Math220-Sum11-HW4

Math220-Sum11-HW4 - Math 220-9809. Summer 2011 Page 1 of 1...

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Unformatted text preview: Math 220-9809. Summer 2011 Page 1 of 1 NAME: DATE: Homework #4: Vector Calculus Due Date— Wednesday, August 3, 2011 Instructions 1. Use this as the cover page of your homework. 2. Clearly identify the problem you are working on. 3. Show all calculations and display answam clearly. Unjustified answers will receive no credit. 4. Write neatly and legibly. 5. Do only the problems that are circled. CHAPTER 16 REVIEW 1151 Exercises 1. A vector field F. a curve C. and a point P are shown.- (a) ls l}. F - dr positive. negative. or zero? Explain. (b) Is div FlP) positive. negative. or zero? Explain. 24) Evaluate the line integral. 2. [CI d5, C is the are of the parabola y = x3 from (0, 0) to (l, I} it y: cosxds. C:x=r, y=3cosr, :=3sinr. OS—réar I. Icydx + (.r + y:) dy, C is the ellipse 4):1 + 9y: = 36 with counterclockwise orientation icy] dx + x: dy. C is the arc of the parabolax = l - yl from (0, -i) to (0. l) E. j} x/xy dx + e"d_v 4- .t2 dz. Cisgivenbyr(r)=t4i+t:j+r3k.051$ 1 L. xydx + y: d)! + yz dz. C is the line segment from (1. 0. — l), to (3. 4, 2') 8. j}. F - dr. where F(x. y) = xyi + .ij and C is given by r(r) =sinri + (l +I)j.05.1$ 1r ‘l'CF - dr, where th._v. :) = e‘i + .rzj + (.i' + y} k and Cisgivenbyr(r) = rzi +r-‘j -tk.0ét~<.l 111. Find the work done by the force field F(.t._v,z) = :i + xj + yk in moving a particle from the point (3. 0.,0) to the point (0. rr/Z. 3) along a ‘5- (a) a straight line " . (b) the helix x = Boos 1‘. y = r. z ——_— 3‘sinr 11—12 Show that F is a conservative vector field. Then find a func- tion f such that F = Vf. ®("—"v) = (1 + xJ’k'T‘ri + (8" + xle'lilj Graphing calculator or computer required 12. F(x._v. :) m sin _\'i + x cos 3'} - sin : 1-: 13—14 Show that F is conservative and use this fact to evaluate f F - dr along the given curve. @(x,y} = (cl-xiv2 - 2x_v")i + (2143' -— 3.1in2 + 4).*"lj. C: r(t) = (t + sin 1Tfll + (2t + cos irrlj. O =E 1‘ $1 14. F(.t._v.:) = c-“i + (.re" + e:_)j + ye: k. C is the line segment from (0. 2. 0) to (4. 0, 3) 15. Verify that Green’s Theorem is true for the line integral 1} xy2 dx — xlydy. where C consists of the parabola y = Jr2 from (—1. I) to (1. 1) and the line segment from (i. 1) to (—1, 1). 16. Use Green's Theorem to evaluate J 1/1 + x1 dx + nya‘y C where C is the triangle with venices (0. 0), (l, 0). and (1. 3). Use Green’s Theorem to evaluate .i'cxly dx — xv: dy, where C is the circle x1 + y2 = 4 with counterclockwise orientation. Find curl F and div F if F(x. y. z) = e." sin y i + e‘-" sin :j + 9“ sin .t k 19. Show that there is no vector field G such that curlG = in + 3yzj — lek 20. Show that. under conditions to be stated on the vector fields F and G, curl(F><G)=FdivG—GdivF+ (G-V)F—(F-V)G 21. If C' is any piecewise-smooth simple closed plane curve and f and g are differentiable functions, show that _|'C f(.\‘) (it + g(y) dy = 0. 22. if f and g are twice differentiable functions. show that V2(fgl =fV39 + ngf-i- 2Vf- V9 23. if f is a harmonic function. that is, Vlf = 0. show that the line integrai Il‘fi. dx — f. d)! is independent of path in any simple region D. 2‘1. (a) Sketch the curve C‘ with parametric equations x==cost y=sinr z=sinr 0$r$21r (it) Find _|'c2.re:-“dx + (2x333 + 23‘ cot :) d_v — ylcselz dz. Computer algebra system required . .mvmz 1152 CHAPTER 16 VECTOR CALCULUS @Find the area of the part of the surface .7 = x1 + 2y that lies above the triangle with vertices (0, O), (l. 0). and (l. 2). 26. (:1) Find an equation of the tangent plane at the point (4, -2, 1) to the parametric surface S given by r(u,v)=uzi-uvj+uzk O$u$3,-3Sv£3 (b) Use a computer to graph the surface S and the tangent plane found in part (a). (C) Set up, but do not evaluate, an integral for the surface area of S. CA3 (d) If 2 2 2 Z . x . y 1+ + , J 1+2' k 1+x2- t+y2 F(.r. y, z) = find F - dS correct to four decimal places. 21—30 Evaluate the surface integral. “5 2 :15, where S is the part of the paraboloid z = x1 + y2 that lies under the plane 2 = 4 23. (.92.: + ylz) dS, where S is the part of the plane 2 = 4 + x + y that lies inside the cylinder Jr2 ~l- y2 = 4 J'J'sF ' dS. where th. y. z) = xzi — 2yj + 3x I: and Sis the sphere x2 + y2 + z2 = 4 with outward orientation 30. 'l'j'sF - (is, where F(x,y,z) = xzi + xyj + zkandSis the part of the paraboloid z = x2 + y2 below the plane 2 = I with upward orientation @Verify that Stokes’ Theorem is true for the vector field F(x,y,z) = xii + yzj + 211:. where S is the part of the paraboloid z = I - x2 - y2 that lies above the xy-plane and S has upward orientation. 32. Use Stokes' Theorem to evaluate fls curl F - (15, where F(Jr.y, z) = xzyzi + yzzj + 212"" k, S is the part of the sphere x2 + y2 + z2 = 5 that lies above the plane 2 = 1, and S is oriented upward. Use Stokes‘ Theorem to evaluate j'c F - dr. where F(x,y. z) = xyi + yzj + zx k, and C is the triangle with vertices (l, 0. 0). (0, i. 0), and {0, 0, l), oriented counter- clockwise as viewed from above. 34. Use the Divergence Theorem to calculate the surface integral F - dS, where F(x, y, z) = fl + y3j + z3 k and S is the surface of the solid bounded by the cylinder x2 -!-y2 = landtheplanesz = 0andz= 2. @ Verify that the Divergence Theorem is true for the vector field F(x,y. z) = xi + yj + z k, where E is the unit ball x1 + y2 + 22 $ 1. 36. Compute the outward flux of xi+yj +zk m. y. z} = through the ellipsoid 4):2 + 9y2 + (if = 36, 37. Let FUJJ) = (311W — 3y)i + (x3: '- 3x)j -+- (x3y + 2z) k Evaluate J'C F - dr, where C is the curve with initial point (0, 0, 2) and terminal point (0, 3, 0) shown in the figure. 38. Let (2x3 + 2er2 —- 2y)l 4- (2y3 + 213): + 2x)j F(x. y) = x2 + y: Evaluate 35C F - dr, where C is shown in the figure. 39. Findj'j'SF - ndS,whereF{x,y,z) = xi + yj + zkandSis the outwardly oriented surface shown in the figure (the bound- ary surface of a cube with a unit comer cube removed). 2' (2.0.2) l2. 2. 0) 40. If the components of F have continuous second partial derivatives and S is the boundary surface of a simple solid region. show that curl F - d3 = 0. 41. Ifa is aconstant vector, r = xi + yj + zit, and Sis an oriented, smooth surface with a simple. closed. smooth. pos- itively oriented boundary curve C. show that 2a-ds=J'C(a><r)-dr S ...
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Math220-Sum11-HW4 - Math 220-9809. Summer 2011 Page 1 of 1...

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