Mathematical Tools for Physics

9—Vector Calculus 1 258

they’re all easy. Take the first of them:

∫y 0 +∆y

y 0

dy

∫z 0 +∆z

z 0

dz

[

vx( 0 )+

(∆x)

∂vx

∂x

( 0 ) + (y−y 0 )

∂vx

∂y

( 0 ) + (z−z 0 )

∂vx

∂z

( 0 ) +

1

2

(∆x)^2

∂^2 vx

∂x^2

( 0 ) +···

]

=vx( 0 )∆y∆z+ (∆x)

∂vx

∂x

( 0 )∆y∆z+

1

2

(∆y)^2 ∆z

∂vx

∂y

( 0 ) +

1

2

(∆z)^2 ∆y

∂vx

∂z

( 0 ) +

1

2

(∆x)^2

∂^2 vx

∂x^2

( 0 )∆y∆z+···

Now look at the second integral, the one that you have to subtract from this one. Before plunging in to the
calculation, stop and look around. What will cancel; what will contribute; what will not contribute? The only
difference is that this is now evaluated atx 0 instead of atx 0 + ∆x. The terms that have∆xin them simply
won’t appear this time. All the rest are exactly the same as before. That means that all the terms in the above
expression that donothave a∆xin them will be canceled when you subtract the second integral. All the terms
thatdohave a∆xwill be untouched. The combination of the two integrals is then

(∆x)

∂vx

∂x

( 0 )∆y∆z+

1

2

(∆x)^2

∂^2 vx

∂x^2

( 0 )∆y∆z+

1

2

(∆x)

∂^2 vx

∂x∂y

( 0 )(∆y)^2 ∆z+···

Two down four to go, but not really. The other integrals are the same except thatxbecomesyandy
becomeszandzbecomesx. The integral over the two faces withyconstant are then

(∆y)

∂vy

∂y

( 0 )∆z∆x+

1

2

(∆y)^2

∂^2 vy

∂y^2

( 0 )∆z∆x+···

and a similar expression for the final two faces. The definition of Eq. ( 9 ) says to add all three of these expressions,
divide by the volume, and take the limit as the volume goes to zero.V = ∆x∆y∆z, and you see that this is a
common factor in all of the terms above. Cancel what you can and you have

∂vx

∂x

( 0 ) +

∂vy

∂y

( 0 ) +

∂vx

∂x

( 0 ) +

1

2

(∆x)

∂^2 vx

∂x^2

( 0 ) +

1

2

(∆y)

∂^2 vy

∂y^2

( 0 ) +

1

2

(∆z)

∂^2 vz

∂z^2

( 0 ) +···