Michael Corral: Vector Calculus

4.5 Stokes’ Theorem 167
Example 4.12. Let f(x,y,z)=zand let C be the curve in 3 parametrized by
x=tsint, y=tcost, z=t, 0≤t≤8π.
Evaluate ∫ f(x,y,z)ds. (Note: C is called a conical helix. See Figure 4.5.1).
C
Solution: Since x ′ (t)=sint+tcost, y ′ (t)=cost−tsint, and z ′ (t)=1, we have
x ′ (t) 2 +y ′ (t) 2 +z ′ (t) 2 = (sin 2 t+2tsintcost+t 2 cos 2 t)+(cos 2 t−2tsintcost+t 2 sin 2 t)+1
= t 2 (sin 2 t+cos 2 t)+sin 2 t+cos 2 t+1
= t 2 +2,
so since f(x(t),y(t),z(t))=z(t)=talong the curve C, then
∫
C
f(x,y,z)ds=
=
∫ 8π
0
∫ 8π
0
t
f(x(t),y(t),z(t))
√
t 2 +2 dt
√
x ′ (t) 2 +y ′ (t) 2 +z ′ (t) 2 dt
( )∣
1
∣∣∣∣∣ 8π
=
3 (t2 +2) 3/2 = 1 (
(64π 2 +2) 3/2 −2 √ 2 ) .
3
0
30
25
t = 8π
20
15
z
10
5
0-25 -20 -15 -10 -5
0
y
5
t = 0
10 15 20 25 30
5
10
15
20
25
0
-25
-20
-15
-10
-5
x
Figure 4.5.1 Conical helix C
Example 4.13. Let f(x,y,z)= xi+yj+2zk be a vector field in 3 . Using the same
curve C from Example 4.12, evaluate ∫ C f·dr.
Solution: It is easy to see that F(x,y,z)= x2
2 + y2
2 + z2 is a potential for f(x,y,z) (i.e.