584 CHAPTER 29. CALCULUS OF VECTOR FIELDS

By the chain rule this equals∫D

Ti j (y (s, t))(

∂y

∂xα

∂xα

∂ s× ∂y

∂xβ

∂xβ

∂ t

)jdsdt.

Summation over repeated indices is used. Remember y = y (t,x) and it is always assumedthe mapping x→ y (t,x) is one to one and so, since on the surface ∂Vt near y, the pointsare functions of (s, t), it follows x is also a function of (s, t). Now by the properties of thecross product, this last integral equals∫

DTi j (x(s, t))

∂xα

∂ s∂xβ

∂ t

(∂y

∂xα

× ∂y

∂xβ

)jdsdt (29.12)

where here x(s, t) is the point of ∂V0 which corresponds with y (s, t) ∈ ∂Vt . Thus

Ti j (x(s, t)) = Ti j (y (s, t)) .

(Perhaps this is a slight abuse of notation because Ti j is defined on ∂Vt , not on ∂V0, but itavoids introducing extra symbols.) Next 29.12 equals∫

DTi j (x(s, t))

∂xα

∂ s∂xβ

∂ tε jab

∂ya

∂xα

∂yb

∂xβ

dsdt

=∫

DTi j (x(s, t))

∂xα

∂ s∂xβ

∂ tεcabδ jc

∂ya

∂xα

∂yb

∂xβ

dsdt

=∫

DTi j (x(s, t))

∂xα

∂ s∂xβ

∂ tεcab

=δ jc︷ ︸︸ ︷∂yc

∂xp

∂xp

∂y j

∂ya

∂xα

∂yb

∂xβ

dsdt

=∫

DTi j (x(s, t))

∂xα

∂ s∂xβ

∂ t∂xp

∂y j

=ε pαβ det(F)︷ ︸︸ ︷εcab

∂yc

∂xp

∂ya

∂xα

∂yb

∂xβ

dsdt

=∫

D(detF)Ti j (x(s, t))ε pαβ

∂xα

∂ s∂xβ

∂ t∂xp

∂y jdsdt.

Now ∂xp∂y j

= F−1p j and also

ε pαβ

∂xα

∂ s∂xβ

∂ t= (xs ×xt)p

so the result just obtained is of the form∫D(detF)F−1

p j Ti j (x(s, t))(xs ×xt)p dsdt =∫D(detF)Ti j (x(s, t))

(F−T )

jp (xs ×xt)p dsdt.

This has transformed the integral over Pt to one over P0, the part of ∂V0 which correspondswith Pt . Thus the last integral is of the form∫

P0

det(F)(T F−T )

ip NpdA

584 CHAPTER 29. CALCULUS OF VECTOR FIELDSBy the chain rule this equalsoy Xa dy Oxg[ritu (s,1) ) (x2 58 Os *Oxg Or oeSummation over repeated indices is used. Remember y = y (t, a) and it is always assumedthe mapping x — y (t,x) is one to one and so, since on the surface OV, near y, the pointsare functions of (s,t), it follows a is also a function of (s,t). Now by the properties of thecross product, this last integral equals)) 2a Oxp Oy oy[rile 52s (52 “Sy, te (29.12)where here x (s,t) is the point of Vo which corresponds with y (s,t) € AV;. ThusTj (@(s,t)) = Ti (y(s,t)).(Perhaps this is a slight abuse of notation because 7;; is defined on OV,;, not on AVo, but itavoids introducing extra symbols.) Next 29.12 equalsAxa IXB OYa AYp~ 1) Fe Gp bith See dep det)) xe Oxp IVa OYp=|, Ti (@(8) Ge Gy cab Oe Bye Ixy OO=6 jca a Dy. dap Ava A= OB Yo PXp PYa PYb= [tlw lot) Ge Gp ee Dxp dy; xq Oxp=E pap det(F )——e,—On)) 2a Oxp Oxp OVe OYa OYp= [til@lo) oe oe Oy; Oxp Axa Oxp_ OXe Oxp OXp"hime eB Gy ar aye etOXpNow %, =F, ; and alsoOXa Ox“PB Os Otso the result just obtained is of the form= (a5 X 21),I (detF) FT) (w(s,1)) (ws x @,), dsdt =I (det F) Ti; (w(s,t)) (F~") ,, (ws x @), dst.This has transformed the integral over P, to one over Po, the part of OV which correspondswith P;. Thus the last integral is of the form‘ -Thn, det (F) (7F ) ip NpdA