Page 410 - Advanced engineering mathematics

390 CHAPTER 12 Vector Integral Calculus

Assuming that neither of these tangent vectors is the zero vector, they both lie in the tangent
plane to the surface at P 0 . Their cross product is therefore normal to this tangent plane. This
leads us to define the normal to the surface at P 0 to be the vector

N(P 0 ) = T u 0 × T v 0

i j k

∂x ∂y ∂z
= (u 0 ,v 0 ) (u 0 ,v 0 ) (u 0 ,v 0 )
∂u ∂u ∂u
∂x (u 0 ,v 0 ) ∂y (u 0 ,v 0 ) ∂z (u 0 ,v 0 )
∂v ∂v ∂v
∂y ∂z ∂z ∂y ∂z ∂x ∂x ∂z ∂x ∂y ∂y ∂x

= − i + − j + − k,
∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v ∂u ∂v
in which all partial derivatives are evaluated at (u 0 ,v 0 ).
To make this vector easier to write, define the Jacobian of two functions f and g to be

∂( f, g) ∂ f/∂u ∂ f/∂v ∂ f ∂g ∂g ∂ f
= = − .
∂(u,v) ∂g/∂u ∂g/∂v ∂u ∂v ∂u ∂v
Then
∂(y, z) ∂(z, x) ∂(x, y)
N(P 0 ) = i + j + k,
∂(u,v) ∂(u,v) ∂(u,v)
with all the partial derivatives evaluated at (u 0 ,v 0 ). This expression is easy to remember with an
observation. Write x, y, z in this order. For the i component of N, delete x, leaving y, z, (in this
order) in the numerator of the Jacobian. For the j component, delete y from x, y, z, but move to
the right, getting z, x in the Jacobian. For the k component, delete z, leaving x, y, in this order.

EXAMPLE 12.16
The elliptical cone has coordinate functions
x = au cos(v), y = au sin(v), z = u
with a and b positive constants. Part of this surface is shown in Figure 12.13.

3

–20 –2 –3
–10 –1
0
1 10
2 20
3

FIGURE 12.13 Elliptical cone z =
au cos(v),y = bu sin(v),z = u.

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October 14, 2010 14:53 THM/NEIL Page-390 27410_12_ch12_p367-424