Transcript 16.7 Surface Integrals

Chapter 16 – Vector Calculus
16.7 Surface Integrals
Objectives:
 Understand integration of
different types of surfaces
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16.7 Surface Integrals
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Surface Integrals

The relationship between surface integrals
and surface area is much the same as the
relationship between line integrals and arc
length.
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16.7 Surface Integrals
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Surface Integrals

Suppose a surface S has a vector equation
r(u, v) = x(u, v) i + y(u, v) j + z(u, v) k
(u, v)  D
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Surface Integrals

In our discussion of surface area in
Section 16.6, we made the approximation
∆Sij ≈ |ru x rv| ∆u ∆v
where:
x
y
z
ru 
i
j k
u
u
u
x y
z
rv  i  j  k
v
v
v
are the tangent vectors at a corner
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Surface Integrals - Equation 2

If the components are continuous and ru and rv
are nonzero and nonparallel in the interior of D,
it can be shown that:
 f ( x, y, z ) dS   f (r(u, v)) | r
u
S
 rv | dA
D
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Surface Integrals

Formula 2 allows us to compute a surface integral by
converting it into a double integral over the parameter
domain D.
◦ When using this formula, remember that
f(r(u, v) is evaluated by writing
x = x(u, v), y = y(u, v), z = z(u, v)
in the formula for f(x, y, z)
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Example 1

Evaluate the surface integral.

1  x 2  y 2 dS ,
S
S is the helicoid with vector equation
r (u, v)  u cos v i  u sin v j  v k , 0  u  1, 0  v   .
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Graphs

Any surface S with equation z = g(x, y)
can be regarded as a parametric surface
with parametric equations
x=x
y=y
z = g(x, y)
◦ So, we have:
 g 
rx  i    k
 x 
 g 
ry  j    k
 y 
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Graphs

Therefore, Equation 2 becomes:
 f ( x, y, z) dS  
S
D
2
 z   z 
f ( x, y, g ( x, y ))       1 dA
 x   y 
2
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Graphs

Similar formulas apply when it is more convenient to
project S onto the yz-plane or xy-plane.

For instance, if S is a surface with equation y = h(x, z)
and D is its projection on the xz-plane, then
 f ( x, y, z) dS  
S
D
 y   y 
f ( x, h( x, z ), z )       1 dA
 x   z 
2
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Example 2 – pg. 1145 # 9

Evaluate the surface integral.
2
x
 yz dS ,
S
S is the part of the plane z  1  2 x  3 y
that lies above the rectangle  0,3   0, 2.
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Oriented Surface

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.
◦ The given choice of n provides S with an orientation.
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Possible Orientations

There are two possible orientations for
any orientable surface.
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Positive Orientation

Observe that n points in the same direction as the
position vector—that is, outward from the sphere.
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Negative Orientation

The opposite (inward) orientation would have been
obtained if we had reversed the order of the parameters
because rθ x rΦ = –rΦ x rθ
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Closed Surfaces

For a closed surface—a surface that is the boundary of a
solid region E—the convention is that:
◦ The positive orientation is the one for which
the normal vectors point outward from E.
◦ Inward-pointing normals give the negative orientation.
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Flux Integral (Def. 8)

If F is a continuous vector field defined on an oriented
surface S with unit normal vector n, then the surface
integral of F over S is:
 F  dS   F  n dS
S
S
◦ This integral is also called the flux of F across S.
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Flux Integral

In words, Definition 8 says that:
◦ The surface integral of a vector field over S is equal to
the surface integral of its normal component over S (as
previously defined).
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Flux Integral


If S is given by a vector function r(u, v), then n is
ru  rv
n
ru  rv
We can rewrite Definition 8 as equation 9:
F

d
S

F

(
r

r
)
dA
u
v


S
D
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Example 3

Evaluate the surface integralS F  dS for the given vector
field F and the oriented surface S. In other words, find
the flux of F across S. For closed surfaces, use the
positive (outward) orientation.
F( x, y, z )  xz i  x j  y k ,
S is the hemisphere x 2  y 2  z 2  25, y  0 oriented
in the direction of the positive y-axis.
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Vector Fields

In the case of a surface S given by a graph
z = g(x, y), we can think of x and y as parameters and
write:
 g g

F  (rx  ry )  ( P i  Q j  R k )    i 
jk 
y
 x


From this, formula 9 becomes formula 10:


g
g
F

d
S


P

Q

R
dA


S
D  x y 
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Vector Fields


g
g
F

d
S


P

Q

R
dA


S
D  x y 
◦ This formula assumes the upward orientation of S.
◦ For a downward orientation, we multiply by –1.
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Example 4

Evaluate the surface integralS F  dS for the given vector
field F and the oriented surface S. In other words, find
the flux of F across S. For closed surfaces, use the
positive (outward) orientation.
F ( x, y , z )  x i  y j  z 4 k ,
S is the part of the cone z  x 2  y 2 beneath
the plane z  1 with downward directions.
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Other Examples

In groups, please work on the following
problems on page 1145:
#’s 12, 14, and 28.
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Example 5 – pg. 1145 # 12

Evaluate the surface integral.
 y dS ,
S
3

2  32
2
S is the surface z   x  y 
3

0  x  1, 0  y  1.
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Example 6 – pg. 1145 # 14

Evaluate the surface integral.
 z dS ,
S
S is the surface x  y  2 z 2 ,
0  y  1, 0  z  1.
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Example 7 – pg. 1145 # 28

Evaluate the surface integralS F  dS for the given vector
field F and the oriented surface S. In other words, find
the flux of F across S. For closed surfaces, use the
positive (outward) orientation.
F ( x, y, z )  xy i  4 x 2 j  yz k ,
S is the surface z  xe y , 0  x  1,
0  y  1, with upward orientation.
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