Chapter 16.1 Gradient,Divergence,and Curl Solutions

Chapter 16.1 Gradient,Divergence,and Curl Solutions

Problem 6

F = xy²i− yz²j + zx²k

divF = _∂x^∂ (xy²) +_∂y^∂ (−yz²) +_∂z^∂ (zx²)

crulF =   

i j k

∂

∂x

∂

∂y

∂

∂z

xy² −yz² zx²   

= 2yzi− 2xzj − 2xyk

Problem 9

Since x = rcosθ, and y = rsinθ, we have r²= x²+ y², and so

∂r

∂x = ^x_r = cosθ

∂r

∂y =^y_r = sinθ

∂

∂xsinθ = _∂x^{∂ y}_r = ^−xy_r3 =−^cosθsinθ_r

∂

∂ysinθ = _∂y^{∂ y}_r = ¹_r−^y_r²3

=^x_r²₃ =^cos_r^2θ

1

∂

∂xcosθ = _∂x^{∂ x}_r = ¹_r−^x_r²3

=^y_r3² = ^sin_r^2θ

∂

∂ycosθ = _∂y^{∂ x}_r = ^−xy_r3 =−^cosθsinθ_r

(The last two derivatives are not needed for this exercise, but will be useful for the next two exercises.) For

F = ri + sinθj we have

divF = ^∂r_∂x+_{∂y sin θ}^∂ = cos θ +^cos_r^2θ

crulF =   

i j k

∂

∂x

∂

∂y

∂

∂z

r sinθ 0   

= (−^sinθcosθ_r − sinθ)k

Problem 12

We use the Maclaurin expansion of F, as presented in the proof of Theorem 1:

F = F0+ F₁x + F₂y + F₃z +· · ·, where

F0= F(0, 0, 0)

F1= _∂x^∂ F(x, y, z)|(0, 0, 0) = (^∂F_∂x¹)i +^∂F_∂x²)j +^∂F_∂x³)k|(0, 0, 0) F2= _∂y^∂ F(x, y, z)|(0, 0, 0) = (^∂F_∂y¹)i +^∂F_∂y²)j + ^∂F_∂y³)k|(0, 0, 0)

F3= _∂z^∂ F(x, y, z)|(0, 0, 0) = (^∂F_∂z¹)i +^∂F_∂z²)j +^∂F_∂z³)k|(0, 0, 0) and where · · · represents terms of degree 2 and higher in x, y, and z. On the top of the box B_a, b, c, we have z = c and ^N = k. On the bottom of the box, we have z = − c and ^N =−k. On both surfaces dS=dxdy.

Thuse

2

(∫ ∫

top +∫ ∫

bottom)F• ^NdS

=∫a

−adx∫b

−bdy(cF3• k − cF3• (−k)) + · · ·

= 8abcF₃• k + · · · = 8abc_∂z^∂ F₃(x, y, z)|(0, 0, 0) +· · ·

where · · · represents terms of degree 4 and higher in a, b, and c. Similar formulas obtain for the two other pairs of faces, and the three formulas combine into

∮

Ba,b,cF• ^NdS = 8abcdivF(0, 0, 0) +· · · It follows that

lima,b,c→0+ 1 8abc

∮

Ba,b,cF• ^NdS = divF(0, 0, 0) .

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