Summary 1: Surface Area
Surface area in rectangular coordinates:
The surface area S determined by a region R in the xy-plane is given by
S   1  f x2  f y2 dxdy or S   1  f x2  f y2 dydx
R
R
Let z=f(x, y).
 
 
f 
f 
 are tangent to
1) At a point P of the surface z=f(x, y), u  1, 0,
, and v   0, 1,
x 
y 


the surface.
2)
i
j
 
u v  1 0
0 1
k
2
2
 f 
f
f
f
 f 
  i
j  1k  1        1  f x2  f y2 ThereT
x
x
y
 x 
 y 
f
y
Therefore 1  f x2  f y2 represents the area of a parallelogram tangent to the surface at
point P.
3)
The area A  xy in the xy  plane correspond s to S  1  f x2  f y2 xy in the
surface z  f ( x, y)
4) S   1  f x2  f y2 dxdy
R
Example. Find the area of the portion of the plane z=x+3y that lies inside the elliptical
x2 y2

1
4
9
Solution. Note: the area of an ellipse is ab
z
z
z  x  3y  f x 
 1 and f y 
 3  1  f x2  f y2  1  1  9  11
x
y
cylinder with equation
 S   1  f x2  f y2 dydx 
R
36  9 x 2
4
2

2


2
11 dydx  2 11 
36  9 x 2
4
2
36  9 x 2
dx  6 11
4
Example. Find the area of the part of the paraboloid z  9  x 2  y 2 that lies above the
plane z=5.
z  9  x 2  y 2 and z  5  5  9  x 2  y 2  x 2  y 2  4
z
z
z  9  x2  y2  f x 
 2 x and f y 
 2 y  1  f x2  f y2  1  4 x 2  4 y 2
x
y
 S  
1  f x2  f y2 dydx  4
R

 /22
2 4 x 2
 
0
1 2

2 3
 /2

1  4r 
2
2 3/ 2
d 
0
0
2
2
1  4 x  4 y dydx  4
0

6
 
1  4r 2 rdrd
0 0
17

17  1
Surface area in cylindrical coordinates:
2
2
 z 
 z 
S   r 2  r 2      drd
 r 
  
R
Consider the surface z=f(x, y) parametrized by the vector space equation
 x  r cos 


 y  r sin   r  x, y. z  r cos  , r sin  , z
 z  g (r , )



r
z
r
z
 cos  , sin  ,
and
  r sin  , r cos  ,
r
r


i
 r r
N

 cos 
r 

  sin 


  sin 


 N 
z

z

j
k
z
sin 

r
z
 r sin  r cos 

z  
z
z 
 r cos  i   cos 
 r sin   j  r cos 2   r sin 2  k
r  

r 
z  
z
z 
 r cos  i   cos 
 r sin   j  rk
r  

r 

r 
2
2

2
2
z
z 
z
z 


 z 
 z 
  sin 
 r cos     cos 
 r sin    r 2  r 2     

r 

r 


 r 
  
2
Problem:
Find the area of the surface given by the part of the paraboloid z  x 2  y 2 in the first
octant bounded by x=0, y=0, and x 2  y 2  4
z
z
z  x2  y2  f x 
 2 x and f y 
 2 y  1  f x2  f y2  1  4 x 2  4 y 2
x
y
 S   1 

f y2 dydx


0
0
R
1

12
 /2

 /22
2 4 x 2
f x2
1  4r 2 3 / 2
2
d 
0
0
1  4 x 2  4 y 2 dydx 
 
1  4r 2 rdrd
0 0

24
17

17  1
Second method: using cylindrical coordinates
2
2
 z 
 z 
S   r 2  r 2      drd where z  r 2 ,r=2
 r 
  
R
 /22
S
 
r  r (2r )  0drd 
2
2
2
0 0

16

2
r 1  4r 2 dr
0
1  4r 
3
2
2 3/ 2
 2
2

0


24
17

17  1
Problem: find the area of the region in the plane z=f(x,y) = 1+2x+2y that lies directly
2
2
above the region of the xy-plane bounded by the parabolas y  x
and x  y .
z
z
z  1  2x  2 y  f x 
 2 and f y 
 2  1  f x2  f y2  1  4  4  3
x
y
 S   1 
 3 dydx  3  dydx  3 
2 4 x 2
f x2

f y2 dydx
R

0
0
1 x
0x
2
1
x  x2

0
1
2
x3 
2 1
 3 x 3 / 2    3    1
3 
 3 3
 3
0
Answer: _____________
Theorem- The area of the surface of z = f(x,y) determined by a region R on the xy plane
 f   f 
S         1 dA
 x   y 
2
2
is given by
R
10. Find the surface of
y=1
z 1 y
2
that is bounded by the planes z=0, x=0, y=x and
answer: ________________