Answers to the Problems in AMBS: Third Quarter

January 30, 2003

Chapter 54

54.1 The tangent plane to the surface x_{3} = f(x_{1}, x_{2}) at the point (x̄_{1}, x̄_{2}) is given by the linearized equation x_{3} = f(x̄_{1}, x̄_{2}) + f′(x̄_{1}, x̄_{2}) [\begin{matrix}x_{1} − x̄_{1} \\ x_{2} − x̄_{2} \end{matrix}], where f′(x_{1}, x_{2}) = ∇f(x_{1}, x_{2}) is the gradient vector. We choose the point (x̄_{1}, x̄_{2}) = (1, 1).

(a) x_{3} = 2 + [2 2] [\begin{matrix}x_{1} − 1 \\ x_{2} − 1 \end{matrix}], i.e., x_{3} = 2x_{1} + 2x_{2}.

(b) x_{3} = 0 + [2 −2] [\begin{matrix}x_{1} − 1 \\ x_{2} − 1 \end{matrix}], i.e., x_{3} = 2x_{1} − 2x_{2}.

(c) x_{3} = 2 + [1 2] [\begin{matrix}x_{1} − 1 \\ x_{2} − 1 \end{matrix}], i.e., x_{3} = x_{1} + 2x_{2} − 1.

(d) x_{3} = 1 + [4 4] [\begin{matrix}x_{1} − 1 \\ x_{2} − 1 \end{matrix}], i.e., x_{3} = 4x_{1} + 4x_{2} − 7.

54.2 (a) f(x) = ∥x∥^{2}x, f(x) − f(y) = ∥x∥^{2}x − ∥y∥^{2}y = (∥x∥^{2} − ∥y∥^{2})x + ∥y∥^{2}(x − y) = ((x + y) · (x − y))x + ∥y∥^{2}(x − y), ∥f(x) − f(y)∥ ≤ |(x+y)·(x−y)|∥x∥+∥y∥^{2}∥x−y∥ ≤ ∥x+y∥∥x−y∥∥x∥+∥y∥^{2}∥x−y∥ = ((∥x + y∥)∥x∥ + ∥y∥^{2})∥x − y∥ ≤ ((∥x∥ + ∥y∥)∥x∥ + ∥y∥^{2})∥x − y∥ ≤ 3∥x − y∥, i.e., L_{f} ≤ 3. If, instead, we argue as in Example 55.10, then we get |\frac{∂f_{i}(x)}{∂x_{j}} | ≤ 5, so that L_{f} ≤ \sqrt{3}\sqrt{3} 5 = 15, which a worse estimate of L_{f}.

(b) ∇f(x) = cos(∥x∥^{2}) [2x_{1} 2x_{2} 2x_{3}] = 2 cos(∥x∥^{2})x. The mean

value theorem gives

|f(x) − f(y)| = |∇f(x̂) · (x − y)| = 2 cos(∥x̂∥^{2})|x̂ · (x − y)|

≤ 2 cos(∥x̂∥^{2})∥x̂∥∥x − y∥ ≤ 2∥x − y∥, L_{f} = 2.

(c) Using (b) we get

∥f(x) − f(y)∥ = \sqrt{(x_{1} − y_{1})^{2} + (x_{2} − y_{2})^{2} + (sin(∥x∥^{2} − sin(∥y∥^{2}))^{2}}

≤ \sqrt{(x_{1} − y_{1})^{2} + (x_{2} − y_{2})^{2} + 4∥x − y∥^{2}} ≤ \sqrt{5}∥x − y∥.

(d) Not Lipschitz continuous. (e)–(f) are too difficult and should be removed.

54.3 is too difficult and should be removed. 54.4 The linearization is f̃(x) = f(x̄) + f′(x̄)(x − x̄).

(a) f̃(x) = 14+[2 4 6] [\begin{matrix}x_{1} − 1 \\ x_{2} − 2 \\ x_{3} − 3 \end{matrix}] = 14+2(x_{1}−1)+4(x_{2}−2)+2(x_{3}−3).

(b) f̃(x) = sin(14) + cos(14) [2 4 6] [\begin{matrix}x_{1} − 1 \\ x_{2} − 2 \\ x_{3} − 3 \end{matrix}]

(c) f̃(x) = [\begin{matrix}14 \\ sin(2) \end{matrix}] + [\begin{matrix}2 & 4 & 6 \\ 0 & cos(2) & 0 \end{matrix}] [\begin{matrix}x_{1} − 1 \\ x_{2} − 2 \\ x_{3} − 3 \end{matrix}]

(d) f̃(x) = [\begin{matrix}14 \\ sin(2) \\ 2 cos(3) \end{matrix}] + [\begin{matrix}2 & 4 & 6 \\ 0 & cos(2) & 0 \\ 2 cos(3) & cos(3) & −2 sin(3) \end{matrix}] [\begin{matrix}x_{1} − 1 \\ x_{2} − 2 \\ x_{3} − 3 \end{matrix}]

54.5 (a) 9x_{2}(2x_{1}^{3} − x_{1}^{2}x_{2} + x_{2}^{3})

(b) 0 54.6 P (x) = f(x̄) + f′(x̄)(x − x̄) + \frac{1}{2}(x − x̄)^{t}f′′(x̄)(x − x̄).

(a)

P (x) = 1 + [\frac{1}{2} \frac{1}{2} \frac{1}{2}] [\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix}] + \frac{1}{2} [x_{1} x_{2} x_{3}] [\begin{matrix}−\frac{1}{4} & −\frac{1}{4} & −\frac{1}{4} \\ −\frac{1}{4} & −\frac{1}{4} & −\frac{1}{4} \\ −\frac{1}{4} & −\frac{1}{4} & −\frac{1}{4} \end{matrix} ] [\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix}]

= 1 + \frac{1}{2}(x_{1} + x_{2} + x_{3}) − \frac{1}{8}(x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + 2x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3})

(b)

P (x) = 0 + [0 0 0] [\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix}] + \frac{1}{2} [x_{1} x_{2} x_{3}] [\begin{matrix}0 & 0 & 0 \\ 0 & 0 & −1 \\ 0 & −1 & 0 \end{matrix}] [\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix}]
= −x_{1}x_{3}

(c) too difficult

P (x) = 1 + [0 0 0] [\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix}] + \frac{1}{2} [x_{1} x_{2} x_{3}] [\begin{matrix}−2 & 0 & 0 \\ 0 & −2 & 0 \\ 0 & 0 & −2 \end{matrix}] [\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \end{matrix}]
= 1 − x_{1}^{2} − x_{2}^{2} − x_{3}^{2}

(e) too difficult

54.7 The problem should be formulated: Linearize f o s, where f(x) = x_{1}x_{2}x_{3}, at t = 1, with ...

We have g(t) = f(s(t)), g′(t) = f′(s(t))s′(t). The linearization is g̃(t) = g(1) + g′(1)(t − 1) = f(s(1)) + f′(s(1))s′(1)(t − 1).

(a) g̃(t) = 1 + [1 1 1] [\begin{matrix}1 \\ 2 \\ 3 \end{matrix}] (t − 1) = 1 + 6(t − 1) = 6t − 5.

(b) g̃(t) = cos(1) sin(1) + [sin(1) cos(1) cos(1) sin(1)] [\begin{matrix}− sin(1) \\ cos(1) \\ 1 \end{matrix}] (t −

1) = cos(1) sin(1) + (− sin^{2}(1) + cos^{2}(1) + cos(1) sin(1))(t − 1).

(c) g̃(t) = 1 + [1 1 1] [\begin{matrix}1 \\ 0 \\ −1 \end{matrix}] (t − 1) = 1.

54.8 On the one hand: f(x) = ∫_{0}^{∞} e^{−xy} dy, f^{(n)}(x) = ∫_{0}^{∞}(−y)^{n}e^{−xy} dy. On the other hand: f(x) = ∫_{0}^{∞} e^{−xy} dy = [\frac{e^{−xy}}{−x} ]_{y=0}^{∞} = x^{−1}, f^{(n)}(x) = (−1)^{n}n!x^{−(n+1)}. We conclude that ∫_{0}^{∞} y^{n}e^{−xy} dy = n!x^{−(n+1)}.

54.9 This should be done by means of a computer program. But we can compute one step by hand as follows. The method of steepest descent with x^{(0)} = [1 1 1] gives

x^{(1)} = x^{(0)} − α∇u(x^{(0)}) = [1 1 1] − α [2 2 4]
= [1 − 2α 1 − 2α 1 − 4α] .

We minimize the function f(α) = u(x^{(1)}) = (1−2α)^{2}+(1−2α)^{2}+2(1−4α)^{2} by solving the equation f′(α) = 0 and find α = 0.3, so that x^{(1)} = [0.4 0.4 −0.2]. This is closer to the minimum, which is located at the origin.

54.10 —

54.11 Should be removed.

54.12 Evaluate f on the lines x_{2} = 0, x_{1} = 0, and x_{1} = x_{2}. This gives f(x_{1}, 0) = 1, f(0, x_{2}) = −1, f(x_{1}, x_{1}) = 0. So f is not well-defined at the origin. Moreover,

∇f(x) = \frac{4x_{1}x_{2}}{∥x∥^{4}} [x_{2} −x_{1}] .

Hence, on the line x_{1} = x_{2} we have ∇f(x_{1}, x_{1}) = x_{1}^{−1} [1 −1], which is arbitrarily large near the origin. Therefore, f is not Lipschitz continuous near the origin.

Chapter 55

55.1 —

55.2 (a) x = [\begin{matrix}cos(t) \\ sin(t) \end{matrix}], 0 ≤ t < 2π.

(b) the same as (a)

(c) empty (no curve)

(d) x = [\begin{matrix}\sqrt{3} cos(t) \\ \sqrt{\frac{3}{2}} sin(t) \end{matrix}], 0 ≤ t < 2π.

55.3 Using Theorem 56.3 we check that ∂u(1, 1, 1)/∂x_{3} = 9 ≠ 0. Hence, the surface can be expressed as x_{3} = g(x_{1}, x_{2}).

55.4 (a) ∇f(x) = [nx_{1}^{n−1}(x_{2}^{n} + x_{3}^{n}) nx_{1}^{n}x_{2}^{n−1} nx_{1}^{n}x_{3}^{n−1}]

(b) ∇f(x) = \frac{[x_{1} x_{2} x_{3}]}{∥x∥} = \frac{x^{t}}{∥x∥}

(c) ∇f(x) = [2x_{1} 2x_{2} 2x_{3}] = 2x^{t}

(d) ∇f(x) = −\frac{x^{t}}{∥x∥^{3}}

(e) ∇f(x) = e^{x_{1}x_{2}x_{3}} [x_{2}x_{3} x_{1}x_{3} x_{1}x_{2}]

55.5 The equation of the tangent plane is ∇f(x̄)(x − x̄) = 0.

(a) n [2 1 1] [\begin{matrix}x_{1} − 1 \\ x_{2} − 1 \\ x_{3} − 1 \end{matrix}] = 0, i.e., 2x_{1} + x_{2} + x_{3} = 4.

(b,c,d) The same curve, namely, ∥x∥ = \sqrt{3}. [1 1 1] [\begin{matrix}x_{1} − 1 \\ x_{2} − 1 \\ x_{3} − 1 \end{matrix}] = 0, i.e.,

x_{1} + x_{2} + x_{3} = 3.

(e) [1 1 1] [\begin{matrix}x_{1} − 1 \\ x_{2} − 1 \\ x_{3} − 1 \end{matrix}] = 0, i.e., x_{1} + x_{2} + x_{3} = 3.

55.6 (a) 3x_{1} + \frac{3}{2}x_{2} − x_{3} = 3

(b) 2x_{1} + 4x_{2} + 6x_{3} = 28
(c) 2πx_{1} − x_{2} = 2π − 2

55.7 x_{1} + 3\sqrt{3}x_{2} = 10, n = [1 3\sqrt{3}].

55.8 Assume that f is differentiable in Q. There are three cases:

(a) There is a point x̂ ∈ Q such that f(x̂) > 0. Since f(x) = 0 on the boundary we conclude that f attains its maximum at an interior point x̄ ∈ Q. At this point we have ∇f(x̄) = 0.

(b) There is a point x̂ ∈ Q such that f(x̂) < 0. Since f(x) = 0 on the boundary we conclude that f attains its minimum at an interior point x̄ ∈ Q. At this point we have ∇f(x̄) = 0.

(c) If f(x) = 0 for every x ∈ Q, then ∇f(x) = 0 for every x ∈ Q.

Chapter 61

61.1 (a) ∇ · F = −1 (b) ∇ × F = (sin(1)e^{2}, −cos(1) − 2sin(1)e^{2}, cos^{2}(1) − sin^{2}(1) + 4) (c) ∇(∇ · F) = (0, −3, 0)

61.2 (∇ × ∇) = 0 and thus (∇ × ∇)u = 0) or (∇ × ∇)u = ∇ × (∇u) = 0, for any sufficiently smooth scalar function u. But ∇ × ∇ × u = ∇ × (∇ × u) is not zero. Recall that the cross product is not associative.

61.4 ∇(rF (r)) = (x_{1}, x_{2})r^{−1}F′(r).

61.5 ∇ · (ω × x) = (∇ × ω) · x − ω · (∇ × x) = 0. By the Divergence Theorem 0 = ∫_{Ω} ∇ · v = ∫_{Γ} n · v. Thus the mass of the fluid inside any subdomain Ω is constant.

61.6 Note that for a rigid transformation we have ∇x̃ = T ∇_{x} with T an orthogonal matrix. Then Δx̃ = ∇x̃ · ∇x̃ = T ∇_{x} · T ∇_{x} = ∇_{x} · T^{t}T ∇_{x} =

∇_{x} · ∇_{x} = Δ_{x}.

61.8 rotu = (u_{x_{2}}, −u_{x_{1}}) and rot(u_{x_{2}}, −u_{x_{1}}) = u_{x_{2}x_{2}} + u_{x_{1}x_{1}}.

61.9 This follows by directly computing the Laplacian.

61.10 This follows by directly computing the Laplacian.

Chapter 63

63.1 (a) 2 sinh 1

(b) 4π\sqrt{2}

(c) 8 (d) \frac{8}{27}(10\sqrt{10} − 1)

(e) Note! Incomplete statement of problem. “0 ≤ t < 2π” should be added. 6

63.2 (a) 2π\sqrt{2}

(b) 2π^{2}\sqrt{2}

(c) −\frac{π\sqrt{2}}{2}

63.3 (a) \frac{1}{2}

(b) 1

(c) \frac{\sqrt{2}}{6}

63.4 (a) 0

(b) 0 (why?)

63.5 (a) 0 (clockwise and counter-clockwise)

(b) 2π (counter-clockwise), −2π (clockwise)

63.6 1

63.7 (a) 2

(b) \frac{7}{3}

(c) 3 − \frac{4}{π}

(d) 3 − \frac{2}{n+1}

63.8 1

63.9 \{x = (x_{1}, x_{2}) ∈ ℝ^{2} : |x − (0, \frac{1}{2})| = \frac{1}{2}\} See the plot below.

63.10 κ = \frac{1}{R}

63.11 −

63.12 (a) κ(x_{1}) = _{\frac{2}{(1+4x_{1}^{2})^{\frac{3}{2}}}}

(b) κ(x_{1}) = _{\frac{6x_{1}}{(1+9x_{1}^{4})^{\frac{3}{2}}}} There is a change of signs in the curvature at the

inflection point x = (x_{1}, x_{2}) = (0, 0). See the plot below.

63.13 Note! Misprint in statement of problem. The third sentence should read “Derive the equilibrium equation y′(x) = \frac{1}{c}s(x) = \frac{1}{c} ∫_{0}^{x} \sqrt{1 + (y′(ξ))^{2}} dξ, with c a constant.”. We study the part of the hanging chain between 0 and x (see the plot below). Letting ρ denote the density (mass per unit length) of the chain, the gravitational force on this part of the chain is ρgs(x). Now letting ψ be the angle the tangent at (x, y(x)) makes with the horizontal axis, PSfrag replacements

x_{1} x_{2}

x_{2} = x_{1}^{2}

x_{1}^{2} + (x_{2} − \frac{1}{2})^{2} = \frac{1}{4}

•

Figure 1: A plot of the circle of curvature \{x = (x_{1}, x_{2}) ∈ ℝ^{2} : |x − (0, \frac{1}{2})| = \frac{1}{2}\} of x_{2} = x_{1}^{2} at x_{1} = 0. (Problem 60.9)

since the chain force is always directed along the tangent horizontal force equilibrium requires that T (0) = T (x) cos ψ and vertical force equilibrium that ρgs(x) = T (x) sin ψ. Dividing these two equilibrium equations gives tan ψ = y′(x) = \frac{ρg}{T (0)}s(x) = \frac{1}{c}s(x), where c = \frac{T (0)}{ρg} can be thought of as a length of the chain acted upon by gravity with a force equal to the tension T (0) at the lowest point of the chain.

Since the arc length between 0 and x is s(x) = ∫_{0}^{x} \sqrt{1 + (y′(ξ))^{2}} dξ, with y′(x) = sinh(\frac{x}{c}) the identity cosh^{2}(\frac{ξ}{c}) − sinh^{2}(\frac{ξ}{c}) = 1 shows that \frac{1}{c}s(x) = \frac{1}{c} ∫_{0}^{x} cosh(\frac{ξ}{c}) dξ = sinh(\frac{x}{c}) = y′(x), and thus, from integration, y(x) = c cosh(\frac{x}{c}) is a solution.

63.14 Note! Misprint in statement of problem. The point (1, 1, 1) does not lie on the given surface. Further, there seem to be many possible such curves. If we consider the point (1, −1, −1), possible directions are α(1, 0, 0) + β(0, 1, 3), where α ∈ ℝ and β ∈ ℝ are arbitrary.

63.15 8 (For a motivation of the formula ds^{2} = ρ^{2}dθ^{2} + dρ^{2}, see the plot below. The expression for the arc length then follows from the fact that dρ = ρ′(θ)dθ).

63.16 L\sqrt{1 + (2πRN)^{2}} (where L is the length of the cylinder, R is the radius of the cylinder, and N is the number of revolutions of the string per unit length (in the axial direction) of the cylinder.)

PSfrag replacements x_{1} x_{2}

x_{2} = x_{1}^{3} κ < 0 κ > 0

•

Figure 2: A plot of the function x_{2} = x_{1}^{3}. The curvature κ(x_{1}) = _{\frac{6x_{1}}{(1+9x_{1}^{4})^{\frac{3}{2}}}}

changes sign at the inflection point x = (x_{1}, x_{2}) = (0, 0). It is negative for x_{1} < 0 and positive for x_{1} > 0. (Problem 60.12 (b))

Chapter 64

64.1 (a) 1

(b) \frac{1}{6}

(c) 2 ln 2 (Use integration by parts.)

(d) 1− \frac{1}{2!2} + \frac{1}{3!3} − \frac{1}{4!4} +... = ∑_{k=1}^{∞} \frac{(−1)^{k+1}}{k!k} (Use the Maclaurin expansion, i.e. Taylor for x̄ = 0.)

64.2 (a) \frac{1}{4}

(b) \frac{1}{4}(e^{2} + 1)

(c) \frac{e}{2} − 1

64.3 (a) ∫_{0}^{1/2} ∫_{1/2}^{1−x_{2}} f(x_{1}x_{2})dx_{1}dx_{2}

(b) ∫_{0}^{1} ∫_{0}^{√\overline{1−x_{2}^{2}}} f(x_{1}x_{2})dx_{1}dx_{2}

(c) ∫_{−1}^{0} ∫_{0}^{x_{1}+1} f(x_{1}x_{2})dx_{2}dx_{1}

(d) ∫_{0}^{1} ∫_{1−x_{2}}^{1} f(x_{1}x_{2})dx_{1}dx_{2} + ∫_{1}^{2} ∫_{1}^{x_{2}−1} f(x_{1}x_{2})dx_{1}dx_{2}

64.4 (a) \frac{11}{60}
(b) \frac{2}{15}

PSfrag replacements

x y

y(x) = c cosh(\frac{x}{c})

•
T (0)
T (x)

T (x) sin ψ T (x) cos ψ

ψ

ρgs(x)

s(x)

Figure 3: A plot of the hanging chain described by the function y(x) = c cosh(\frac{x}{c}) (with c = \frac{1}{2}). Force equilibrium for the part of the chain between 0 and x requires that T (0) = T (x) cos ψ (horizontal equilibrium) and ρgs(x) = T (x) sin ψ (vertical equilibrium). (Problem 60.13)

(c) \frac{7}{2}

(d) 1

64.5 (a) e − 1

(b) (1 − e^{−2})(2 − 5e^{−1})
(c) \frac{3}{4}

(d) \frac{π}{4} (Use integration by parts to obtain: ∫ \sqrt{a − x^{2}}dx = \frac{1}{2}x\sqrt{a − x^{2}}−

\frac{\sqrt{a}}{2} arccos(\frac{x}{\sqrt{a}}))

64.6 The volume of the metal removed is \frac{4π}{3} (a^{3} − (a^{2} − b^{2})^{3/2}). (Use polar coordinates.)

64.7 \frac{2π}{5} a_{1}a_{2} (Use elliptic coordinates.)

64.8 e^{2} − 1

64.9 \frac{9π}{4}

64.10 The area within the cardoid is \frac{3π}{2} . (Use polar coordinates.)

64.11 (a) \frac{e^{2}−3}{2}
(b) e^{2}(2e − 1)

PSfrag replacements x_{1}

x_{2}
θ
△θ
ρ
ρ △ θ
△s
△ρ

Figure 4: A plot of the relation between small changes △ρ and △θ in ρ and θ, and the corresponding change in position △s. From the figure and old Pythagoras we get (△s)^{2} ≈ ρ^{2}(△θ)^{2} + (△ρ)^{2}, which in the infinitesimal limit turns into ds^{2} = ρ^{2}dθ^{2} + dρ^{2}. (Problem 60.15)

(c) \frac{1}{2}(1, 1)

64.12 (a) 2(1 − \frac{2}{e})

(b) \frac{1}{15}(25\sqrt{5} − 4\sqrt{2} − 31)
(c) 2ln2 − \frac{5}{2} (Division of polynomials.)

64.13 (a) \frac{4π}{3} (Complete the squares and take x_{1} = r cos(θ) + 1 and x_{2} = r sin(θ) + 1.)

(b) \frac{1}{9^{4}2} (Complete the squares and take x_{1} = r cos(θ) + \frac{1}{3} and x_{2} =

\sqrt{3}r sin(θ).)
(c) −π

Chapter 65

65.2 \frac{40π}{3} (26\sqrt{26} − 125)

65.6 (a) \sqrt{6} (Since x = s(y) = My is a linear transformation, the area scale from the parameter domain to the surface S is constant, equal to |\frac{∂s}{∂y_{1}} × \frac{∂s}{∂y_{2}} | = |(1, 0, 1) × (0, 1, 2)| = \sqrt{6}, where \frac{∂s}{∂y_{1}} = (1, 0, 1) and \frac{∂s}{∂y_{2}} = (0, 1, 2) are the columns of M. Therefore, the area of S is \sqrt{6} times the area of Q, and since Q = {y = (y_{1}, y_{2}) ∈ ℝ^{2} : 0 ≤ y_{1} ≤ 1, 0 ≤ y_{2} ≤ 1} is the unit square in ℝ^{2} with area (1 − 0)(1 − 0) = 1, the area of S is equal to \sqrt{6}. But we can equivalently view S as the parallelogram spanned by the two vectors v = (1 − 0)\frac{∂s}{∂y_{1}} = (1, 0, 1) and w = (1 − 0)\frac{∂s}{∂y_{2}} = (0, 1, 2) with area equal to |v × w|, which obviously yields the same result. See the plot below.)

PSfrag replacements x_{1} x_{2} x_{3}

(1, 0, 1) (0, 1, 2)

Figure 5: A plot of the surface S, a parallelogram spanned by the vectors (1, 0, 1) and (0, 1, 2). (Problem 62.6)

(b) \frac{3\sqrt{6}}{2}

65.10 Note! Parameter domain is not specified. \frac{4}{\sqrt{3}} (for 0 ≤ y_{1} ≤ 1, 0 ≤ y_{2} ≤ π)

65.11 Note! n denotes the outward unit normal to S.

\frac{1}{2}

65.15 4π^{2}ab

Chapter 66

66.1 Hint: Use the fact that E = \frac{1}{2}m(r, θ, ϕ)v^{2} = \frac{1}{2}m(r, θ, ϕ)(ωr)^{2}

66.2

66.3 (1) 1

(2) \frac{(e^{2}−1)^{2}}{4}

(3) 4π

66.4 (a) \frac{1}{24}
(b) \frac{10π}{24}

66.5 (1) 2π log(2) + π^{2} Hint: Use polar coordinates for x_{1} and x_{2}

(2) \frac{m}{2}
(3) \frac{7m}{12}
66.6 (1) \frac{8π}{e}

(2) 2

(3) \frac{n}{2}

66.7 (1) 0 (2) π

(3) \frac{4π}{5}

66.8

66.9 π^{n/2} Hint: Use the fact that

∫_{−∞}^{∞} ∫_{−∞}^{∞} e^{−||x||^{2}}dx_{1}dx_{2} = ∫_{−∞}^{∞} ∫_{−∞}^{∞} e^{−(x_{1}^{2}+x_{2}^{2})}dx_{1}dx_{2} =

∫_{−∞}^{∞} ∫_{−∞}^{∞} e^{−x_{1}^{2}} · e^{−x_{2}^{2}}dx_{1}dx_{2} = ∫_{−∞}^{∞} e^{−x_{1}^{2}}dx_{1} · ∫_{−∞}^{∞} e^{−x_{2}^{2}}dx_{2}

66.10 \frac{7m}{6}

66.11 Hint: Use the fact that {(x, y) : x ∈ E_{y}, 0 < y < ∞} = {(x, y) : |f(x)| > y, 0 < y < ∞} = {(x, y) : 0 < y < |f(x)|, x ∈ R^{n}}

Chapter 67

67.1 (64.4) Gauss’ theorem applied to vw yields

∫_{Ω} \frac{∂vw}{∂x_{i}} dx = ∫_{Γ} vwn_{i}ds.

On the other hand the product rule for differentiation yields

∫_{Ω} \frac{∂vw}{∂x_{i}} dx = ∫_{Ω} \frac{∂v}{∂x_{i}} wdx + ∫_{Ω} v \frac{∂w}{∂x_{i}} dx.

The result follows by equating the right hand sides. (64.5) Follows immediately by the definition. (64.6) Componentwise it reads, using (64.4),

∫_{Ω} \frac{∂v}{∂x_{i}} \frac{∂w}{∂x_{i}} dx = ∫_{Γ} v \frac{∂w}{∂x_{i}} n_{i}ds − ∫_{Ω} v \frac{∂^{2}w}{∂x_{i}^{2}} dx.

(64.8) Follows by interchanging the roles of v and w in the right hand side of (64.6) and equating the two right hand sides.

67.3 (a) ∫_{Γ} u · nds = 0 by Gauss’ theorem.

(b) Nothing.

67.4

67.5 For the mapping \overline{x}_{i} = x_{i} + a_{i}x_{i}, i = 1, 2, the Jacobian J becomes 1 + a_{1} + a_{2} + a_{1}a_{2}. For area preserving maps J = 1. For small deformations the term a_{1}a_{2} can be neglected from which we conclude that divu = a_{1} + a_{2} is almost zero.

67.6 (a) Here the divergence theorem applies and since div u = 0 we conclude

that ∫_{Γ} u · nds = 0.

(b) ∫_{Γ} u · nds = 2π. Here the divergence theorem does not apply since div u is not defined in the origin. Instead compute the curve integral by using polar coordinates.

67.7 Γ and \overline{Γ} with outward normals n and −\overline{n}, respectively, close a domain Ω in the plane to which the divergence theorem applies. We get

0 = ∫_{Ω} div udx = ∫_{Γ} u · nds + ∫\overline{Γ} u · (−\overline{n})ds.

Consequently

∫_{Γ} u · nds = ∫\overline{Γ} u · \overline{n}ds.

67.8 (a) 0. Hint: Close the curve by adding the line segments from (0, 0) to

(1, 0) and from (0, 0 to (0, 1), respectively.

(b) log 5. Hint: Close the curve by adding the line segment from (1, 0) to (1, 2).

67.9 ∇×u = 0, thus there exists a potential φ such that u = ∇φ. An integration gives φ(x) = x_{1} exp(x_{1}x_{2}) + C, where C is an arbitrary constant.

67.10 Note! Misprint in statement of problem. Reads: v = −\frac{1}{2π} log(x − \overline{x}) Should read: v = −\frac{1}{2π} log(∥x − \overline{x}∥) Let Ω_{ϵ} = \{∥x∥ < R : ∥x − \overline{x}∥ > ϵ\} On Ω_{ϵ} we have Δv = 0, c.f. exercise 58.9. Further, since w and ∇w · n vanish for ∥x∥ > R, for some R large enough, (64.16) reduces to

∫_{Ω_{ϵ}} vfdx = − ∫_{∂Ω_{ϵ}} v∇w · nds + ∫_{∂Ω_{ϵ}} w∇v · nds.

In fact, since ∥x − \overline{x}∥ = ϵ and n = −\frac{x−\overline{x}}{∥x−\overline{x}∥} on ∂Ω_{ϵ}, we get

∫_{Ω_{ϵ}} vfdx = \frac{1}{2π} ∫_{∂Ω_{ϵ}}(log ϵ)∇w · nds + \frac{1}{2πϵ} ∫_{∂Ω_{ϵ}} wds.

Using polar coordinates in the curve integrals, a passage to the limit, letting ϵ → 0 in the integrals, gives

∫_{Ω_{ϵ}} vfdx → ∫_{Ω} vfdx,

\frac{1}{2π} ∫_{∂Ω_{ϵ}}(log ϵ)∇w · nds → 0

and

\frac{1}{2πϵ} ∫_{∂Ω_{ϵ}} wds → w(\overline{x}).

Hence,

w(\overline{x}) = ∫_{Ω} vfdx.

67.11 Analogous arguments as in 64.10. Let Ω̃_{ϵ} = {x ∈ Ω_{ϵ} : x_{2} > 0}. (64.16) takes the form

∫Ω̃_{ϵ} vfdx = −\frac{1}{2π} ∫_{∂Ω̃_{ϵ}} log ϵ∇w · nds + \frac{1}{2πϵ^{2}} ∫_{∂Ω̃_{ϵ}} w · nds + ∫_{\{x_{2}=0\}} wg · nds.

By noting that

\frac{1}{2πϵ^{2}} ∫_{∂Ω_{ϵ}} w · nds → \frac{1}{2}w(\overline{x}),

since the curve integral only takes half a round for points \overline{x} = (\overline{x}_{1}, 0) the result now follows.

67.12 Follows immediately from (64.16).

Chapter 68

68.3 j = 0, 1 and 2 gives the values 4π, 2π and 0, respectively. Compute the integrand and use polar coordinates. For the case j = 2 one can also compute the divergence wihich is zero and apply the divergence theorem. Compare with excercise 64.6.

68.4 Dismissed.

68.5 (1) 4π

(2) 0

(3) 4π

(4) 0.

68.6 A direct computation gives div F = (α + 2)∥x∥^{α−1}.

68.7 An application of the divergence theorem gives

I = ∫ ∫ ∫_{V} divFdx_{1}dx_{2}dx_{3}.

To minimize the volume integral corresponds to solving the inequality divF ≤ 0, from which the set V of integration is determined.

divF = x_{1}^{2} − 2x_{1} + x_{2}^{2} − 4x_{2} + 4x_{3}^{2} + 8x_{3} + 5 ≤ 0.

A completion of the squares gives:

\frac{(x_{1} − 1)^{2}}{4} + \frac{(x_{2} − 2)^{2}}{4} + (x_{3} + 1)^{2} ≤ 1,

which is recognized as an ellipsoid centered at (1,2,-1) with half axes 2, 2 och 1, respectively. An introduction of ellipsoidal coordinates

x = 1 + 2r sin θ cos φ,
y = 2 + 2r sin θ sin φ,
z = −1 + r cos θ,

with the Jacobian J = 4r^{2} sin θ finally yields

I = ∫_{0}^{2π} ∫_{0}^{π} ∫_{0}^{1}(4r^{2} − 4)4r^{2} sin θ drdθdφ = −128π/15.

68.8 Close the surface with a lid Γ̃ from the bottom, where Γ̃ : x_{1}^{2} + x_{2}^{2} ≤ 4, x_{3} = 0. By choosing downward normal n = (0, 0, −1) from Γ̃ we can apply the Gauss theorem.

∫_{Γ} F · nds = ∫_{Ω} div F dx − ∫Γ̃ F · nds,

where Ω is the solid upper half sphere with radius 2. The surface integral vanishes on Γ̃ since F · n = (f_{1}, f_{2}, 0) · (0, 0, −1) = 0 here. Moreover taking the divergence of F gives div F = x_{1}. A standard use of spherical coordinates gives

∫_{Γ} F · nds = 0,

by periodicity.

68.9 k. Compare with exercise 65.5 (4) for instance. Each point x_{i} inside the closed surface Γ contributes with the value 1, since F is normalized with 1/4π.

Chapter 69

69.2 We have ∇·(u×a) = a·∇×u−u·∇×a, but a is constant ⇒ ∇×a = 0 ⇒

∫_{Ω} ∇ · (u × a) dx = ∫_{Ω} a · (∇ × u) dx = a · ∫_{Ω} ∇ × u dx.  (1)

Also (u × a) · n = (n × u) · a ⇒

∫_{Γ}(u × a) · n ds = ∫_{Γ}(n × u) · a ds = a · ∫_{Γ}(n × u) ds  (2)

Now (1), (2) and the divergence theorem gives

a · (∫_{Ω} ∇ × u dx − ∫_{Γ}(n × u) ds) = 0.

And this holds for all constant vectors a ⇒

∫_{Ω} ∇ × u dx = ∫_{Γ}(n × u) ds.

69.3 (u_{1}, u_{2}) = (−v_{2}, v_{1}) ⇒ ∇ · u = ∇ × v and u · n = n × v.

69.5 14π. Hint: u = \frac{(−x_{2},x_{1},x_{3})}{(x_{1}^{2}+x_{2}^{2})∥x∥} conservative.

69.6 We have ∇ × (va) = v∇ × a + (∇v) × a = (∇v) × a (a constant ⇒ ∇ × a = 0), and (∇v) × a · n = n × (∇v) · a ⇒ ∫_{S}(∇ × (va)) · n ds = a · ∫_{S} n × (∇v) ds. Also ∫_{Γ}(va) · ds = a · ∫_{Γ} v ds, and Stoke’s theorem gives that a · (∫_{S} n × (∇v) ds − ∫_{Γ} v ds) = 0, but this is true for all constant vectors a ⇒ ∫_{S} n × (∇v) ds = ∫_{Γ} v ds.

69.7 ∇ × (x_{2}, 2x_{3}, 3x_{1}) = −(2, 3, 1), s(x_{1}, x_{2}) = (x_{1}, x_{2}, \sqrt{1 − x_{1}^{2} − x_{2}^{2}}) ⇒ s′_{,1} = (1, 0, −x_{1}/\sqrt{1 − x_{1}^{2} − x_{2}^{2}}), s′_{,2} = (0, 1, −x_{2}/\sqrt{1 − x_{1}^{2} − x_{2}^{2}}) ⇒ s′_{,1} × s′_{,2} = (x_{1}/\sqrt{1 − x_{1}^{2} − x_{2}^{2}}), x_{2}/\sqrt{1 − x_{1}^{2} − x_{2}^{2}}), 1) ⇒

|s′_{,1} × s′_{,2}| = \frac{1}{1 − x_{1}^{2} − x_{2}^{2}}

n = \frac{s′_{,1} × s′_{,2}}{|s′_{,1} × s′_{,2}|} = (\frac{x_{1}}{1 − x_{1}^{2} − x_{2}^{2})^{3/2}} , \frac{x_{2}}{1 − x_{1}^{2} − x_{2}^{2})^{3/2}} , \frac{1}{1 − x_{1}^{2} − x_{2}^{2}} )

⇒ (∇ × (x_{2}, 2x_{3}, 3x_{1})) · n = ^{\frac{−2x_{1}−3x_{2}−\sqrt{1−x_{1}^{2}−x_{2}^{2}}}{(1−x_{1}^{2}−x_{2}^{2})^{3/2}}}

69.8 We have \frac{1}{2} ∫_{Γ} u · ds = \frac{1}{2} ∫_{Ω}(∇ × u) · n ds by Stoke’s theorem. Further ∇ × u = ∇ × (−x_{2}, x_{1}, 0) = (0, 0, 2) and n = (0, 0, 1) ⇒ (∇ × u) · n = 2 ⇒ \frac{1}{2} ∫_{Ω} 2 ds = ∫_{Ω} ds = A(Ω).

Chapter 70

70.1 Using the formula ∇ × u = (\frac{∂u_{3}}{∂x_{2}} − \frac{∂u_{2}}{∂x_{3}} , \frac{∂u_{1}}{∂x_{3}} − \frac{∂u_{3}}{∂x_{1}} , \frac{∂u_{2}}{∂x_{1}} − \frac{∂u_{1}}{∂x_{2}} ) we find that

(a) ∇ × u = (0 − 0, 0 − 0, 0 − 0) = (0, 0, 0) so that u = ∇ϕ for some potential ϕ. Integrating ∇ϕ = (x_{1}, x_{2}, x_{3}), we obtain ϕ(x) = \frac{1}{2}(x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) + C = \frac{1}{2}∥x∥^{2} + C, where C is an arbitrary constant.

(b) ∇ × u = (1 − 0, 1 − 0, 1 − 0) = (1, 1, 1) ≠ (0, 0, 0) so that u is not a potential field!

(c) ∇ × u = (0 − 0, −1 − (−1), 2x_{2} − 2x_{2}) = (0, 0, 0) so that u = ∇ϕ for some ϕ. Integrating we find that ϕ(x) = x_{1}x_{2}^{2} − x_{1}x_{3} + x_{3}^{3} + C.

70.2 Hint: Verify by direct differentiation (under the integral sign) that ∇×ψ = u. At some point, use the fact that ∇ · u = 0.

70.3 We find for u(x) = \frac{(−x_{2},x_{1},0)}{||x||^{2}} that ∇ × u = (\frac{∂u_{3}}{∂x_{2}} − \frac{∂u_{2}}{∂x_{3}} , \frac{∂u_{1}}{∂x_{3}} − \frac{∂u_{3}}{∂x_{1}} , \frac{∂u_{2}}{∂x_{1}} − \frac{∂u_{1}}{∂x_{2}} ) = (0, 0, 0) for (x_{1}, x_{2}) ≠ (0, 0), while for s(t) = (r cos(t), r sin(t), 0) we have ∫_{Γ} u · ds = ∫_{0}^{2π} \frac{1}{r^{2}} (−r sin(t), r cos(t), 0) · (−r sin(t), r cos(t), 0) dt = ∫_{0}^{2π} dt = 2π, just as in the two dimensional counter example.