS , F = dS , S is the ellipsoid x2 y z + + = 25 x3 + y , e2 z , y z , S S is the outward oriented surface of the solid bounded by x + y = 4, z = −1, z = 6.6 6.7 6.8 Verify that the Stokes’ Theorem is true for the given vector field F and surface S a F = y, z , − x , S is the part of z = x + y − 9, x + y ≤ 9, oriented downward b F = y, z , x , S is the part of x + y + z = −1, x ≤ 0, y ≤ 0, z ≤ 0, oriented downward Verify that the Divergence Theorem is true for the given vector field F on the domain D a F = xy, y, − xz , D is the cube [0,1]3 b F = x , xy, xz , D is the region bounded by z = x + y 2and z = c F = xy, zy, xz , D is the region bounded by x + y = 4, z = −1 and z = d F = ax, ay, az , a > 0, and D is the ball x + y + z ≤ a Use Stockes’ Theorem to evaluate ∫∫ curl F ⋅ n dS S a F = e yz , y z , z , S is the part of the hemisphere x + y + z = 9, x ≥ 0, that lies x inside the cylinder y + z = , oriented in the direction of the positive x-axis 87 Calculus – Chapter 6: Surface Integrals Nguyen Van Ho F = xyz , xy, x yz , S consists of the top and the four sides but not the bottom of the b cube [0, 1]3 , oriented outward ANSWERS  c + ; b + ln ; 6.1 a 6.2 a I = 16  ; b I= 6.3 a 0; b 16; 6.4 a 13  / 20 (Hint: Add the disk oriented downward) b 16 c 6.5 a 12  a ; b 6.8 a −4  ; b 23 16 69 − 128 ; − = 280 105 840  a5 ; 4560  ; d 1/3; c 144  e c 62  / c 4 a ; 144  ; f d ( ) +  a3  /12 1/2 ; REFERENCES [1] James Stewart, CALCULUS, Fourth edition, Brooks/Cole, 1999 [2] George B Thomas, Jr., CALCULUS, Pearson Education, Inc., 2006 88 A COURSE IN CALCULUS NGUYỄN VĂN HỘ CONTENTS CHAPTER 1.1 1.2 1.3 VECTORS AND GEOMETRY OF SPACE Vectors A Vectors B Dot Product C Cross Product D Scalar Triple Product Equations of Lines and Planes A Equations of Lines B Equations of Planes Cylinders and Quadratic Equations 1.3 3 10 10 11 12 Cylindrical and Spherical Coordinates A Cylindrical Coordinates B Spherical Coordinates Exercises 13 13 14 15 CHAPTER VECTOR FUNCTIONS 18 2.1 Vector Functions and Space Curves 18 2.2 Derivatives of Vector Functions 19 2.3 Integrals of Vector Functions 21 2.4 Arc Length of Space Curves 22 2.5 Curvature 23 2.6 Normal and Binormal Vectors 25 2.7 Motions in Space Velocity and Acceleration 26 Exercises CHAPTER 3.1 3.2 3.3 3.4 3.5 27 DOUBLE INTEGRALS Definitions and Properties Iterated Integrals Change of Variables in Double Integrals Double Integrals in Polar Coordinates Application of Double Integrals Area and Volume Density and Mass Moment and Center of Mass 29 29 31 32 34 35 35 36 37 A COURSE IN CALCULUS NGUYỄN VĂN HỘ Moment of Inertia Surface Area Exercises CHAPTER 4.1 4.2 4.3 4.4 4.5 40 TRIPLE INTEGRALS Definitions and Properties Iterated Integrals Change of Variables in Triple Integrals Triple Integrals in Cylindrical Coordinates Application of Triple Integrals 4.5.1 Volume 4.5.2 Density and Mass 4.5.3 Moments, Center of Mass, and Moment of Inertia Exercises CHAPTER 5.1 LINE INTEGRALS 5.2 Line Integrals of Vector Fields 5.3 The Fundamental Theorem 5.4 Green’s Theorem 5.5 Curl 5.6 Divergence 5.7 Vector Form of Green’s Theorem Exercises 6.1 SURFACE INTEGRALS Surface Integrals 6.2 Surface Integrals of Vector Fields 6.3 Stokes’ Theorem 6.4 The Divergence Theorem Exercises REFERENCES 42 42 43 44 45 46 46 47 47 49 Line Integrals CHAPTER 38 38 50 50 53 56 59 62 65 66 68 70 70 75 81 84 86 88
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