Calculus 2nd Edition Rogawski

SECTION 17.5 Surface Integrals of Vector Fields 1001
EXAMPLE 4 Let v = 〈 x 2 + y 2 , 0,z 2〉 be the velocity field (in centimeters per second)
of a fluid in R 3 . Compute the flow rate through the upper hemisphere S of the unit sphere
centered at the origin.
Solution We use spherical coordinates:
x = cos θ sin φ, y = sin θ sin φ, z = cos φ
The upper hemisphere corresponds to the ranges 0 ≤ φ ≤ π 2
and 0 ≤ θ ≤ 2π. By Eq. (2)
in Section 17.4, the upward-pointing normal is
n = T φ × T θ = sin φ 〈 cos θ sin φ, sin θ sin φ, cos φ 〉
We have x 2 + y 2 = sin 2 φ, so
v = 〈 x 2 + y 2 , 0,z 2〉 = 〈 sin 2 φ, 0, cos 2 φ 〉
v · n = sin φ 〈 sin 2 φ, 0, cos 2 φ 〉 · 〈cos
θ sin φ, sin θ sin φ, cos φ 〉
∫∫
S
v · dS =
= sin 4 φ cos θ + sin φ cos 3 φ
∫ π/2 ∫ 2π
φ=0
θ=0
(sin 4 φ cos θ + sin φ cos 3 φ)dθ dφ
The integral of sin 4 φ cos θ with respect to θ is zero, so we are left with
∫ π/2 ∫ 2π
φ=0
θ=0
sin φ cos 3 φ dθ dφ = 2π
= 2π
∫ π/2
φ=0
cos 3 φ sin φ dφ
( ) ∣
− cos4 φ ∣∣∣
π/2
4
φ=0
= π 2 cm3 /s
Since n is an upward-pointing normal, this is the rate at which fluid flows across the
hemisphere from below to above.
Electric and Magnetic Fields
The laws of electricity and magnetism are expressed in terms of two vector fields, the
electric field E and the magnetic field B, whose properties are summarized in Maxwell’s
four equations. One of these equations is Faraday’s Law of Induction, which can be
formulated either as a partial differential equation or in the following “integral form”:
∫
E · ds = − d ∫∫
B · dS 5
C dt S
FIGURE 12 The positive direction along the
boundary curve C is defined so that if a
pedestrian walks in the positive direction
with the surface to her left, then her head
points in the outward (normal) direction.
The tesla (T) is the SI unit of magnetic field
strength. A one-coulomb point charge
passing through a magnetic field of 1 T at
1 m/s experiences a force of 1 newton.
In this equation, S is an oriented surface with boundary curve C, oriented as indicated in
Figure 12. The line integral of E is equal to the voltage drop around the boundary curve
(the work performed by E moving a positive unit charge around C).
To illustrate Faraday’s Law, consider an electric current of i amperes flowing through
a straight wire. According to the Biot-Savart Law, this current produces a magnetic field
B of magnitude B(r) = µ 0|i|
T, where r is the distance (in meters) from the wire and
2πr
µ 0 = 4π · 10 −7 T-m/A.At each point P , B is tangent to the circle through P perpendicular
to the wire as in Figure 13(A), with direction determined by the right-hand rule: If the
thumb of your right hand points in the direction of the current, then your fingers curl in
the direction of B.