MATH 550, PROBLEM 18 FROM PAGE 429
Problem 18. Find the surface area of the helicoid parameterized by X:D→
R3,D=(0,1) ×(0,2πn), na positive integer, where X(r, θ)=hrcos θ, r sin θ, θi,
(r, θ)∈D.
Solution:
DX(r, θ)=
cos θ−rsin θ
sin θrcos θ
01
.
According to the formula (6) on page 425 the surface area should be computed by
the rule
Z2πn
0Z1
0kDX(r, θ)ˆ
e1×DX(r, θ )ˆ
e2kdr dθ.
Computing the cross product we get
DX(r, θ)ˆ
e1×DX(r, θ )ˆ
e2=
cos θ
sin θ
0
×
−rsin θ
rcos θ
1
=
sin θ
−cos θ
r
.
Hence kDX(r, θ)ˆ
e1×DX(r, θ )ˆ
e2k=√1+r
2
. Thus the surface area of the helicoid
is
Z2πn
0Z1
0p1+r
2dr dθ =2πn Z1
0p1+r
2dr.
This last integral is done by a trigonometric substitution followed by integration
by parts, or by look up in the table of integrals. Thus
Zp1+r
2dr =r
2p1+r
2+1
2ln |r+p1+r
2
|+C.
Therefore
Z1
0p1+r
2dr =√2
2+ln(1 + √2)
2,
and hence the area is πn[√2 + ln(1 + √2)].
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