16.5

Surface Integrals

( ) = u v

S S

area S d dudv r r

Parametrization of a surface: , ( , ), ( , ), ( , )u v x u v y u v z u vr

where ( , ) lie in some region in the planeu v uv

Review :

area element of a surface u vd dudv r r

Ex: Parametrization of a sphere :

, sin( )cos( ), sin( )sin( ), cos( ) ,

a

a a a

r

2area element: sin( )d a r r 2( recall volume element is sin( ) )dV

1 2

1 2

1 2

If in the region and

we have , then

with and both traversed counter clockwise

x y

C C

C C

Q P Pdx Qdy Pdx Qdy

C C

Theorem : between

2Area of a sphere of radius is 4a a

2 2 2 2 2 2

Compute the surface area of that portion of the

the sphere 2 cut out by the cone , 0.x y z z x y z

Example :

2Recall: sin( )a r r

2Surface area : 1 sin( ) S S

d d d a d d r r

2 sin( )a d d

2

2

0 /4

sin( )a d d

2

/4 04

2sin( ) 2 2cosd d

22 2 2 4 2 2

2

2 2a

Surface area of the graph of a function: ( , ).z f x y

, , , ( , )u v u v f u vr

1, 0, u ufr

0, 1, v vfr

u v r r 1 0

0 1

u

v

f

f

i j k

u vf f i j k

2 2Surface area element of ( , ) : 1 x yz f x y d f f dxdy

2 2 ( ) 1 x yarea S f f dxdy

2 2 2 2 2

Find the surface area of the portion of the cone

lying above the disc ( 1) 1.z x y x y

Example :

2 2 ( , )z f x y x y

2 2 2 2

1 2

2x

xf x

x y x y

2 2

y

yf

x y

2 22 2

2 2 2 21 1x y

x yf f

x y x y

2 2

2 2

22

x y

x y

2 2 ( ) 1 x yarea S f f dxdy

( ) 2 2 ( ) 2R

area S dxdy area R

2 2 ( ) 1 x yarea S f f dxdy

Surface area of a level surface : ( , , ) cF x y z

assume that we can solve for z, i.e. ( , ) with ( , , ( , )) cz h x y F x y h x y

differentiate implicity with respect to : + 0x z

hx F F

x

= , = yx

x y

z z

FFh h

F F

2 2Surface area element of ( , ) : 1 x yz h x y d h h dxdy

2 2

1 yx

z z

FF

F F

2 2 21= x y z

z z

FF F F

F F

Surface area element of ( , , ) c : z

FF x y z d dxdy

F

( ) zS R

Farea S d dxdy

F

where R is the projection of S onto the xy plane.

(we need to assume S projects uniquely, i.e. no folding)

2 2 2 2F x y z a

2 ,2 ,2F x y z

2 2 24 4 4F x y z 2 2 24 x y z

24 2a a 2F a 2

2z

F a a

F z z

2 2 2 2

22 2 2 2

Intersection between

and

2

x y z a z a h

x y a a h ah h

1

R

a dAz

2 2 2

1

R

a dAa x y

2 2 2 2

Find the surface area of a spherical cap of height h

in a sphere of radius a: , . x y z a a h z a

Example :

( ) zR

Farea S dA

F

R is the projection of the cap into the xy plane.

A disc whose boundary is a circle.

2 2 2

1Surface Area

R

a dAa x y

22 2

2 20 0

ah hr

a drda r

22 2

2 2

00

ah h

a d a r

2 22 2a a ah h a

Surface area of a spherical cap of height h

in a sphere of radius a is equal to 2 ah

2 2 2 R: 2x y ah h

2 22 2a a ah h a

2 a a h a 2 ah

2

(if 2 , we get the whole

sphere with area 2 4 )

h a

ah a

2 2 1 if ( , )

and R is the projection of S onto the xy plane

x y

S R

Gd G f f dxdy z f x y

= where , ( , ), ( , ), ( , )

describes S in parametric form.

u v

S R

Gd G dudv u v x u v y u v z u v r r r

= where ( , , ) c describes S implicitly,

and R is the projection of S onto the xy plane

zS R

FGd G dxdy F x y z

F

General surface integral , , where is a surface in 3-space.S

G x y z d S

2 The surface S: lies in the first quadrant and is bounded by

0, 4, 0, 3. It has mass density equal to the distance to the yz plane.

What is its total mass?

y x

y y z z

Example :

mass density x total mass =S

xd

cannot project into the xy plane (why not?)

2 , 0x zf x f

2instead project into the xz plane (or yz plane): ( , )y f x z x

2 2 21 1 4 x zd f f dxdz x dxdz

2total mass 1 4 S

x x dxdz 3 2

2

0 0

1 4 x x dxdz

3 2

3/22

00

1 21 4

8 3dz x

3/2117 1

4

if 0 y 4, then 0 2 x

Let , , be the velocity field of a fluid.x y zv

Volume of fluid flowing through

an element of surface area per unit time

is approximated by

comp d nv v n

Flux through a surfaceRecall: Outward flux across a

curve C in the plane is C

ds F n

The total volume of fluid flowing through

the surface S per unit time is called the

flux through S

if the vector field is denoted by ,

then flux =

( or other common notation )

S

S

d

dS

F

F n

F n

flux = S

d v n

A surface is if there exists a continuous unit normal vector

function defined at each point on the surface

S orientable

n

orientable

surface with

and n n

upward

orientation

n

downward

orientation

n

non-orientable

surface since

becomes n n

1

To find , define the surface by , , , then S g x y z c gg

n n

In a flux integral, we need to choose a normal S

d F n n.

2 2 4g x y z

2 ,2 ,1g x y

2 24 4 1g x y

2 2

1 12 ,2 ,1

4 4 1g x y

g x y

n2 20 4

0 2,0 2

z x y

r

2 2

3 3 3 2

0 0

cos sin 4r r rdrd

5 4

2 23 3 2

5 40

0

cos sin 2r rr d

2

3 3325

0

cos sin 4 d

8

2 2 2 21 1

2 2 Let , ,z and S the paraboloid 4 ,0 4.

Compute the flux of through S, where is the upward pointing normal.

x y z x y z Example : F

F n

(upward pointing normal,

positive z coordinate)

2 2

3 3

0 0

why is cos sin 0?d d

2 2

2 2

1 1

2 2

2 ,2 ,1 , ,z

4 4 1

x yx y

x y

F n

3 3

2 2

4 4 1

x y z

x y

F n

2 21 x yd f f dxdy

2 21 4 4x y dxdy

3 32 2

2 2

flux =

14 4 1

R

x y

R

d

x y zf f dxdy

x y

F n

3 3 ( ) R

x y z dxdy

3 3 2 2 ( 4 ) R

x y x y dxdy