Graphics Reference
In-Depth Information
9.9.9. Theorem.
(1) K(
p
) = k
1
k
2
.
(2) H(
p
) = (k
1
+ k
2
)/2.
Proof.
Part (1) follows from equations (9.45) and the fact that the determinant of
a linear transformation is independent of a basis. Part (2) follows from a similar prop-
erty for the trace function.
To get a little feel for this new definition of Gauss curvature, we rework Examples
9.9.1-9.9.3.
9.9.10. Example.
Let
S
be a sphere of radius r about a point
c
. Then the Gauss map
is defined by
pc
pc
-
-
1
r
()
=
(
)
np
=
pc
-
and
1
)
¢
()
=¢
()
()
¢
()
(
)
=
(
o
D
np
g
0
n
g
0
g
0
.
r
Therefore, a matrix for D
n
(
p
) is
1
Ê
ˆ
0
Á
Á
˜
˜
r
,
1
0
Ë
¯
r
so that K = 1/r
2
and H =-1/r. We can also see that k
1
= k
2
=-1/r.
9.9.11. Example.
If
S
is a plane, then the Gauss map is constant, so that D
n
(
p
) =
0 and the matrix for D
n
(
p
) is zero. It follows that K = H = 0.
9.9.12. Example.
Let S be the cylinder of radius r about the z-axis defined by
2
2 2
+=.
xy r
The Gauss map is defined by
1
()
=
(
)
np
p
-
p e
∑
3
r
and
1
)
¢
()
=¢
()
-∑¢
()
D
np
()
¢
()
(
g
0
)
=
(
n
o
g
0
(
g
0
e
g
0
)
.
3
r
