This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 2209809. 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} = (clxiv2  2x_v")i + (2143' — 3.1in2 + 4).*"lj. C: r(t) = (t + sin 1Tfll + (2t + cos irrlj. O =E 1‘ $1 14. F(.t._v.:) = c“i + (.re" + e:_)j + ye: k.
C is the line segment from (0. 2. 0) to (4. 0, 3) 15. Verify that Green’s Theorem is true for the line integral
1} xy2 dx — xlydy. where C consists of the parabola y = Jr2
from (—1. I) to (1. 1) and the line segment from (i. 1)
to (—1, 1). 16. Use Green's Theorem to evaluate
J 1/1 + x1 dx + nya‘y
C where C is the triangle with venices (0. 0), (l, 0). and (1. 3). Use Green’s Theorem to evaluate .i'cxly dx — xv: dy,
where C is the circle x1 + y2 = 4 with counterclockwise
orientation. Find curl F and div F if F(x. y. z) = e." sin y i + e‘" sin :j + 9“ sin .t k
19. Show that there is no vector ﬁeld G such that curlG = in + 3yzj — lek 20. Show that. under conditions to be stated on the vector ﬁelds
F and G, curl(F><G)=FdivG—GdivF+ (GV)F—(FV)G 21. If C' is any piecewisesmooth simple closed plane curve and f and g are differentiable functions, show that
_'C f(.\‘) (it + g(y) dy = 0. 22. if f and g are twice differentiable functions. show that
V2(fgl =fV39 + ngfi 2Vf V9 23. if f is a harmonic function. that is, Vlf = 0. show that the line
integrai Il‘ﬁ. dx — f. d)! is independent of path in any simple
region D. 2‘1. (a) Sketch the curve C‘ with parametric equations x==cost y=sinr z=sinr 0$r$21r (it) Find _'c2.re:“dx + (2x333 + 23‘ cot :) d_v — ylcselz dz. Computer algebra system required . .mvmz 1152 CHAPTER 16 VECTOR CALCULUS @Find the area of the part of the surface .7 = x1 + 2y that lies
above the triangle with vertices (0, O), (l. 0). and (l. 2). 26. (:1) Find an equation of the tangent plane at the point
(4, 2, 1) to the parametric surface S given by r(u,v)=uziuvj+uzk O$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 xyplane 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 2ads=J'C(a><r)dr
S ...
View
Full
Document
 Summer '08
 Helton
 Math, Multivariable Calculus

Click to edit the document details