sol08

�= xˆi is a gradient field and f = Clearly F Then �

�

�

20.3 4b.

x 2

2

�. is a scalar potential for F

�󰀚· dR �= f(1, 0)−f(−1, 0)=0F

C

7.󰀍20.4 2a.

ˆ, with t from 0 to 1. Then �= atˆLet P =(a, b, c) and R i + btˆj+ ctk

ˆ,�=(abct2 +abt+act+a)ˆF i+(abct2 +abt+bct+b)ˆj+(abct2 +act+bct+c)kso

�

1

0

�

�·󰀂dR dt = F

dt

�

1

(3abct2 + 2abt + 2act + 2bct + a + b + c)dt =

0

= abc + ab + ac + bc + a + b + c

Now if we put x, y, z instead of a, b, c, we obtain the scalar potential f(x, y, z)= xyz + xy + xz + yz + x + y + z. It is easy to verify that

�. f is indeed a scalar potential for F8.󰀍� 20.5 1.

Let f be the scalar potential we want to find. Then �

f(x, y, z) =(ye xy + ze xz)dx + a(y, z)=

= exy + e xz + a(y, z).󰀛

Hence󰀎

xe xy + 2 = fy = xe xy +

Hence

a(y, z)=2y + b(z),

f(x, y, z)= exy + e xz + 2y + b(z).

so

b(z)= c󰀛

∂a(y, z)

,∂y󰀚

so

∂a(y, z)󰀎

=2. ∂y

xe xz = fz = xe xz + b�(z),

We can take b(z) =0, andobtain a scalar potential󰀎

f(x, y, z)= exy +󰀎e xz + 2y.

The given curve C goes from P(0, 0, 1)to Q(1, 1, 0). Hence

�

�󰀚· dR �= f(Q)−f(P)= e + 1 F

C

3