Page 108 - Vector Analysis

104 CHAPTER 4. Vector Calculus

thus using (4.2) we find that

żb ˝ γ)(t)}γ 1(t)}Rn dt ´ ε ă ( ˝ γ)}γ 1}Rn, P) ´ ε
(f L (f 4

a

kk

ÿ inf ÿ sup f (ξ)ℓ(γ([ti´1, ti]))
ď f (ξ)ℓ(γ([ti´1, ti])) ď
i=1 ξPγ([ti´1,ti])
i=1 ξPγ([ti´1,ti])

ď ( ˝ γ)}γ 1}Rn, P) + ε ă żb ˝ γ)(t)}γ 1(t)}Rn dt + ε.
U (f 4 (f

a

Since ε ą 0 is chosen arbitrary, we conclude (4.1). ˝

Example 4.24. Let C be the upper half part of the circle centered at the origin with radius

ż

R ą 0 in the xy-plane. Evaluate the line integral y ds.

First, we parameterize C by C

γ(t) = (R cos t, R sin t) t P [0, π] .

Then

ż żπ żπ

y ds = R sin t››(´R sin t, R cos t)››R2dt = R2 sin t dt = 2R2 .
C0 0

Example 4.25. Find the mass of a wire lying along the first octant part of the curve of
intersection of the elliptic paraboloid z = 2 ´ x2 ´ 2y2 and the parabolic cylinder z = x2
between (0, 1, 0) and (1, 0, 1) if the density of the wire at position (x, y, z) is ϱ(x, y, z) = xy.

Note that we can parameterize the curve C by

?
γ(t) = (t, 1 ´ t2, t2) t P [0, 1] .

Therefore, the mass of the curve can be computed by

ż ϱ ds = ż1 ? ´ t2 ››(1, ? ´t , 2t)››R3 dt = ż1 ? ´ t2 a1 ´ t2 + t2 + 4t2(1 ´ t2) dt
t1 1´ t1 ?
C 0 t2 0
1 ´ t2

ż 1 1 ż 1 ? π

= a ´ (1 ´ 2t2)2 dt = 2 ´ u2du = 1ż 4 2 cos2 θ dθ
t2
0 4 ´1 4´
1[ ]ˇθ= π
θ 4

= + sin(2θ ) ˇ π π1
4 = +.

4 2 ˇθ=´ π 84
4