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Unformatted text preview: 205 2.5. INTEGRATION ALONG CURVES
The arclength integral
2π s (C ) = 0 √
2 − 2 cos θ dθ can be rewritten, multiplying and dividing the integrand by
√
2π
√
1 − cos2 θ
√
=2
dθ
1 + cos θ
0 √ 1 + cos θ , as which suggests the substitution
u = 1 + cos θ
du = − sin θ
since the numerator of the integrand looks like sin θ . However, there is a
√
pitfall here: the numerator does equal sin2 θ , but this equals sin θ only
when sin θ ≥ 0, which is to say over the ﬁrst half of the curve, 0 ≤ θ ≤ π ;
for the second half, it equals − sin θ . Therefore, we break the integral into
two
√ 2π 2
0 √ √ √
sin θ dθ
√
−2
1 + cos θ
0
0
√
√
=2
−u−1/2 du − 2 1 − cos θ dθ = π 2π 2 2 √
=2 2 2 π
2 sin θ dθ
√
1 + cos θ u−1/2 du 0 u−1/2 du 0 √
= 4 2u1/2 2
0 = 8.
We note one technical point here: strictly speaking, the parametrization of
the cycloid is not regular: while it is continuously diﬀerentiable, the
velocity vector is zero at the ends of the arch. To get around this problem,
we can think of this as an improper integral, taking the limit of the
→
arclength of the curve − (θ ), ε ≤ θ ≤ 2π − ε as ε → 0. The principle here
p
(similar, for example, to the hypotheses of the Mean Value Theorem) is
that the velocity can vanish at an endpoint of an arc in Theorem 2.5.1, or
more generally that it can vanish at a set of isolated points of the curve20
and the integral formula still holds, provided we don’t “back up” after that.
20 With a little thought, we see that it can even vanish on a nontrivial closed interval. ...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
 Fall '08
 ALL
 Calculus

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