Section 16.2

1. Problem Let T(x, y) be a temperature | x | ≤ 10, 0 ≤ y ≤ 10. Determine, and sketch the T = 0, 20, 40, 60 level sets. If the level is empty,state that.

(a)T = x² + y²

(c)T = 10(x + y)

Solution (a)

(c)

2. Problem The position vector R = x i + y j to any point in (x, y) in the x, y plane constitutes a vector field. Draw the R vector (to scale) at eight points that are equally spaced around the circle r = 1, and at eight points that are equally spaced around the circle r = 3.

Solution Easy.

3. Problem Consider the vector field v = x i − y j in 0 ≤ x ≤ ∞, 0 ≤ y ≤ ∞.

(a) On a single graph, sketch the v vectors at the 16 points (m, n), where m and n are integers such that 0 ≤ m ≤ 3 and 0 ≤ n ≤ 3. (Here it will help to assign a convenient scale, as we have done in Fig.4, so that the arrow representations of the vectors will not lie on top of each other.)

(b) Determine the curves along which v has constant magnitude and those along which v has constant direction. Are these results in agreement with your sketch of the field in part (a)?

Solution Constant magnitude: x² + y² = const.

Constant direction:

Section 16.3

1. Problem Work out div v for the following vector fields, each of which is defined over all of 3-space, say. Further, evaluate div v at P = (3, −1, 4), and verify that v is C¹ in some region R containing P. (t is the time.)

(a) v = a i + b j + c k (a,b,c constants)

(e) v = x y i − 2 (x² + z²) k

Solution

(a) v = (a, b, c) (constant). Therefore ∇·v = 0.
(e) v = ( x y, 0, − 2 (x² + z²) ). ∇·v = y − 4z .

2. Problem Determine the divergence div v of the fluid flow field v given by (3) in Section 16.2 as

v =

⎛
⎝

U a² (y² − x²)

(x² + y²)²

− 2 x y U a²

(x² + y²)²

⎞
⎠

Solution

∇·v

2 a² U x (x² − 3 y²)

(x² + y²)³

2 a² U x ( − x² + 3 y²)

(x² + y²)³

(1)

(2)

3. Problem Make up a vector field v(x, y, z) such that

(a) ∇·v = 0 everywhere

(e) ∇·v > 0 in | x | > 1 and ∇·v < 0 in | x | < 1

Solution (a)

v = (1, 1, 1)

for instance.

(e)

v = (

x³

− x, 1, 1)

for instance.

4. Problem This exercise is to promote understanding of the limit definition (3) as

div v(P) ≡

lim
B → 0

⌠
(⎜)
⌡

n ·v dA

Specifically, we ask you to evaluate div v at P = (0, 0, 0), say, by actually evaluating (∫_S n ·v dA)/V and taking the limit as B → 0 take B to be cube | x | ≤ ϵ, | y | ≤ ϵ, | z | ≤ ϵ.

Show that your result agrees with that obtained (much more readily) from (7) as

div v =

∂v_x

∂x

∂v_y

∂y

∂v_z

∂z

NOTE: Do not merely mimic our steps (4)-(6). Rather, actually compute ∫n ·v dA on each of the six faces, add the results, divide by V = 8 ϵ³, and take the limit as ϵ→ 0.

(a) v = 2 i − j + 4 k

(b) v = 3 i + 4 j − 2 k

(d) v = (x + 1) sin y j

(e) v = x i + 2 y j − 4 z³ k

(f) v = (x² − 4 x + y z²) i + j

Solution (a) At x = −ϵ, n = (−1, 0, 0) so v·n = − 2 and

⌠
⌡

x = − ϵ

n·v dS = (2ϵ)² ×( − 2).

At x = ϵ, n = (1, 0, 0) so v·n = 2 and

⌠
⌡

x = ϵ

n·v dS = (2ϵ)² ×(2).

Repeat this for y = ±ϵ and z = ±ϵ. Thus,

⌠
(⎜)
⌡

n ·v dA

= 0.

Needless to say, div v = 0.

⌠
⌡

x = −ϵ

n·v dy dz = 5 ϵ

⌠
⌡

ϵ

−ϵ

⌠
⌡

ϵ

−ϵ

e^z dz = 10 ϵ² (e^ϵ − e^−ϵ).

At x = ϵ, n = (1, 0, 0) so v·n = 5 ϵe^z and

⌠
⌡

x = ϵ

n·v dy dz = 10 ϵ² (e^ϵ − e^−ϵ).

At y = −ϵ, n = (0, −1, 0) and v·n = 0 so

⌠
⌡

y = −ϵ

n·v dx dz = 0.

Repeat this for y = ϵ and z = ±ϵ. All the surface integrals are 0.

Thus,

⌠
(⎜)
⌡

n ·v dS = 20 ϵ² (e^ϵ − e^−ϵ),

and

⌠
(⎜)
⌡

n ·v dS

20 ϵ² (e^ϵ − e^−ϵ)

8 ϵ³

e^ϵ − e^−ϵ

As ϵ→ 0,

lim
ϵ→ 0

e^ϵ − e^−ϵ

lim
ϵ→ 0

e^ϵ + e^−ϵ

= 5.

(L'Hospital's theorem)

On the other hand,

div v|_(0,0,0) = 5 e^z |_(0,0,0) = 5.

For other problems, follow the procedure above.

5. Problem (Invariance property of div v) Let a C¹ vector field v be represented in terms of some partial Cartesian x, y frame as v = v_x(x, y) i + v_y(x, y) j. (We limit ourselves to the two-dimensional case merely to reduce the algebra; the story is essentially identical for three-dimensional fields.) Then, as derived in this section,

div v =

∂v_x

∂x

∂v_y

∂y

As emphasized in the text, this number will be the same at a given point P = (x, y), independent of the choice of the location and orientation of the reference coordinate system. In other words, it should be equally true that

div v =

∂v_x′

∂x′

∂v_y′

∂y′

for any values of a, b and α (see the accompanying figure - α is the angle of rotation).

Here we ask you to show that this is true, namely, that

∂v_x

∂x

∂v_y

∂y

∂v_x′

∂x′

∂v_y′

∂y′

Solution You can prove this invariance (i.e. divergence of v is invariant in any (Cartesian) coordinate systems) if you recall in tensor algebra that ∇·v is a scalar. To prove this, one can start with

v_i′,i′

(β_i′jv_j)_,i′

β_i′jv_j,i′

β_i′jv_j,k x_k,i′

β_i′jvj,kβ_ki′

β_i′jβ_i′jv_j,k

δ_kj v_j,k

v_j,j

Alternatively, note that

∂

∂x

∂x′

∂x

∂

∂x′

∂y′

∂x

∂

∂y′

cosθ

∂

∂x′

− sinθ

∂

∂y′

∂

∂y

∂x′

∂y

∂

∂x′

∂y′

∂y

∂

∂y′

sinθ

∂

∂x′

+ cosθ

∂

∂y′

v_x = cos θv_x′ − sin θv_y′,

and

v_y = sin θv_x′ + cos θv_y′.

Thus,

∂v_x

∂x

(cosθ

∂

∂x′

− sinθ

∂

∂y′

)(cos θv_x′ − sin θv_y′)

cos² θ

∂v_x′

∂x′

− cos θsin θ

∂v_y′

∂x′

− cos θsin θ

∂v_x′

∂y′

+ sin²θ

∂v_y′

∂y′

and

∂v_y

∂y

(sin θ

∂

∂x′

+ cos θ

∂

∂y′

)(sin θv_x′ + cos θv_y′)

sin²θ

∂v_x′

∂x′

+ cos θsin θ

∂v_y′

∂x′

+ cos θsin θ

∂v_x′

∂y′

+ cos² θ

∂v_y′

∂y′

Therefore,

∂v_x

∂x

∂v_y

∂y

∂v_x′

∂x′

∂v_y′

∂y′

Section 16.4

1. Problem Work out grad u for the following scalar fields, each of which is defined over all of 3-space. Further, evaluate grad u at (9, 4, −1).

(a) u = 6 x

Solution

(a) ∇u = (6, 0, 0).

2. Problem Evaluate the directional derivative [(du)/(ds)], at the designated point P, in the direction of the given vector v.

(a) u = x² + y² + z², P = (2, 1, 5), v = i

(e) u = √{x (y + z)}, P = (1, 2, 3), v = i − j + k

Solution

(a) ∇u|_{(2, 1, 5)} = (2 x, 2 y, 2 z) |_{(2, 1, 5)} = (4, 2, 10) and n = (1, 0, 0). Therefore,

∂u

∂n

( =

) = n ·∇u = 4.

(e) ∇u|_{(8, −1, 2)} = (2, 0, 2 z) |_{(8, − 1, 2)} = (2, 0, 4) and n = (0, 2/√5, 1/√5). Therefore,

∂u

∂n

( =

) = n ·∇u =

√5

3. Problem For the temperature field T(x, y) given in Fig.3, evaluate the following quantities directly from the figure, with the help of suitable difference quotients, and compare your results with the exact values obtained from the expression t = 100 x y.

(a) ∇T at (0, 4.1)

(b) ∇T at (1, 1)

(d) ∇T at (0.5, 0.2)

(e) ∇T at (1.6, 0.5)

(f) ∇T at (1.5, 1)

(g) [(dT)/(ds)] at (0.4, 1) in the direction of - i - j

(h) [(dT)/(ds)] at (1, 1) in the direction of - i - 2 j

(i) [(dT)/(ds)] at (1, 0.2) in the direction of i + j

(j) [(dT)/(ds)] at (0.5, 0.2) in the direction of - j

(k) [(dT)/(ds)] at (1.6, 0.5) in the direction of i - j

(l) [(dT)/(ds)] at (1.5, 1) in the direction of - i - 2 j

Solution Straightforward.

4. Problem Let the electrical potential (i.e., the voltage) be given by V(x, y, z) = 3 x² y − x z. If a positive charge is placed at P = (x, y, z), in what direction will the charge begin to move ? NOTE: It is known, from electric field theory, that such a charge will begin to move in the direction of maximum rate voltage drop.

(a) P = (2, 3, − 1)

Solution

− ∇V |_{(2, 3, − 1)} = ( −6 x y + z, − 3 x², x)|_{(2, 3, −1)} = (−37, −12, 2).

Therefore, it is likely to move to the direction of (−37, −2, 2).

5. Problem (Convective derivative) Let v(x, y, z, t) = v_x(x, y, z, t) i + v_y(x, y, z, t) j + v_z(x, y, z, t) k be a fluid velocity field, and consider some scalar property of the flow, such as the temperature field T(x, y, z, t). If we swim along any desired path according to x = x(t), y = y(t), z = z(t), then the timewise rate of change of T that we observe is, by chain differentiation,

T(x(t), y(t), z(t)) =

∂T

∂t

∂T

∂x

∂T

∂y

∂T

∂z

(3)

(a) If we choose to drift with the fluid, then dx/dt = v_x, dy/dt = v_y, dz/dt = v_z.

In this case, show that (3) becomes

∂T

∂t

+ (v ·∇) T.

(4)

This is often called the convective derivative because it is the derivative obtained when we drift, or convect, with the fluid. The special notation D/Dt, suggested by Sir G. G. Stokes (1819 - 1903), is often used:

D( )

∂( )

∂t

+ (v ·∇)( ).

(5)

(b) To understand the nature of contributions of each of the two terms on the right-hand side of (4) apply (4) to three simple cases: T = 2 t, v = 0; T = 3 x, v = i; T = 2 t + 3 x, v = i. Include whatever words of explanation (and sketches) seem appropriate.

D( )

∂( )

∂t

+ (divv)( ) ?

Solution (a)

∂T

∂t

∂T

∂x

u +

∂T

∂y

v +

∂T

∂z

(6)

∂T

∂t

+ ∇T ·q.

(7)

(b) For T = 2 t, q = (0, 0, 0),

= 2

For T = 2 t + 3 x, q = (U, 0, 0),

= 2 + 3 U.

q ·∇T ≠ (∇·q) T.

Section 16.5

1. Problem Work out curl v for the following vector fields. Further evaluate curl v at (3, 4, −1).

(a) v = a i + b j + c k (a,b,c constants)

(e) v = x y i − 2 (x² + z²) k

Solution

(a) v = (a, b, c). ∇×v = 0.

(e) v = (x y, 0, − 2 (x² + y²)), ∇×v = (0, 4 x, −x). At (3, 4, −1), ∇×v = (0, 12, −3).

Section 16.6

1. Problem Evaluate ∇² u and ∇×∇u in each case.

Solution

2. Problem Evaluate ∇·∇×v and ∇×( ∇×v) in each case.

(a) v = x i + y j + z k

Solution (a) v = (x, y, z). ∇×v = 0. Therefore, ∇·∇×v = 0 and ∇×(∇×v) = 0.

3. Problem Are the parentheses needed in (14), i.e., in ∇×( ∇×v) ? Explain.

curl curv v = curl² v = ∇×(∇×v) = ∇(∇·v) − ∇² v.

Solution

Yes as (∇×∇) ×v is identically 0 (∇×∇ = 0).

4. Problem ∇² u = 0 is called Laplace's Equation. Show that the following are solutions of Laplace's equation.

(a) x y + 3 z

(b) x² − y² − 2 x z

(d) y³ − 3 x²y

(e) ln (x² + y²) for x² + y² ≠ 0

(f) [1/(x² + y² + z²)] for x² + y² + z² ≠ 0

Solution

Easy.

5. Problem If v is any vector, and R = x i + y j + z k, show that (v ·∇) R = v.

Solution

(v ·∇) R

(v_x

∂

∂x

+ v_y

∂

∂y

+ v_z

∂

∂z

)(x,y,z)

(8)

(v_x, v_y, v_z)

(9)

(10)

6. Problem Derive the following equations [as we done in the text for (4)].

(f) equation(14)

Solution (a),(b)(c),(d),(e) … Shown in class.

(f)

∇×(∇×v)

⎛
⎝

∇·v

⎞
⎠

∇−

⎛
⎝

∇·∇

⎞
⎠

(11)

∇

⎛
⎝

∇·v

⎞
⎠

− ∇² v.

(12)

7 Problem The following scalar equations occur in fluid mechanics, solid mechanics, and electromagnetics. Re-express them more concisely in terms of vector and vector differential operator notation. Here x, y, z subscripts denote the respective components of a given vector, not partial derivatives. To illustrate what we are after, consider the equations

∂u

∂x

= F_x,

∂u

∂y

= F_y,

∂u

∂z

= F_z.

(13)

If we multiply these equations by i, j, k, respectively, and add them, we obtain

∂u

∂x

i +

∂u

∂y

j +

∂u

∂z

k = F_x i + F_y j + F_z k

or in more concise vector notation, ∇u = F.

(a)

∂σ

∂t

∂

∂x

(σv_x) +

∂

∂y

(σv_y) +

∂

∂z

(σv_z) = 0.

(b)

∂² u_x

∂t²

= μ

⎛
⎝

∂² u_x

∂x²

∂² u_x

∂y²

∂² u_x

∂z²

⎞
⎠

+ (λ+ μ)

∂Θ

∂x

+ F_x

∂² u_y

∂t²

= μ

⎛
⎝

∂² u_y

∂x²

∂² u_y

∂y²

∂² u_y

∂z²

⎞
⎠

+ (λ+ μ)

∂Θ

∂y

+ F_y

∂² u_z

∂t²

= μ

⎛
⎝

∂² u_z

∂x²

∂² u_z

∂y²

∂² u_z

∂z²

⎞
⎠

+ (λ+ μ)

∂Θ

∂z

+ F_z

where c, μ, λ are constants.

(c)

∂E_z

∂y

−

∂E_y

∂z

∂B_x

∂t

= 0

∂E_x

∂z

−

∂E_z

∂x

∂B_y

∂t

= 0

∂E_y

∂x

−

∂E_x

∂y

∂B_z

∂t

= 0

∂σ

∂t

+ ∇·(σv) = 0.

Solution (b)

∂² u

∂t²

= μ∆u + (μ+ λ) ∇Θ + F.

(c)

∇×E +

∂B

∂t

= 0.

8. Problem The differential operator ∇⁴ ≡ ∇² ∇² is known as the biharmonic (or bi - Laplacian) operator, and the partial differential equation ∇⁴u = 0 is known as the biharmonic equation.

(a) If u = u(x, y), show that

∇⁴ u = u_xxxx + 2 u_xxyy + u_yyyy

where the subscripts denote partial derivatives.

(b) Write out ∇⁴ u, in Cartesian coordinates, for the three dimensional case u = u(x, y, z).

Solution (a)

∇⁴ u

(

∂²

∂x²

∂²

∂y²

)(

∂²u

∂x²

∂²u

∂y²

)

(14)

∂⁴ u

∂x⁴

+ 2

∂⁴ u

∂x² ∂y²

∂⁴ u

∂y⁴

(15)

(b)

∇⁴ u

(

∂²

∂x²

∂²

∂y²

∂²

∂z²

)(

∂²u

∂x²

∂²u

∂y²

∂²u

∂z²

)

(16)

∂⁴ u

∂x⁴

∂⁴ u

∂y⁴

∂⁴ u

∂z⁴

+ 2

∂⁴ u

∂x² ∂y²

+ 2

∂⁴ u

∂y² ∂z²

+ 2

∂⁴ u

∂z² ∂x²

(17)

10. Problem Given that

∇×v = −

∂w

∂t

(18)

∇×w =

∂v

∂t

(19)

∇·v = 0,

(20)

∇·w = 0,

(21)

show that v and w satisfy the wave equations

c² ∇² v =

∂² v

∂t²

and c² ∇² w =

∂² w

∂t²

HINT: One of equations (1) to (14) will be useful. Also, note that [(∂)/(∂t)]∇·u = ∇·[(∂u)/(∂t)] and [(∂)/(∂t)]∇×u = ∇×[(∂u)/(∂t)].

Solution From (14),

∇² v

∇(∇·v) − ∇×(∇×v)

(22)

0 − ∇×(1/c)

∂w

∂t

(23)

(1/c)

∂

∂t

(∇×w )

(24)

(1/c)

∂

∂t

(

∂v

∂t

)

(25)

−

c²

∂² v

∂t²

(26)

Similar procedure for w.

Section 16.8

1. Problem In each case, verify the divergence theorem by working out ∫_V ∇· vdV and ∫_Sn· v dA and showing that the results are equal.

(d) v = j + x² z k, V: the cube 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1

(g) v = x y j, V: the pentahedron with vertices at (0, 0, 0), (1, 0, 0), (0, 2, 0), (0, 0, 1), (1, 0, 1), (1, 2, 0).

Solution (d) LHS: ∇·v = x² so

⌠
⌡

∇·v dV

⌠
⌡

1

0

x² dx

⌠
⌡

1

0

⌠
⌡

1

0

(27)

(28)

RHS: On the surface, z = 0, n = (0, 0, − 1) and v = (0, 1, 0) so n·v = 0. On z = 1, n = (0, 0, 1) and v = (0, 1, x²) so n ·v = x².

⌠
⌡

n·v dS

⌠
⌡

1

0

⌠
⌡

1

0

x² dx dy

(29)

(30)

On x = 0, n = (−1, 0, 0) so n·v = 0. On x = 1, n = (1, 0, 0) so n·v = 0. On y = 0, n = (0, −1, 0) and v = (0, 1, x² z) so n ·v = −1.

⌠
⌡

n·v dS

⌠
⌡

1

0

⌠
⌡

1

0

(−1) dx dz

(31)

−1.

(32)

On y = 1, n = (0, 1, 0) and v = (0, 1, x² z) so n·v = 1.

⌠
⌡

n·v dS

⌠
⌡

1

0

⌠
⌡

1

0

1 dx dz

(33)

(34)

Adding the above surface integrals yields

+ ( − 1) + (1) =

(g) LHS: ∇·v = x so that

⌠
⌡

∇·v dV

⌠
⌡

1

0

⌠
⌡

2

0

⌠
⌡

1 − y/2

0

x dz dy dx

(35)

⌠
⌡

1

0

⌠
⌡

2

0

(x −

x y

) dy dx

(36)

⌠
⌡

1

0

x dx

(37)

(38)

RHS: On y = 0, v = (0, 0, 0) so n·v = 0. On x = 0, v = (0, 0, 0) so n·v = 0. On x = 1, v = (0, y, 0) and n = (1, 0, 0) so n·v = 0. On z = 0, n·v = 0. On y + 2 z = 2, the equation of the plane can be written as z = 1 − y/2. Therefore,

√

1 + z_x² + z_y²

dx dy

(39)

⎛
√

1 +

dx dy

(40)

√5

dx dy.

(41)

The normal, n, is

n = (0,

√5

Thus,

n·v =

x y

√5

and

⌠
⌡

n·v dS

⌠
⌡

1

0

⌠
⌡

2

0

x y

√5

(42)

(43)

2. Problem Let S be a piecewise smooth orientable closed surface enclosing a region of volume V. Show that

(a) ∫_S n d A = 0 HINT: Show that ∫_S n ·a d A = 0 for every constant vector a.

(b) ∫_S n ·(x i) d A = V

(d) ∫_S n ·(x i + y j + zk) d A = 3 V

Solution

(a)

⌠
(⎜)
⌡

n·a dA

⌠
⌡

∇·a dV

(44)

(45)

Choose a = (1,0,0) so that

⌠
(⎜)
⌡

n·v dA

⌠
(⎜)
⌡

n_x dA

(46)

(47)

By choosing a = (0,1,0) and (0,0,1) yields

⌠
(⎜)
⌡

n·vdA

⌠
(⎜)
⌡

n_y dA

(48)

(49)

and

⌠
(⎜)
⌡

n_z dA = 0

which is equivalent to

⌠
(⎜)
⌡

n dA = 0.

(b)

⌠
(⎜)
⌡

n·(x i) dA

⌠
⌡

∇·(xi) dV

(50)

⌠
⌡

1 dV

(51)

(52)

(c)

⌠
(⎜)
⌡

n·(x i + y j) dA

⌠
⌡

∇·(x i + yj) dV

(53)

⌠
⌡

(1 + 1) dV

(54)

2V.

(55)

(d) Same as (c).

3. Problem Verify Green's first identity below in each case.

⌠
⌡

( ∇u ·∇v + u ∇² v ) d V =

⌠
(⎜)
⌡

∂v

∂n

d A.

(a) u = 2x³,v = xy², V: the rectangular prism 0 ≤ x ≤ 2, 0 ≤ y ≤ 3, 0 ≤ z ≤ 1

Solution

(a) LHS: ∇u = (6x², 0,0), ∇v = (y²,2xy,0) and ∇² v = 2x. Therefore,

∇u ·∇v + u ∇² v

6x²y² + 2x³(2x)

(56)

4 x⁴ + 6 x² y².

(57)

⌠
⌡

(4x⁴ + 6x²y²) dV

⌠
⌡

2

0

⌠
⌡

3

0

⌠
⌡

1

0

dz (4x⁴ + 6x²y²)

(58)

1104

(59)

RHS: On z = 0,

∂v

∂n

= −

∂v

∂z

= 0.

On z = 1,

∂v

∂n

∂v

∂z

= 0.

On x = 0,

∂v

∂n

= −

∂v

∂x

= − y².

Thus,

∂v

∂n

= 2x³ ( − y²) = 0.

On x = 2,

∂v

∂n

∂v

∂x

= y².

Thus,

∂v

∂n

= 2x³ (y²) = 16 y².

Therefore,

⌠
⌡

16 y² dy dz = 144.

On y = 0,

∂v

∂n

= −

∂v

∂y

= − 2xy = 0.

On y = 3,

∂v

∂n

∂v

∂y

= 2xy = 6x.

Thus,

∂v

∂n

= 12 x⁴.

Therefore,

⌠
⌡

12 x⁴ dx dz =

384

Adding the above integrals together yields

144 +

384

1104

4. Problem Verify the two-dimensional divergence theorem below in each case.

⌠
⌡

∇·v d A =

⌠
(⎜)
⌡

n ·v ds.

(a) v = x i + y j R: the rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ b

Solution

(a) LHS: ∇·v = 2 and

⌠
⌡

∇·v dS = 2ab.

RHS: On y = 0, ds = dx, n = (0, − 1). Thus, n·v = − y = 0. On x = a, ds = dy, n = (1,0). Thus, n·v = x = a.

⌠
⌡

b

0

a dy = ab.

On y = b, ds = − dx, n = (0,1). Thus, n·v = y = b.

⌠
⌡

0

a

b ( − dx) = ab.

On x = 0, ds = − dy, n = ( − 1,0). Thus, n·v = − x = 0. Therefore,

⌠
(⎜)
⌡

n·v dl = 2ab.

5. Problem In each case evaluate I = ∫¹₀ ∫¹₀ ∫¹₀ f dx dy dz directly. Then determine a vector field v such that ∇·v = f, convert I to a surface integral by using the divergence theorem, and evaluate the surface integral.

(a) f = x² y z

Solution

(a)

I =

⌠
⌡

1

0

⌠
⌡

1

0

⌠
⌡

1

0

x² y z dx dy dz =

Choose

v = (

x³ y z

, 0, 0),

for instance, so that ∇·v = x² y z. On z = 0, v = 0. On z = 1,

v = (

x³ y

, 0, 0),

n = (0, 0, 1) and n ·v = 0. On x = 0, v = 0. On x = 1,

v = (

y z

, 0, 0),

n = (1, 0, 0) and

n·v =

y z

Thus,

⌠
⌡

n·v dS =

⌠
⌡

dy dz =

On y = 0, v = 0. On y = 1,

v = (

y³ z

, 0, 0),

n = (0, 1, 0) and n ·v = 0.

6. Problem Derive the following results from results in this section.

(a)

⌠
⌡

∇² u d V =

⌠
⌡

∂u

∂n

d A

(b)

⌠
⌡

[ ||∇u|| ² + u∇² u] d V =

⌠
⌡

∂u

∂n

d A

CAUTION: Most author write (∇u)² for ∇ u ·∇ u, in place of our ||∇u|| ².

Solution

(a)

⌠
(⎜)
⌡

∂u

∂n

dA =

⌠
(⎜)
⌡

n ·∇u dA =

⌠
⌡

∇·∇u dA.

(b)

⌠
(⎜)
⌡

∂u

∂n

⌠
(⎜)
⌡

u n ·∇u dA

(60)

⌠
⌡

∇·( u ∇u) dV

(61)

⌠
⌡

(∇u ·∇u + u ∇·∇u) dV.

(62)

8. Problem (Alternative proof of divergence theorem) An alternative approach to the proof of the divergence theorem is to express the volume integral as an iterated integral, and carry out one integration. To illustrate the procedure, consider the simpler two - dimensional case below

⌠
⌡

∇·v d A =

⌠
(⎜)
⌡

n ·v ds,

and suppose that R is convex in the x direction, i.e., each crosshatched silver running from x_L(y) to x_R(y) (see sketch) lies entirely within R. Now

⌠
⌡

∇·v d A =

⌠
⌡

⎛
⎝

∂v_x

∂x

∂v_y

∂y

⎞
⎠

d A =

⌠
⌡

∂v_x

∂x

d x d y +

⌠
⌡

∂v_y

∂y

dx dy .

The first term becomes

⌠
⌡

∂v_x

∂x

d x d y =

⌠
⌡

y₂

y₁

⌠
⌡

x_R(y)

x_L(y)

∂v_x

∂x

d x d y

⌠
⌡

right

v_x( ds cos α) −

⌠
⌡

left

v_x (ds cos α)

⌠
⌡

right

v_x(i ·n) d s −

⌠
⌡

left

v_x( − i·n)ds

⌠
⌡

right

(v_x i)·n d s +

⌠
⌡

left

(v_xi)·nds

⌠
⌡

(v_xi)·nds.

We now state the problem: assuming that R is convex in the direction as well, show that

⌠
⌡

∂v_y

∂y

dx dy =

⌠
⌡

(v_yi)·nds,

and hence infer (47) from (8.1) and (8.2). NOTE: Extension to nonconvex regions is not difficult but will not be considered.

Solution Omitted.

10. Problem (Continuity equation) Let S₁ be a completely permeable plane surface within a steady fluid flow field. Let the velocity field be v(x, y, z), and let the mass density field be σ(x,y,z). Then set of streamlines that pass through the set points on the edge of S₁ form a "streamtube," as sketched in the accompanying figure. Let S₂ be another plane cross section of the streamtube an arbitrary distance downstream S₁, and let the areas of S₁, S₂ be A₁, A₂, respectively. Recalling the continuity equation (29), namely ∇·(σv) = 0 (since the field is steady, so that ∂σ/ ∂t = 0), integrate this equation over a control volume V which is bounded by S₁, S₂, and the streamtube surface S₃ between S₁ and S₂, apply the divergence theorem and thus show that

⌠
⌡

S₁

n ·(σv) d A +

⌠
⌡

S₂

n ·(σv) d A = 0

(63)

If σ and v are constant over S₁ and S₂, and S₁,S₂ are normal to the flow, show that (1) reduces to the simpler form

σ₁ A₁ V₁ = σ₂ A₂ V₂

(64)

often given in fluid mechanics texts, where ||v|| = V₁ on S₁ and ||v|| = V₂ on S₂.

Solution ∇·(σq) = 0. Therefore,

⌠
⌡

∇·(σq) dV

⌠
(⎜)
⌡

n·(σq) dA

(65)

(66)

⌠
⌡

S₁

n·(σq) dA +

⌠
⌡

S₂

n·(σq) dA +

⌠
⌡

S₃

n·(σq) dA = 0.

(67)

Note that over the surface, S₃, n·q = 0. Therefore,

⌠
⌡

S₁

n·(σq) dA +

⌠
⌡

S₂

n·(σq) dA = 0.

If σ and q are constant on S₁ and S₂, this becomes

n₁ ·σ₁ q₁

⌠
⌡

S₁

dA + n₂ ·σ₂ q₂

⌠
⌡

S₂

dA = 0

− σ₁ V₁ A₁ + σ₂ V₂ A₂ = 0.

Section 16.9

2. Problem Verify Stole's theorem. S is a plane surface with straight edges. The vertices and orientation of C are given.

(e) v = x² y z j, C: (0, 1, 0) to (1, 1, 0) to (1, 0, 1) to (0, 0, 1) to (0, 1, 0)

Solution (e) Surface integral:

n = (0,

√2

)

(note the direction).

∇×v = 2 x y z |_{z = 1 − y} = 2 x y (1 − y).

Therefore,

√

1 + z_x² + z_y²

(68)

√2 dx dy.

(69)

Thus,

⌠
⌡

n·∇×v d A

−

⌠
⌡

1

0

⌠
⌡

1

0

2 x y (1 − y) dx dy

(70)

−

(71)

Line integral:

v ·dr = x² y z dy.

This is 0 for (0, 1, 0)→ (1, 1, 0), (1, 0, 1)→ (0, 0, 1),and (0, 0, 1)→ (0, 1, 0). Thus,

⌠
⌡

v·dr

⌠
⌡

(1, 1, 0)→ (1, 0, 1)

x² y z dy

(72)

⌠
⌡

0

1

1² y (1 − y) dy

(73)

−

(74)

3. Problem Verify Stole's theorem.

(c) v = y² j − x y² z k. S is the surface of a unit cube with corners at (0, 0, 0), (1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 0, 1), (1, 1, 0), (0, 1, 1), (1, 1, 1), excluding the face x = 1. Take n = k on the z = 1 face.

Solution (c) Surface integral: ∇×v = (−2 x y z, y² z, 0). On z = ±1, n = (0, 0, ±1) and n ·(∇×v) = 0. On x = 0, n = (−1, 0, 0) and n ·(∇×v) = 2 x y z |_{x = 0} = 0. On y = 0, n = (0, − 1, 0) and n·(∇×v) = − y² z|_{y = 0} = 0. On y = 1, n = (0, 1, 0) and n ·(∇×v) = y² z|_{y = 1} = z. Thus,

⌠
⌡

n·∇×v dS

⌠
⌡

1

0

z dz

⌠
⌡

1

0

(75)

(76)

Line integral:

v·dR = y² dy − x y² z dz

but x = 1 so v·dR = y² dy − y² z dz.

(1, 0, 0)→ (1, 0, 1): x = 1, y = 0 and dy = 0 so v·dR = 0.

(1, 0, 1) → (1, 1, 1): x = 1, z = 1 so v·dR = y² dy. Thus,

⌠
⌡

1

0

y² dy =

(1, 1, 1) → (1, 1, 0): x = 1, y = 1 so v ·dR = − z dz. Thus,

⌠
⌡

0

1

− z dz =

(1, 1, 0)→ (1, 0, 0): x = 1, z = 0 so v·dR = y² dy. Thus,

⌠
⌡

0

1

y² dy = −

Therefore,

⌠
(⎜)
⌡

v·dR = 0 + 1/3 + 1/2 − 1/3 =

6. Problem Given the field F = y² i and the contour shown, the following conflicting calculations were put forward in exam papers

⌠
(⎜)
⌡

F ·dR =

⌠
(⎜)
⌡

y² dx =

⌠
⌡

4

0

[4 − (x − 2)²]dx +

⌠
⌡

0

4

0dx =

but

⌠
(⎜)
⌡

F ·dR =

⌠
⌡

n ·∇×F d A =

⌠
⌡

2 y d A = 2

⌠
⌡

4

0

y (y dx) = 2

⌠
⌡

4

0

[4 − (x − 2)² ] d x =

Find, and correct, the error.

Solution The area integration (∫_S 2 y dA) is wrong. ∇×F = (0, 0, −2 y). The integral range can be expressed as x − 2 = r cos θ and y = r sin θ with dA = r dr dθ. Therefore,

⌠
⌡

2 y dA

⌠
⌡

2

0

⌠
⌡

π

0

2 r sin θr dr dθ

(77)

⌠
⌡

2

0

r² dr

⌠
⌡

π

0

sin θdθ

(78)

(79)

8. (Heuristic proof of Stole's theorem) Our proof of Stole's theorem was limited to the case where S is flat. This exercise is to indicate how to prove the theorem for a general surface S that is not necessarily flat. We use the limit definition of the curl,

curlv (P) ≡

lim
B → 0

⎧
⎨
⎩

⌠
⌡

S′

n×vdA

⎫
⎬
⎭

(80)

given in Exercise 6 of section 16.5. Let B be a right circular cylinder of a small height h, and let the base of B be a plane region of area A, bounded by a simple closed curve C′, oriented so that the unit normal ν to the base, and C′, are in accordance with the right - hand rule as seen in the figure.

Dotting both sides of (1) with ν, and recalling that A ·B ×C = A ×B ·C, show that

ν·curlv =

lim
A→ 0

⎧
⎨
⎩

⌠
⌡

C′

v·dR

⎫
⎬
⎭

(81)

or, in differential form,

ν·curlvdA ≈

⌠
⌡

C′

v·dR

(82)

as A → 0. Now take the surface S in the Stole's theorem and partition it with a large number of curves as C′, as suggested in the figure, and number them as C′₁, C′₂, …. Write down (3) for each curve (C′₁, C′₂, …), realizing that ν is the normal n to S. Add these equations and, noting internal cancellation analogous to that which occurred in our proof of the divergence theorem, show that Stole's theorem follows.

(Note that S′ in (1) is the closed surface of the infinitesimal body B, whereas the S in the figure is above, is the open cap - like surface in Stole's theorem.)

Solution Will be shown in class.

9. Problem (Faraday's law) Faraday's law (that the emf around a closed curve equals the negative of the time rate of change of the magnetic flux through the curve) may be expressed as

⌠
(⎜)
⌡

E ·dR = −

⌠
⌡

B·n d A = −

⌠
⌡

∂B

∂t

·n d A

(83)

for every (fixed) surface S with closed boundary C in the field, where E is the electric field intensity, B is the magnetic flux density, t is the time, and where the relative orientation of C and n are the same as in Stokes' theorem. Apply Stokes' theorem to the line integral, use the arbitrariness and C to deduce (heuristically) that the relation

∇×E = −

∂B

∂t

(84)

holds at all points in the field. Equation (5) is one of the Maxwell's equations for time - varying fields: for steady state it is reduced to ∇×E = 0.

Solution

⌠
(⎜)
⌡

E·dR

⌠
⌡

n·∇×E dA

(85)

−

⌠
⌡

n·

∂B

∂t

dA.

(86)

Thereby,

∇×E = −

∂B

∂t

10. Problem (Surfaces with holes) Here we indicate a modest extension of Stokes' theorem to the case where S has one or more in it. It should suffice to consider the surface S shown in left - hand figure, which for simplicity, has only one hole and lies in the plane of the paper. Thus, the boundary C of S is not a simple closed curve, as called for in Stokes' theorem

(a) Slit the region as shown in the middle figure, and define the contours C₁, C₂, C₃, C₄ enclosing the region S′. Since and C = C₁ + C₂ + C₃ + C₄ do satisfy the requirements given in Stokes' theorem, that theorem can now be applied. Let the gap between C₃ and C₄ tend to zero, show that we obtain

⌠
⌡

n·∇×vdA =

⌠
(⎜)
⌡

C₁

v·dR +

⌠
(⎜)
⌡

C₂

v·dR

(87)

in accordance with the right - hand figure, where n is direction out of the paper, toward the reader. NOTE: Since Green's theorem is but a special case of Stokes' theorem, a result analogous to (6) follows immediately for Green's theorem

⌠
⌡

⎛
⎝

∂Q

∂x

−

∂P

∂y

⎞
⎠

d A =

⌠
(⎜)
⌡

C₁

Pdx + Qdy +

⌠
(⎜)
⌡

C₂

P dx + Q dy.

(88)

Solution (a) Shown in class.

11. Problem Verify Stokes' theorem (6) above.

(a) v = y³ i, S is the region S₁ shown, and n = k

Solution

(a)v = (y³, 0, 0), ∇×v = (0, 0, − 3 y²) and n ·∇×v = − 3 y². Therefore,

⌠
⌡

n·∇×v dS

⌠
⌡

(− 3 y²) dx dy

(89)

⌠
⌡

2

−2

⌠
⌡

2

−2

(−3 y²) dx dy −

⌠
⌡

1

0

⌠
⌡

0

−1

(−3 y²) dx dy

(90)

− 63.

(91)

The line integral is along (2, −2) → (2, 2) → (−2, 2) → (−2, −2) → (2, −2).

⌠
(⎜)
⌡

outside

y³ dx +

⌠
(⎜)
⌡

inside

y³ dx

⌠
⌡

2

− 2

(−2)³ dx +

⌠
⌡

− 2

2

(2)³ dx +

⌠
⌡

− 1

0

(0)³ d +

⌠
⌡

0

− 1

(1)³ dx

(92)

−63.

(93)

13. Problem If C is a piecewise smooth simple closed curve in the x, y plane, oriented counterclockwise, show that the area A enclosed by C is given by the line integral

A =

⌠
(⎜)
⌡

(x dy − y dx).

(94)

Solution

⌠
(⎜)
⌡

(x dy − y dx)

⌠
⌡

(

∂

∂x

(

x) −

∂

∂y

( −

y))dA

(95)

⌠
⌡

(96)

(97)

14. Problem Use (8) in Exercise 13 to compute A if C is the boundary of the

Solution (c) Let x = 3 cos θ and y = 3 sin θ.

A =

⌠
(⎜)
⌡

(x dy − y dx)

⌠
⌡

2π

0

(cos² θ+ sin² θ) dθ

(98)

9 π.

(99)

15. Problem Show that (33) (see below) does not hold for the case where P = − y/(x² + y²), Q = x/(x² + y²), and S is the unit disk x² + y² ≤ 1. Explain why this failure does not violate Green's theorem.

⌠
⌡

⎛
⎝

∂Q

∂x

−

∂P

∂y

⎞
⎠

dA =

⌠
(⎜)
⌡

P dx + Q dy

Solution Because

∂Q

∂x

−

∂P

∂y

fails to exist at (0, 0).

Section 16.10

2. Problem Show whether or not v is irrotational: if it is, find its potential Φ such that v = ∇Φ. In what domain D is your result valid?

(a) v = i − 2 j − 8 k

(g) v = 2 x z i + 3 y j + x² k

Solution (a) Yes as ∇×v = 0. Solving for

∂Φ

∂x

(100)

∂Φ

∂y

−2,

(101)

∂Φ

∂z

−8

(102)

yields Φ = x − 2 y − 8 z for instance.

(g) Yes as ∇×v = 0. Solving for

∂Φ

∂x

2 x z,

(103)

∂Φ

∂y

3 y,

(104)

∂Φ

∂z

x²

(105)

yields

Φ = x² z +

y²

for instance.

3. Problem Can functions f(x, y, z), g(x, y, z) be found such as irrotational? If so, find one such f and g. If not, why?

(a) v = x y i − z j + f k

Solution (a)

∇×v = (1 +

∂f

∂y

, −

∂f

∂x

, − x)

which cannot be zero because of −x. No such functions exist.

5. Problems Consider ∫_C v ·dR, where C is the path x = sin τ, y = cos τ, z = 2 τ from τ = 0 to τ = π. Show that v is irrotational, and use this fact to evaluate the integral two ways: by path simplification and by the potential method. Sketch the simplified path.

(c)v = 5 x^3/2 e^2y i + (4 x^5/2 e^2y + 5 y² j

Solution

∂Φ

∂x

= 5 x^3/2 e^2y

and

∂Φ

∂y

= 4 x^5/2 e^2y + 5 y²

yields

Φ = 2 x^5/2 e^2y +

y³

for instance. Thereby,

⌠
⌡

v ·dR

Φ|_(0,1,0)^{(0, − 1, 2 π)}

(106)

(− 5/3) − (5/3)

(107)

−

(108)

6. Problem Repeat Exercise 5, but where C is path x = cos³ τ, y = τ², z = 3 τ from τ = 0 to τ = π.

(a) v = x i + y j + z k.

Solution (a) Yes as ∇×v = 0. Φ = (x² + y² + z²)/2 for instance.

⌠
⌡

v·dR

Φ|_{(1, 0, 0)}^{( −1, π², 3 π)}

(109)

(π⁴ + 9 π²).

(110)

7. Problem (Exact differentials) An expression P dx + Q dy + R dz, where P, Q, and R are functions of x, y, z defined in some domain D, is called a first-order differential form in the three variables. If there exists a function f such that P dx + Q dy + R dz = df in D, then the form is said to be an exact differential. Assuming that P, Q, R are C¹ in D, show that for P dx + Q dy + R dz to be an exact differential it is necessary and sufficient that ∇×(P i + Q j + R k) = 0 in D. Further, show that f is the scalar potential of P i + Q j + R k.

Solution If ∇×(P, Q, R) = 0 then, (P, Q, R) is irrotational and there exists a potential, f, such that

P =

∂f

∂x

, Q =

∂f

∂y

,R =

∂f

∂z

Therefore,

P dx + Q dy + R dz =

∂f

∂x

dx +

∂f

∂y

dy +

∂f

∂z

dz.

10. Problem (Solenoidal fields) Let v be a vector field defined in a domain D. If ∇·v = 0 at each point in D, then v is said to be solenoidal in D. We have the following companion to theorem 16.10.1:

THEOREM 16.10.2 Solenoidal Field

If v is C¹ and solenoidal in a simply connected domain D, then there exists a vector field w in D such that v = ∇×w

Prove this theorem, for the case where D is the prism x₁ < x < x₂, y₁ < y < y₂, z₁ < z < z₂. HINT: we need to show that the equation v = ∇×w does admit a solution w. To do this, start with the three scalar equations

∂w_z

∂y

−

∂w_y

∂z

= v_x,

∂w_x

∂z

−

∂w_z

∂x

= v_y,

∂w_y

∂x

−

∂w_x

∂y

= v_z.

(111)

There is enough leeway (1) to set one of the components of w, say w_x, equal to zero. Then the latter two equations in (1) become

∂w_z

∂y

= − v_y, and

∂w_y

∂x

= v_z.

(112)

which can be integrated to give w_y to within an arbitrary additive function A(y, z) and w_z to within an arbitrary additive function B(y, z). Putting these results into the first equation in (1), show that it is possible to choose A(y, z) = B(y, z) = 0 so that

w = 0 i +

⎡
⎣

⌠
⌡

x

x₀

v_z(ξ,y,z) ∂ξ

⎤
⎦

j +

⎡
⎣

⌠
⌡

y

y₀

v_x(x₀, η, z) ∂η−

⌠
⌡

x

x₀

v_y(ξ, y, z) ∂ξ

⎤
⎦

(113)

where x₀, y₀ are any constants such that x₁ ≤ x₀ ≤ x₂ and y₁ ≤ y₀ ≤ y₂. Show that to the right - hand side of (3) we can add ∇f, the gradient of an arbitrary scalar function f that is C² in D. NOTE: Observe the pattern that emerges in this section: If v is irrotational (∇×v = 0), v is expressible as the gradient of scalar potential Φ(v = ∇Φ). Analogously, if v is solenoidal ( ∇·v = 0), v is expressible as the curl of a vector potential w (v = ∇×w).

The notions of irrotational and solenoidal fields are complementary in the sense that C¹ text field v, defined in a bounded simply connected domain D, can be split as v = v₁ + v₂, where v₁ is irrotational in D and v₂ is solenoidal in D, a fact that we state without proof.

Solution Explained in class. Just follow the hint.

11. Problem Show that v is solenoidal (see Exercise 10), and determine a vector potential w. (As noted in Exercise 10, w is uniquely determined.)

(a) v = a i + b j + c k

(d) v = x y i − y z k

Solution (a) ∇·v = 0. w = (0, c x, a y − b x) for instance.

(d) ∇·v = y + 0 − y. Use (b) in Problem 9.

(0,

⌠
⌡

x

0

( − y z) d ξ,

⌠
⌡

y

0

(0) (η) dη)

(114)

(0, − x y z, 0)

(115)

for instance.

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On 01 Oct 2022, 10:50.