VC.10
Surface Area Calculations and
Surface Integrals (Day 2)
Example 1: Surface Area
C o n sid e r th e tw o -d im e n sio n al su rfac e in x yz -sp ac e d e sc rib e d b y th e
x
e q u atio n f( x , y )  sin  y  c o s   . Fin d th e su rfac e are a o f th e su rfac e
2
g iv e n th e b o u n d s 0  x  4 an d 0  y  2  :
First, w e com e up w ith a param eterizatio n of the surface:
x(u , v )  u
y(u , v )  v
u
z(u , v )  sin ( v ) co s  
2
0  u 4
0  v  2
The Surface Area Conversion Factor
N ow w e can think of this as a m ap from a uv-rectangle to a xyz-surface.
(u , v )

 u 
 u , v , sin ( v ) cos   
 2 

The Surface Area Conversion Factor
W e can also consider how a sm all uv-rectangle m aps into xyz-space:
A s in p reviou s ch ap ters, w e'll relate a sm all ch an g e in area on th e u vrectan g u lar reg ion relates to a ch an g e in su rface area on th e xyz-su rface .
N otice th at u v-rectan g les of fixed area m ap in to little xyz-su rfaces
of varyin g su rface area.
The Surface Area Conversion Factor
Im agine a tiny rectangle in the uv-plane :
  x(u , v )  y(u , v )  z(u , v ) 
,
,



v

v
v


dv
du
  x(u , v )  y(u , v )  z(u , v ) 
,
,



u

u
u


T h e c h an g e in su rfac e are a th at re su lts w ith th e x yz -su rfac e c an b e
ap p ro x im ate d b y th e are a o f th e p aralle lo g ram g e n e rate d b y th e
tan g e n t v e c to rs g iv e n b y tak in g th e p art ial d e riv ativ e o f th e m ap
(x (u ,v ),y(u ,v ),z (u ,v )) w ith re sp e c t to u an d v .
The Surface Area Conversion Factor
  x (u , v )  y(u , v )  z(u , v )    x (u , v )  y(u , v )  z(u , v ) 
,
,
,
,




u

u

u

v

v
v

 

  x(u , v )  y(u , v )  z(u , v ) 
,
,



v

v
v


dv
du
  x(u , v )  y(u , v )  z(u , v ) 
,
,



u

u
u


N o w re call th at th e are a o f a p aralle lo g ram in 3 D -sp ace can b e q u ick ly
co m p u te d b y fin d in g th e m ag n itu d e o f th e cro ss p ro d u ct o f its
g e n e ratin g v e cto rs. T h is is e q u iv ale n t to th e le n g th o f th e n o rm al v e c to r
sh o w n ab o v e .
The Surface Area Conversion Factor
  x (u , v )  y(u , v )  z(u , v )    x (u , v )  y(u , v )  z(u , v ) 
,
,
,
,




u

u

u

v

v
v

 

  x(u , v )  y(u , v )  z(u , v ) 
,
,



v

v
v


dv
du
A longer normal vector means
that the uv-rectangle has much
more surface area in xyz-space
(more curvature, really)
  x(u , v )  y(u , v )  z(u , v ) 
,
,



u

u
u


  x(u , v )  y(u , v )  z(u , v )    x(u , v )  y(u , v )  z(u , v ) 
SA xyz (u , v )  
,
,
,
,


u
u
u
v
v
v

 

The Surface Area Conversion Factor
G iv e n a p aram e te rizatio n o f a su rface (x (u ,v ),y(u ,v ),z(u ,v )), yo u can
fin d th e su rface are a o f th is su rface fo r u 1  u  u 2 an d v 1  v  v 2 b y
in te g ratin g th e fo llo w in g :
  x(u , v )  y(u , v )  z(u , v )    x(u , v )  y(u , v )  z(u , v ) 
SA xyz (u , v )  
,
,
,
,



u

u

u

v

v

v

 

If you w rite T(u , v )  ( x(u , v ), y(u , v ), z(u , v )), you can w rite th is a b it
m ore con cisely as th e follow in g :
S A x yz (u , v ) 

T
u

T
v
 T T   T T 





 u v   u v 
The Surface Area Conversion Factor
G iv e n a p aram e te rizatio n o f a su rface (x (u ,v ),y(u ,v ),z(u ,v )), yo u can
fin d th e su rface are a o f th is su rface fo r u 1  u  u 2 an d v 1  v  v 2 b y
in te g ratin g th e fo llo w in g :
v 2 u2

dA 
R

S A x yz (u , v ) d u d v
v 1 u1
L e t T (u ,v )  ( x (u , v ), y(u , v ), z(u , v )) b e a m ap fro m u v -sp ace to x yz - sp ace :
T h e n S A x yz (u , v ) 

T
u

T
v
 T T   T T 






u

v

u

v

 

Example 1 Revisited: Surface Area
x
Fin d th e su rface are a o f th e su rface f( x , y )  sin  y  co s   g iv e n th e
2
b o u n d s 0  x  4 an d 0  y  2  :

 u 
Let T(u , v )   x(u , v ), y(u , v ), z(u , v )    u , v , sin ( v ) c os    for 0  u  4 an d 0  v  2  .
 2 

 T T   T T 






u

v

u

v

 

M ath e m at ica calcu late s S A xyz (u , v ) 

1
4
3(7  co s(u ))  (3  5co s(u ))co s( 2 v )
2 4
U se M ath e m atica ag ain :
  SA
0 0
xyz
(u , v ) d u d v  2 8 .2 9 8
Example 2: Surface Integrals
C o n sid e r th e p aram e te rize d su rface b e lo w fo r 0  u  4 an d 0  v  2  :
T (u , v ) 
 x(u , v ), y(u , v ), z(u , v ) 

 u 
  u , v , sin ( v ) co s   
 2 

T h e su rface is m ad e of a m ixtu re of variou s m etals of varyin g
2
d en sity d escrib ed b y g (x,y,z)  xy  z g / cm . Fin d th e m ass of th e su rface.
Use M athem atica and the previously com pu ted SA xyz (u, v ) :
2 4
  g( x(u , v ), y(u , v ), z(u , v ))SA
0 0
xyz
(u , v ) d u d v  1 7 4 .2 2 1 g
Summary: Surface Integrals:
G iv e n a p aram e te rizatio n o f a su rface (x (u ,v ),y(u ,v ),z(u ,v )), yo u can
fin d th e su rface in te g ral o f th e fu n ctio n g (x,y,z) fo r u 1  u  u 2 an d
v 1  v  v 2 w ith re sp e ct to su rface are a b y in te g ratin g th e fo llo w in g :
v 2 u2
 g( x , y , z )d A    g( x(u , v ), y(u , v ), z(u , v )) S A
R
x yz
(u , v ) d u d v
v 1 u1
L e t T (u ,v )  ( x (u , v ), y(u , v ), z(u , v )) b e a m ap fro m u v -sp ace to x yz - sp ace :
T h e n S A x yz (u , v ) 

T
u

T
v
 T T   T T 






u

v

u

v

 

Example 3: Surface Area Enclosed in a
Curve Plotted on a Surface (Bonus Material!)
Fin d th e are a o f th e e llip se (2 co s(t),3 s in (t)) p lo tte d o n th is sam e su rface :
T (u , v )   x(u , v ), y(u , v ), z(u , v ) 

 u 
  u , v , sin ( v ) co s   
 2 

T h e fille d in e llip se is g iv e n b y:
(2s co s( t ), 3s sin ( t ))
fo r 0  t  2  an d 0  s  1 .
T o m ap it on to th e su rface, let u( s , t )  2s cos( t ) an d v( s , t )  3s sin ( t ), an d
p lot  x(u( s , t ), v( s , t )), y(u( s , t ), v( s , t )), z( u( s , t ), v( s , t ))  .
Example 3: Surface Area Enclosed in a
Curve Plotted on a Surface (Bonus Material!)
Fin d th e are a o f th e e llip se (2 co s(t),3 s in (t)) p lo tte d o n th is sam e su rface :
T (u , v )   x(u , v ), y(u , v ), z(u , v ) 

 u 
  u , v , sin ( v ) co s   
 2 

S o w e are h av e a se co n d ch an g e o f
v ariab le s w ith u( s , t )  2s co s( t ) an d
v ( s , t )  3s sin ( t ) !
2 1

ellip se
SA xyz (u , v ) d u d v 

0 0
SA xyz (u( s , t ), v( s , t )) A uv ( s , t ) d s d t
Example 3: Surface Area Enclosed in a
Curve Plotted on a Surface (Bonus Material!)
Fin d th e are a o f th e e llip se (2 co s(t),3 s in (t)) p lo tte d o n th is sam e su rface :
T (u , v )   x(u , v ), y(u , v ), z(u , v ) 

 u 
  u , v , sin ( v ) co s   
 2 

U sin g u( s , t )  2s co s( t ) an d v ( s , t )  3s sin ( t ),
A uv ( s , t ) 

u
v
s
u
s
v
t
t
2 co s( t )
3 sin ( t )
 2s sin ( t )
3s co s( t )
 6s
Example 3: Surface Area Enclosed in a
Curve Plotted on a Surface (Bonus Material!)
Fin d th e are a o f th e e llip se (2 co s(t),3 s in (t)) p lo tte d o n th is sam e su rface :
T (u , v )   x(u , v ), y(u , v ), z(u , v ) 

 u 
  u , v , sin ( v ) co s   
 2 

Le ttin g M ath e m atica p o lish th is o n e o ff, w e g e t:
2 1

0 0
2 1
S A xyz (u( s , t ), v ( s , t )) A u v ( s , t ) d s d t 

6 s S A xyz (u( s , t ), v ( s , t )) d s d t
0 0
 2 2 .0 6 6 7
Today’s Material Did Not Reference Vector
Fields Acting on a Surface!
Let R be a solid in three dimensions with boundary surface (skin)
C with no singularities on the interior region R of C. Then the
net flow of the vector field Field(x,y,z) ACROSS the closed
surface is measured by:
Ò
 Fie ld( x , y , z )  o u te rn o rm al d A
C

 d iv Fie ld d x
dy dz
R
Le t Fie ld( x , y , z )  m ( x , y , z )  n( x , y , z )  p( x , y , z ).
W e d e fin e th e d iv e rg e n ce o f th e v e cto r fi e ld as:
d iv Fie ld( x , y , z ) 
m
x

n
y

p
z
 D[m[ x , y , z ], x ]  D[n[ x , y , z ], y ]  D[p[ x , y , z ], z ]
Today’s Material Did Not Reference Vector
Fields Acting on a Surface!
The Divergence Theorem is great for a closed surface, but it is not useful at all
when your surface does not fully enclose a solid region. In this situation, we
will need to compute a surface integral. For a parameterized surface, this is
pretty straightforward:
 Fi e ld( x , y , z )  o u te rn o rm al d A
C

t2
 
t1
s2
s1
Fie ld( x( s , t ), y( s , t ), z( s , t ))  n o rm al( s , t ) d s d t