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Surface Integrals of
Vector Fields

Just as we did with line integrals we now need to move on to surface integrals of vector fields
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The same thing will hold true with surface integrals
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Let’s start off with a surface that has two sides (while this may seem strange, recall that the
Mobius Strip is a surface that only has one side!) that has a tangent plane at every point (except
possibly along the boundary)
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This means that every surface will have two sets of normal
vectors
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There is one convention that we will make in regards to certain kinds of oriented surfaces
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A surface S is closed if it is the boundary of some solid
region E
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We say that the closed
surface S has a positive orientation if we choose the set of unit normal vectors that point outward
from the region E while the negative orientation will be the set of unit normal vectors that point
in towards the region E
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In order to work with surface integrals of vector fields we will need to be able to write down a
formula for the unit normal vector corresponding to the orientation that we’ve chosen to work
with
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First, let’s suppose that the function is given by z = g ( x, y )
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Now,
recall that ∇f will be orthogonal (or normal) to the surface given by f ( x, y, z ) = 0
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The only potential problem is that it might not
be a unit normal vector
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In or case this is,

∇f
n=
∇f

In this case it will be convenient to actually compute the gradient vector and plug this into the
formula for the normal vector
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First, notice that the component of the normal vector in the z-direction (identified by the k in the
normal vector) is always positive and so this normal vector will generally point upwards
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This will be important when we are working with a closed surface and we want the positive
orientation
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Remember that the “positive”
orientation must point out of the region and this may mean downwards in places
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Again, remember that we always have that option
when choosing the unit normal vector
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We could just as easily done the above work for
surfaces in the form y = g ( x, z ) (so f ( x, y, z ) = y − g ( x, z ) ) or for surfaces in the form

x = g ( y, z ) (so f ( x, y, z ) = x − g ( y, z ) )
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But if the vector is normal to the tangent plane at a point then it will also be normal to the surface
at that point
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In order to guarantee that it is a unit normal vector we
will also need to divide it by its magnitude
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If it doesn’t then we can always take the negative of this vector and
that will point in the correct direction
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Okay, now that we’ve looked at oriented surfaces and their associated unit normal vectors we can
actually give a formula for evaluating surface integrals of vector fields
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This is sometimes called the flux of
F across S
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We will need to be careful with each of the following
formulas however as each will assume a certain orientation and we may have to change the
normal vector to match the given orientation
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In this case let’s also
assume that the vector field is given by F = P i + Q j + R k and that the orientation that we are
after is the “upwards” orientation
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If we’d needed the “downward”
orientation then we would need to change the signs on the normal vector
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So, we really need to be careful here when using this
formula
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When we’ve been given a surface that is not in parametric form there are in fact 6 possible
integrals here
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Given each form of the surface there will be two possible unit normal vectors and we’ll need to
choose the correct one to match the given orientation of the surface
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Notice as well that because we are using the unit normal vector the messy square root will always
drop out
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All we’ll need to work with is the numerator of the unit vector
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It should also be noted that the square root is nothing
more than,

( gx ) + ( g y )
2

2

+ 1 = ∇f

so in the following work we will probably just use this notation in place of the square root when
we can to make things a little simpler
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In this case the surface integral is,

∫∫ F idS = ∫∫ F in dS
S

S

⌠⌠ ⎛ ru × rv
= ⎮⎮ F i⎜⎜
⌡⌡ ⎝ ru × rv

⎞
⎟⎟ ru × rv dA
⎠

D

= ∫∫ F i( ru × rv ) dA
D

Again note that we may have to change the sign on ru × rv to match the orientation of the surface
and so there is once again really two formulas here
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Note as well that there are even times when we will used the definition,

∫∫ F idS = ∫∫ F in dS ,
S

S

directly
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Let’s now work a couple of examples
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Assume that S has positive
orientation
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This
means that we have a closed surface
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Let’s first get a sketch of S so we can get a feel for what is going on and in which direction we
will need to unit normal vectors to point
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Also note that in
order for unit normal vectors on the paraboloid to point away from the region they will all need to
point generally in the negative y direction
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Since S is composed of the two surfaces we’ll need to do the surface integral on each and then
add the results to get the overall surface integral
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In this case we
have the surface in the form y = g ( x, z ) so we will need to derive the correct formula since the
one given initially wasn’t for this kind of function
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First
define,

f ( x , y , z ) = y − g ( x, z ) = y − x 2 − z 2

We will next need the gradient vector of this function
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However, as noted above we need the normal vector point in the negative y
direction to make sure that it will be pointing away from the enclosed region
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We don’t really need to divide this by the
magnitude of the gradient since this will just cancel out once we actually do the integral
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Also, the dropping of the minus sign is not a typo
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S1 : The Paraboloid
Okay, here is the surface integral in this case
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In this case D is the disk of radius 1 in the xz-plane and so it makes sense to use
polar coordinates to complete this integral
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x = r cos θ

z = r sin θ

0 ≤ θ ≤ 2π

0 ≤ r ≤1

Note that we kept the x conversion formula the same as the one we are used to using for x and let
z be the formula that used the sine
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Here is the evaluation of this integral
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This one is actually
fairly easy to do and in fact we can use the definition of the surface integral directly
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Now we want the unit normal vector to point away from the enclosed region and since it must

also be orthogonal to the plane y = 1 then it must point in a direction that is parallel to the y-axis,
but we already have a unit vector that does this
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It also points in the correct direction for us to use
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In this case since we are using the definition directly we won’t get the canceling
of the square root that we saw with the first portion
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Here is the value of the surface integral
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Here is the surface integral that we
were actually asked to compute
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Assume that S has the positive
orientation
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To help us visualize this here is a sketch of the surface
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Now, in order for the unit normal vectors on the sphere to point away from
enclosed region they will all need to have a positive z component
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On the other hand, the unit normal on the bottom of the disk must point in the negative z direction
in order to point away from the enclosed region
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In this case since the surface is a sphere we will need to
use the parametric representation of the surface
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0 ≤ θ ≤ 2π

0≤ϕ ≤

π

2

Next, we need to determine rθ × rϕ
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rθ (θ , ϕ ) = −3sin ϕ sin θ i + 3sin ϕ cos θ j
rϕ (θ , ϕ ) = 3cos ϕ cos θ i + 3cos ϕ sin θ j − 3sin ϕ k
i

j

rθ × rϕ = −3sin ϕ sin θ

3sin ϕ cos θ

k
0

3cos ϕ cos θ

3cos ϕ sin θ

−3sin ϕ

= −9sin 2 ϕ cos θ i − 9sin ϕ cos ϕ sin 2 θ k − 9sin 2 ϕ sin θ j − 9sin ϕ cos ϕ cos 2 θ k
= −9sin 2 ϕ cos θ i − 9sin 2 ϕ sin θ j − 9sin ϕ cos ϕ ( sin 2 θ + cos 2 θ ) k
= −9sin 2 ϕ cos θ i − 9sin 2 ϕ sin θ j − 9sin ϕ cos ϕ k

Note that we won’t need the magnitude of the cross product since that will cancel out once we
start doing the integral
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Therefore we will need to use the
following vector for the unit normal vector
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Okay, next we’ll need

F ( r (θ , ϕ ) ) = 3sin ϕ cos θ i + 3sin ϕ sin θ j + 81cos 4 ϕ k

Remember that in this evaluation we are just plugging in the x component of r (θ , ϕ ) into the
vector field etc
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F ( r (θ , ϕ ) )i(rθ ×rϕ ) = 27 sin 3 ϕ cos 2 θ + 27 sin 3 ϕ sin 2 θ + 729sin ϕ cos5 ϕ

= 27 sin 3 ϕ + 729sin ϕ cos5 ϕ
Now we can do the integral
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In this case we are looking at
the disk x 2 + z 2 ≤ 9 that lies in the plane z = 0 and so the equation of this surface is actually
z = 0
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This also means that we can use the definition of the surface integral here with

n = −k
We need the negative since it must point away from the enclosed region
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Here is surface integral that we were asked to look
at
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If
v is the velocity field of a fluid then the surface integral

∫∫ v idS
S

represents the net amount of fluid flowing through S

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