ࡱ>  79456y] RtbjbjVyyf @]zzzz&&&&8^bT&FZ  h"2B4B4B4B4B4B4B$HuJzXB}i^XB0zzZ 000 zl:2B&&zzzz2B0\0a3?,2B>&&i% ANVector Analysis Chapter 1 Algebra of Vectors I. Elementary properties of vectors Consider a Cartesian coordinate system, in which we define unit vectors, i, j, and k having the directions of positive x, y, and z axes, respectively. Any vector v whose projections on these axes are vx, vy, and vz, can be written as v = vxi + vyj + vzk or v = [vx, vy, vz] The number vx, vy, and vz are called the scalar components of v in the x, y, and z directions. If, when the initial point of v coincides with the origin, the angle measured to v from the positive x, y, and z axes are denoted by a, b, and g, respectively, there follows vx = vcosa, vy = vcosb, vz = vcosg, where v = | v | =gQL] Equation.2  is the magnitude or length of v, the number cosa, cosb, and cosg are called the direction cosines of v. It is clearly that the direction cosines satisfy the equation  EMBED MSDraw  cos2a + cos2b + cos2g = 1. Addition of vectors: If a = a1i + a2j + a3k, b = b1i + b2j + b3k, then a + b = (a1 + b1)i + (a2 + b2)j + (a3 + b3)k = b + a ca = ca1i + ca2j + ca3k A pair of nonparallel two vectors in a plane forms a basis of the plane. That is, any vector of the plane can be uniquely determined by a linear combination of the basis. Similarly a basis for three-space is obtained from three noncoplanar vectors. Prove that the diagonals of a parallelogram bisect each other. Solution: Let  EMBED Equation.2 , then  EMBED Equation.3   EMBED Equation.3  m(a + b) = (1 - n)a + nb ( (m + n - 1)a + (m - n)b = 0 Since a and b are independent, m + n - 1 = 0, and m - n = 0. We get m = n = 1/2, and point O is the midpoint of the diagonals.  EMBED MSDraw  If  EMBED Equation.3 .  EMBED Equation.3   EMBED MSDraw  If  EMBED Equation.3 .  EMBED Equation.3  II. Inner product (Dot product or Scalar product) Definition: a ( b = abcosq, where q, 0 ( q ( p, is the angle between a and b. In components a ( b = a1b1 + a2b2 + a3b3 Properties: a ( b = b ( a a ( a = a2 Theorem: The inner product of two nonzero vectors is zero if and only if the two vectors are perpendicular. gQL] MSDraw  Projection of a vector a in the direction of a vector b is p = acosq = a EMBED Equation.2 = a ( eb, where eb is the unit vector in the direction of b. An orthonormal basis for 3-space is {a, b, c}, a = b = c = 1, any vector v in the space can be written as v = (v ( a)a + (v ( b)b + (v ( c)c. Find the straight line which passes through the two points P(x1, y1, z1), Q(x2, y2, z2). Solution: Assume any point in the plane is X(x, y, z), then gQL] Equation.2  ( [(x - x1)i + (y - y1)j + (z - z1)k] = t[(x2 - x1)i + (y2 - y1)j + (z2 - z1)k] gQL] Equation.2  is the equation of the line. Find a plane which passes through the point P(x0, y0, z0) and is normal to the vector N = ai + bj + ck. Solution: Assume any point in the plane is X(x, y, z),  EMBED Equation.3  (ai + bj + ck) ( [(x - x0)i + (y - y0)j + (z - z0)k] = 0 a(x - x0) + b(y - y0) + c(z - z0) = 0 is the equation of the plane. gQL] MSDraw  gQL] MSDraw  Prove that the triangle inscribed in a semicircle is a right triangle. Solution: Let  EMBED Equation.2 , then  EMBED Equation.3  [Exercises] 1. Find all vectors v such that (v ( v)v = 169v and 4i ( v = 3j ( v = k ( v. 2. Find the angle between the planes 2x - y + 2z = 1 and x - y = 2. 3. Find the scalar and vector projections of -7i + 14j + 7k along 2i + 3j - 6k. 4. Find the magnitude of the scalar component of the force vector F = i + 2j + 2k in the direction of the straight line with equations x = y = 2z. [Answers] 1. v = ( 3i ( 4j ( 12k 2. 45( 3. -2, (-4i - 6j + 12k)/7 4. 8/3 III. Outer product (Cross product, Vector product) Definition: The outer product of two vectors a and b is a vector v = a ( b The magnitude of v is v = absinq, 0 ( q ( p, q is the angle between a and b. The direction of v is the direction of advance of a right-hand screw as vector a is turned into vector b, and it is perpendicular to both a and b. In components, a = a1i + a2j + a3k, b = b1i + b2j + b3k, v = v1i + v2j + v3k 5"i ( i = j ( j = k ( k = 0, i ( j = k, j ( k = i, k ( i = j, j ( i = -k, k ( j = -i, i ( k = -j 4"v1 = a2b3 - a3b2, v2 = a3b1 - a1b3, v3 = a1b2 - a2b1 or v = EMBED Equation.3  gQL] MSDraw  Properties: 1. a ( b = -b ( a 2. (a ( b) ( c ( a ( (b ( c) in general 3. If a ( b = 0, (1) then one of the two vectors is zero. (2) a and b has the same or opposite direction. 4. | a ( b | = absinq is the area of the parallelogram formed by the vectors a and b, as shown. gQL] MSDraw  Find the distance d of point O to the line gQL] Equation.2 . Solution: EMBED Equation.3  Find the distance d between two lines which is not at the same plane. gQL] MSDraw  Solution: Suppose the unit vectors along the two lines are e1, e2 respectively, and P, Q are points on the two lines respectively, then d =EMBED Equation.2 \* mergeformat Find the distance of two lines L1: EMBED Equation.3 . Solution: (1)L1veTTϑpv1 = 4i - 3j - k L2veTTϑpv2 = 3i - 4j - 2k -P(4m + 11,-3m - 5,-m - 7)pL1 Nvޞ Q(3n - 5,-4n + 4,-2n + 6)pL2 Nvޞ NPQvݍ►pNjke}vݍ GR  EMBED Equation.3 (v1 = 0 ( [(3n - 4m - 16)i + (-4n + 3m + 9)j + (-2n + m + 13)k] ( (4i - 3j - k) = 0 4(3n - 4m - 16) - 3(-4n + 3m + 9) - (-2n + m + 13) = 0 ( 26n - 26m = 104& & (  EMBED Equation.3 (v2 = 0 ( [(3n - 4m - 16)i + (-4n + 3m + 9)j + (-2n + m + 13)k] ( (3i - 4j - 2k) = 0 3(3n - 4m - 16) - 4(-4n + 3m + 9) - 2(-2n + m + 13) = 0 ( 26n - 29m = 110& & ( ( - (: 3m = -6 ( m = -2 NeQ(_n = 2 m, nNV_P(3, 1, -5), Q(1, -4, 2) Njke}vݍ►p EMBED Equation.3  (2)-S+TL1Ns^LL2vs^bEpa(x - 11) + b(y + 5) + c(z + 7) = 0 GR EvlTϑ ( L1 ( (ai + bj + ck) ( (4i - 3j - k) = 0 ( 4a - 3b - c = 0 EvlTϑ ( L2 ( (ai + bj + ck) ( (3i - 4j - 2k) = 0 ( 3a - 4b - 2c = 0 a: b: c = EMBED Equation.3 = 2: 5: (-7) s^bEp2(x - 11) + 5(y + 5) - 7(z + 7) = 0 ( 2x + 5y - 7z - 46 = 0 L2NޞA(-5, 4, 6) GRA0Rs^bEvݍsSpiQjke}vݍ vQ 0. Solution: r = rer ( EMBED Equation.2 = asinq er + r eq = asinq er + a(1 - cosq) eq EMBED Equation.2 \* mergeformat 2. Frenet formula We have a unit tangent vector T = EMBED Equation.2 , and define  EMBED Equation.2 = kN, k ( 0 (2.2) where k is the magnitude of  EMBED Equation.2  and is called the curvature, and N is a unit vector normal to T and is called the normal or principal normal to the curve. The reciprocal of the curvature, r = EMBED Equation.2 , is called the radius of curvature. In addition to the two perpendicular vector T and N, a third unit vector known as the binormal vector, is definede by B = T ( N then  EMBED Equation.2  Since  EMBED Equation.2  is perpendicular to B, and from the right hand side of the above equation,  EMBED Equation.2  is perpendicular to T,  EMBED Equation.2  can be written in the form  EMBED Equation.2 = -tN (2.3) where t is the torsion or twisting of the curve.  EMBED Equation.2 = -tN ( T + B ( kN = -kT + tB (2.4) The three formulas, equations (2.2), (2.3), and (2.4) are called Frenet formulas. In general, the position vector r is a function of t, to find k and t, we differentiate r with respect to t:  EMBED Equation.3  Given a space curve r(t) = a cost i + a sint j + ct k, find (1)curvature (2)torsion (3)unit tangent, normal, and binormal vectors.  EMBED Equation.3  Substituting into a, we have N = -costi - sintj  EMBED Equation.3  Show that for a curve y = y(x) in the x-y plane, k(x) = EMBED Equation.2  Solution: r = x i + y j v = i + y( j = v T, ( v = EMBED Equation.2  a = y(j = EMBED Equation.2 T + kv2N v ( a = kv3 B ( y(k = kv3 B ( k = EMBED Equation.2  [Exercise] The position vector of a particle P is r(t) = t i + t2 j + 2t3/3 k. Find (1) the velocity, acceleration, and speed of P, (2) the tangential and centripetal components of acceleration, (3) the curvature of the curve C traversed by P, and (4) the minimum radius of curvature of C. [Answer] (1) v = i + 2t j + 2t2 k, a = 2j + 4t k, v = 1 + 2t2 (2) at = 4t, an = 2 (3) k = 2/(1 + 2t2)2 (4) rmin = 1/2 III. Gradient Definition: For a given scalar function f(x, y, z) the gradient of f is the vector function grad f = (f = EMBED Equation.2  where ( = EMBED Equation.2 , read del, is a differential operator. Then  EMBED Equation.2  where T is the unit tangent vector to the curve described by r(s). Since (f ( T = | (f | cosq, where q is the angle between (f and T, then  EMBED Equation.2 = | (f | cosq; so the component of (f in the direction of T is the directional derivative of f in that direction. It readily follows from this result that (f is in the direction in which  EMBED Equation.2  has its maximum values; this maximum value is equal to | (f |. Consider a surface S in space given by f(x, y, z) = constant. As long as we move along this surface on any curve C, f has constant value and consequently  EMBED Equation.2 = 0. Thus  EMBED Equation.2 ( (f ( T where r(s) is the position vector describing the curve C. Since (f is completely determined by f and C is any curve on the surface, (f must be normal to the surface. Find a unit normal vector n of the cone of revolution z2 = 4(x2 + y2) at point P(1, 0, 2). Solution: The cone is the level surface f = 0 of f(x, y, z) = 4(x2 + y2) - z2, Thus (f = 8x i + 8y j - 2z k and at P (f|(1, 0, 2)= 8i - 4k n = EMBED Equation.2  Find the directional derivative of f(x, y, z) = 4x2y + 3y2 + 2xz2 at the point (1, -1, 0) in the direction of the vector b = -3i + 4j. Solution: (f = (8xy + 2z2) i + (4x2 + 6y) j + 4xz k (f|(1, -1, 0)= -8i - 2j eb = EMBED Equation.2  (f ( eb = 16/5 [Exercise] The temperature at any point in space is given by T = xy + yz + zx. (1) Find the direction cosines of the direction in which the temperature changes most rapidly with distance from the point (1, 1, 1), and determine the maximum rate of change. (2) Find the derivative of T in the direction of the vector 3i - 4k at the point (1, 1, 1). [Answer] (1)((i + j + k)/ EMBED Equation.3  (2)-2/5 IV. Divergence of a vector field Definition: Let v(x, y, z) be a differentiable vector function and v = v1i + v2j + v3k. Then the function div v = (( v = EMBED Equation.2  is called the divergence of v. For a small rectangular solid with edges (x, (y, and (z. In the y direction the outflow of v through the right-hand face is v2(x, y + (y, z)(x(z, through the left-hand face is -v2(x, y, z) (x(z, the total outflow in the y direction is the sum [v2(x, y + (y, z) - v2(x, y, z)](x(z The difference in these values of v2 is  EMBED Equation.2 (y Thus the net outflow from the two faces is  EMBED Equation.2 (x(y(z Similarly, the two faces in the x and z direction contribute respectively  EMBED Equation.2 (x(y(z and  EMBED Equation.2 (x(y(z After we divide by the volume (x(y(z, we have (( v = EMBED Equation.2  as the net outflow per unit volume. If F = x i + y2z j + xz3 k, find (( F. Solution: (( F = 1 + 2yz + 3xz2 Find (( r, given r = xi + yj + zk. Solution: (( r = 1 + 1 + 1 = 3 V. Curl of a vector field Definition: Let v(x, y, z) = v1i + v2j + v3k be a differentiable vector function. Then the function curl v = ( ( v = EMBED Equation.2  is called the curl of v or the rotation of v. Let w = w1i + w2j + w3k represent the angular velocity of a rigid body, the velocity field is v = w ( r = (w1i + w2j + w3k) ( (xi + yj + zk). The curl of v is ( ( v = EMBED Equation.2 = 2w Hence, the curl of the velocity field in the case of the rotation of a rigid body has the direction of the axis of rotation, and its magnitude equals twice the angular speed w of the rotation. VI. Vector identities ((f1f2) = f1(f2 + f2(f1 ( ( (fF) = (f ( F + f( ( F ( ( (fF) = (f ( F + f( ( F (f(u) = EMBED Equation.2 (u ( ( (F ( G) = G ( (( ( F) - F ( (( ( G) ( ( (F ( G) = (G ( ()F - (F ( ()G + (( ( G)F - (( ( F)G ((F ( G) = (F ( ()G + (G ( ()F + F ( (( ( G) + G ( (( ( F) ( ( (( ( F) = ((( ( F) - (2F ( ( r = 3 ( ( r = 0 ( ( ((f) = 0 ( ( (( ( F) = 0 ( ( ((f1 ( (f2) = 0 Tensor notations: a ( b = ai bi a ( b = eijk aj bk (f = (i f (2f = ( ( ((f) = (i (i f ( ( F = (i Fi ( ( F = eijk (j Fk where eijk = EMBED Equation.2  and eijkeips = djp dks - djs dkp where dij = EMBED Equation.2  Prove (A ( B) ( C = (A ( C)B - (B ( C)A. Solution: (A ( B) ( C = eijk (A ( B)j Ck = eijk ejlm Al Bm Ck = ejki ejlm Al Bm Ck = (dkl dim - dkm dil) Al Bm Ck = Ak Bi Ck - Ai Bk Ck = (A ( C)B - (B ( C)A Prove ( ( (F ( G) = (G ( ()F - (F ( ()G + (( ( G)F - (( ( F)G. Solution: ( ( (F ( G) = eijk (j (F ( G)k = eijk eklm (j (Fl Gm) = ekij eklm (j (Fl Gm) = (dil djm - dim djl) (j (Fl Gm) = (j (Fi Gj) - (j (Fj Gi) = (j Fi Gj + Fi (j Gj - (j Fj Gi - Fj (j Gi = (G ( ()F + (( ( G)F - (( ( F)G - (F ( ()G Prove ( ( ((f) = 0. Solution: ( ( ((f) = eijk (j ((f)k = eijk (j (k f Since (j (k f = (k (j f and eijk = -eikj ( eijk (j (k f = 0 Prove (v ( ()v = EMBED Equation.2 (|v|2 - v ( (( ( v). Solution: v ( (( ( v) =eijk vj (( ( v)k = eijk vj (eklm (l vm) = ekij eklm vj (l vm = (dil djm - dim djl)vj (l vm = vj (i vj - vj (j vi = EMBED Equation.2 (i (vj vj) - vj (j vi = EMBED Equation.2 ((v ( v) - (v ( ()v = EMBED Equation.2 (|v|2 - (v ( ()v ( (v( ()v =  EMBED Equation.2 (|v|2 - v ( (( ( v) [Exercise] Show that ((v ( v) = 2v ( (v + 2v ( (( ( v). Chapter 3 Vector Integral Calculus I. Line integrals 1. Definition A line integral of a vector function F(r) over a curve C is  EMBED Equation.3  In terms of components  EMBED Equation.3  If x, y, and z are function of t, we have  EMBED Equation.3  Find the value of the line integral  EMBED Equation.3 , when F = -yi + xyj and C is the circular arc as shown from A to B. Solution: We may represent C by r = costi + sintj 0 ( t ( p/2 and F = -sinti + costsintj  EMBED Equation.3  Find the value of the line integral  EMBED Equation.3 , when f = x2 + x3y and C is the line y = 2x from (0, 0) to (2, 4). Solution: We may represent C by r = xi + yj ( dr = dxi + dyj ds = | dr | = EMBED Equation.3  and f = x2 + x3y = x2 + x3(2x) = x2 + 2x4  EMBED Equation.3  [Exercise] Evaluate the line integral  EMBED Equation.3 , with F = 5zi + xyj + x2zk from point A(0, 0, 0) to point B(1, 1, 1) along (1) C: the straight line r = ti + tj + tk. (2) C: the parabolic arc r = ti + tj + t2k. [Solution](1) F = 5ti + t2j + t3k, dr = (i + j + k)dt  EMBED Equation.3 = 37/12 (2) F = 5t2i + t2j + t4k, dr = (i + j + 2tk)dt  EMBED Equation.3 =7/3 2. Conservative fields A vector field F is said to be conservative if there can be found some scalar f such that F = (f, i.e.,  EMBED Equation.3 . Then f is called a potential function or simply potential for F, and F ( dr = (f ( dr = EMBED Equation.3 = df The line integral from A to B along a curve C is  EMBED Equation.3 = f(B) - f(A) This shows that the value of line integral is simply the difference of the value of f at the two endpoints of C and is independent of the path C. Hence the line integral around a closed curve of a conservative field is zero. Since ( ( ((f) = 0 ( ( ( F = 0 There follows a fact that if and only if ( ( F = 0, the vector field F is conservative. Find the line integral of  EMBED Equation.3 , with F = 2xi + 2yj + 4zk from point A(0, 0, 0) to point B(2, 2, 2) along C: the straight line r = ti + tj + tk. Solution: 5" ( ( F = 0 4"the line integral is independent of the path, and the potential is  EMBED Equation.3  f = x2 + y2 + 2z2 and  EMBED Equation.3 = f(2, 2, 2) - f(0, 0, 0) = 16  EMBED MSDraw  II. Surface integrals Definition: the flux of F through the surface S to be the surface integral  EMBED Equation.3 , where n is the unit normal vector to S. From the figure shown, ds | cosg | = dxdy, g is the angle between the normal of ds and z-axis. n ( k = cosg  EMBED Equation.3  Similarly, if a and b are the angle between the normal of ds and x and y axes respectively,  EMBED Equation.3   EMBED MSDraw  When the surface S is parametrized by r(u,v)  EMBED Equation.3  If F = n, the surface integral is area of the surface itself.  EMBED MSDraw  Given F = x2i + 3y2k and S is the portion of the plane x + y + z = 1 in the first octant, evaluate  EMBED Equation.3 . Solution: Let f = x + y + z - 1 the unit normal vector n of S is  EMBED Equation.3  Calculate the surface integral of the vector function F = xi + yj over the portion of the surface of the unit sphere S: x2 + y2 + z2 = 1 above the xy-plane, z ( 0. Solution: (1)Let f = x2 + y2 + z2, the unit normal vector of S is  EMBED Equation.3  Let x = rcosq, y = rsinq, we have  EMBED Equation.2  (2)Let x = sinqcosf, y = sinqsinf, z = cosq, then r = sinqcosfi + sinqsinfj +cosqk, F = sinqcosfi + sinqsinfj, and  EMBED Equation.3  III. Volume integral Definition: the volume integral of a function f over the volume V is  EMBED Equation.3  In Cartesian coordinate system, dv = dxdydz. If f = 1, the volume integral is volume of V. Find the volume integral of f(x, y, z) = x + yz over the box bounded by the coordinate planes, x = 1, y = 2, and z = 1 + x.  EMBED MSDraw   EMBED Equation.3   EMBED MSDraw  Find the volume of the region of space above the xy plane and beneath the plane z = 2 + x + y, bounded by the planes y = 0, x = 0, and the surface y = 1 - x2.  EMBED Equation.3  IV. Divergence theorem (Gauss theorem) Divergence theorem: Let V be a closed bounded region in space whose boundary is a piecewise smooth oriented surface S. Let F be a vector function that is continuous first partial derivative in some domain containing V. Then  EMBED Equation.3   EMBED Equation.3  Solution: (1) Let f = x2 + y2 + z  EMBED Equation.3  [Exercises] 1. Evaluate  EMBED Equation.3 , S is the lateral surface of the portion of the cylinder x2 + y2 = 1 for which 0 ( z ( 1. [Solution] (1) Let f = x2 + y2 - 1, then n = xi + yj  EMBED Equation.3  2. Evaluate EMBED Equation.3 , with S: x2 + y2 + z2 = 4. [Solution] (1)n = (xi + yj + zk)/2  EMBED Equation.3  (2)( ( (7xi - zk) = 7 - 1 = 6  EMBED Equation.3  V. Stokes theorem Stokes theorem: Let S be a piecewise smooth oriented surface in space and let the boundary of S be a piecewise smooth simple closed curve C. Let F be a continuous vector function that has continuous first partial derivative in a domain in space containing S. Then  EMBED Equation.3  [Note] The positive direction along C is defined as the direction along which an observer, traveling on the positive side of S, would proceed in keeping the enclosed area to his left. Find the surface integral  EMBED Equation.3 , with F = yi + zj + xk and S: the paraboloid z = 1 - (x2 + y2), z ( 0.  EMBED Equation.3   EMBED Equation.3  Evaluate  EMBED Equation.3 , where C is the circle x2 + y2 = 4, z = -3, oriented counterclockwise as seen by a person standing at the origin, and F = yi + xz3j - zy3k. Solution: (1) Let x = 2cosq, y = 2sinq F ( dr = (2sinqi - 54cosqj + 24sin3qk) ( (-2sinqi + 2cosqj)dq = (-4sin2q - 108cos2q)dq  EMBED Equation.3 = -112p (2) As surface S bounded by C we can take the plane circular disk x2 + y2 ( 4 in the plane z = -3. Then n = k, and ( ( F| z=-3 = (9y2 - 27x)i - 28k, (( ( F) ( n = -28  EMBED Equation.3 = -28(p ( 22) = -112p VI. Green s theorem If F = F1i + F2j is a vector function that is continuously differentiable in a domain in the xy-plane containing a simply connected bounded close region S whose boundary C is a piecewise smooth simple closed curve. Then the Stokes theorem can be reduced to  EMBED Equation.3  which is the Greens theorem. Evaluate  EMBED Equation.3 , C is a closed curve from (0, 0) to (1, 1) along y = x2 and back to (0, 0) along y = x. Solution:  EMBED MSDraw  (1)  EMBED Equation.2  (2)  EMBED Equation.2  Evaluate the line integral  EMBED Equation.2 F(dr counterclockwise around the boundary C of the region R, where F = xsinyi - ysinxj, R is the rectangle: 0 ( x ( p, 0 ( y ( p/2. Solution: (1) On C1: y = 0, dy = 0, F = 0( EMBED Equation.2 ; On C2: x = p, dx = 0, F = psinyi ( EMBED Equation.2 ; On C3: y = p/2, dy = 0, F = xi - EMBED Equation.2 ; On C4: x = 0, dx = 0, F = 0 ( EMBED Equation.2 .  EMBED Equation.2 .  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Times New Roman4- 2 ( 2 3cos 2  ) 2 `( 2 sini 2 %)Symbol- 2 "p 2 0 p 2 p 2 p 2 0p 2 &pTimes New Roman4- 2  x 2 Zx 2 ^dx 2 dx &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39q@I^I  1 2 CompObj|fObjInfo~Equation Native \_975248073Fޢޢ+(2cos2x) 2 +(2sin2x) 2 dx 01 +" = 1+4 2 dx 01 +" = 1+4 2L EllOle PIC LMETA @CompObj f E  .1   &  & MathType "-@VE  Times New Roman%- 2 n]d 2 Td 2 ndr 2 d 2 `ar 2 dY d 2  d Times New Roman9- 2 WrW 2  rWTimes New Romanp>- 2 nr 2 `e 2 d eSymbol- 2 q 2 |q 2 d q 2 `b= 2 `?+ & "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo!Equation Native _974834646$F`ߢ`ߢOle r΀tI0kI drd=drde r +rde r dL66  .1  @1&@1PIC #&LMETA CompObj%'fObjInfo( & MathType "-Ct  "-t}-}!!-6F-6N!(2(-2(}y(-}(((/Times New Roman-- 2 ,( 2 sini 2 ) 2  [ 2 i ( 2  cos 2 )] 2 cos 2 )( 2 -,cos 2 4/)Times New Roman-- 2 a 2  a 2 d 2 a 2 8a 2 !d 2  %a 2 /dSymbol- 2 q 2 q 2 Wq 2 !q 2 "q 2 \.q 2 ~0q Symbol- 2 upw 2 pw 2 3'pw Times New Roman-- 2 t2p 2 t2p 2 0p 2 2p 2 td2p 2 t2p 2 G0p 2 2p 2 z&0p 2 &2pTimes New Roman-- 2  1 2 2 2 w2 2 (2 2 *1Symbol- 2  + 2  - 2 = 2 P- 2 #= 2 +-Symbol- 2 ? 2  2 % &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native _9752480900+F⢜⢜Ole PIC *-Ld(fI@jI s= (asin) 2 +[a(1"cos)] 2 d 02 +" = 2a 2 "2a 2 cos  d 02 +" =a 2(1"cos)  d 02 +" =a2sin2d 02 +" =4a("cos2) 02 =8aLWETlWE  .1   & & MathType "-@META CompObj,.fObjInfo/Equation Native @Times New Roman- 2 nTd 2 fdsTimes New Roman9- 2 nr & "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qr$̩II drds _9752480972F墜墜Ole PIC 14LMETA LElE  .1  &@ & MathType "-@$Times New Roman- 2 nTd 2 dsTimes New Romany- 2 nT & "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qr$0II dTdsLElCompObj35fObjInfo6Equation Native @_975248108)L9F0碜0碜Ole PIC 8;LMETA CompObj:<jE  .1  &@ & MathType "-@$Times New Roman- 2 nTd 2 dsTimes New Romany- 2 nT & "Systems- FMicrosoft {s边 3.0 DS Equation Equation.39q (yJ}JyJ dTdsLElE  .ObjInfo=Equation Native <_975248117@F=ꢜ=ꢜOle PIC ?BLMETA CompObjACfObjInfoD1  &` & MathType "-@JTimes New Roman- 2 pe1Symbol- 2 Tk &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39q     !#$%&'(*/1234568=?@ABCDFKMNOPQRSTUVWXYZ[]`adghijklmnopqrstuvwxyz}rxI|I 1LElE   .1  &` & MathType "-@Equation Native 8_1006978747GFKKOle PIC FILMETA HCompObjHJfObjInfoKEquation Native 0b  95YUTimes New Roman1- 2 nTd 2 ds 2 n4d 2 ds 2 n d 2 ( ds 2 nMd 2 ds 2 nmd 2 dsTimes New RomanA- 2 nB 2 `(T 2 `N 2 n T 2 `<N 2 `T 2 n N 2 `T 2 n-NSymbol- 2 `= 2 `r 2 ` = 2 ` 2 `+ 2 `  2 `= 2 `-Times New Roman1- 2 `( 2 ` ) & "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qwII dBds=dds(TN)=dTdsN+TdNds=NN+TdNds=TdNds_975248138>aNF`.`.Ole PIC MP LMETA "LElE  .1  &@ & MathType "-@0Times New Roman-- 2 nTd 2 dsTimes New Roman-- 2 nB & "Systems-N FMicrosoft Equation 3.0 DS Equation Equation.39qr$II dBdsxLElCompObjOQ)fObjInfoR+Equation Native ,@_975165244Z~UFOle -PIC TW.LMETA 0CompObjVX7fE  .1  &@ & MathType "-@0Times New Roman-- 2 nTd 2 dsTimes New Roman-- 2 nB & "Systems-N FMicrosoft Equation 3.0 DS Equation Equation.39q$,fIDjI dBds LElE  .1  &@ObjInfoY9Equation Native :@_975165231\FOle ;PIC [^<LMETA >CompObj]_EfObjInfo`G & MathType "-@0Times New Roman-- 2 nTd 2 dsTimes New Roman-- 2 nB & "Systems-N FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native H@_975248154cF;;Ole IPIC beJL$,fIDjI dBds LE lE  .1  `&  & MathType "-@<n  9META LCompObjdf\fObjInfog^Equation Native _Times New Roman=- 2 nTd 2 ds 2 n@d 2  ds 2 n d 2 . ds 2 nMd 2 dsTimes New RomanET- 2 nN 2 `4B 2 `T 2 n B 2 `HT 2 `B 2 n TSymbol- 2 `= 2 ` 2 ` = 2 `( 2 `+ 2 ` Times New Roman=- 2 `( 2 ` ) & "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qrΜtI0kI dNds=dds(BT)=dBdsT+BdTds FMicrosoft Equation 3.0 DS Equation Equation.39q@kI$fI "r=drdt=drdsdsdt_975161763tSjF@@Ole bCompObjikcfObjInfoleEquation Native f_1268386564oFH H Ole {CompObjnp|f=vT"r=2vT+vdTdt=2vT+vdTdsdsdt=2vT+v 2 N"r=2vT+2vdTdt+ddt(v 2 )N+v 2 dNdt=2vT+2vvN+ddt(v 2 )N+v 3 ("T+B)=(2v" 2 v 3 )T+[v2v+ddt(v 2 )]N+v 3 B"r"r=v 3 B!=|"r"r||"r| 3 (2.5)("r"r)""r= 2 v 6 !=("r"r"r)|"r"r| 2 FMicrosoft Equation 3.0 DS Equation Equation.39q SolutiObjInfoq~Equation Native _1268386782mtF  Ole on:v="r="asinti+acostj+ck=vT!v= a 2 +c 2 ,2v=0T=vv="asint a 2 +c 2 i+acost a 2 +c 2 j+c a 2 +c 2 ka="r="acosti"asintj=2vT+v 2 N=(a 2 +c 2 )N,"r=asinti"acostj"r"r=ijk"asintacostc"acost"asint0=v 3 B!acsinti"accostj+a 2 k=v 3 Bv 3 = a 2 c 2 +a 4 =a a 2 +c 2 !=aa 2 +c 2 FMicrosoft Equation 3.0 DS EqCompObjsufObjInfovEquation Native _975165786yF V  V uation Equation.39q  =("r"r"r)|"r"r| 2 =a 2 ca 2 (a 2 +c 2 )=ca 2 +c 2 B=TN=1 a 2 +c 2 ijk"asintacostc"cost"sint0=1 a 2 +c 2 (csinti"ccostj+ak)L` Ole PIC x{LMETA CompObjz|j  .1  &< & MathType "-DD@@@@ @$Symbol- 2 c 2 c 2 + 2 c 2 w c 2 B =Times New RomanA- 2 y 2 y 2  y 2  dy 2  dxTimes New RomanD- 2 T( 2 ) 2 ( 2 6) Times New RomanA- 2 $/=Times New RomanD- 2 1 Times New RomanA- 2 $G2p 2 $3p 2 $V2p & "Systems- FMicrosoft {s边 3.0 DS Equation Equation.39qׄ yJ}JpyJ 2yab(1+2y 2 ) 3/2 (2y=dydx)L3ObjInfo}Equation Native _975165798wSF V Ole PIC LMETA CompObjjObjInfo3 J .1  &`d & MathType` "-Hy  "-yd-dd&d&dZTimes New Roman-- 2  1 Times New RomanE- 2 T2pSymbol- 2  + 2 !cTimes New Roman,- 2 Sy & "Systems- FMicrosoft {s边 3.0 DS Equation Equation.39q, ~JpyJ~J  1+2y 2Equation Native H_936423460FOle PIC LL``  .1  @&` & MathType MT Extrap- 2 Mi&Times New Roman!META hCompObjmObjInfoEquation Native ,- 2 `@v &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29q ~JpyJ~J 2vLC _9752481777F`c`cOle PIC LMETA C   .1  ` &  & MathType "-NhhNMM@@@yNNuu@3@ Symbol- 2 c 2 c 2 = 2 c 2 c 2 + 2 cTimes New Roman-- 2 y 2 v 2 y 2 y Times New Roman1- 2 3p 2 92p 2 j 3p 2 ) 2pTimes New Roman-- 2 1 2 @( 2 ) Times New Roman1- 2  /= &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qrΜII 2yv 3 =2y(1+2y 2 ) 3/2CompObjfObjInfoEquation Native _1268398849FOle PIC LMETA 4CompObjfX@?L] ]   .1  @ &  & MathType "-@Dko Symbol- 2 dW 2 { 2 d 2  2 d 2    #%&'()*+,-./0124789:=?@ABCEJLMNOPRWYZ[\]_dfghijlopsuvwxyz{|}~Times New Roman- 2 ddfi 2 Xx 2 dfi 2 y 2 d fi 2  zTimes New Roman)- 2 `ii 2 `j 2 ` kSymbol- 2 `=+ 2 `+ & "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39q  "f"xi+"f"yj+"f"zkLg ObjInfo Equation Native  _936424165F@@Ole  PIC LMETA CompObjmObjInfog  { .1  @@ &  & MathType "-@#6 Symbol- 2 d 2 W 2 d 2 : 2 d 2 MTimes New Roman)- 2 4x 2 #y 2  zTimes New Roman,- 2 `8ii 2 `?j 2 ` kSymbol- 2 `+ 2 `+ & "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29q`(yJ}JyJ xiEquation Native |_975248271F}}Ole !PIC "L+yj+zkLA| A  .1  @`&  & MathType "-@v  META $CompObj3fObjInfo5Equation Native 68-  T8Times New Roman<- 2 nTdfi 2 rds 2 dfi 2 x 2 ndx 2 ds 2 d fi 2  y 2 nA dy 2 S ds 2 dXfi 2 Rz 2 ndz 2 dsSymbol- 2 `h= 2 `+ 2 `& + 2 d 2  2 d  2 6  2 dk 2  & "Systems-o# FMicrosoft Equation 3.0 DS Equation Equation.39qr ^I8I dfds="f"xdxds+"f"ydyds+"f"zdzds="f"drds="f"T.L{Ehl{E  .1  @&_975248326F}}Ole ;PIC <LMETA >h & MathType "-@Times New RomanE- 2 p`dfi 2 xds &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qr$ vI|I dfdsCompObjDfObjInfoFEquation Native G@_975248333FOle HPIC ILMETA KhCompObjQjL{Ehl{E  .1  @& & MathType "-@Times New RomanE- 2 p`dfi 2 xds &  "Systems- FMicrosoft {s边 3.0 DS Equation Equation.39q yJ}JxyJ dfdsL{EhlObjInfoSEquation Native T<_936425182F00Ole UPIC VLMETA XhCompObj^mObjInfo`{E  .1  @& & MathType "-@Times New RomanE- 2 p`dfi 2 xds &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29q yJ}JxyJ dfdsL{Ehl{E  .1  @& & MathType "-@Equation Native a<_1268398454F88Ole bPIC cLMETA ehCompObjkfObjInfomEquation Native nTimes New RomanE- 2 p`dfi 2 xds &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39q  dfds="f"drds=0!"f"drdsLH C .1  @ & & MathType0 "-hh@T_1082930219FppOle qPIC rLMETA t  "- -||v#-#<j-<r0-0<w-<Symbol- 2  2  2 0= 2  - 2 `I + 2 ` - 2 = 2 -Times New RomanX- 2 fi 2 fi Times New RomanI- 2 (J 2 ,8 2 ,8 2 F)JTimes New RomanX- 2 `k ( 2 ` ) Times New RomanI- 2 `1p 2  0p 2 2p 2 S 2p 2 72pTimes New RomanX- 2  8 2  4 2 `8 2 ` 4 2 ^2 2 ,5 2 q1 2 ,5Times New RomanI- 2  ii 2 z k 2 ii 2 k & "Systems- FMicrosoft Equation 3.0 DS EqCompObjfObjInfoEquation Native ,_936426463Fuation Equation.39q zI I "f"f (1,0,2) =8i"4k 8 2 +("4) 2 =2 5  i"1 5  kOle PIC LMETA CompObjmLy y  \ .1  ` &@  & MathType "-Jh hJn n@m=cTimes New Roman- 2 pb 2 b 2 `ii 2 `jSymbol- 2 `= 2 `P- 2 `N+Times New Roman- 2 p3 2 z5 2 p4 2 5 &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qObjInfoEquation Native p_1007069837FPS!PS!Ole TtzJuJzJ bbab=-35i+45j FMicrosoft Equation 3.0 DS Equation Equation.39q0hmI`xI  3  ;2 3 CompObjfObjInfoEquation Native L_975248642F`$`$Ole PIC LMETA 4CompObjjL @   .1  @ &  & MathType "-@G Symbol- 2 nW 2  2 n 2 + 2 n 2 Times New Roman9- 2 nv 2 Fx 2 nv 2 y 2 ncv 2 z Times New RomanDg- 2 1p 2 \2p 2  3pSymbol- 2 `+ 2 `i+ &  "Systems- FMicrosoft {s边 3.0 DS Equation Equation.39qׄtzJuJzJ v 1 x+v 2 y+v 3 zLObjInfoEquation Native _936452423Fm'm'Ole PIC LMETA CompObjmObjInfo  .1  @& & MathType "-@nSymbol- 2 nW 2 Times New Roman - 2 nv 2 hy Times New Roman-- 2 2p &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qh4 ~JpyJ~J v 2 yEquation Native P_936452522Fp(p(Ole PIC LL  .1  @& & MathType "-@nSymbol- 2 nW 2 Times New Roman - 2 nv 2 hy META CompObjmObjInfoEquation Native PTimes New Roman-- 2 2p &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qh4 ~JpyJ~J v 2 y_936452610F,,Ole PIC LMETA L  .1  @&` & MathType "-@ASymbol- 2 qW 2 Times New Romana- 2 qv 2 Ex Times New Roman-- 2 1p &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qh4yJ}JxyJ v 1 CompObjmObjInfoEquation Native P_936452621$F//xL  .1  @&` & MathType "-@WSymbol- 2 qW Ole PIC LMETA CompObjm  "$%&'()*+,-./02567:<=>?@ABCDEFGHIJKLMORSTUXZ[\]^`eghijklmnopqrstuvwxyz{|}~2 Times New Roman-- 2 qv 2 Yz Times New Roman- 2 3p &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qObjInfoEquation Native P_936452736F00Ole  h4yJ}JxyJ v 3 zL @   .1  @ &  & MathType "-@PIC  LMETA  4CompObjmObjInfoG Symbol- 2 nW 2  2 n 2 + 2 n 2 Times New Roman9- 2 nv 2 Fx 2 nv 2 y 2 ncv 2 z Times New RomanDg- 2 1p 2 \2p 2  3pSymbol- 2 `+ 2 `i+ &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qׄtzJuJzJ v 1 x+v 2 y+v 3 zEquation Native _1268399818FТ3Т3Ole  PIC !LL  x(    .1  @& & MathType "- v   :  2XHHXXXTimes New RomanMETA #xCompObj1fObjInfo3Equation Native 4- 2 ii 2 6j 2 kSymbol- 2  2  2   2  2  2 Times New Roman- 2 Jx 2 ry 2 z 2 hv 2 hv 2 hv Times New RomanE- 2 R1p 2 p2p 2 3p & "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39q0 Times New Roman~i~j~k""x""y""zv 1 v 2 v 3 _1267966013 F66Ole 8PIC   9LMETA ;L7 L(7  W .1  @& & MathType "-   m  ]  O XHHXzzTimes New RomanY- 2 eii 2 5 j 2 kSymbol- 2 > 2  2   2   2  2 f 2 hnw 2 h?w 2 hOw 2 h6 w 2 hw 2 hwTimes New RomanY- 2 x 2  y 2 _z 2 h6z 2 hy 2 h x 2 h z 2 hy 2 hx Times New Roman- 2 2p 2 n3p 2 ~ 3p 2 W 1p 2 1p 2 2pSymbol- 2 h"- 2 h - 2 h- & "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj  NfObjInfoPEquation Native Q>_936539076F6868" ijk""x""y""z 2 z" 3 y 3 x" 1 z 1 y" 2 xL{EhlOle VPIC WLMETA YhCompObj_m{E  .1  @& & MathType "-@Times New Roman1- 2 p`dfi 2 edu &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29q 0J}JJ dfduLqp 8q 7 .1  &@ObjInfoaEquation Native b<_1082930295FC;C;Ole cPIC dLMETA fxCompObjfObjInfo & MathTypeTimes New Roman- 2 1 2 J 1 2 F 2 2 f 3 2 R2 2 r3 2 b1 2 3 2 1 2 2 2 1 2 J 1 2 F 3 2 N 2 2 R2 2 Z1 2 V3 2 3 2 2 2 1 2 0 2  ( 2 ,` 2 P,` 2 ) 2  ( 2  ,` 2  ,` 2  ) 2  ( 2 ,` 2  ,` 2 )( 2 ,` 2 ,` 2 ) 2  ( 2 ,` 2 P,` 2 ) 2  ( 2  ,` 2  ,` 2  ) 2  ( 2 ,` 2 ,` 2 )( 2 ,` 2 ,` 2 )Times New Roman- 2 ii 2 ji 2 k 2 ii 2 ji 2 kSymbol- 2 = 2 - 2 = 2 ; 2 ; 2 ; 2 f; 2 z;Times New Roman-2   otherwiseii & "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39q ׼DII 1,(i,j,k)=(1,2,3),(2,3,1),(3,1,2)"1,(i,j,k)=(1,3,2)Equation Native _936540778F0Q>0Q>Ole PIC !L,(2,1,3),(3,2,1)0,otherwise{ L   4 .1   &@META pCompObj "mObjInfo#Equation Native l & MathTypeTimes New Romang- 2 1 2 0Times New Roman- 2 ii 2 gjiSymbol- 2 = 2 ; 2 ; 2 &;Times New Romang-2  otherwiseii & "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qVP(yJ}JyJ 1i=j0otherwise{LEl_936541336&F??Ole PIC %(LMETA dE  .1  &@ & MathType "-@#Times New Roman- 2 pQ1 2 V2 &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qCompObj')mObjInfo*Equation Native 4_936550828-FBB0J}JJ 12LElE  .1  &@ & MathType "-@Ole PIC ,/LMETA dCompObj.0m#Times New Roman- 2 pQ1 2 V2 &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29q0J}JJ 12ObjInfo1Equation Native 4_936550896+@4FPEPEOle PIC 36LMETA dCompObj57mObjInfo8LElE  .1  &@ & MathType "-@#Times New Roman- 2 pQ1 2 V2 &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29q0J}JJ 12LElE  .Equation Native 4_936550981;FxGxGOle PIC :=LMETA dCompObj<>mObjInfo?Equation Native 41  &@ & MathType "-@#Times New Roman- 2 pQ1 2 V2 &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29q0J}JJ 12LElE  .1  &@ & 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"Systems- "-'' FMicrosoft DrawingMSDrawMSDraw9q  ]&"MrEdMicrosoft Draw]&MrEdLCompObjXObjInfoOle10Native_974212196Fqq+ Times New Roman < =&MrEd`xT&  x` h   "- &  & MrEd%p &  & MrEd% &  & MrEd%p@ &  & MrEd%@ &  & MrEd%pq &  & MrEd%@@ &  & MrEd% &  & MrEd%&MrEdYh- YidU&MrEdph hdl&MrEd@ @p<&MrEd@ p<&MrEd@q @Xu|<&MrEdpP Ll &  & MrEd% &  & MrEd%  &  & MrEd%  &  & MrEd%@  &  & MrEd%QEA! &  & MrEd%_8@ &MrEd&Times New Roman{-. & 2 F(y & .&MrEd. & 2 :~z & .&MrEd5. & 2 T7x & .&MrEdTimes New Roman{-. & 2 +dx & .&MrEdSL. & 2 S'Ndy & .&MrEd>Times New Roman{-. & 2 f@k & .&MrEdEy. & 2 {n & .&MrEdi\Symbol-. & 2 i !^g & .&MrEdBTimes New Roman{-. & 2 BC@ds & . &  & MrEd%@ &  & MrEd%@@ &  & MrEd%@h &  & MrEd%$  &  & MrEd% &  & MrEd%@_, &  & MrEd%@^S &  & MrEd%h &  & MrEd%h "System- "-''      !I$I$Ole CompObjfObjInfoEquation Native  FMicrosoft Equation 3.0 DS Equation Equation.39q F"ds S +"+" orF"nds S +"+" FMicrosoft Equation 3.0 DS Eq_974222977FOle CompObjfObjInfouation Equation.39q\ ds=dxdy|n"k|F"nds S +"+" =F"ndxdy|n"k| S +"+" FMicrosoft Equation 3.0 DS EqEquation Native _974223138 F Ole CompObjfuation Equation.39q8,1 F"nds S +"+" =F"ndydz|n"i| S +"+" =F"ndzdx|n"j| S +"+"ObjInfoEquation Native _1082931263F  Ole PIC LMETA  CompObj&XObjInfo(L '   k]&"MrEdMicrosoft Draw]&MrEd+ өQq < =&MrEd& 0  ev       !"#$%'+./0123458:;<=>?@ABCDEFGHIJKLMNOPQRSTVYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~ &MrEd P "--  PP &MrEd0P 0PP0 &  & MrEd% &MrEd@ @@&MrEd> => &  & MrEd%0&MrEdTsTimes New Roman-. & 2 uP & .&MrEd. & 2 Q & .&MrEd+. & 2 +5)R & .&MrEd1|a. & 2 |1zS & .&MrEdTimes New Roman-. &2 ^2(u,v) & .&63&MrEd>#. &2 > %(u+ u,v+ v) & .&MrEd"Symbol-. & 2 i$D & .&MrEd+. & 2 -D & . & &_&MrEdTTimes New Roman-. &2  nV(u,v+ v) & .&MrEdcWSymbol-. & 2 c;YD & . & &IQ&MrEd AHTimes New Roman-. &2   sJ(u,v+ v) & .&MrEd ISymbol-. & 2 1KD & . & "Systems- "-'' FMicrosoft DrawingMSDrawMSDraw9q FMicrosoft Equation 3.0 DS Equation Equation.39q \ /PQ="r"udu,/PS="r"vdvOle10Nativez$_974224427F``Ole )CompObj*f1%& 2  n ]&"MrEdMicrosoft Draw]&MrEd+ Times New Roman < =&MrEdo& 0 o   &MrEd P "--  PL&MrEd0P 0PT4 &  & MrEd% &MrEd@ @<&MrEd> =B{ &  & MrEd%0&MrEdMsTimes New Roman{-. & 2 uP & .&MrEd. & 2 ~Q & .&MrEd+ . & 2 +5)R & .&MrEd1|f. & 2 |1zS & .&_&MrEdW. &2  Y( , + ) & .&MrEdcWSymbol-. & 2 c<YD & .&MrEdRTimes New Roman{-. & 2 *Tu & .&MrEd 5O. & 2 tQv & .&MrEdU. & 2  `Wv & . & &m&MrEdyTimes New Roman{-. &2 yewL( , ) & .&MrEduKTimes New Roman{-. & 2 usu & .&MrEdeu. & 2 ues&v & . & &Oo&MrEdWyTimes New Roman}-. & 2 W( & .&MrEd~Times New Roman}-. & 2 ~Du & .&MrEdTimes New Roman}-. & 2 "+ & .&MrEd4Symbol-. & 2 [D & .&MrEd>|nTimes New Roman}-. & 2 |>zu & .&MrEdnTimes New Roman}-. & 2 n, & .&MrEdxTimes New Roman}-. & 2 xv_v & .&MrEdTimes New Roman}-. & 2 7}+ & .&MrEd}GSymbol-. & 2 }n{D & .&MrEdHwsTimes New Roman}-. & 2 wHu v & .&MrEdx|Times New Roman}-. & 2 |xz0) & . & &Y&MrEd. & 2 *x( & .&MrEdTimes New Roman}-. & 2 Qu & .&MrEd!VTimes New Roman}-. & 2 !+ & .&MrEdZSymbol-. & 2 Z3D & .&MrEdTimes New Roman}-. & 2  mu & .&MrEdTimes New Roman}-. & 2 A, & .&MrEd!Times New Roman}-. & 2 `v & .&MrEd/QTimes New Roman}-. & 2 / ) & . &  "System- "-''''ObjInfo,Equation Native -'_1139171156FOle 6ds=/PQ/PS="r"u"r"vdudvandF"ds S +"+" =F" S +"+" "r"u"r"vdudvLkCPIC 7LMETA 9CompObjUXObjInfoWk z]&"MrEdMicrosoft Draw]&MrEd+ өQq < =&MrEd& 0  N   "- & %% & MrEd% & %l%k & MrEd% & %% & MrEd% &  & MrEd% &  & MrEd% &  & MrEd%&MrEdgTimes New Roman8-. & 2 g ecx & .&MrEdPS{. & 2 SPQy & .&MrEd&. & 2 jz & .&MrEdOTimes New Roman)-. & 2 ?Q1 & .&MrEd =. & 2  w1 & .&MrEdW . & 2 WFU1 & . "Systems- "-'' FMicrosoft DrawingMSDrawMSDraw9qOle10NativeX _974224940FডডOle CompObjf   ]&"MrEdMicrosoft Draw]&MrEd+ өQq < =&MrEd& 0  l   "- &  & MrEd% &  & MrEd% &  & MrEd%&MrEdTTimes New Roman-. & 2 T Rcx & .&MrEdqF. & 2 FqD2y & .&MrEd&. & 2 jz & .&MrEdOTimes New Roman-. & 2 ?Q1 & .&MrEd =. & 2  w1 & .&MrEdW . & 2 WFU1 & . &  & MrEd% "- &  & MrEd% "- &  & MrEd% "- &  & MrEd% "- &  & MrEd% "- &  & MrEd% "- &  & MrEd%  &  & MrEd%  &  & MrEd%N &  & MrEd%N &  & MrEd% &  & MrEd%  "System- "-''MrEd FMicrosoft Equation 3.0 DS Equation Equation.39q9 , F"ds S +"+" FMicrosoft Equation 3.0 DS EqObjInfoEquation Native U_974227648F--Ole CompObjfObjInfoEquation Native {_983430912F::uation Equation.39q_Xl n="f|"f|=1 3  (i+j+k)F"ds S +"+" =F"ndxdy|n"k| S +"+" =x 2 +3y 2  3  dxdy1/ 3 S +"+" =(x 2 +3y 2 )dxdy S +"+" =(x 2 +3y 2 ) 01"x +" dy 01 +" dx=[x 2 (1"x)+(1"x) 3 ] 01 +" dx=13Ole CompObjfObjInfoEquation Native  FMicrosoft Equation 3.0 DS Equation Equation.39q\fIdjI n="f|"f|=2xi+2yj+2zk (2x) 2 +(2y) 2 +(2z) 2 =xi+yj+zkF"ds S +"+" =F"ndxdy|n"k| S +"+" =(x 2 +y 2 )dxdyz S +"+" =4x 2 +y 2  1"(x 2 +y 2 )  dy 0 1"x 2 +" dx 01 +"LE+E+ ; .1  `@'&' & MathType "-@M@ Times New Roman- 2 8_1082932156F`HOle PIC LMETA 4 2 `4 2 1 Times New Roman!- 2 62p 2 x^0p 2 x2p 2 x0p 2 x2p 2 2p 2 x0p 2 x2p 2 x0p 2 x,2pTimes New Roman- 2 ]sini 2 cos 2  cos 2  sini 2 e( 2 cos 2  ) 2 J!sini Times New Roman!- 2 x?/= 2 x/= 2 xg/= 2 x/=Symbol- 2 f 2 f 2  f 2 f 2 f 2 q 2 f 2 =#f 2 $f 2 <&q Symbol- 2 xpw 2 xpw 2 xpw 2 xpwSymbol- 2 ) 2 ) 2 ) 2 )$Symbol- 2 *= 2 -Times New Roman- 2 d 2 Dd 2 #d 2 |%d &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjfObjInfoEquation Native _1269261360eFΪΪ װzII F"ds S +"+" =4r 2  1"r 2 "(x,y)"(r,) 01 +" 0/2 +" drd=4r 2  1"r 2 r 01 +" 0/2 +" drd=4sin 2 cos 0/2 +" 0/2 +" cossindd=4(1"cos 2 )sin 0/2 +" 0/2 +" dd=4"cos+cos 3 3() 0/20/2 +" d=4(1"13) 0/2 +" d=43=2  FMicrosoft Equation 3.0 DS Equation Equation.39qOle CompObjfObjInfoEquation Native e "r""r"=ijkcoscoscossin"sin"sinsinsincos0=sin 2 cosi+sin 2 sinj+sincoskF""r""r"dd S +"+" =(sincosi+sinsinj)" 02 +" 02 +" (sin 2 cosi+sin 2 sinj+sincos)dd=sin 3 dd 02 +" 02 +" =sin(1"cos 2 )dd 02 +" 02 +" ="cos+cos 3 3() 02 d 02 +" =1"13()d 02 +" =23 02 =43 FMicrosoft Equation 3.0 DS Equation Equation.39q_975247000FۭۭOle CompObjfObjInfoEquation Native  X_1082835555F 鰣 鰣Ole  PIC  L                       " % & ' ( ) * + , - . / 0 1 2 3 4 5 6 7 8 9 : ; < = > ? @ A B C D E F G H I L O P Q R S T U V W X Y Z [ \ ] ^ a c d e f g h i j k l m n o p q r s t u v w x y z { | } ~  r<^IoI fdv V +"+"+"L D c D  i]&"MrEdMicrosoft Draw]&MrEd+ Times New Roman < =META  CompObj! XObjInfo# Ole10Native$ d &MrEdn7& 0 n    "- & %p3%pR4 & MrEd%p@p & %@ _%@ " & MrEd%p@@ & %/%/ & MrEd%p@/ &  & MrEd% &  & MrEd%  &  & MrEd%   &  & MrEd%  &  & MrEd%p &  & MrEd% p &  & MrEd%@ &  & MrEd%  &  & MrEd%@&MrEdvTimes New Romanq-. & 2 v7x & .&MrEd,W%. & 2 ,'y & .&MrEd). & 2 U+z & . "Systems- "-'' FMicrosoft DrawingMSDrawMSDraw9q`   ]&"MrEdMicrosoft Draw]&MrEd\+ Times New Roman < =&MrEdq8& 0 q    "- &  & MrEd% &  & MrEd%  &  & MrEd%   &  & MrEd%  "- &  & MrEd%p &  & MrEd% p "- &  & MrEd%@ &  & MrEd%  &  & MrEd%@&MrEdyTimes New Roman{-. & 2 y:x & .&MrEdFq4. & 2 F6y & .&MrEd). & 2 U+z & . "- &  & MrEd%p p@ &  & MrEd%@p@ &  & MrEd%p@ "- &  & MrEd%0 &  & MrEd%p q &  & MrEd%@@ &  & MrEd%@ - &  & MrEd%@U &  & MrEd%p &  & MrEd%_p &  & MrEd%0 &  & MrEd%0 "System- "-''' FMicrosoft Equation 3.0 DS Equation Equation.39qN#H Solution:(x+yz)dz_1272952701FooOle J CompObjK fObjInfoM Equation Native N ?_1082835783F}}Ole _ PIC ` L 01+x +" dx 01 +" dy 02 +" =xz+yz 2 2() 01+x dx 01 +" dy 02 +" =[x(1+x)+y2(1+x) 2 ]dx 01 +" dy 02 +" =(x 2 2+x 3 3)+y2(1+x) 3 3[] 01 dy 02 +" =(56+76y)dy 02 +" =56y+712y 2  02 =4L D c D  ]&"MrEdMicrosoft Draw]&MrEd+ Times New Roman < =&MrEdn7& 0 n  META b h CompObj XObjInfo Ole10Native D   "- & %p3%pR4 & MrEd%p@p & %@ _%@ " & MrEd%p@@ & %/%/ & MrEd%p@/&MrEdvTimes New Roman0-. & 2 v7x & .&MrEd,W%. & 2 ,'y & .&MrEd). & 2 U+z & . &  & MrEd%pP &  & MrEd%pP0P&MrEd0- O/0 &  & MrEd% &  & MrEd%0P0@&MrEd0 ?/0&Ntq &MrEdV|. & 2 |Vzy & .&MrEdTimes New Roman--. & 2 =1 & .&MrEdSymbol-. & 2 O- & .&MrEd!LTimes New Roman,-. & 2 !x & .&MrEdLiTimes New Roman0-. & 2 L2 & . &  "Systems- "-'' FMicrosoft DrawingMSDrawMSDraw9q@   ]&"MrEdMicrosoft Draw]&MrEd+ Times New Roman < =&MrEdna(7&  an    "- & %px%pxR & MrEd%p,px & % @_% @" & MrEd%@ @ & %/%/ & MrEd%/&MrEdvTimes New Roman{-. & 2 v7x & .&MrEd=. & 2 ]?y & .&MrEdi. & 2 igKz & .&MrEdy7- 7;u &  & MrEd%8 &  & MrEd%@&MrEd @&MrEdV . & 2 Vy & .&MrEdTimes New Roman{-. & 2 =1 & .&MrEdSymbol-. & 2 O - & .&MrEd!LTimes New Roman{-. & 2 !x & .&MrEdLiTimes New Roman{-. & 2 L2 & . "- &  & MrEd%ph &  & MrEd%ph8 &  & MrEd%p/p@ &  & MrEd%p@@ &  & MrEd%p@ "System- "-''xxxxxxxxxxxxxx'' FMicrosoft Equation 3.0 DS Equation Equation.39q_1272952832j/F@@Ole  CompObj fObjInfo Nп Solution:dz 02+x+y +" dy 01"x 2 +" dx 01 +" =(2+x+y)dy 01"x 2 +" dx 01 +" =(2+x)y+Equation Native  _1272953732FOle  CompObj fy 2 2[] 01"x 2 dx 01 +" =[(2+x)(1"x 2 )+(1"x 2 ) 2 2]dx 01 +" =(x 4 2"x 3 "3x 2 +x+52)dx 01 +" =x 5 10"x 4 4"x 3 +x 2 2+52x() 01 =110"14"1+12+52=3720 FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo Equation Native  _1271006182F  Ole  N{п F"ds S +"+" =""Fdv V +"+"+" . FMicrosoft Equation 3.0 DS Equation Equation.39qCompObj fObjInfo Equation Native  2_1272980717 F`+`+D Evaluate[xyi+xzj+(1"z"yz)k]"ds S +"+" ,withS:z=1"x 2 "y 2 ,ze"0. FMicrosoft Equation 3.0 DS Equation Equation.39qOle  CompObj  fObjInfo  Equation Native  0pD n=2xi+2yj+k 4x 2 +4y 2 +1  F"nds S +"+" =F"ndxdy|n"k| S +"+" =(2x 2 y+2xyz+1"z"y                        # $ ' * + , - . / 0 1 2 3 4 5 6 7 8 9 : ; < = > ? @ A B E H K N O P Q R S T U V W X Y Z [ \ ] ^ _ ` a b c d e f g h i j k l m n o p q r s v y z { ~ z)dxdy S +"+" =[xyi+xzj+(1"z"yz)k]"2xi+2yj+k 4x 2 +4y 2 +1  dxdy2xi+2yj+k 4x 2 +4y 2 +1  "k S +"+" =(2x 2 y+2xyz+1"z"yz)dxdy S +"+" =[2x 2 y+2xy(1"x 2 "y 2 )+1"(1"x 2 "y 2 )"y(1"x 2 "y 2 )]dxdy S +"+" =[3x 2 y+2xy(1"x 2 "y 2 )+(x 2 +y 2 )"y+y 3 ]dxdy S +"+" =[3r 3 cos 2 sin+2r 2 sincos(1"r 2 )+r 2 "rsin+r 3 sin 3 ]rdrd 01 +" 02 +" =(35 02 +" cos 2 sin+16sincos+14"13sin+15sin 3 )d=("15cos 3 "124cos2+14+13cos"15cos+115cos 3 ) 02 =2(2)F"nds S +"+" =""Fdv V +"+"+" "F"nds S 1  +"+" =("1)dv V +"+"+" "F"("k)dxdy S 1  +"+" ="dzdxdy 01"x 2 "y 2 +" " 1"y 2  1"y 2 +" "11 +" +(1"z"yz) z=0 dxdy S 1  +"+" ="(1"r 2 )rdrd 01 +" 02 +" +="(12"14)d 02 +" +=2 FMicrosoft Equation 3.0 DS Eq_1082836628{|FOle  CompObj  fObjInfo! uation Equation.39q|dII [x 2 i"(1+2x)j+zk]"ds S +"+" FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native " _1144604181F@ģ@ģOle % CompObj& fObjInfo( Equation Native ) \_1082836828FǣǣOle C l@IdI F"ndydz|n"i| S +"+" =[x 3 "y(1+2x)]dydz|x| "11 +" 01 +" =[x 3 "y(1+2x)]dydzx "11 +" 01 +" +[x 3 "y(1+2x)]dydz"x "11 +" 01 +" =0(2)""Fdv V +"+"+" "F"kdxdy S 1  +"+" "F"("k)dxdy S 2  +"+" =(2x+1)dzdxdy 01 +" " 1"y 2  1"y 2 +" "11 +" "1dxdy S 1  +"+" +0=(2rcos+1)rdrd 01 +" 02 +" "=( 02 +" 23cos+12)d"="=0 FMicrosoft Equation 3.0 DS Equation Equation.39qXzI,I (7xi"      !"#$'(-)*+,/.023N456978;:=<>?@ABCEDGFHIJMKLOTdPQRSUYVWXZ][\^_`abceh{fgijknlmorpqstuxvwyz|}~ne 6jn?$5@s0` CZJq*4\aMa(;+KRsAb ],`YF( f8X0"Dd\B R S AN? ^2!:3*p:QsSh`!`!:3*p:QsS.``Љ-P.xT=hA~fM3wG$Iу$E h%I.sB촳IZR8& 1yf y0l,Դ1P] 8Q g֭< wUT`B9-F{rpv\V=Rw4fI"^(WhoS9;){LZӂ"n|:^1W̽!жw+Xw֖\Zw1y\qkS k={7qEܧss8kam۾h*>̥qݟ7&'B(n` .&lwu)qo0eO8]͒=dO W&ϝ[>9*Y+ 3|HD #f0#^9r)uN˥\>1q8`w7W[7sJ3z1J{+ >`S0h/%$rjp_ NUv#$VYJP&/򿺀DdxB S S AO? _2nGt8#4HFJ`!BGt8#4HFz. xUMLAl mM5%5F!WhBb MBw~cRoE;EIsM~z ?p=L@rMc6,u,u,HZG,",D⺣rQBz%m.xu1|4byu,Ә^`'-um1sL ^c~8]Gk寅MDjw >+T3ϣY%:ml6஘[ tm/8^4pW<(3EY OGD=Q%.pJm]Iez͐Яᠻ^:_}=ֽO8'pMr4 |RwuOї7 t7C{ C#Ɉy-rɁ!wb:T8nS|Pw^2o9x[|/Nߪ|kJzo֗$OVShA }H܉W>5\Mcn$k}/hB3~y0G2@-DIj{3|RA9-+KyG l,جxvb>v?8(_Ԯ6.yFL4BAq߯:)2\H!F!~僈Wޫ-g=ttAf";+~3EŞh4,₶qkP?V2 R694$_M# A׳p-SqĦժz~aaOLm&o$E pMtcP[m&DdD 0B z S Av? b27W8AURb `!7W8AURbPkkxK@߽ҦeA?Z"ঃ,VPhliҭ B7A]ANEH &wy "@ȂxB|EPt10t],l,d;øvLf0$?]Xxz٭t yI(M;Nu4ǭaKC x~_A~ ~"hzz=jցٰPO oUf2&}<'^찷⓸P}|&%Veg%~,&G)W|W2.%◐3+&*Z?U31WsE*9r=r XI p+ 2rX6GpPDd B u S Aq? c2nSpL`!nSpL @hxS+Q>yy{x E(IJbգCy)j Kg ؾXX*K%eW;w+?os=w AЕF1 PQnH56 ԝ@+dUź ON Vv֗ԋ*]XL!',PsB2u*K;{ǃDB.m |*2?յGv'dOC1eɹK|k}!:$M}"S੿ŠZѹDo?_|>PY(y1殡XMcGlSg'L 7LfSG99`)J1[yC1weAa}yE_';ȿBj?ԵbbS(or=1̥x͋xDdB v S Ar? d2C܍eWkj `!C܍eWkj  @xx]RMKQ?cID@ @T ƱMM?ЦE)#pӦ7sϽs;A| cʛDBX[ sHljb$Z2JO$z1 ɥ&l)$d :,CFY HIRr[rso'!Ӽt~~vӾͻ;9ps\<L{*V-(|?*Fķ?)?@?Nv(7?3c+>;x/·p5'}ՌwbQgQ}E!/ zK>EN%!^`;fy9L$aV,P:a6Dd+0B w S As? e2")IOnuDgʠ-|7`!t")IOnuDgʠ- MkBxcdd``gd``beV dX,XĐ  A?d@zjx|K2B* R6Uf: P 27)?"](6Z4 CiF0n548?ķAͤ|! ~ Ay pClb ;>qڄ #pI 8 3j&R2v4nhR̶F&&\bt3i_DdHB T S AP? f2{g>};*lgWm`!Og>};*lg>@@" xcdd``gd``beV dX,XĐ ɁفURcgb x@e깡jx|K2B* R6Uf: P 27)?"](6br`)>F0W}I?~br< &1BMtgI0ae\nrfTeL%ōAzRag!(J 0ww021)W2IY,DdXB U S AQ? g2dL1pr@$@~`!8L1pr@$~ X+xOHQߟy3i\HpŨHad mjkCBt)NBEz"TNs Z3f~~3~h"QRA::"@ċGbւ&Pa2PG٣9TD@ĥw, @WQO$2}c9ɭzju!{җsoZ,]H`{lKxUe M ɦs3|o4;t(juCwbLzCa yn{UGvzoW9ZkT&*}؞_/_}[G:ySN-OYqXwuݯu@a=?zQƙj`N[/gd$o1#?b=V{A{|~#_Ow0v3VQ8΋OM .˜SrKͽD?D9ոyC.EYG0+R9Nx a.l찭Nvv߼; N) ^Ӧ'uCm?eS_oS22Z܇,u?>\5| u>x\_+E\o?̷3n"6'0x;WS} "M _TېgҠ7?66Dd+0B # S A? 2Z w1+df|x`!tZ w1+df MkBxcdd``gd``beV dX,XĐ  A?d@zjx|K2B* R6Uf: P 27)?"](6Z4 CiF0n548?ķAͤ|! ~ Ay pClb ;>qڄ #pI 8 3j&R2v4nhR̶F&&\bt3|`ZDd&B [ S AU? 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Equation Native  e_1272956962%>1Fףף"ds S +"+" FMicrosoft Equation 3.0 DS Equation Equation.39qN  Solution:(1)n=2xi+2yj+k 4x 2 +4y Ole  CompObj02 fObjInfo3 Equation Native  2 +1  ,"F="i"j"k("F)"n="2x"2y"1 4x 2 +4y 2 +1  ,n"k=1 4x 2 +4y 2 +1  ("F)"ds S +"+" =("2x"2y"1)dx " 1"y 2  1"y 2 +" dy "11 +" =("2rcos"2rsin"1)rdr 01 +" d 02 +" =("23cos"23sin"12)d 02 +" =" FMicrosoft Equation 3.0 DS Equation Equation.39qN{ (2)("F)"ds S +"+" =F"dr C +" ,wher_12729562506F`أڣOle  CompObj57 fObjInfo8 Equation Native  _1272957081;FۣۣOle  CompObj:< feC:x 2 +y 2 =1Letx=cos,y=sinF"dr=(sini+0j+cosk)"("sini+cosj)d="sin 2 dF"dr C +" ="sin 2 d 02 +" ="1"cos22d 02 +" ="2+sin24 02 =" FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo= Equation Native  Q_12729571159C@FޣޣOle  N5 , F"dr C +" FMicrosoft Equation 3.0 DS Equation Equation.39qNP F"dr C +" =("4sin 2 CompObj?A fObjInfoB Equation Native  _1272958055HEF6ࣜ6ࣜ"108cos 2 )d 02 +" FMicrosoft Equation 3.0 DS Equation Equation.39qN ("F)"ds S +"+" ="28Ole  CompObjDF fObjInfoG Equation Native  ds S +"+" FMicrosoft Equation 3.0 DS Equation Equation.39qNj , ("F)"n=("F)"k="F 2 "x""F _1272958036JFC㣜C㣜Ole  CompObjIK fObjInfoL Equation Native  _974298403*`OFQ棜Q棜Ole  CompObjNP f1 "y("F)"ds S +"+" ="F 2 "x""F 1 "ydx S +"+" dyandF"dr C +" =F 1 dx+F 2 dy C +" hence"F 2 "x""F 1 "ydx S +"+" dy=F 1 dx+F 2 dy C +" FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoQ Equation Native  _1086373764TF磜@^飜Ole  y2 [(xy+y 2 )dx+x 2 dy] C +"L_ 3 P_ 3  ]&"MrEdMicrosoft Draw]&MrEdPIC SV LMETA  $CompObjUW XObjInfoX + Times New Roman < =&MrEd& 0  P    "- & %% & MrEd% & % ?3% 4 & MrEd%   &  & MrEd% &MrEd0-  0&MrEdTimes New Roman-. & 2  ax & .&MrEdP{C. & 2 PEy & .&MrEdF~Times New Roman-. &2 F-D(1,1) & . 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( & MathType "-$m&Times New Roman - 2 `{[ 2 `( 2 `) 2 ` ] 2 `( 2 `Y) 2 `( 2 `)Times New Roman- 2 ` x 2 `x 2 `3x 2 `N dx 2 `X x 2 `%xdx 2 `3x 2 `x 2 `"x 2 `$dx 2 `.x 2 ` dxSymbol- 2 `` 2 `o+ 2 ` + 2 `$+ 2 `*` 2 `+ 2 `+ 2 `k"= 2 `#-Symbol- 2 ? 2  Times New Roman - 2 2p 2 2p 2 : 2p 2 (2p 2 80p 2 81p 2 2p 2 2p 2 81p 2 80pTimes New Roman- 2 `P2 2 p;%1 2 $20 &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qz yJ}JxyJ [xx 2 +(x 2 ) 2 ]dx+x 2 (2xdx)+ 01  (xx+x 2 )dx+x 2 dx=-120 10 LG#ElG#E  ._974298502bF@Ole D PIC adE LMETA G 1   & & MathType  "-! "-s!-s%f\f\f "-  Times New Roman- 2 8Sp 2 8yb 2 8ybTimes New RomanE- 2 `Vx 2 `&x 2 `M y 2 ` dxdy 2 `x 2 ` y 2 ``dxdySymbol- 2 ? 2  2 Symbol- 2 `c- 2 `4+ 2 `V= 2 `- 2 `= 2 `-Times New Roman- 2 `[ 2 `( 2 ` )] 2 `C( 2 `) 2 `2 2 `\2 2 `2 2 pz1 2 20 Times New RomanE- 2 8@0p 2 8b1p &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qh  S CompObjce\ fObjInfof^ Equation Native _ _953964169iF0N+"+" [2x"(x+2y)]dxdy=(x"2y)dxdy y y  +" 01 +" ="120L{hq{  .Ole d PIC hke LMETA g hCompObjjlm m1  @@& & MathType Times New Romanh- 2 CSymbol- 2 i? &  "Systems- FMicrosoft Equation 2.0 媩 DS Equation Equation.29qObjInfomo Equation Native p @_1082840654upFhhOle q $(yJ}JyJ  C  L*E8l*E  .1  &&& & MathType "-$m&Times New Roman - 2 `{PIC orr LMETA t tCompObjqs fObjInfot  [ 2 `( 2 `) 2 ` ] 2 `( 2 `Y) 2 `( 2 `)Times New Roman- 2 ` x 2 `x 2 `3x 2 `N dx 2 `X x 2 `%xdx 2 `3x 2 `x 2 `"x 2 `$dx 2 `.x 2 ` dxSymbol- 2 `` 2 `o+ 2 ` + 2 `$+ 2 `*` 2 `+ 2 `+ 2 `k"= 2 `#-Symbol- 2 ? 2  Times New Roman - 2 2p 2 2p 2 : 2p 2 (2p 2 80p 2 81p 2 2p 2 2p 2 81p 2 80pTimes New Roman- 2 `P2 2 p;%1 2 $20 &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qL ? @ A B C D E F G H I J K L M N O P Q R S T U V W X Y Z [ \ ] ^ _ ` a b c d e f i k l m n o p q r s t u v w x y z { | } ~  & MathType "-$m&Times New Roman - 2 `{[ 2 `( 2 `) 2 ` ] 2 `( 2 `Y) 2 `( 2 `)Times New Roman- 2 ` x 2 `x 2 `3x 2 `N dx 2 `X x 2 `%xdx 2 `3x 2 `x 2 `"x 2 `$dx 2 `.x 2 ` dxSymbol- 2 `` 2 `o+ 2 ` + 2 `$+ 2 `*` 2 `+ 2 `+ 2 `k"= 2 `#-Symbol- 2 ? 2  Times New Roman - 2 2p 2 2p 2 : 2p 2 (2p 2 80p 2 81p 2 2p 2 2p 2 81p 2 80pTimes New Roman- 2 `P2 2 p;%1 2 $20 &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39q\ zI\I F"dr Equation Native  x_1082841783F  Ole  PIC  LC +" =F"dr C 1  +" +F"dr C 2  +" +F"dr C 3  +" +F"dr C 4  +" =" 2 2L*E8lMETA  tCompObj& fObjInfo( Equation Native ) ,*E  .1  &&& & MathType "-$m&Times New Roman - 2 `{[ 2 `( 2 `) 2 ` ] 2 `( 2 `Y) 2 `( 2 `)Times New Roman- 2 ` x 2 `x 2 `3x 2 `N dx 2 `X x 2 `%xdx 2 `3x 2 `x 2 `"x 2 `$dx 2 `.x 2 ` dxSymbol- 2 `` 2 `o+ 2 ` + 2 `$+ 2 `*` 2 `+ 2 `+ 2 `k"= 2 `#-Symbol- 2 ? 2  Times New Roman - 2 2p 2 2p 2 : 2p 2 (2p 2 80p 2 81p 2 2p 2 2p 2 81p 2 80pTimes New Roman- 2 `P2 2 p;%1 2 $20 &  "Systems- FMicrosoft Equation 3.0 DS Equation Equation.39qzI,I (2)F"dr C +" =("F 2 "x""F 1 "y)dxdy S +"+" =("ycosx"xcosy)dxdy 0 +" 0/2 +" =("ysinx"x 2 2cosy) 0 dy 0/2 +" =" 2 2cosydy 0/2 +" =" 2 2siny 0/2 =" 2 2. FMicrosoft DrawingMSDraw_1082834907Fp p Ole 6 CompObj7 XObjInfo9 MSDraw9q P6 e]&"MrEdMicrosoft Draw]&MrEd+ Times New Roman < =&MrEd!& 0  9y Ole10Native:  _1086373182F  Ole g PIC h L  "- &  & MrEd%p &  & MrEd%0 &  & MrEd%pp0 &  & MrEd%p00 &  & MrEd%p0P &  & MrEd%P &  & MrEd%0PP &  & MrEd%P &  & MrEd%0 &  & MrEd% &  & MrEd%/P0 &  & MrEd% &  & MrEd%-W0P &  & MrEd%p0&MrEdTimes New Roman~-. & 2 Xy & .&MrEd} . & 2 }L{x & .&MrEd:`D. & 2 :Fz & .&MrEd'j>Symbol-. & 2 '@D & .&MrEdE. & 2 ]GD & .&MrEd*. & 2 *((D & .&MrEd$ATimes New Roman~-. & 2 $"y & .&MrEdy2. & 2 y:4x & .&MrEd9. & 2 ?;z & . 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