Advanced book on Mathematics Olympiad

Real Analysis 581

∮

C

Pdx+Qdy+Rdz=

∫∫

S

(

∂Q

∂x

−

∂P

∂y

)

dxdy+

(

∂R

∂y

−

∂Q

∂z

)

dydz

+

(

∂P

∂z

−

∂R

∂x

)

dzdx.

Writing the parametrization with coordinate functions−→v 1 (s) = (x(s), y(s), z(s)),
−→v
2 (t)=(x′(t), y′(t), z′(t)), the linking number ofC 1 andC 2 (with the factor 41 πignored)
becomes
∮

C 1

∮

C 2

(x′−x)(dz′dy−dy′dz)+(y′−y)(dx′dz−dz′dx)+(z′−z)(dy′dx−dx′dy)

((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )^3 /^2

The 1-formPdx+Qdy+Rdz, which we integrate onC=C 1 ∪C′ 1 ,is
∮

C 2

(x′−x)(dz′dy−dy′dz)+(y′−y)(dx′dz−dz′dx)+(z′−z)(dy′dx−dx′dy)

((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )^3 /^2

.

Note that here we integrate against the variablesx′,y′,z′, so this expression depends
only onx, y, andz. Explicitly,

P(x, y, z)=

∮

C 2

−(y′−y)dz′+(z′−z)dy′

((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )^3 /^2

,

Q(x,y,z)=

∮

C 2

(x′−x)dz′−(z′−z)dx′

((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )^3 /^2

,

R(x, y, z)=

∮

C 2

−(x′−x)dy′+(y′−y)dx′

((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )^3 /^2

.

By the general form of Green’s theorem, lk(C 1 ,C 2 )=lk(C 1 ′,C 2 )if

∂Q

∂x

−

∂P

∂y

=

∂R

∂y

−

∂Q

∂z

=

∂P

∂z

−

∂R

∂x

= 0.

We will verify only∂Q∂x−∂P∂y=0, the other equalities having similar proofs. The part of
it that containsdz′is equal to
∮

C 2

− 2 ((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )−^3 /^2

+ 3 (x′−x)^2 ((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )−^5 /^2

+ 3 (y′−y)^2 ((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )−^5 /^2 dz′

=

∮

C 2

((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )−^3 /^2

+ 3 (z′−z)^2 ((x′−x)^2 +(y′−y)^2 +(z′−z)^2 )−^5 /^2 dz′