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12.5 Surface Integrals 393

where D is the set of points in the x, y - plane over which the surface is defined. We now
recognize that this area is actually the integral of the length of the normal vector:

area of = N(x, y) dx dy. (12.9)
D
This is analogous to the formula for the length of a curve as the integral of the length of the
tangent vector. More generally, if is given by coordinate functions x(u,v), y(u,v), z(u,v) for
(u,v) varying over some set D in the u,v - plane, then

area of = N(u,v) du dv.
D

12.5.4 Surface Integrals
The line integral of f (x, y, z) over C with respect to arc length is
b

2
2
2

f (x, y, z)ds = f (x(t), y(t), z(t)) x (t) + y (t) + z (t) dt.
C a
We want to lift this idea up one dimension to integrate a function over a surface instead of over
a curve. To do this, imagine that the coordinate functions are functions of two variables u and
b
v,so ···dt will be replaced by ···du dv. The differential element of arc length ds for
a D
C will be replaced by the differential element of surface area on , which by equation (12.9) is
dσ = N(u,v) du dv.
Let be a smooth surface with coordinate functions x(u,v), y(u,v), z(u,v) for (u,v) in D.

Let f be continuous on . Then the surface integral of f over is denoted f (x, y, z)dσ

and is defined by

f (x, y, z)dσ = f (x(u,v), y(u,v), z(u,v)) N(u,v) du dv.
D
If is piecewise smooth, then the line integral of f over is the sum of the line integrals over
the smooth pieces.
If is given by z = S(x, y) for (x, y) in D, then

2 2
∂S ∂S

f (x, y, z)dσ = f (x, y, S(x, y)) 1 + + dx dy.
D ∂x ∂y
EXAMPLE 12.20

We will compute the surface integral xyz dσ over the part of the surface

1
2
x = u cos(v), y = u sin(v), z = u sin(2v)
2
corresponding to (u,v) in D : 1 ≤ u ≤ 2,0 ≤ v ≤ π.
First we need the normal vector. The components of N(u,v) are:
∂(y, z) sin(v) u cos(v)
2
= 2 = u [sin(v)cos(2v) − cos(v)sin(2v)],
∂(u,v) u sin(2v) u cos(2v)

2
∂(z, x) u sin(2v) u cos(2v) 2
= =−u [sin(v)sin(2v) + cos(v)cos(2v)],
∂(u,v) cos(v) −u sin(v)
and

∂(x, y) cos(v) −u sin(v)
= = u.
∂(u,v) sin(v) u cos(v)
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October 14, 2010 14:53 THM/NEIL Page-393 27410_12_ch12_p367-424