Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Surface Integrals of
Vector Fields

Just as we did with line integrals we now need to move on to surface integrals of vector fields
...
The same thing will hold true with surface integrals
...

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)
...
This means that every surface will have two sets of normal
vectors
...

There is one convention that we will make in regards to certain kinds of oriented surfaces
...
A surface S is closed if it is the boundary of some solid
region E
...
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
...

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
...

First, let’s suppose that the function is given by z = g ( x, y )
...
Now,
recall that ∇f will be orthogonal (or normal) to the surface given by f ( x, y, z ) = 0
...
The only potential problem is that it might not
be a unit normal vector
...
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
...


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
...

This will be important when we are working with a closed surface and we want the positive
orientation
...
Remember that the “positive”
orientation must point out of the region and this may mean downwards in places
...
Again, remember that we always have that option
when choosing the unit normal vector
...
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 ) )
...

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
...
In order to guarantee that it is a unit normal vector we
will also need to divide it by its magnitude
...
If it doesn’t then we can always take the negative of this vector and
that will point in the correct direction
...

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
...
This is sometimes called the flux of
F across S
...
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
...
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
...
If we’d needed the “downward”
orientation then we would need to change the signs on the normal vector
...
So, we really need to be careful here when using this
formula
...

When we’ve been given a surface that is not in parametric form there are in fact 6 possible
integrals here
...

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
...

Notice as well that because we are using the unit normal vector the messy square root will always
drop out
...
All we’ll need to work with is the numerator of the unit vector
...
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
...
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
...

Note as well that there are even times when we will used the definition,

∫∫ F idS = ∫∫ F in dS ,
S

S

directly
...

Let’s now work a couple of examples
...
Assume that S has positive
orientation
...
This
means that we have a closed surface
...

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
...
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
...

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
...
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
...
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
...
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
...
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
...
Also, the dropping of the minus sign is not a typo
...


S1 : The Paraboloid
Okay, here is the surface integral in this case
...
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
...


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
...

Here is the evaluation of this integral

---