Surface Area and Surface Integrals

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

the surface area is given by

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

– Let’s check it out in maple

),(),,(),,(),( vuzvuyvuxvur

u vRr r dudv

rrr ,sin,cos

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

and

),(,,,, yxfyxzyxr

),(,1,0),(,0,1 yxfryxfr yyxx

yxyx ffrr ,1,0,0,1

1)()(1,, 22 yxyx ffff

Alternative Notation• If we want to find the surface area of a function,

z = f(x,y), than we can use the following

• Example– Find the surface area of the plane

z = 6 – 3x – 2y that lies in the first octant

2 2( ) ( ) 1x yRf f dxdy

• We can calculate the surface area over any given region

• Example– Find the surface area of the function z = xy

between the two cylinders2 2

2 2

1

4

x y

x y

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

• To calculate a surface integral of g over the surface D if the surface is defined parametrically we have

• 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

( ( , )) u vRg r u v r r dudv

• To calculate a surface integral of g over the surface D if the surface is given by z = f(x,y) we can use

• 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

2 2( , , ( , )) ( ) ( ) 1x yRg x y f x y f f dxdy

Surface Integrals of Vector Fields• Recall that a line integral of a vector field could

be interpreted as work done by the force field on a particle moving along the path

• If the vector field    represents the flow of a fluid, then the surface integral of   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

F

F

F

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 that is perpendicular to the surface

F

Surface Integrals of Vector Fields• Let represent a unit normal vector to the surface• Than in order to find the component of that is

perpendicular we can use our dot product

– This is 0 if and are perpendicular– Positive if and are in the same direction– Negative if and are in opposite directions

• Given some fluid flow , integrating will determine the total flux of fluid through a surface– It will be positive if it is in the same direction as – Negative if it is in the opposite direction of

F n

Fn

F

n

F n

F

n

F

n

F

nn

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

and we get the following

• This can be simplified!

( ( , )) u vRg r u v r r dA

u vRF n r r dA

Surface Integrals of Vector Fields• The formula for a unit normal vector given our

surface parameterization is

• Inserting that into our surface integral

we get

u v

u v

r rnr r

u vu vR

u v

r rF r r dAr r

u vRF n r r dA

r

Surface Integrals of Vector Fields• We can cancel scalars

to get• Example– The surface will be the parabaloid , 0 ≤ z

≤ 1 with the vector field– Should our integral be positive or negative?– How can we tell?

u vu vR

u v

r rF r r dAr r

u vRF r r dA

2 2z x y , ,0F x y

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,

• 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

1 2 1andn n n

• 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

• Since is a normal vector to the surface we can rewrite the integral as

• Now if our surface is given by a function z = f(x,y) then where f(x,y,z) = f(x,y) - z and our

integral becomes

• Let’s try our previous example again with this method

u vRF r r dA

u vr r

R

F n dA

( , , )n f x y z

( , , )R

F f x y z dA

Surface Integrals of Vector Fields• Calculate the flux of of the surface

S which is a hemisphere given by the following

• 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

2, ,F x y z

2 2 2

2 2

1

1

x y z

x y

Relationship between Surface Integrals and Line Integrals

• To calculate a line integral we use

which summed up the components of the vector field that were tangent to the path given by

• To calculate a surface integral we use

which sums up the components of the vector field that are in the normal direction given by

( ( )) where '( ) , '( )C

F r t dr dr x t dt y t dt

dr

u vRF r r dA

u vr r