Page 411 - Advanced engineering mathematics

12.5 Surface Integrals 391

Since
x
2 2
y
2
z = + .
a b
this surface is a “cone” with major axis the z - axis. Planes z = k parallel to the x, y - plane
√
intersect this surface in ellipses. We will write the normal vector P 0 at P 0 = (a 3/4,b/4,1/2)
obtained when u = u 0 = 1/2 and v = v 0 = π/6. Compute the Jacobian components:
∂(y, z) ∂y ∂z ∂z ∂y
= −
∂(u,v) ∂u ∂v ∂u ∂v (1/2,π/6)
√
=[b sin(v)(0) − bu cos(v)] (1/2,π/6) =− 3b/4,
∂(z, x) ∂z ∂x ∂x ∂z
= −
∂(u,v) ∂u ∂v ∂u ∂v
(1/2,π/6)
=[−au sin(v) − a cos(v)(0)] (1/2,π/6) =−a/4,

∂(x, y) ∂x ∂y ∂y ∂x
= −
∂(u,v) ∂u ∂v ∂u ∂v
(1/2,π/6)
=[a cos(v)bu cos(v) − b sin(v)(−au sin(v))] (1/2,π/6) = ab/2.
Then
√
3b a ab
N(P 0 ) =− i − j + k.
4 4 2
We frequently encounter the case that a surface is given by an equation z = S(x, y), with
u = x and v = y as parameters. In this case

∂(y, z) ∂(y, z) 0 1 ∂S
= = =− ,
∂(u,v) ∂(x, y) ∂S/∂x ∂S/∂y ∂x

∂(z, x) ∂(z, x) ∂S/∂x ∂S/∂y ∂S
= = =− ,
∂u,v ∂(x, y) 1 0 ∂y
and

∂(x, y) ∂(x, y) 10
= = = 1.
∂(u,v) ∂(x, y) 01
Now the normal vector at P 0 : (x 0 , y 0 ) is
∂S ∂S
N(P 0 ) =− (x 0 , y 0 )i − (x 0 , y 0 )j + k
∂x ∂y
∂z ∂z
=− (x 0 , y 0 )i − (x 0 , y 0 )j + k.
∂x ∂y

EXAMPLE 12.17
Let be the hemisphere given by

2
z = 4 − x − y .
2
√
We will find the normal vector at P 0 : (1, 2,1). Compute
∂z ∂z √
=−1 and =− 2.
√
√
∂x (1, 2) ∂y (1, 2)
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

October 14, 2010 14:53 THM/NEIL Page-391 27410_12_ch12_p367-424