14.7 Surface Integrals
Suppose f is a function of three variables whose domain include a surface S . After
dividing S into patches Sij with area ∆Sij and choosing a point Pij∗ from each Sij , we
dene the surface integral of f over the surface S as
¨
f (x, y, z) dS := lim
kP k→0
S
m X
n
X
f (Pij∗ )∆Sij
i=1 j=1
If S is given by r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k, (u, v) ∈ D, then
¨
¨
f (x(u, v), y(u, v), z(u, v))|ru × rv | dA
f (x, y, z) dS :=
S
D
Example 1. Evaluate
˜
lies in the rst octant.
S
y dS , where S is the part of the plane 3x + 2y + z = 6 that
If a thin sheet has the shape of a surface S and the density at the point (x, y, z) is
ρ(x, y, z), then the total mass of the sheet is
¨
ρ(x, y, z) dS
m=
S
Example 2. Find the mass of a thin funnel in the shape of a cone z = x2 + y 2 inside
the cylinder x2 + y 2 = 2x, if its density is a function ρ(x, y, z) = x2 + y 2 + z 2 .
p
Oriented Surfaces
If it is possible to choose a unit normal vector n at every such point (x, y, z) so that
n varies continuously over S , then S is called an oriented surface and the given
choice of n provides S with an orientation. There are two possible orientations for
any orientable surface.
When S is a smooth orientable surface given in parametric form by a vector function
r(u, v), then it is automatically supplied with the orientation of the unit normal vector
n=
ru × rv
|ru × rv |
and the opposite orientation is given by −n.
Convention: For a closed surfaces (is the boundary of a solid region), the positive
orientation is outward.
A Möbius strip is an example of non-orientable surfaces.
Surface Integrals of Vector Fields
Suppose that S is an oriented surface with unit normal vector n and a uid with density
ρ(x, y, z) and velocity eld v(x, y, z) ows through S . If we divide S into small patches
Sij with area ∆Sij and choose a point Pij∗ from each Sij , then the mass of uid per
unit time crossing Sij in the direction of n is approximately equal to ρv · n∆Sij .
By forming Riemann sum and taking limit, we get the rate of ow through S :
¨
ρv · n dS
S
Denition. If F is a continuous vector eld dened on an oriented surface S with
unit normal vector n, then the surface integral of F over S is
¨
¨
F · dS :=
S
F · n dS
S
This integral is also called the ux of F across S .
If S is given by a vector function r(u, v) with (u, v) ∈ D, then
¨
¨
F · n dS =
S
Thus
F·
ru × rv
|ru × rv | dA
|ru × rv |
D
¨
¨
F · dS =
S
F · (ru × rv ) dA
D
Example 3. Evaluate
centered at the origin.
˜
S
F · dS where F = z i + y j + x k and S is the unit sphere
An application of surface integrals occurs in the study of heat ow. Suppose the
temperature at a point (x, y, z) in a body is u(x, y, z). Then the heat ow is dened
as the vector eld
F = −K ∇u
where K is a constant called the conductivity of the substance. The rate of heat
ow across the surface in the body is then given by the surface integral
¨
¨
F · dS = −K
S
∇u · dS
S
Example 4. The temperature at the point (x, y, z) in a substance with conductivity
K = 10 is u(x, y, z) = x2 + y 2 + z 2 . Find the rate of heat ow inward across the sphere
x2 + y 2 + z 2 = 4.