Solucao impares Calculo 6ed CAP.16

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17 VECTOR CALCULUS

ET 16

17.1 Vector Fields

ET 16.1

1. F({> |) =1₂(i + j)

All vectors in this eld are identical, with length1 2 and

direction parallel to the line| = {.

3. F({> |) = | i +1₂j

The length of the vector| i +1₂j is t

|2+1₄. Vectors are tangent to parabolas opening about the{-axis.

5. F({> |) = | is +{ j {2₊_|2

The length of the vectors| i + { j {2₊_|2 is 1.

7. F({> |> }) = k

All vectors in this eld are parallel to the}-axis and have length 1.

9. F({> |> }) = { k

At each point ({> |> }), F({> |> }) is a vector of length |{|. For{ A 0, all point in the direction of the positive }-axis, while for{ ? 0, all are in the direction of the negative }-axis. In each plane { = n, all the vectors are identical.

269

TX.10

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11. F({> |) = h|> {i corresponds to graph II. In the rst quadrant all the vectors have positive {- and |-components, in the second quadrant all vectors have positivecomponents and negative |-components, in the third quadrant all vectors have negative {-and|-components, and in the fourth quadrant all vectors have negative {-components and positive |-components. In addition, the vectors get shorter as we approach the origin.

13. F({> |) = h{ 2> { + 1i corresponds to graph I since the vectors are independent of | (vectors along vertical lines are identical) and, as we move to the right, both the{- and the |-components get larger.

15. F({> |> }) = i + 2 j + 3 k corresponds to graph IV, since all vectors have identical length and direction.

17. F({> |> }) = { i + | j + 3 k corresponds to graph III; the projection of each vector onto the {|-plane is { i + | j, which points away from the origin, and the vectors point generally upward because their}-components are all 3.

19.

The vector eld seems to have very short vectors near the line| = 2{. ForF({> |) = h0> 0i we must have |2 2{| = 0 and 3{| 6{2= 0. The rst equation holds if| = 0 or | = 2{, and the second holds if { = 0 or | = 2{. So both equations hold [and thus F({> |) = 0] along the line| = 2{. 21. i({> |) = {h{| i({> |) = i{({> |) i + i|({> |) j = ({h{|· | + h{|)i + ({h{|· {) j = ({| + 1)h{|i + {2h{|j 23. i({> |> }) = i{({> |> }) i + i|({> |> }) j + i}({> |> }) k = s { {2₊_|2₊_}2 i + | s {2₊_|2₊_}2j + } s {2₊_|2₊_}2k 25. i({> |) = {2 | i({> |) = 2{ i j. The length ofi({> |) is4{2+ 1. When{ 6= 0, the vectors point away from the|-axis in a slightly downward direction with length that increases as the distance from the|-axis increases.

27.We graphi along with a contour map of i.

The graph shows that the gradient vectors are perpendicular to the level curves. Also, the gradient vectors point in the direction in whichi is increasing and are longer where the level curves are closer together.

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SECTION 17.2 LINE INTEGRALS ET SECTION 16.2 ¤ 271 29.i({> |) = {2+|2 i({> |) = 2{ i + 2| j. Thus, each vector i({> |) has the same direction and twice the length of

the position vector of the point ({> |), so the vectors all point directly away from the origin and their lengths increase as we move away from the origin. Hence,i is graph II.

31.i({> |) = ({ + |)2 i({> |) = 2({ + |) i + 2({ + |) j. The {- and |-components of each vector are equal, so all vectors are parallel to the line| = {. The vectors are 0 along the line | = { and their length increases as the distance from this line increases. Thus,i is graph II.

33.Atw = 3 the particle is at (2> 1) so its velocity is V(2> 1) = h4> 3i. After 0.01 units of time, the particle’s change in location should be approximately 0=01 V(2> 1) = 0=01 h4> 3i = h0=04> 0=03i, so the particle should be approximately at the point (2=04> 1=03).

35. (a) We sketch the vector eldF({> |) = { i | j along with several approximate ow lines.The ow lines appear to be hyperbolas with shape similar to the graph of| = ±1@{, so we might guess that the ow lines have equations | = F@{.

(b) If{ = {(w) and | = |(w) are parametric equations of a ow line, then the velocity vector of the ow line at the point ({> |) is {0(w) i + |0(w) j. Since the velocity vectors coincide with the vectors in the vector eld, we have {0₍_{w) i + |}0₍_{w) j = { i | j g{@gw = {, g|@gw = |. To solve these differential equations, we know}

g{@gw = { g{@{ = gw ln |{| = w + F { = ±hw + F ₌_Dhw_{for some constant}_{D, and}

g|@gw = | g|@| = gw ln ||| = w + N | = ±hw + N ₌_Ehw_{for some constant}_{E. Therefore}

{| = Dhw_Ehw₌_{DE = constant. If the ow line passes through (1> 1) then (1) (1) = constant = 1 {| = 1}

| = 1@{, { A 0.

17.2 Line Integrals

ET 16.2

1.{ = w3and| = w, 0 w 2, so by Formula 3 U F|3gv = U2 0 w3 t_g{ gw 2 +g|_gw2gw =U₀2w3s(3w2)2+ (1)2gw =U₀2w39w4+ 1gw = ₃₆1 ·2₃9w4+ 13@2l2 0= 1 54(1453@2 1) or541 145145 1 3.Parametric equations forF are { = 4 cos w, | = 4 sin w, ₂ w ₂. Then

U

F{|4gv =

U@2

@2(4 cosw)(4 sin w)4

s

(4 sin w)2+ (4 cosw)2gw =U_@2@2 45cosw sin4ws16(sin2w + cos2w) gw = 45U_@2@2 (sin4w cos w)(4) gw = (4)61₅sin5w@2_@2=2 · 4₅6 = 1638=4

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5. If we choose{ as the parameter, parametric equations for F are { = {, | ={ for 1 { 4 and U F {2_|3_{_{g| =}U4 1 k {2_{· (}_{{ )}3_{l 1 2{g{ = 1 2 U4 1 {3₁_g{ = 1₂1₄{4 {4₁= 1₂64 4 1₄+ 1=243₈ 7. F = F₁+F₂ OnF1: { = {, | = 0 g| = 0 g{, 0 { 2. OnF2: { = {, | = 2{ 4 g| = 2 g{, 2 { 3. Then U F{| g{ + ({ |) g| = U F1{| g{ + ({ |) g| + U F2{| g{ + ({ |) g| =U₀2(0 + 0)g{ +U₂3[(2{2 4{) + ({ + 4)(2)] g{ =U₂3(2{2 6{ + 8) g{ = 17₃

9. { = 2 sin w, | = w, } = 2 cos w, 0 w . Then by Formula 9, U F{|} gv = U 0(2 sinw)(w)(2 cos w) t_g{ gw 2 +g|_gw2+g}_gw2gw

=U₀4w sin w cos ws(2 cosw)2+ (1)2+ (2 sinw)2gw =U₀2w sin 2ws4(cos2w + sin2w) + 1 gw =25U₀w sin 2w gw = 251₂w cos 2w +1₄sin 2w

0

integrate by parts with x = w, gy = sin 2w gw

=25₂ 0=5

11. Parametric equations forF are { = w, | = 2w, } = 3w, 0 w 1. Then U F{h|}gv = U1 0 wh(2w)(3w) 12+ 22+ 32gw =14U₀1wh6w2gw =14k₁₂1h6w2l1 0= 14 12 (h6 1). 13. U_F{2|} g} =U₀1(w3)2(w)w2· 2w gw =U₀12w9gw = 1₅w101 0=15 15. OnF₁: { = 1 + w g{ = gw, | = 3w g| = 3 gw, } = 1 g} = 0 gw, 0 w 1. OnF2: { = 2 g{ = 0 gw, | = 3 + 2w g| = 2 gw, } = 1 + w g} = gw, 0 w 1. Then U F({ + |}) g{ + 2{ g| + {|} g} =U_F 1({ + |}) g{ + 2{ g| + {|} g} + U F2({ + |}) g{ + 2{ g| + {|} g} =U₀1(1 +w + (3w)(1)) gw + 2(1 + w) · 3 gw + (1 + w)(3w)(1) · 0 gw +U₀1(2 + (3 + 2w)(1 + w)) · 0 gw + 2(2) · 2 gw + (2)(3 + 2w)(1 + w) gw =U₀1(10w + 7) gw +U₀1(4w2+ 10w + 14) gw =5w2+ 7w1 0+ ₄ 3w3+ 5w2+ 14w 1 0= 12 +613 =973

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SECTION 17.2 LINE INTEGRALS ET SECTION 16.2 ¤ 273 17. (a) Along the line{ = 3, the vectors of F have positive |-components, so since the path goes upward, the integrand F · T is

always positive. ThereforeU_F

1F · gr =

U

F1F · T gv is positive.

(b) All of the (nonzero) eld vectors along the circle with radius 3 are pointed in the clockwise direction, that is, opposite the direction to the path. SoF · T is negative, and thereforeU_F

2F · gr = U F2F · T gv is negative. 19.r(w) = 11w4i + w3j, so F(r(w)) = (11w4)(w3)i + 3(w3)2j = 11w7i + 3w6j and r0(w) = 44w3i + 3w2j. Then U FF · gr = U1 0 F(r(w)) · r0(w) gw = U1 0(11w7· 44w3+ 3w6· 3w2)gw = U1 0(484w10+ 9w8)gw = 44w11+w91₀= 45. 21.U_FF · gr =U₀1 sinw3> cos(w2)> w4· 3w2> 2w> 1gw

=U₀1(3w2sinw3 2w cos w2+w4)gw = cos w3 sin w2+1₅w51

0=65 cos 1 sin 1

23.F(r(w)) = (hw)hw2i + sinhw2j = hww2i + sinhw2j, r0(w) = hwi 2whw2j. Then ] FF · gr = ] 2 1 F(r(w)) · r 0₍_{w) gw =}] 2 1 k hww2 hw_{+ sin}_hw2 ·2whw2l gw = ] 2 1 k h2ww2_2whw2_sin_hw2l_{gw 1=9633} 25.{ = w2, | = w3, } = w4so by Formula 9, U F{ sin(| + }) gv = U5 0(w2) sin(w3+w4) s (2w)2+ (3w2)2+ (4w3)2gw =U₀5w2sin(w3+w4)4w2+ 9w4+ 16w6gw 15=0074

27.We graphF({> |) = ({ |) i + {| j and the curve F. We see that most of the vectors starting on F point in roughly the same direction asF, so for these portions of F the tangential component F · T is positive. Although some vectors in the third quadrant which start onF point in roughly the opposite direction, and hence give negative tangential components, it seems reasonable that the effect of these portions ofF is outweighed by the positive tangential components. Thus, we would expect U

FF · gr =

U

FF · T gv to be positive.

To verify, we evaluateU_FF · gr. The curve F can be represented by r(w) = 2 cos w i + 2 sin w j, 0 w 3₂ , soF(r(w)) = (2 cos w 2 sin w) i + 4 cos w sin w j and r0(w) = 2 sin w i + 2 cos w j. Then

U

FF · gr =

U3@2

0 F(r(w)) · r0(w) gw

=U₀3@2[2 sin w(2 cos w 2 sin w) + 2 cos w(4 cos w sin w)] gw = 4U₀3@2(sin2w sin w cos w + 2 sin w cos2w) gw

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29. (a)U_FF · gr =U₀1Ghw21> w5H· 2w> 3w2gw =U₀12whw21+ 3w7gw =khw21+3₈w8l1 0= 11 8 1@h (b)r(0) = 0, F(r(0)) = h1> 0; r1 2 = G 1 2>212 H , F r1 2 = G h1@2_> 1 42 H ; r(1) = h1> 1i, F(r(1)) = h1> 1i.

In order to generate the graph with Maple, we use the PLOT command (not to be confused with the plot command) to dene each of the vectors. For example,

v1:=PLOT(CURVES([[0,0],[evalf(1/exp(1)),0]]));

generates the vector from the vector eld at the point (0> 0) (but without an arrowhead) and gives it the name v1. To show everything on the same screen, we use the display command. In Mathematica, we use ListPlot (with the

PlotJoined -A True option) to generate the vectors, and then Show to show everything on the same screen. 31. { = hwcos 4w, | = hwsin 4w, } = hw, 0 w 2 .

Theng{

gw =hw( sin 4w)(4) hwcos 4w = hw(4 sin 4w + cos 4w), g|

gw =hw(cos 4w)(4) hwsin 4w = hw(4 cos 4w + sin 4w), and g}gw =hw, so v g{ gw 2 + g| gw 2 + g} gw 2

=s(hw)2[(4 sin 4w + cos 4w)2+ (4 cos 4w + sin 4w)2+ 1] =hws16(sin24w + cos24w) + sin24w + cos24w + 1 = 32hw

Therefore U F{3|2} gv = U2 0 (hwcos 4w)3(hwsin 4w)2(hw) (3 2hw)gw =U₀232h7wcos34w sin24w gw = _5,632,705172,704 2 (1 h14) 33. We use the parametrization{ = 2 cos w, | = 2 sin w, ₂ w ₂. Then

gv =tg{ gw 2 +g|_gw2gw =s(2 sin w)2+ (2 cosw)2gw = 2 gw, so p =U_Fn gv = 2nU_@2@2 gw = 2n(), { = 1 2n U F{n gv =21 U@2 @2(2 cosw)2 gw = 21 4 sinw@2_@2=4,| = _2n1 U_F|n gv =₂1 U_@2@2 (2 sinw)2 gw = 0. Hence ({> |) =4> 0. 35. (a){ = 1 p ] F{({> |> }) gv , | = 1 p ] F|({> |> }) gv, } = 1 p ] F}({> |> }) gv where p = U F({> |> }) gv. (b)p =U_Fn gv = nU₀2s4 sin2w + 4 cos2w + 9 gw = n13U₀2gw = 2n13, { = 1 2n13 ] 2 0 2n13 sinw gw = 0, | = 1 2n13 ] 2 0 2n13 cosw gw = 0, } = 1 2n13 ] 2 0 n13 (3w) gw = 3 2 22= 3. Hence ({> |> }) = (0> 0> 3).

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SECTION 17.2 LINE INTEGRALS ET SECTION 16.2 ¤ 275 37.From Example 3,({> |) = n(1 |), { = cos w, | = sin w, and gv = gw, 0 w

L{=U_F|2({> |) gv =U₀sin2w [n(1 sin w)] gw = nU₀(sin2w sin3w) gw =1₂nU₀(1 cos 2w) gw nU₀(1 cos2w) sin w gw

Letx = cos w, gx = sin w gw in the second integral

=nk₂ +U₁1(1 x2)gxl=n₂ 4₃

L|=U_F{2({> |) gv = nU₀cos2w (1 sin w) gw = n₂U₀(1 + cos 2w) gw nU₀cos2w sin w gw =n₂ 2₃, using the same substitution as above.

39.Z =U_FF · gr =U₀2hw sin w> 3 cos wi · h1 cos w> sin wi gw =U₀2(w w cos w sin w + sin w cos w + 3 sin w sin w cos w) gw =U₀2(w w cos w + 2 sin w) gw =1₂w2 (w sin w + cos w) 2 cos w2₀

integrate by parts in the second term

= 22 41.r(w) = h1 + 2w> 4w> 2wi, 0 w 1, Z =U_FF · g r =U₀1h6w> 1 + 4w> 1 + 6wi · h2> 4> 2i gw =U₀1(12w + 4(1 + 4w) + 2(1 + 6w)) gw =U₀1(40w + 6) gw =20w2+ 6w1 0= 26

43.LetF = 185 k. To parametrize the staircase, let { = 20 cos w, | = 20 sin w, } =₆90w = 15w, 0 w 6 Z =U_FF · gr =U6 0 h0> 0> 185i · 20 sin w> 20 cos w>15 gw = (185)15 U6 0 gw = (185)(90) 1=67 × 104ft-lb

45. (a)r(w) = hcos w> sin wi, 0 w 2, and let F = hd> ei. Then

Z =U_FF · g r =U₀2hd> ei · h sin w> cos wi gw =U₀2(d sin w + e cos w) gw =d cos w + e sin w2

0

=d + 0 d + 0 = 0 (b) Yes.F ({> |) = n x = hn{> n|i and

Z =U_FF · g r =U2

0 hn cos w> n sin wi · h sin w> cos wi gw =

U2

0 (n sin w cos w + n sin w cos w) gw =

U2

0 0gw = 0.

47.The work done in moving the object isU_FF · gr =U_FF · T gv. We can approximate this integral by dividing F into 7segments of equal length v = 2 and approximating F · T, that is, the tangential component of force, at a point ({_l> |_l)on each segment. SinceF is composed of straight line segments, F · T is the scalar projection of each force vector onto F. If we choose ({_l> |_l)to be the point on the segment closest to the origin, then the work done is

U

FF · T gv 7

S

l = 1

[F({_l> |_l)· T({_l> |_l)] v = [2 + 2 + 2 + 2 + 1 + 1 + 1](2) = 22. Thus, we estimate the work done to be approximately 22 J.

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17.3 The Fundamental Theorem for Line Integrals

ET 16.3

1. F appears to be a smooth curve, and since i is continuous, we know i is differentiable. Then Theorem 2 says that the value ofU_Fi · gr is simply the difference of the values of i at the terminal and initial points of F. From the graph, this is 50 10 = 40.

3. C(2{ 3|)@C| = 3 = C(3{ + 4| 8)@C{ and the domain of F is R2which is open and simply-connected, so by Theorem 6F is conservative. Thus, there exists a function i such that i = F, that is, i_{({> |) = 2{ 3| and

i|({> |) = 3{ + 4| 8. But i{({> |) = 2{ 3| implies i({> |) = {2 3{| + j(|) and differentiating both sides of this equation with respect to| gives i_|({> |) = 3{ + j0(|). Thus 3{ + 4| 8 = 3{ + j0(|) so j0(|) = 4| 8 and j(|) = 2|2_{8| + N where N is a constant. Hence i({> |) = {}2_{3{| + 2|}2_{8| + N is a potential function for F.}

5. C(h{sin|)@C| = h{cos| = C(h{cos|)@C{ and the domain of F is R2. HenceF is conservative so there exists a function i such thati = F. Then i{({> |) = h{sin| implies i({> |) = h{sin| + j(|) and i|({> |) = h{cos| + j0(|). But i|({> |) = h{cos| so j0(|) = 0 j(|) = N. Then i({> |) = h{sin| + N is a potential function for F.

7. C(|h{+ sin|)@C| = h{+ cos| = C(h{+{ cos |)@C{ and the domain of F is R2. HenceF is conservative so there exists a functioni such that i = F. Then i{({> |) = |h{+ sin| implies i({> |) = |h{+{ sin | + j(|) and

i|({> |) = h{+{ cos | + j0(|). But i|({> |) = h{+{ cos | so j(|) = N and i({> |) = |h{+{ sin | + N is a potential

function forF.

9. C(ln | + 2{|3)@C| = 1@| + 6{|2=C(3{2|2+{@|)@C{ and the domain of F is {({> |) | | A 0} which is open and simply connected. HenceF is conservative so there exists a function i such that i = F. Then i_{({> |) = ln | + 2{|3implies i({> |) = { ln | + {2_|3₊_{j(|) and i}

|({> |) = {@| + 3{2|2+j0(|). But i|({> |) = 3{2|2+{@| so j0(|) = 0 j(|) = N and i({> |) = { ln | + {2_|3₊_{N is a potential function for F.}

11. (a)F has continuous rst-order partial derivatives and C

C|2{| = 2{ = CC{({2)onR2, which is open and simply-connected. Thus,F is conservative by Theorem 6. Then we know that the line integral of F is independent of path; in particular, the value ofU_FF · gr depends only on the endpoints of F. Since all three curves have the same initial and terminal points, U

FF · gr will have the same value for each curve.

(b) We rst nd a potential functioni, so that i = F. We know i_{({> |) = 2{| and i_|({> |) = {2. Integrating i{({> |) with respect to {, we have i({> |) = {2| + j(|). Differentiating both sides with respect to | gives i|({> |) = {2+j0(|), so we must have {2+j0(|) = {2 j0(|) = 0 j(|) = N, a constant. Thusi({> |) = {2| + N. All three curves start at (1> 2) and end at (3> 2), so by Theorem 2,

U

FF · gr = i(3> 2) i(1> 2) = 18 2 = 16 for each curve.

13. (a)i_{({> |) = {|2impliesi({> |) =1₂{2|2+j(|) and i_|({> |) = {2| + j0(|). But i_|({> |) = {2| so j0(|) = 0 j(|) = N, a constant. We can take N = 0, so i({> |) = 1

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SECTION 17.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS ET SECTION 16.3 ¤ 277 (b) The initial point ofF is r(0) = (0> 1) and the terminal point is r(1) = (2> 1), so

U

FF · gr = i(2> 1) i(0> 1) = 2 0 = 2.

15. (a)i_{({> |> }) = |} implies i({> |> }) = {|} + j(|> }) and so i_|({> |> }) = {} + j_|(|> }). But i_|({> |> }) = {} so j|(|> }) = 0 j(|> }) = k(}). Thus i({> |> }) = {|} + k(}) and i}({> |> }) = {| + k0(}). But

i}({> |> }) = {| + 2}, so k0(}) = 2} k(}) = }2+N. Hence i({> |> }) = {|} + }2(takingN = 0). (b)U_F F · gr = i(4> 6> 3) i(1> 0> 2) = 81 4 = 77.

17. (a)i_{({> |> }) = |2cos} implies i({> |> }) = {|2cos} + j(|> }) and so i_|({> |> }) = 2{| cos } + j_|(|> }). But i|({> |> }) = 2{| cos } so j|(|> }) = 0 j(|> }) = k(}). Thus i({> |> }) = {|2cos} + k(}) and

i}({> |> }) = {|2sin} + k0(}). But i}({> |> }) = {|2sin}, so k0(}) = 0 k(}) = N. Hence i({> |> }) = {|2_cos_{} (taking N = 0).}

(b)r(0) = h0> 0> 0i, r() = 2> 0> soU_F F · gr = i(2> 0> ) i(0> 0> 0) = 0 0 = 0.

19.HereF({> |) = tan | i + { sec2| j. Then i({> |) = { tan | is a potential function for F, that is, i = F so F is conservative and thus its line integral is independent of path. Hence

U

Ftan| g{ + { sec2| g| =

U

FF · g r =i

2>₄ i(1> 0) = 2 tan₄ tan 0 = 2.

21.F({> |) = 2|3@2i + 3{s| j, Z =U_FF · g r. Since C(2|3@2)@C| = 3 | = C(3{s| )@C{, there exists a function i such thati = F. In fact, i{({> |) = 2|3@2 i({> |) = 2{|3@2+j(|) i|({> |) = 3{|1@2+j0(|). But

i|({> |) = 3{ | so j0(|) = 0 or j(|) = N. We can take N = 0 i({> |) = 2{|3@2. Thus

Z =U_FF · g r = i(2> 4) i(1> 1) = 2(2)(8) 2(1) = 30.

23.We know that if the vector eld (call itF) is conservative, then around any closed path F,U_FF · gr = 0. But take F to be a circle centered at the origin, oriented counterclockwise. All of the eld vectors that start onF are roughly in the direction of motion alongF, so the integral around F will be positive. Therefore the eld is not conservative.

25. From the graph, it appears thatF is conservative, since around all closed paths, the number and size of the eld vectors pointing in directions similar to that of the path seem to be roughly the same as the number and size of the vectors pointing in the opposite direction. To check, we calculate

C

C|(sin|) = cos | = CC{(1 +{ cos |). Thus F is conservative, by Theorem 6.

27.SinceF is conservative, there exists a function i such that F = i, that is, S = i{,T = i|, andU = i}. SinceS , T and U have continuous rst order partial derivatives, Clairaut’s Theorem says that CS@C| = i{|=i|{=CT@C{,

CS@C} = i{}=i}{=CU@C{, and CT@C} = i|}=i}| =CU@C|. THE FUNDHE FUND

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29. G = {({> |) | { A 0, | A 0} = the rst quadrant (excluding the axes). (a)G is open because around every point in G we can put a disk that lies in G.

(b)G is connected because the straight line segment joining any two points in G lies in G. (c)G is simply-connected because it’s connected and has no holes.

31. G =({> |) | 1 ? {2+|2? 4=the annular region between the circles with center (0> 0) and radii 1 and 2. (a)G is open.

(b)G is connected.

(c)G is not simply-connected. For example, {2+|2= (1=5)2is simple and closed and lies withinG but encloses points that are not inG. (Or we can say, G has a hole, so is not simply-connected.)

33. (a)S = | {2₊_|2,CS_C| = | 2_{2 ({2+|2)2 andT = { {2₊_|2, CT_C{ = | 2_{2 ({2+|2)2. Thus CS C| =CTC{. (b)F1:{ = cos w, | = sin w, 0 w , F2: { = cos w, | = sin w, w = 2 to w = . Then

]

F1

F · gr =]

0

( sin w)( sin w) + (cos w)(cos w) cos2w + sin2w gw = ] 0 gw = and ] F2 F · gr =] 2gw =

Since these aren’t equal, the line integral ofF isn’t independent of path. (Or notice thatU_F

3F · gr =

U2

0 gw = 2 where

F3is the circle{2+|2= 1, and apply the contrapositive of Theorem 3.) This doesn’t contradict Theorem 6, since the

domain ofF, which is R2except the origin, isn’t simply-connected.

17.4 Green's Theorem

ET 16.4

1. (a) Parametric equations forF are { = 2 cos w, | = 2 sin w, 0 w 2. Then K

F({ |) g{ + ({ + |) g| =

U₂

0 [(2 cosw 2 sin w)(2 sin w) + (2 cos w + 2 sin w)(2 cos w)] gw

=U₀2(4 sin2w + 4 cos2w) gw =U₀24gw = 4w2₀ = 8

(b) Note thatF as given in part (a) is a positively oriented, smooth, simple closed curve. Then by Green’s Theorem, K F({ |) g{ + ({ + |) g| = UU G k C C{({ + |) C|C ({ |) l gD =UU_G[1 (1)] gD = 2UU_GgD = 2D(G) = 2(2)2= 8 3. (a) F1: { = w g{ = gw, | = 0 g| = 0 gw, 0 w 1. F2: { = 1 g{ = 0 gw, | = w g| = gw, 0 w 2. F3: { = 1 w g{ = gw, | = 2 2w g| = 2 gw, 0 w 1. Thus K F{| g{ + {2|3g| = K F1+ F2+ F3 {| g{ + {2_|3_g| =U₀10gw +U₀2w3gw +U₀1(1 w)(2 2w) 2(1 w)2(2 2w)3gw = 0 +1₄w42₀+2₃(1 w)3+8₃(1 w)61₀= 410₃ =2₃

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SECTION 17.4 GREEN’S THEOREM ET SECTION 16.4 ¤ 279 (b)K_F{| g{ + {2|3g| =UU_Gk_C{C ({2|3)_C|C ({|)lgD =U₀1U₀2{(2{|3 {) g| g{ =U₀11₂{|4 {||=2{ |=0 g{ = U1 0(8{5 2{2)g{ =43 23 = 23

5. The regionG enclosed by F is given by {({> |) | 0 { 2> { | 2{}, so U F{|2g{ + 2{2| g| = UU G k C C{(2{2|) C|C ({|2) l gD =U₀2U_{2{(4{| 2{|) g| g{ =U₀2{|2|=2{_|={ g{ =U₀23{3g{ = 3₄{42₀= 12 7.U_F| + h{g{ + (2{ + cos |2)g| =UU_Gk_C{C (2{ + cos |2)_C|C | + h{lgD =U₀1U_||2 (2 1) g{ g| =U₀1(|1@2 |2)g| =1₃ 9.U_F|3g{ {3g| =UU_G k C C{({3)C|C (|3) l gD =UU_G(3{2 3|2)gD =U₀2U₀2(3u2)u gu g =3U₀2gU₀2u3gu = 3(2)(4) = 24

11.F({> |) = { + |3> {2+| and the region G enclosed by F is given by {({>|) | 0 { >0 | sin{}. F is traversed clockwise, so F gives the positive orientation.

U FF · g r = U F _{{ + |}3_{g{ +}_{2₊| g| = UU G k C C{ {2₊s_| C C| { + |3l_gD =U₀U₀sin {(2{ 3|2)g| g{ = U₀2{| |3|=sin {_|=0 g{

=U₀(2{ sin { sin3{) g{ = U₀(2{ sin { (1 cos2{) sin {) g{ =2 sin{ 2{ cos { + cos { 1₃cos3{₀ [integrate by parts in the rst term] =2 2 +2₃=4₃ 2

13.F({> |) = h{+{2|> h| {|2and the regionG enclosed by F is the disk {2+|2 25. F is traversed clockwise, so F gives the positive orientation.

U FF · g r = U F(h{+{2|) g{ + (h| {|2)g| = UU G k C C{(h| {|2)C|C (h{+{2|) l gD =UU_G(|2 {2)gD =UU_G({2+|2)gD =U₀2U₀5(u2)u gu g =U₀2gU₀5u3gu = 21₄u45₀= 625₂

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15. HereF = F₁+F₂where

F1can be parametrized as{ = w, | = 1, 1 w 1, and

F2is given by{ = w, | = 2 w2, 1 w 1.

Then the line integral is K F1+F2 |2_h{_{g{ + {}2_h|_{g| =}U1 1[1· hw+w2h · 0] gw +U₁1 [(2 w2)2hw(1) + (w)2h2w2(2w)] gw =U₁1 [hw (2 w2)2hw 2w3h2w2]gw = 8h + 48h1 according to a CAS. The double integral is

]] G CT C{ CSC| gD =] 1 1 ] 2{2 1

(2{h| 2|h{)g| g{ = 8h + 48h1, verifying Green’s Theorem in this case.

17. By Green’s Theorem,Z =U_FF · gr =U_F{({ + |) g{ + {|2g| =UU_G(|2 {) g| g{ where F is the path described in the question andG is the triangle bounded by F. So

Z =U1 0 U1{ 0 (|2 {) g| g{ = U1 0 ₁ 3|3 {| | = 1{ | = 0 g{ = U1 0 ₁ 3(1 {)3 {(1 {) g{ =₁₂1(1 {)41₂{2+1₃{3₀1=1₂+1₃₁₂1=₁₂1

19. LetF1be the arch of the cycloid from (0> 0) to (2> 0), which corresponds to 0 w 2, and let F2be the segment from (2> 0) to (0> 0), so F2is given by{ = 2 w, | = 0, 0 w 2. Then F = F1 F2is traversed clockwise, soF is oriented positively. ThusF encloses the area under one arch of the cycloid and from (5) we have

D = K_F | g{ =U_F₁| g{ +U_F₂| g{ =U₀2(1 cos w)(1 cos w) gw +U₀20 (gw) =U₀2(1 2 cos w + cos2w) gw + 0 =w 2 sin w +1₂w +1₄sin 2w2

0 = 3

21. (a) Using Equation 17.2.8 [ ET 16.2.8], we write parametric equations of the line segment as{ = (1 w){1+w{2, | = (1 w)|1+w|2, 0 w 1. Then g{ = ({2 {1)gw and g| = (|2 |1)gw, so U F{ g| | g{ = U1 0 [(1 w){1+w{2](|2 |1)gw + [(1 w)|1+w|2]({2 {1)gw =U₀1({₁(|₂ |₁) |₁({₂ {₁) +w[(|₂ |₁)({₂ {₁) ({₂ {₁)(|₂ |₁)])gw =U₀1({1|2 {2|1)gw = {1|2 {2|1

(b) We apply Green’s Theorem to the pathF = F1 F2 · · · Fq, whereFlis the line segment that joins ({l> |l)to ({l+1> |l+1)forl = 1, 2, = = =, q 1, and Fqis the line segment that joins ({q> |q)to ({1> |1). From (5),

1 2

U

F{ g| | g{ =

UU

G gD, where G is the polygon bounded by F. Therefore

area of polygon =D(G) =UU_G gD = 1₂U_F{ g| | g{ =1₂U_F 1{ g| | g{ + U F2{ g| | g{ + · · · + U Fq1{ g| | g{ + U Fq{ g| | g{ To evaluate these integrals we use the formula from (a) to get

D(G) =1

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SECTION 17.4 GREEN’S THEOREM ET SECTION 16.4 ¤ 281 (c)D =1₂[(0· 1 2 · 0) + (2 · 3 1 · 1) + (1 · 2 0 · 3) + (0 · 1 (1) · 2) + (1 · 0 0 · 1)]

= 1₂(0 + 5 + 2 + 2) =9₂

23.We orient the quarter-circular region as shown in the gure. D = 1 4d2so{ = 1 d2_@2 L F { 2_{g| and | =} 1 d2_@2 L F| 2_g{. HereF = F1+F2+F3whereF1:{ = w, | = 0, 0 w d; F2:{ = d cos w, | = d sin w, 0 w ₂; and

F3:{ = 0, | = d w, 0 w d. Then K F{2g| = U F1{ 2_{g| +}U F2{ 2_{g| +}U F3{ 2_{g| =}Ud 0 0gw + U@2 0 (d cos w)2(d cos w) gw + Ud 0 0gw

=U₀@2d3cos3w gw = d3U₀@2(1 sin2w) cos w gw = d3sinw 1₃sin3w@2₀ =2₃d3 so{ = 1 d2_@2 L F{ 2_{g| =} 4d 3. K F|2g{ = U F1| 2_{g{ +}U F2| 2_{g{ +}U F3| 2_{g{ =}Ud 0 0gw + U_@2 0 (d sin w)2(d sin w) gw + U_d 0 0gw

=U₀@2(d3sin3w) gw = d3U₀@2(1 cos2w) sin w gw = d31₃cos3w cos w@2

0 =23d3, so| = 1 d2_@2 L F| 2_{g{ =} 4d 3. Thus ({> |) = 4d 3> 4d 3 .

25.By Green’s Theorem,1₃K_F|3g{ = 1₃UU_G(3|2)gD =UU_G|2 gD = L{and

1 3 K F{3g| = 13 UU G(3{2)gD = UU G{2 gD = L|.

27.SinceF is a simple closed path which doesn’t pass through or enclose the origin, there exists an open region that doesn’t contain the origin but does containG. Thus S = |@({2+|2)andT = {@({2+|2)have continuous partial derivatives on this open region containingG and we can apply Green’s Theorem. But by Exercise 17.3.33(a) [ ET 16.3.33(a)],

CS@C| = CT@C{, soK_FF · gr =UU_G0gD = 0.

29.Using the rst part of (5), we have thatUU_Ug{ g| = D(U) =U_CU{ g|. But { = j(x> y), and g| = Ck

Cxgx + CkCygy, and we orientCV by taking the positive direction to be that which corresponds, under the mapping, to the positive direction alongCU, so ] CU{ g| = ] CVj(x> y) Ck Cxgx + CkCygy = ] CVj(x> y) CkCxgx + j(x> y) CkCygy

=±UU_V_CxC j(x> y)Ck_Cy_CyC j(x> y)Ck_CxgD [using Green’s Theorem in thexy-plane] =±UU_V_CxCjCk_Cy +j(x> y)_{Cx Cy}C2k Cj_CyCk_Cx j(x> y)_{Cy Cx}C2kgD [using the Chain Rule] =±UU_V_CxC{C|_Cy C{_Cy_CxC|gD [by the equality of mixed partials] = ±UU_VC({>|)_C(x>y)gx gy

The sign is chosen to be positive if the orientation that we gave toCV corresponds to the usual positive orientation, and it is negative otherwise. In either case, sinceD(U) is positive, the sign chosen must be the same as the sign of C({> |)

C(x> y). ThereforeD(U) = ]] Ug{ g| = ]] V C({> |)_{C(x> y)}gxgy.

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17.5 Curl and Divergence

ET 16.5

1. (a) curlF = × F = i j k C@C{ C@C| C@C} {|} 0 {2| = ({2 0) i (2{| {|) j + (0 {}) k ={2i + 3{| j {} k (b) divF = · F = C C{({|}) + CC|(0) +C}C ({2|) = |} + 0 + 0 = |} 3. (a) curlF = × F = i j k C@C{ C@C| C@C} 1 { + |} {| } = ({ |) i (| 0) j + (1 0) k = ({ |) i | j + k (b) divF = · F = C C{(1) +C|C ({ + |}) + CC} {| }=} 1 2} 5. (a) curlF = × F = i j k C@C{ C@C| C@C} { s {2₊_|2₊_}2 | s {2₊_|2₊_}2 } s {2₊_|2₊_}2 = 1 ({2+|2+}2)3@2 [(|} + |}) i ({} + {}) j + ({| + {|) k] = 0 (b) divF = · F = C C{ # { s {2₊_|2₊_}2 $ + C C| # | s {2₊_|2₊_}2 $ + C C} # } s {2₊_|2₊_}2 $ ={ 2₊_|2₊_}2_{2 ({2+|2+}2)3@2 + {2₊_|2₊_}2_|2 ({2+|2+}2)3@2 + {2₊_|2₊_}2_}2 ({2+|2+}2)3@2 = 2{2+ 2|2+ 2}2 ({2+|2+}2)3@2 = 2 s {2₊_|2₊_}2 7. (a) curlF = × F = i j k C@C{ C@C| C@C} ln{ ln({|) ln({|}) = {} {|} 0 i |} {|} 0 j + | {| 0 k = 1 |> 1 {> 1 { (b) divF = · F = C C{(ln{) + CC|(ln({|)) + CC}(ln({|})) = 1 {+{|{ +{|}{| = 1 {+ 1 |+ 1 }

9. If the vector eld isF = S i + T j + U k, then we know U = 0. In addition, the {-component of each vector of F is 0, so S = 0, hence CS_C{ =CS

C| =CSC} = CUC{ = CUC| =CUC} = 0.T decreases as | increases, so CTC| ? 0, but T doesn’t change in the{- or }-directions, so CT C{ =CTC} = 0. (a) divF = CS C{ +CTC| +CUC} = 0 + CTC| + 0? 0 (b) curlF = CU C| CTC} i + CS C} CUC{ j + CT C{ CSC| k = (0 0) i + (0 0) j + (0 0) k = 0

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SECTION 17.5 CURL AND DIVERGENCE ET SECTION 16.5 ¤ 283 11.If the vector eld isF = S i + T j + U k, then we know U = 0. In addition, the |-component of each vector of F is 0, so

T = 0, hence CT_C{ = CT

C| = CTC} = CUC{ =CUC| =CUC} = 0.S increases as | increases, so CSC| A 0, but S doesn’t change in the{- or }-directions, so CS C{ = CSC} = 0. (a) divF = CS C{ +CTC| +CUC} = 0 + 0 + 0 = 0 (b) curlF = CU C| CTC} i + CS C} CUC{ j + CT C{ CSC| k = (0 0) i + (0 0) j + 0 CS C| k = CS_C|k SinceCS

C| A 0, CSC|k is a vector pointing in the negative }-direction.

13.curlF = × F = i j k C@C{ C@C| C@C} |2_}3 ₂_{|}3 ₃_{|2_}2 = (6{|}2 6{|}2)i (3|2}2 3|2}2)j + (2|}3 2|}3)k = 0

andF is dened on all of R3with component functions which have continuous partial derivatives, so by Theorem 4, F is conservative. Thus, there exists a function i such that F = i. Then i{({> |> }) = |2}3implies

i({> |> }) = {|2_}3₊_{j(|> }) and i}

|({> |> }) = 2{|}3+j|(|> }). But i|({> |> }) = 2{|}3, soj(|> }) = k(}) and i({> |> }) = {|2_}3₊_{k(}). Thus i}

}({> |> }) = 3{|2}2+k0(}) but i}({> |> }) = 3{|2}2sok(}) = N, a constant. Hence a potential function forF is i({> |> }) = {|2}3+N.

15.curlF = × F = i j k C@C{ C@C| C@C} 2{| {2+ 2|} |2 = (2| 2|) i (0 0) j + (2{ 2{) k = 0, F is dened on all of R3,

and the partial derivatives of the component functions are continuous, soF is conservative. Thus there exists a function i such thati = F. Then i_{({> |> }) = 2{| implies i({> |> }) = {2| + j(|> }) and i_|({> |> }) = {2+j_|(|> }). But i|({> |> }) = {2+ 2|}, so j(|> }) = |2} + k(}) and i({> |> }) = {2| + |2} + k(}). Thus i}({> |> }) = |2+k0(}) but

i}({> |> }) = |2sok(}) = N and i({> |> }) = {2| + |2} + N. 17.curlF = × F = i j k C@C{ C@C| C@C} |h{ _h{ ₂_} = (0 0) i (0 0) j + (h{ h{)k = 2h{k 6= 0, soF is not conservative.

19.No. Assume there is such aG. Then div(curl G) = C

C{({ sin |) + CC|(cos|) + CC}(} {|) = sin | sin | + 1 6= 0, which contradicts Theorem 11.

21.curlF = i j k C@C{ C@C| C@C} i({) j(|) k(}) = (0 0) i + (0 0) j + (0 0) k = 0. Hence F = i({) i + j(|) j + k(}) k is irrotational.

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For Exercises 23–29, letF({> |> }) = S1i + T1j + U1kandG({> |> }) = S2i + T2j + U2k. 23. div(F + G) = divhS1+S2> T1+T2> U1+U2i = C(S1+S2) C{ +C(T1 +T₂) C| +C(U1 +U₂) C} = CS1 C{ +CSC{2 +CTC|1 +CTC|2 +CUC}1 +CUC}2 = CS1 C{ +CTC|1 +CUC}1 + CS2 C{ +CTC|2 +CUC}2 = divhS₁> T₁> U₁i + divhS₂> T₂> U₂i = div F + div G

25. div(iF) = div(i hS₁> T₁> U₁i) = divhiS₁> iT₁> iU₁i = C(iS1)

C{ +C(iT1 ) C| +C(iU1 ) C} = i CS_C{1 +S₁Ci C{ + i CT_C|1 +T₁Ci C| + i CU_C}1 +U₁ Ci C} =i CS1 C{ +CTC|1 +CUC}1 +hS1> T1> U1i · Ci C{> CiC|> CiC} =i div F + F · i 27. div(F × G) = · (F × G) = C@C{ C@C| C@C} S1 T1 U1 S2 T2 U2 = C C{ T1 U1 T2 U2 C|C S1 U1 S2 U2 +C}C S1 T1 S2 T2 = T1CU2 C{ +U2 CTC{1 T2CUC{1 U1CTC{2 S1CU2 C| +U2CSC|1 S2CUC|1 U1CSC|2 + S1CT_C}2 +T2CS_C}1 S2CT_C}1 T1CS_C}2 = S2 CU1 C| CTC}1 +T₂ CS1 C} CUC{1 +U₂ CT1 C{ CSC|1 S1 CU2 C| CTC}2 +T1 CS2 C} CUC{2 +U1 CT2 C{ CSC|2 =G · curl F F · curl G 29. curl(curlF) = × ( × F) = i j k C@C{ C@C| C@C} CU1@C| CT1@C} CS1@C} CU1@C{ CT1@C{ CS1@C| = C2_T 1 C|C{ C 2_S 1 C|2 C 2_S 1 C}2 +C 2_U 1 C}C{ i + C2_U 1 C}C| C 2_T 1 C}2 C 2_T 1 C{2 +C 2_S 1 C{C| j + C2_S 1 C{C} C 2_U 1 C{2 C 2_U 1 C|2 +C 2_T 1 C|C} k Now let’s consider grad(divF) 2F and compare with the above. (Note that2F is dened on page 1102 [ ET 1066].)

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SECTION 17.5 CURL AND DIVERGENCE ET SECTION 16.5 ¤ 285 grad(divF) 2F = C2_S 1 C{2 + C2_T 1 C{C| +C 2_U 1 C{C} i + C2_S 1 C|C{+C 2_T 1 C|2 + C2_U 1 C|C} j + C2_S 1 C}C{+C 2_T 1 C}C| +C 2_U 1 C}2 k C2_S 1 C{2 +C 2_S 1 C|2 +C 2_S 1 C}2 i + C2_T 1 C{2 +C 2_T 1 C|2 +C 2_T 1 C}2 j + C2_U 1 C{2 + C2_U 1 C|2 + C2_U 1 C}2 k = C2_T 1 C{C|+C 2_U 1 C{C} C 2_S 1 C|2 C 2_S 1 C}2 i + C2_S 1 C|C{+C 2_U 1 C|C} C 2_T 1 C{2 C 2_T 1 C}2 j + C2_S 1 C}C{+C 2_T 1 C}C| C 2_U 1 C{2 C 2_U 2 C|2 k

Then applying Clairaut’s Theorem to reverse the order of differentiation in the second partial derivatives as needed and comparing, we have curl curlF = grad div F 2F as desired.

31. (a)u = s{2+|2+}2=s { {2₊_|2₊_}2i + | s {2₊_|2₊_}2 j + } s {2₊_|2₊_}2 k = { i +| j + } k s {2₊_|2₊_}2 = r u (b) × r = i j k C C{ C|C C}C { | } = C C|(}) CC}(|) i + C C}({) CC{(}) j + C C{(|) CC|({) k = 0 (c) 1 u = # 1 s {2₊_|2₊_}2 $ = 1 2s{2+|2+}2(2{) {2₊_|2₊_}2 i 1 2s{2+|2+}2 (2|) {2₊_|2₊_}2 j 1 2s{2+|2+}2 (2}) {2₊_|2₊_}2 k = { i+| j + } k ({2+|2+}2)3@2 = ru3 (d) ln u = ln({2+|2+}2)1@2=1₂ ln({2+|2+}2) = { {2₊_|2₊_}2i + _{2₊_||2₊_}2j +_{2₊_|}2₊_}2 k = { i +| j + } k {2₊_|2₊_}2 = _ur2

33.By (13),K_Fi(j) · n gv =UU_Gdiv(ij) gD =UU_G[i div(j) + j · i] gD by Exercise 25. But div(j) = 2j. HenceUU_Gi2j gD =K_Fi(j) · n gv UU_Gj · i gD.

35.Leti({> |) = 1. Then i = 0 and Green’s rst identity (see Exercise 33) says UU G2j gD = K F(j) · n gv UU G0 · j gD UU G2j gD = K Fj · n gv. But j is harmonic on G, so 2_{j = 0} K Fj · n gv = 0 and K FGnj gv = K F(j · n) gv = 0.

37. (a) We know that$ = y@g, and from the diagram sin = g@u y = g$ = (sin )u$ = |w × r|. But v is perpendicular to bothw and r, so that v = w × r.

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(b) From (a),v = w × r = i j k 0 0 $ { | } = (0· } $|) i + (${ 0 · }) j + (0 · | { · 0) k = $| i + ${ j (c) curlv = × v = i j k C@C{ C@C| C@C} $| ${ 0 = C C|(0) CC}(${) i + C C}($|) CC{(0) j + C C{(${) CC|($|) k = [$ ($)] k = 2$ k = 2w

39. For any continuous functioni on R3, dene a vector eldG({> |> }) = hj({> |> })> 0> 0i where j({> |> }) =U₀{i (w> |> }) gw. Then divG = C

C{(j({> |> })) + CC|(0) +C}C (0) = C{C U{

0 i(w> |> }) gw = i({> |> }) by the Fundamental Theorem of

Calculus. Thus every continuous functioni on R3is the divergence of some vector eld.

17.6 Parametric Surfaces and Their Areas

ET 16.6

1. S (7> 10> 4) lies on the parametric surface r(x> y) = h2x + 3y> 1 + 5x y> 2 + x + yi if and only if there are values for x andy where 2x + 3y = 7, 1 + 5x y = 10, and 2 + x + y = 4. But solving the rst two equations simultaneously gives x = 2, y = 1 and these values do not satisfy the third equation, so S does not lie on the surface.

T(5> 22> 5) lies on the surface if 2x + 3y = 5, 1 + 5x y = 22, and 2 + x + y = 5 for some values of x and y. Solving the rst two equations simultaneously givesx = 4, y = 1 and these values satisfy the third equation, so T lies on the surface. 3. r(x> y) = (x + y) i + (3 y) j + (1 + 4x + 5y) k = h0> 3> 1i + x h1> 0> 4i + y h1> 1> 5i. From Example 3, we recognize

this as a vector equation of a plane through the point (0> 3> 1) and containing vectors a = h1> 0> 4i and b = h1> 1> 5i. If we

wish to nd a more conventional equation for the plane, a normal vector to the plane isa × b = i j k 1 0 4 11 5 = 4i j k

and an equation of the plane is 4({ 0) (| 3) (} 1) = 0 or 4{ | } = 4.

5. r(v> w) = v> w> w2 v2, so the corresponding parametric equations for the surface are{ = v, | = w, } = w2 v2. For any point ({> |> }) on the surface, we have } = |2 {2. With no restrictions on the parameters, the surface is} = |2 {2, which we recognize as a hyperbolic paraboloid.

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SECTION 17.6 PARAMETRIC SURFACES AND THEIR AREAS ET SECTION 16.6 ¤ 287 7.r(x> y) = x2+ 1> y3+ 1> x + y, 1 x 1, 1 y 1.

The surface has parametric equations{ = x2+ 1,| = y3+ 1,} = x + y, 1 x 1, 1 y 1. In Maple, the surface can be graphed by entering plot3d([uˆ2+1,vˆ3+1,u+v],u=-1..1,v=-1..1);. In

Mathematica we use the ParametricPlot3D command. If we keepx constant atx₀,{ = x2₀+ 1, a constant, so the corresponding grid curves must be the curves parallel to the|}-plane. If y is constant, we have | = y3₀+ 1, a constant, so these grid curves are the curves parallel to the{}-plane. 9.r(x> y) = x cos y> x sin y> x5.

The surface has parametric equations{ = x cos y, | = x sin y, } = x5_, _{1 x 1, 0 y 2. Note that if x = x}

0is constant then} = x5₀is constant and{ = x0cosy, | = x0siny describe a circle in{, | of radius |x0|, so the corresponding grid curves are

circles parallel to the{|-plane. If y = y0, a constant, the parametric equations become{ = x cos y0,| = x sin y0,} = x5. Then | = (tan y0){, so these are the grid curves we see that lie in vertical

planes| = n{ through the }-axis.

11.{ = sin y, | = cos x sin 4y, } = sin 2x sin 4y, 0 x 2, ₂ y ₂. Note that ify = y₀is constant, then{ = sin y₀is constant, so the

corresponding grid curves must be parallel to the|}-plane. These are the vertically oriented grid curves we see, each shaped like a “gure-eight.” Whenx = x0is held constant, the parametric equations become{ = sin y, | = cos x₀sin 4y,

} = sin 2x0sin 4y. Since } is a constant multiple of |, the corresponding grid curves are the curves contained in planes } = n| that pass through the {-axis.

13.r(x> y) = x cos y i + x sin y j + y k. The parametric equations for the surface are { = x cos y, | = x sin y, } = y. We look at the grid curves rst; if we xy, then { and | parametrize a straight line in the plane } = y which intersects the }-axis. If x is held constant, the projection onto the{|-plane is circular; with } = y, each grid curve is a helix. The surface is a spiraling ramp, graph I.

15.r(x> y) = sin y i + cos x sin 2y j + sin x sin 2y k. Parametric equations for the surface are { = sin y, | = cos x sin 2y, } = sin x sin 2y. If y = y0is xed, then{ = sin y0is constant, and| = (sin 2y0) cosx and } = (sin 2y0) sinx describe a

circle of radius|sin 2y₀|, so each corresponding grid curve is a circle contained in the vertical plane { = sin y₀parallel to the

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|}-plane. The only possible surface is graph II. The grid curves we see running lengthwise along the surface correspond to holdingx constant, in which case | = (cos x0) sin 2y, } = (sin x0) sin 2y } = (tan x0)|, so each grid curve lies in a plane} = n| that includes the {-axis.

17. { = cos3x cos3y, | = sin3x cos3y, } = sin3y. If y = y0is held constant then} = sin3y0is constant, so the

corresponding grid curve lies in a horizontal plane. Several of the graphs exhibit horizontal grid curves, but the curves for this surface are neither circles nor straight lines, so graph III is the only possibility. (In fact, the horizontal grid curves here are members of the family{ = d cos3x, | = d sin3x and are called astroids.) The vertical grid curves we see on the surface correspond tox = x₀held constant, as then we have{ = cos3x₀ cos3y, | = sin3x0 cos3y so the corresponding grid curve lies in the vertical plane| = (tan3x₀){ through the }-axis.

19. From Example 3, parametric equations for the plane through the point (1> 2> 3) that contains the vectors a = h1> 1> 1i and b = h1> 1> 1i are { = 1 + x(1) + y(1) = 1 + x + y, | = 2 + x(1) + y(1) = 2 + x y,

} = 3 + x(1) + y(1) = 3 x + y.

21. Solving the equation for| gives |2= 1 {2+}2 | =1 {2+}2. (We choose the positive root since we want the part of the hyperboloid that corresponds to| 0.) If we let { and } be the parameters, parametric equations are { = {, } = }, | =1 {2+}2.

23. Since the cone intersects the sphere in the circle{2+|2= 2,} =2and we want the portion of the sphere above this, we can parametrize the surface as{ = {, | = |, } =s4 {2 |2where{2+|2 2.

Alternate solution: Using spherical coordinates, { = 2 sin ! cos , | = 2 sin ! sin , } = 2 cos ! where 0 !  4 and

0 2.

25. Parametric equations are{ = {, | = 4 cos , } = 4 sin , 0 { 5, 0 2.

27. The surface appears to be a portion of a circular cylinder of radius 3 with axis the{-axis. An equation of the cylinder is |2₊_}2_{= 9, and we can impose the restrictions 0}_{{ 5, | 0 to obtain the portion shown. To graph the surface on a}

CAS, we can use parametric equations{ = x, | = 3 cos y, } = 3 sin y with the parameter domain 0 x 5,₂ y 3₂ . Alternatively, we can regard{ and } as parameters. Then parametric equations are { = {, } = }, | = 9 }2, where 0 { 5 and 3 } 3.

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SECTION 17.6 PARAMETRIC SURFACES AND THEIR AREAS ET SECTION 16.6 ¤ 289 31. (a) Replacing cosx by sin x and sin x by cos x gives parametric equations

{ = (2 + sin y) sin x, | = (2 + sin y) cos x, } = x + cos y. From the graph, it appears that the direction of the spiral is reversed. We can verify this observation by noting that the projection of the spiral grid curves onto the{|-plane, given by { = (2 + sin y) sin x, | = (2 + sin y) cos x, } = 0, draws a circle in the clockwise direction for each value ofy. The original equations, on the other hand, give circular projections drawn in the counterclockwise direction. The equation for} is identical in both surfaces, so as} increases, these grid curves spiral up in opposite directions for the two surfaces.

(b) Replacing cosx by cos 2x and sin x by sin 2x gives parametric equations { = (2 + sin y) cos 2x, | = (2 + sin y) sin 2x, } = x + cos y. From the graph, it appears that the number of coils in the surface doubles within the same parametric domain. We can verify this observation by noting that the projection of the spiral grid curves onto the{|-plane, given by { = (2 + sin y) cos 2x, | = (2 + sin y) sin 2x, } = 0 (where y is constant), complete circular revolutions for 0 x while the original surface requires 0 x 2 for a complete revolution. Thus, the new surface winds around twice as fast as the original surface, and since the equation for} is identical in both surfaces, we observe twice as many circular coils in the same }-interval.

33.r(x> y) = (x + y) i + 3x2j + (x y) k.

rx=i + 6x j + k and ry=i k, so rx× ry=6x i + 2 j 6x k.

Since the point (2> 3> 0) corresponds to x = 1, y = 1, a normal vector to the surface at (2> 3> 0) is 6 i + 2 j 6 k, and an equation of the tangent plane is6{ + 2| 6} = 6 or 3{ | + 3} = 3.

35.r(x> y) = x2i + 2x sin y j + x cos y k r(1> 0) = (1> 0> 1). rx= 2x i + 2 sin y j + cos y k and ry= 2x cos y j x sin y k,

so a normal vector to the surface at the point (1> 0> 1) is rx(1> 0) × ry(1> 0) = (2 i + k) × (2 j) = 2 i + 4 k.

Thus an equation of the tangent plane at (1> 0> 1) is 2({ 1) + 0(| 0) + 4(} 1) = 0 or { + 2} = 1.

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37. The surfaceV is given by } = i({> |) = 6 3{ 2| which intersects the {|-plane in the line 3{ + 2| = 6, so G is the triangular region given by({> |)0 { 2> 0 | 3 3₂{. By Formula 9, the surface area ofV is

D(V) = ]] G v 1 + C} C{ 2 + C} C| 2 gD =UU_Gs1 + (3)2+ (2)2gD =14UU_GgD =14D(G) =141₂· 2 · 3= 314= 39. } = i({> |) = 2₃({3@2+|3@2)andG = {({> |) | 0 { 1> 0 | 1}. Then i{={1@2,i|=|1@2and

D(V) =UU_Gt1 + ({ )2+|2gD =U₀1U₀11 +{ + | g| g{ =U₀1 k 2 3({ + | + 1)3@2 l|=1 |=0g{ = 2 3 U1 0 k ({ + 2)3@2 ({ + 1)3@2 l g{ =2₃k2₅({ + 2)5@22₅({ + 1)5@2l1 0= 4 15(35@2 25@2 25@2+ 1) =154(35@2 27@2+ 1) 41. } = i({> |) = {| with 0 {2+|2 1, so i{=|, i|={ D(V) =UU_Gs1 +|2+{2gD =U₀2U₀1u2+ 1u gu g =U₀2k1₃(u2+ 1)3@2lu=1 u=0 g =U₀21₃22 1g =2₃ 22 1 43. } = i({> |) = |2 {2with 1 {2+|2 4. Then

D(V) =UU_Gs1 + 4{2+ 4|2gD =U₀2U₁21 + 4u2u gu g =U₀2gU₁2 u1 + 4u2gu =2 0 k 1 12(1 + 4u2)3@2 l2 1= 6 1717 55

45. A parametric representation of the surface is{ = {, | = 4{ + }2,} = } with 0 { 1, 0 } 1. Hencer{× r}= (i + 4 j) × (2} j + k) = 4 i j + 2} k.

Note: In general, if | = i({> }) then r{× r}=Ci

C{i j + CiC} k and D (V) = ]] G v 1 + Ci C{ 2 + Ci C} 2 gD. Then D(V) =U1 0 U1 0 17 + 4}2g{ g} =U₀117 + 4}2g} = 1₂}17 + 4}2+17₂ ln2} +4}2+ 171 0= 21 2 +174 ln2 +21 ln17 47. rx=h2x> y> 0i, ry=h0> x> yi, and rx× ry= y2> 2xy> 2x2. Then

D(V) =UU_G|rx× ry| gD =U₀1U₀2 y4+ 4x2y2+ 4x4gy gx =U₀1U₀2 s(y2+ 2x2)2gy gx

=U₀1U₀2(y2+ 2x2)gy gx =U₀11₃y3+ 2x2yy=2_y=0gx =U₀18₃+ 4x2gx =8₃x +4₃x31₀= 4

49. } = i({> |) = h{2|2with{2+|2 4.

D(V) =UU_Gt1 +2{h{2|22+2|h{2|22gD =UU_Gs1 + 4({2+|2)h2({2+|2)gD

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SECTION 17.6 PARAMETRIC SURFACES AND THEIR AREAS ET SECTION 16.6 ¤ 291 51. (a)D(V) = ]] G v 1 + C} C{ 2 + C} C| 2 gD =] 6 0 ] 4 0 v 1 + 4{ 2_{+ 4}_|2 (1 +{2+|2)4g| g{.

Using the Midpoint Rule withi({> |) = v 1 + 4{ 2_{+ 4}_|2 (1 +{2+|2)4,p = 3, q = 2 we have D(V) S3 l = 1 2 S m = 1i

{l> |mD = 4 [i(1> 1) + i(1> 3) + i(3> 1) + i(3> 3) + i(5> 1) + i(5> 3)] 24=2055

(b) Using a CAS we haveD(V) = ] 6 0 ] 4 0 v 1 + 4{2+ 4|2

(1 +{2+|2)4 g| g{ 24=2476. This agrees with the estimate in part (a) to the rst decimal place.

53.} = 1 + 2{ + 3| + 4|2, so D(V) = ]] G v 1 + C} C{ 2 + C} C| 2 gD = ] 4 1 ]1 0 s 1 + 4 + (3 + 8|)2g| g{ = ] 4 1 ]1 0 s 14 + 48| + 64|2g| g{. Using a CAS, we have

U4 1 U1 0 s 14 + 48| + 64|2g| g{ =45₈ 14 +15₁₆ln115 + 314515₁₆ln35 +145 or 45₈ 14 +15₁₆ln115 + 370 35 +70 .

55. (a){ = d sin x cos y, | = e sin x sin y, } = f cos x {2

d2 +| 2

e2 +} 2

f2 = (sinx cos y)2+ (sinx sin y)2+ (cosx)2

= sin2x + cos2x = 1

and since the ranges ofx and y are sufcient to generate the entire graph, the parametric equations represent an ellipsoid.

(b)

(c) From the parametric equations (withd = 1, e = 2, and f = 3), we calculate

rx= cosx cos y i + 2 cos x sin y j 3 sin x k and ry= sin x sin y i + 2 sin x cos y j. So

rx× ry= 6 sin2x cos y i + 3 sin2x sin y j + 2 sin x cos x k, and the surface area is given by

D(V) =U₀2U₀|rx× ry| gx gy =U₀2U₀

s

36 sin4x cos2y + 9 sin4x sin2y + 4 cos2x sin2x gx gy 57.To nd the regionG: } = {2+|2implies} + }2= 4} or }2 3} = 0. Thus } = 0 or } = 3 are the planes where the

surfaces intersect. But{2+|2+}2= 4} implies {2+|2+ (} 2)2= 4, so} = 3 intersects the upper hemisphere. Thus (} 2)2= 4 {2 |2or} = 2 +s4 {2 |2. ThereforeG is the region inside the circle {2+|2+ (3 2)2= 4, that is,G =({> |) | {2+|2 3. D(V) = ]] G t 1 + [({)(4 {2 |2)1@2]2+ [(|)(4 {2 |2)1@2]2gD = ] 2 0 ] 3 0 u 1 + u2 4 u2u gu g = ]2 0 ] 3 0 2u gu 4 u2g = ]2 0 k 2(4 u2₎1@2lu= 3 u=0 g =U₀2(2 + 4) g = 22₀ = 4

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59. LetD(V₁)be the surface area of that portion of the surface which lies above the plane} = 0. Then D(V) = 2D(V₁). Following Example 10, a parametric representation ofV1is{ = d sin ! cos , | = d sin ! sin ,

} = d cos ! and |r!× r| = d2sin!. For G, 0 ! ₂ and for each xed!,{ ₂1d2+|2 1₂d2or

d sin ! cos 1 2d

2

+d2sin2! sin2 (d@2)2impliesd2sin2! d2sin! cos 0 or

sin! (sin ! cos ) 0. But 0 ! ₂, so cos sin ! or sin₂ + sin ! or ! ₂ ₂ !. HenceG =(!> ) | 0 ! ₂,! ₂ ₂ !. Then

D(V1) =U₀@2U_{! (@2)}(@2) !d2sin! g g! = d2U₀@2( 2!) sin ! g! =d2[( cos !) 2(! cos ! + sin !)]@2₀ =d2( 2) ThusD(V) = 2d2( 2).

Alternate solution: Working on V1we could parametrize the portion of the sphere by{ = {, | = |, } =sd2 {2 |2. Then|r{× r|| = v 1 + { 2 d2_{2_|2 + |2 d2_{2_|2 = d s d2_{2_|2 and D(V1) = ]] 0 ({ (d@2))2_{+ |}2_(d@2)2 d s d2_{2_|2gD = ] @2 @2 ] d cos 0 d d2_u2u gu g =U_@2@2 d(d2 u2)1@2lu = d cos u = 0 g = U@2 @2d2[1 (1 cos2)1@2]g =U_@2@2 d2(1 |sin |) g = 2d2U₀@2(1 sin ) g = 2d2₂ 1 ThusD(V) = 4d2₂ 1= 2d2( 2). Notes:

(1) Perhaps working in spherical coordinates is the most obvious approach here. However, you must be careful in setting upG.

(2) In the alternate solution, you can avoid having to use|sin | by working in the rst octant and then multiplying by 4. However, if you set upV1as above and arrived atD(V1) =d2, you now see your error.

17.7 Surface Integrals

ET 16.7

1. The faces of the box in the planes{ = 0 and { = 2 have surface area 24 and centers (0> 2> 3), (2> 2> 3). The faces in | = 0 and | = 4 have surface area 12 and centers (1> 0> 3), (1> 4> 3), and the faces in } = 0 and } = 6 have area 8 and centers (1> 2> 0), (1> 2> 6). For each face we take the point S_lmto be the center of the face andi({> |> }) = h0=1({+|+}), so by Denition 1,

UU

Vi({> |> }) gV [i(0> 2> 3)](24) + [i(2> 2> 3)](24) + [i(1> 0> 3)](12)

+ [i(1> 4> 3)](12) + [i(1> 2> 0)](8) + [i(1> 2> 6)](8)

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SECTION 17.7 SURFACE INTEGRALS ET SECTION 16.7 ¤ 293 3.We can use the{}- and |}-planes to divide K into four patches of equal size, each with surface area equal to1₈ the surface

area of a sphere with radius50, so V = 1₈(4)502= 25. Then (±3> ±4> 5) are sample points in the four patches, and using a Riemann sum as in Denition 1, we have

UU

Ki({> |> }) gV i(3> 4> 5) V + i(3> 4> 5) V + i(3> 4> 5) V + i(3> 4> 5) V

= (7 + 8 + 9 + 12)(25) = 900 2827 5.} = 1 + 2{ + 3| so C}

C{ = 2and C}C| = 3. Then by Formula 4, UU V{2|} gV = ]] G{ 2_|}vC} C{ 2 + C} C| 2 + 1gD =U₀3U₀2{2|(1 + 2{ + 3|)4 + 9 + 1g| g{ =14U₀3U₀2({2| + 2{3| + 3{2|2)g| g{ =14U₀31₂{2|2+{3|2+{2|3|=2_|=0g{ =14U₀3(10{2+ 4{3)g{ =1410₃{3+{43₀= 17114

7.V is the part of the plane } = 1 { | over the region G = {({> |) | 0 { 1> 0 | 1 {}. Thus UU V|} gV = UU G|(1 { |) s (1)2+ (1)2+ 1gD =3U₀1U₀1{| {| |2g| g{ =3U₀11₂|21₂{|21₃|3|=1{_|=0 g{ =3U₀11₆(1 {)3g{ = ₂₄3(1 {)4l1 0 = 3 24 9.r(x> y) = x2i + x sin y j + x cos y k, 0 x 1, 0 y @2, so

rx× ry= (2x i + sin y j + cos y k) × (x cos y j x sin y k) = x i + 2x2siny j + 2x2cosy k and

|rx× ry| =

s

x2_{+ 4}_x4_sin2_{y + 4x}4_cos2_{y =}s_x2_{+ 4}_x4_(sin2_{y + cos}2_{y) = x}_{1 + 4}_x2_(since_{x 0). Then by}

Formula 2, UU

V|} gV =

UU

G(x sin y)(x cos y) |rx× ry| gD =

U@2 0

U1

0(x sin y)(x cos y) · x

1 + 4x2gx gy

=U₀1x31 + 4x2gxU₀@2siny cos y gy letw = 1 + 4x2 x2= 1₄(w 1) and1₈gw = x gx =U₁51₈ ·1₄(w 1)w gwU₀@2siny cos y gy = ₃₂1 U₁5w3@2wgwU₀@2siny cos y gy

= ₃₂1k2₅w5@22₃w3@2l5 1 ₁ 2sin2y @2 0 =321 2 5(5)5@223(5)3@225 +23 ·1 2(1 0) = 485 5 +₂₄₀1

11.V is the portion of the cone }2={2+|2for 1 } 3, or equivalently, V is the part of the surface } =s{2+|2over the regionG =({> |) | 1 {2+|2 9. Thus ]] V{ 2_}2_{gV =}]] G{ 2₍_{2₊_|2₎ y x x w#_s { {2₊_|2 $2 + # | s {2₊_|2 $2 + 1gD = ]] G{ 2₍_{2₊_|2₎ v {2₊_|2 {2₊_|2 + 1gD = ]] G 2{2({2+|2)gD =2 ] 2 0 ] 3 1 (u cos )2(u2)u gu g =2U₀2cos2 gU₁3u5gu =21₂ +1₄sin 22₀ 1₆u6₁3=2 () ·1₆(36 1) =364 2 3

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13. Using{ and } as parameters, we have r({> }) = { i + ({2+}2)j + } k, {2+}2 4. Then r{× r}= (i + 2{ j) × (2} j + k) = 2{ i j + 2} k and |r{× r}| =4{2+ 1 + 4}2=s1 + 4({2+}2). Thus UU V| gV = UU {2_+}2₄ ({2+}2)s1 + 4({2+}2)gD =U₀2U₀2u21 + 4u2u gu g =U₀2gU₀2u21 + 4u2u gu = 2U₀2u21 + 4u2u gu letx = 1 + 4u2 u2=1₄(x 1) and1₈gx = u gu = 2U₁171₄(x 1)x ·1₈gx =₁₆1U₁17(x3@2 x1@2)gx =₁₆1 k 2 5x5@223x3@2 l17 1 = 1 16 k 2 5(17)5@223(17)3@225+23 l = 60 39117 + 1

15. Using spherical coordinates and Example 17.6.10 [ ET 16.6.10] we haver(!> ) = 2 sin ! cos i + 2 sin ! sin j + 2 cos ! k and|r_!× r| = 4 sin !. ThenUU_V({2} + |2}) gV =U₀2U₀@2(4 sin2!)(2 cos !)(4 sin !) g! g = 16 sin4!@2

0 = 16.

17. V is given by r(x> y) = x i + cos y j + sin y k, 0 x 3, 0 y @2. Then rx× ry= i × ( sin y j + cos y k) = cos y j sin y k and |rx× ry| =

s cos2y + sin2y = 1, so UU V(} + {2|) gV = U@2 0 U3

0(siny + x2cosy)(1) gx gy =

U@2

0 (3 siny + 9 cos y) gy

= [3 cos y + 9 sin y]@2₀ = 0 + 9 + 3 0 = 12

19. F({> |> }) = {| i + |} j + }{ k, } = j({> |) = 4 {2 |2, andG is the square [0> 1] × [0> 1], so by Equation 10 UU VF · gS = UU G[{|(2{) |}(2|) + }{] gD = U1 0 U1 0[2{2| + 2|2(4 {2 |2) +{(4 {2 |2)]g| g{ =U₀11₃{2+11₃{ {3+34₁₅g{ = 713₁₈₀

21. F({> |> }) = {}h|i {}h|j + } k, } = j({> |) = 1 { |, and G = {({> |) | 0 { 1> 0 | 1 {}. Since V has downward orientation, we have

UU VF · gS = UU G[{}h|(1) ({}h|)(1) + }] gD = U1 0 U1{ 0 (1 { |) g| g{ =U₀11₂{2 { +1₂g{ = 1₆

23. F({> |> }) = { i } j + | k, } = j({> |) =s4 {2 |2andG is the quarter disk

({> |)0 { 2> 0 | 4 {2. V has downward orientation, so by Formula 10, UU VF · gS = UU G k { ·1 2(4 {2 |2)1@2(2{) (}) ·12(4 {2 |2)1@2(2|) + | l gD = ]] G # {2 s 4 {2 |2 s 4 {2 |2·s | 4 {2 |2 +| $ gD =UU_G{2(4 ({2+|2))1@2gD = U₀@2U₀2(u cos )2(4 u2)1@2u gu g

=U₀@2cos2 gU₀2u3(4 u2)1@2gu letx = 4 u2 u2= 4 x and 1₂gx = u gu =U₀@21₂+1₂cos 2gU₄01₂(4 x)(x)1@2gx =1₂ +1₄sin 2@2 0 1 2 k 8x 2₃x3@2l0 4= 4 1 2 16 +16 3 =4₃

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SECTION 17.7 SURFACE INTEGRALS ET SECTION 16.7 ¤ 295 25.LetV₁be the paraboloid| = {2+}2, 0 | 1 and V₂the disk{2+}2 1, | = 1. Since V is a closed

surface, we use the outward orientation.

OnV1:F(r({> })) = ({2+}2)j } k and r{× r}= 2{ i j + 2} k (since the j-component must be negative on V1). Then UU V1F · gS = UU {2_{+ }}2₁ [({2+}2) 2}2]gD = U₀2U₀1(u2+ 2u2cos2) u gu g =U₀2 1₄(1 + 2 cos2) g = ₂ +₂= OnV₂:F(r({> })) = j } k and r_}× r_{=j. ThenUU_V 2F · gS = UU {2_{+ }}2₁ (1)gD = . HenceUU_VF · gS = + = 0.

27.HereV consists of the six faces of the cube as labeled in the gure. On V₁: F = i + 2| j + 3} k, r|× r}=i andUU_V₁F · gS =U₁1 U₁1 g| g} = 4; V2:F = { i + 2 j + 3} k, r}× r{=j andUU_V 2F · gS = U1 1 U1 12g{ g} = 8; V3:F = { i + 2| j + 3 k, r{× r|=k andUU_V₃F · gS =U₁1 U₁1 3g{ g| = 12; V4:F = i + 2| j + 3} k, r}× r|=i andUU_V 4F · gS = 4; V5:F = { i 2 j + 3} k, r{× r} =j andUU_V 5F · gS = 8; V6:F = { i + 2| j 3 k, r|× r{ =k andUU_V₆F · gS =U₁1 U₁1 3g{ g| = 12. HenceUU_VF · gS =S6_l=1UU_V lF · gS = 48.

29.HereV consists of four surfaces: V₁, the top surface (a portion of the circular cylinder|2+}2= 1);V₂, the bottom surface (a portion of the{|-plane); V3, the front half-disk in the plane{ = 2, and V4, the back half-disk in the plane{ = 0. OnV1: The surface is} =s1 |2for 0 { 2, 1 | 1 with upward orientation, so

]] V1 F · gS =] 2 0 ] 1 1 % {2₍₀₎_|2 # s | 1 |2 $ +}2 & g| g{ =] 2 0 ] 1 1 # |3 s 1 |2 + 1 | 2 $ g| g{ =U₀2ks1 |2+1₃(1 |2)3@2+| 1₃|3l|=1 |=1g{ = U2 0 43g{ =83

OnV2: The surface is} = 0 with downward orientation, so UU V2F · gS = U2 0 U1 1 }2_{g| g{ =}U2 0 U1 1(0)g| g{ = 0

OnV₃: The surface is{ = 2 for 1 | 1, 0 } s1 |2, oriented in the positive{-direction. Regarding | and } as parameters, we haver|× r} =i and

UU V3F · gS = U1 1 U 1|2 0 {2g} g| = U1 1 U 1|2 0 4g} g| = 4D (V3) = 2

OnV₄: The surface is{ = 0 for 1 | 1, 0 } s1 |2, oriented in the negative{-direction. Regarding | and } as parameters, we use (r|× r}) =i and

UU V4F · gS = U1 1 U 1|2 0 {2g} g| = U1 1 U 1|2 0 (0)g} g| = 0 ThusUU_VF · gS = 8₃+ 0 + 2 + 0 = 2 +8₃.

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31. } = {| C}@C{ = |, C}@C| = {, so by Formula 4, a CAS gives UU V{|} gV = U1 0 U1 0 {|({|) s |2₊_{2_{+ 1}_{g{ g| 0=1642.}

33. We use Formula 4 with} = 3 2{2 |2 C}@C{ = 4{, C}@C| = 2|. The boundaries of the region 3 2{2 |2 0 are t 3 2 { t 3 2 and

3 2{2 | 3 2{2, so we use a CAS (with precision reduced to seven or fewer digits; otherwise the calculation may take a long time) to calculate

]] V{ 2_|2_}2_{gV =}] 3@2 3@2 ] 3 2{2 3 2{2{ 2_|2₍₃_2{2_|2₎2s₁₆_{₂_{+ 4}_|₂_{+ 1}_{g| g{ 3=4895}

35. IfV is given by | = k({> }), then V is also the level surface i({> |> }) = | k({> }) = 0. n = i_{|i({> |> })|}({> |> }) = k{i + j k}k

k2_{_{+ 1 +}_k2_} , andn is the unit normal that points to the left. Now we proceed as in the

derivation of (10), using Formula 4 to evaluate ]] VF · gS = ]] VF · n gV = ]] G (S i + T j + U k) Ck C{i j + CkC}k v Ck C{ 2 + 1 + Ck C} 2 v Ck C{ 2 + 1 + Ck C} 2 gD

whereG is the projection of V onto the {}-plane. Therefore ]] VF · gS = ]] G S Ck_C{ T + U Ck_C} gD. 37. p =UU_VN gV = N · 41₂d2= 2d2N; by symmetry P_{} =P_|} = 0, and

P{|=UU_V}N gV = NU₀2U₀@2(d cos !)(d2sin!) g! g = 2Nd31₄cos 2!@2₀ =Nd3. Hence ({> |> }) =0> 0>1₂d. 39. (a)L}=UU_V({2+|2)({> |> }) gV (b)L} =UU_V({2+|2) 10s{2+|2 gV = UU 1 {2_{+ |}2₁₆ ({2+|2) 10s{2+|2 2gD =U₀2U₁42 (10u3 u4)gu g = 224329₁₀ =4329₅ 2

41. The rate of ow through the cylinder is the uxUU_Vv · n gV =UU_Vv · gS. We use the parametric representation r(x> y) = 2 cos x i + 2 sin x j + y k for V, where 0 x 2, 0 y 1, so rx=2 sin x i + 2 cos x j, ry=k, and the

outward orientation is given byrx× ry= 2 cosx i + 2 sin x j. Then UU Vv · gS = U2 0 U1 0

y i + 4 sin2_{x j + 4 cos}2_{x k}_{· (2 cos x i + 2 sin x j) gy gx}

=U₀2U₀12y cos x + 8 sin3xgy gx = U₀2cosx + 8 sin3xgx =sinx + 81₃(2 + sin2x) cos x2₀ = 0kg@s

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SECTION 17.8 STOKES’ THEOREM ET SECTION 16.8 ¤ 297 43.V consists of the hemisphere V₁given by} =sd2 {2 |2and the diskV₂given by 0 {2+|2 d2,} = 0.

OnV₁: E = d sin ! cos i + d sin ! sin j + 2d cos ! k,

T!× T=d2sin2! cos i + d2sin2! sin j + d2sin! cos ! k. Thus

UU

V1E · gS =

U2 0

U@2

0 (d3sin3! + 2d3sin! cos2!) g! g

=U₀2U₀@2(d3sin! + d3sin! cos2!) g! g = (2)d31 +1₃= 8₃d3 OnV₂: E = { i + | j, and r_|× r_{=k soUU_V

2E · gS = 0. Hence the total charge is t = %0

UU

VE · gS = 83d3%0.

45.Nx = 6=5(4| j + 4} k). V is given by r({> ) = { i +6 cos j +6 sin k and since we want the inward heat ow, we user_{× r=6 cos j 6 sin k. Then the rate of heat ow inward is given by

UU V(N x) · gS = U2 0 U4 0 (6=5)(24) g{ g = (2)(156)(4) = 1248.

47.LetV be a sphere of radius d centered at the origin. Then |r| = d and F(r) = fr@ |r|3=f@d3({ i + | j + } k). A parametric representation forV is r(!> ) = d sin ! cos i + d sin ! sin j + d cos ! k, 0 ! , 0 2. Then r!=d cos ! cos i + d cos ! sin j d sin ! k, r =d sin ! sin i + d sin ! cos j, and the outward orientation is given byr!× r=d2sin2! cos i + d2sin2! sin j + d2sin! cos ! k. The ux of F across V is

UU

VF · gS =

U 0

U2

0 _df3(d sin ! cos i + d sin ! sin j + d cos ! k)

·d2_sin2_{! cos i + d}2_sin2_{! sin j + d}2_sin_{! cos ! k}_{g g!}

= f d3 U 0 U2 0 d3

sin3! + sin ! cos2!g g! = fU₀U₀2sin! g g! = 4f Thus the ux does not depend on the radiusd.

17.8 Stokes' Theorem

ET 16.8

1.BothK and S are oriented piecewise-smooth surfaces that are bounded by the simple, closed, smooth curve {2+|2= 4, } = 0 (which we can take to be oriented positively for both surfaces). Then K and S satisfy the hypotheses of Stokes’ Theorem, so by (3) we knowUU_KcurlF · gS =U_FF · gr =UU_ScurlF · gS (where F is the boundary curve).

3.The paraboloid} = {2+|2intersects the cylinder{2+|2= 4in the circle{2+|2= 4,} = 4. This boundary curve F should be oriented in the counterclockwise direction when viewed from above, so a vector equation ofF is

r(w) = 2 cos w i + 2 sin w j + 4 k, 0 w 2. Then r0₍_{w) = 2 sin w i + 2 cos w j,}

F(r(w)) = (4 cos2_{w)(16) i + (4 sin}2_{w)(16) j + (2 cos w)(2 sin w)(4) k = 64 cos}2_{w i + 64 sin}2_{w j + 16 sin w cos w k>}

and by Stokes’ Theorem, UU VcurlF · gS = U FF · gr = U2 0 F(r(w)) · r0(w) gw = U2

0 (128 cos2w sin w + 128 sin2w cos w + 0) gw

= 1281₃cos3w +1₃sin3w2₀ = 0