Surface Area and Surface
Integrals
Surface Area
• Given some surface in 3 space, we want to
calculate its surface area
• Just as before, a double integral can be used to
calculate the area of a surface
• We are going to look at how to calculate the
surface area of a parameterized surface over a
given region
• Given the vector parameterization

r (u, v)  x(u, v), y(u, v), z(u, v)
the surface area is given by

R
ru  rv dudv
Let’s take a look at where this comes from
• Example
– Find the surface area of a cone with a height of 1
– The parameterization is
r cos , r sin  , r
– Let’s check it out in maple
Alternative Notation
• If we want to find the surface area of a function,
z = f(x,y), than we can simplify the cross product
• Then

r  x, y, z  x, y, f ( x, y)

rx  1,0, f x ( x, y )
and

ry  0,1, f y ( x, y )
 
rx  ry  1,0, f x  0,1, f y 
 f x , f y ,1  ( f x )  ( f y )  1
2
2
Alternative Notation
• If we want to find the surface area of a function,
z = f(x,y), than we can use the following

R
( f x )  ( f y )  1 dxdy
2
2
• Example
– Find the surface area of the plane
z = 6 – 3x – 2y that lies in the first octant
• We can calculate the surface area over any
given region
• Example
– Find the surface area of the function z = xy
between the two cylinders
x  y 1
2
2
x  y 4
2
2
Surface Integrals
• A surface integral involves integrating a
function over some surface in 3 space
• We have calculated integrals of functions over
regions in the xy plane and over 3 dimensional
figures, now we want to integrate over a 2
dimensional surface in 3 space
• Thus if the function represents a density, the
surface integral would calculate the total mass
of the 2D plate that has the shape of the surface
Surface Integrals
• To calculate a surface integral of g over the
surface D if the surface is defined parametrically
we have

R
g ( r (u, v )) ru  rv dudv
• Example
– Calculate the surface integral of f(x,y) = xy over the
cone of radius 1 in the first octant from the previous
example
Surface Integrals
• To calculate a surface integral of g over the surface
D if the surface is given by z = f(x,y) we can use

R
g ( x, y, f ( x, y )) ( f x )  ( f y )  1 dxdy
2
2
• Example
– Find the surface integral of the function g(x,y,z) = xyz
over the plane z = 6 – 3x – 2y that lies in the first octant
Surface Integrals of Vector Fields
• Recall that a line integral of a vector field F
could be interpreted as work done by the force
field on a particle moving along the path
• If the vector field F represents the flow of a
fluid, then the surface integral of F will
represent the amount of fluid flowing through
the surface (per unit time)
• In this case the amount of fluid flowing through
the per unit time is called the flux
• Surface integrals of a vector field are sometimes
referred to as flux integrals
Surface Integrals of Vector Fields
• The term flux comes from physics
• It is used to denote the rate of transfer of:
– Fluid
• liquid flow density
– Particles
• Electromagnetism
– Energy across a surface
• Total charge of a surface
Surface Integrals of Vector Fields
• Imagine water flowing through a surface
– If the flow of water is perpendicular to the surface a
lot of water will flow through and the flux will be
large
– If the flow of water is parallel to the surface then no
water will flow through the surface and the flux will
be zero
• In order to calculate the flux we must add up the
component of F that is perpendicular to the
surface
Surface Integrals of Vector Fields
• Let n represent a unit normal vector to the surface
• Than in order to find the component of F that is
perpendicular we can use our dot product
F n
– This is 0 if F and n are perpendicular
– Positive if F and n are in the same direction
– Negative if F and n are in opposite directions
• Given some fluid flow F, integrating F  n will
determine the total flux of fluid through a surface
– It will be positive if it is in the same direction as n
– Negative if it is in the opposite direction of n
Surface Integrals of Vector Fields
• Now we must sum over our surface so we will
combine our dot product with our formula for a
surface integral from before

R
g ( r (u, v )) ru  rv dA
and we get the following
F

n
r

r



R
• This can be simplified!
u
v
dA
Surface Integrals of Vector Fields
• The formula for a unit normal vector given our
surface parameterization r is
ru  rv
n
ru  rv
• Inserting that into our surface integral
  F  n  r  r
R
we get
u
v
dA

ru  rv 
F

r

r
dA


u
v
R
r

r
u
v 

Surface Integrals of Vector Fields
• We can cancel scalars

ru  rv 
F

r

r
dA


u
v
R
r

r
u
v 

to get
• Example
F

r

r
dA





R
u
v
2
2
z

x

y
– The surface will be the parabaloid
,0≤z
≤ 1 with the vector field F  x,  y,0 
– Should our integral be positive or negative?
– How can we tell?
Surface Integrals of Vector Fields
• In order for a surface to have an orientation the surface
must have two sides
• Thus every point will have two normal vectors,
n1 and n2  n1
• The set we choose determines the orientation which is
described as the positive orientation
• You should be able to choose a normal vector in a way
so that if it varies in a continuous way over the surface,
when you return to the initial position it still points in the
same direction
• The Möbius band is not orientable
– No matter where you start to construct a continuous unit
normal field, moving the vector continuously around the
surface will return it to the starting point with a direction
opposite to the one it had when it started.
Surface Integrals of Vector Fields
• As mentioned before, a surface integral over a vector
field is positive if the normal of the surface and flow are
in the same direction, negative if they are in opposite
directions and 0 if they are perpendicular
• How do we know which normal to use for a surface?
• A surface is closed if it is the boundary of some solid
region
– For example the surface of a sphere is closed
– A closed surface has a positive orientation if we choose the set
of normal vectors that point outward from the region
– A closed surface has a negative oritenation if we choose the set
of normal vectors that point inward toward the region
– This convention is only used for closed surfaces
– The surface in our previous example was not closed so this
does not apply
Surface Integrals of Vector Fields
• In order to calculate our surface integral we use
  F   r  r   dA
u
R
v
• Since ru  rv is a normal vector to the surface we can
rewrite the integral as
  F  n  dA
R
• Now if our surface is given by a function z = f(x,y) then
n  f ( x, y, z) where f(x,y,z) = f(x,y) - z and our
integral becomes
  F  f ( x, y, z)  dA
R
• Let’s try our previous example again with this method
Surface Integrals of Vector Fields
• Calculate the flux of F  x, y, z  of the surface S
which is a hemisphere given by the following
2
x  y  z 1
2
2
2
x2  y2  1
• In this case we have a closed bounded region so our
surface has a positive orientation that is pointing
outwards
– Should we expect our integral to be positive or negative?
– In order to calculate this integral we will have to break S into 2
separate regions
Relationship between Surface Integrals
and Line Integrals
• To calculate a line integral we use
 F (r (t ))  dr where dr  x '(t )dt, y '(t )dt 
C
which summed up the components of the vector
field that were tangent to the path given by dr
• To calculate a surface integral we use
R F   ru  rv  dA
which sums up the components of the vector field
that are in the normal direction given by ru  rv

