of the cross product of the sides of a parallelogram gives the area of the parallelogram
and consequently one can express the element of surface area as
dS =|∂r
∂u
du ×∂r
∂v
dv |=|∂r
∂u
×∂r
∂v
|dudv (7 .53)
Using the vector identity
(A×B)·(C×D) = (A·C)(B·D)−(A·D)(B·C)
with A=C =
∂r
∂u and
B =D =∂r
( ∂v one finds
∂r
∂u ×
∂r
∂v
)
·
(
∂r
∂u ×
∂r
∂v
)
=
(
∂r
∂u ·
∂r
∂u
)(
∂r
∂v ·
∂r
∂v
)
−
(
∂r
∂u ·
∂r
∂v
)
·
(
∂r
∂u ·
∂r
∂v
)
Define the quantities
E=∂r
∂u
·∂r
∂u
=
(
∂x
∂u
) 2
+
(
∂y
∂u
) 2
+
(
∂z
∂u
) 2
F=
∂r
∂u ·
∂r
∂v =
∂x
∂u
∂x
∂v +
∂y
∂u
∂y
∂v +
∂z
∂u
∂z
∂v
G=
∂r
∂v ·
∂r
∂v =
(
∂x
∂v
) 2
+
(
∂y
∂v
) 2
+
(
∂z
∂v
) 2
(7 .54)
then one can write
dS =|
∂r
∂u ×
∂r
∂v |dudv =
√
EG −F^2 dudv (7 .55)
Alternatively one can write
∂r
∂u ×
∂r
∂v =
∣∣
∣∣
∣∣
ˆe 1 ˆe 2 ˆe 3
∂x
∂u
∂y
∂u
∂z
∂x ∂u
∂v
∂y
∂v
∂z
∂v
∣∣
∣∣
∣∣
=
(
∂y
∂u
∂z
∂v
−∂y
∂v
∂z
∂u
)
ˆe 1 −
(
∂x
∂u
∂z
∂v
−∂z
∂u
∂x
∂v
)
ˆe 2 +
(
∂x
∂u
∂y
∂v
−∂x
∂v
∂y
∂u
)
ˆe 3
and the magnitude of this cross product is given by
∣∣
∣∣∂r
∂u
×∂r
∂v
∣∣
∣∣=
√(
∂y
∂u
∂z
∂v
−∂y
∂v
∂z
∂u
) 2
+
(
−∂x
∂u
∂z
∂v
+∂z
∂u
∂x
∂v
) 2
+
(
∂x
∂u
∂y
∂v
−∂x
∂v
∂y
∂u
) 2
(7 .56)
Expanding the equation (7.56) one finds that the element of surface can be repre-
sented by the equation (7.55). To find the area of the surface one need only evaluate
the double integral
Surface Area =
∫β
α
∫δ
γ
√
EG −F^2 dv du (7 .57)
