Recall : The area of the parallelogram determined by two 3-vectors
is the length of cross product of the two vectors.
A parametrized surface defined by
(u,v) = (x(u,v), y(u,v), z(u,v)) for
(u,v) in D can be partitioned into
“approximate parallelograms”
each of which can be determined
by u(u,v) and v(u,v) at a point
on the surface. The area of the
whole surface is approximately
equal to the sum of the areas of
the “approximate parallelograms”
each equal to ||u(u,v)  v(u,v)||
The exact surface area is the limit
at a point on the surface.
of the approximate surface area as
the number of approximate
parallelograms in the partition goes
||u(u,v)  v(u,v)|| du dv
to infinity, and this limit is
D
Recall (from Section 4.2):
b
||c  (t)|| dt .
The arc length of the path c(t) from c(a) to c(b) is
a
The definition of surface area is a generalization of arc length:
The area of the parametrized surface defined by
(u,v) = (x(u,v), y(u,v), z(u,v)) for (u,v) in D is
||u(u,v)  v(u,v)|| du dv
D
Example Consider the part of the sphere x2 + y2 + z2 = 1 which
lies above the xy plane and is bounded by the elliptical cylinder
3x2 + 4y2 = 3 .
(a) Write the part of the sphere above the xy plane as a
function z = f(x,y).
z = f(x,y) = (1 – x2 – y2)1/2
(b) Let D be the region inside the ellipse 3x2 + 4y2 = 3 on the xy plane.
Describe D as a y-simple region.
– 1  x  1 , – (3 – 3x2)1/2/2  y  (3 – 3x2)1/2/2
(c) Parametrize the part of the sphere which lies above the xy plane and
is bounded by the elliptical cylinder.
(u,v) = ( u , v , (1 – u2 – v2)1/2 )
for
– 1  u  1 , – (3 – 3u2)1/2/2  v  (3 – 3u2)1/2/2
(d) Find the surface area of the part of the sphere which lies above the
xy plane and is bounded by the elliptical cylinder.
u(u,v) = ( 1 , 0 , – u(1 – u2 – v2)–1/2 )
v(u,v) = ( 0 , 1 , – v(1 – u2 – v2)–1/2 )
u(u,v)  v(u,v) = ( u(1 – u2 – v2)–1/2 , v(1 – u2 – v2)–1/2 , 1 )
1
||u(u,v)  v(u,v)|| = —————
(1 – u2 – v2)1/2
1
||u(u,v)  v(u,v)|| du dv =
D
–1
(3 – 3u2)1/2/2
1
————— dv du =
(1 – u2 – v2)1/2
– (3 – 3u2)1/2/2
(3 – 3u2)1/2/2
1
————— dv du =
(1 – u2 – v2)1/2
1
–1
– (3 – 3u2)1/2/2
(3 – 3u2)1/2/2
1
dv
———————— ——— du =
(1 – (v/1 – u2)2)1/2 1 – u2
1
–1
– (3 – 3u2)1/2/2
1
v
arcsin –––––– du =
(1 – u2)1/2
Recall:
1
d
f /(x)
— arcsin[f(x)] = —————
1 – [f(x)]2
dx
(3 – 3u2)1/2/2
–1
v = – (3 – 3u2)1/2/2
(3 – 3u2)1/2/2
1
v
arcsin ––––––
du =
2
1/2
(1 – u )
–1
v = – (3 – 3u2)1/2/2
1
3
arcsin ––
2

 3
arcsin ––––
2
–1
1


3
–1

 
3
4
du = 
3
du =
Recall that the line integral of a vector field F over a path c is defined
to be b
F(c(t)) • c  (t) dt
F • ds .
and is often denoted by
c
a
In an analogous manner, the surface integral of a vector field F with
parametrization (u,v) of the surface is defined to be
F((u,v)) • [u(u,v)  v(u,v)] du dv
D
F • dS .
and is often denoted by

Observe that
c  (t)
in the definition of the line integral, then the line
(1) If F = ———
|| c  (t) ||
integral is the length of the path.
Observe that
c  (t)
in the definition of the line integral, then the line
(1) If F = ———

|| c (t) ||
integral is the length of the path.
u(u,v)  v(u,v)
(2) If F = ————————– in the definition of the surface
|| u(u,v)  v(u,v) ||
integral, then the surface integral is the area of the surface.
Recall Green’s Theorem (Theorem 1 on page 522) which says that if
F = P(x,y)i + Q(x,y)j is a vector field defined on D, then
Q
P
— – —
x
y
D
curl(F) • k
dA =
F • ds
D
Recall Green’s Theorem (Theorem 1 on page 522) which says that if
F = P(x,y)i + Q(x,y)j is a vector field defined on D, then
Q
P
— – —
x
y
D
curl(F) • k
dA =
F • ds
D
Suppose a surface S can be either defined as a function or parametrized
over a region D satisfying the conditions of Green’s Theorem.
Stokes’ Theorem (Theorem 5 on page 533 and Theorem 6 on page 538)
says that if F is a vector field in R3 defined on the surface S and the
boundary of the surface is the path S, then
curl(F) • [u(u,v)  v(u,v)] du dv =
D
F • ds
S