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Math 32B Discussion Session
Week 7 Notes
February 21 and 23, 2017
In last week’s notes we introduced surface integrals, integrating scalar-valued functions
over parametrized surfaces. As with our previous integrals, we used a transformation
(namely, the parametrization) to rewrite our integral over a more familiar domain, and
picked up a fudge factor along the way. This week we want to integrate vector fields over
surfaces.
Surface Integrals of Vector Fields
3 3
Suppose we have a surface S ⊂ R and a vector field F defined on R , such as those seen in
the following figure:
Wewanttomakesense of what it means to integrate the vector field over the surface. That
is, we want to define the symbol Z
S F·dS.
When defining integration of vector fields over curves we set things up so that our integral
would measure the work done by the vector field along the curve. This doesn’t lift very
nicely to surfaces — which direction in the surface should be considered positive? Instead
of measuring the extent to which F is pushing along the surface, we’ll measure the extent to
which F is pushing through S. So instead of work, we want the surface integral of a vector
field to measure the flux of the vector field through the surface.
We can measure the pointwise contribution to flux in much the same way we measured
the pointwise contribution to work for line integrals. We focus on a fixed point p in our
surface. At this point, the flux through S is the same as the flux through the tangent plane
to S at p, as indicated here:
1
The blue vector in this figure is the vector associated to p by the vector field F. Only that
part of this vector which is perpendicular to the tangent plane is making any contribution
to the flux. That is, the blue vector can be decomposed as a sum of two vectors, one of
which is perpendicular to the tangent plane, and one of which is in the tangent plane. The
vector that is in the tangent plane is not pushing through the surface, and thus makes no
contribution to the flux. All of the flux data of our blue vector is encoded in its projection
onto the line normal to the tangent plane. The pointwise contribution to flux is then the
magnitude of this vector, so we see that
pointwise contribution to flux = kF·nk,
where n is the unit normal vector to S at p pointing in the positive direction. For this reason
we make the following definition.
Definition. Let S ⊂ R3 be a surface and suppose F is a vector field whose domain contains
S. We define the vector surface integral of F along S to be
ZZSF·dS:=ZZS(F·n)dS,
where n(P) is the unit normal vector to the tangent plane of S at P, for each point P in S.
The situation so far is very similar to that of line integrals. When integrating scalar
valued functions we pick up a strange fudge factor, and when integrating vector fields we
compute the dot product of our vector field with some distinguished unit vector field. Just
as in the line integral case, the fudge factor and the distinguished vector field are related in
way that greatly simplifies the computational difficulty of integrating vector fields.
Theorem 1. Let G(u,v) be an oriented parametrization of an oriented surface S with
parameter domain D. Assume that G is one-to-one and regular, except possibly at the
boundary of D. ThenZZ ZZ
S F·dS = DF(G(u,v))·N(u,v)dudv.
Our lone application of this theorem will be a fluid flow example. If a fluid is flowing
with velocity vector field v, its flow rate across S, measured in units of volume per unit
time, is given by ZZ
flow rate across S = S v · dS.
Example. (Section 17.5, Exercise 26) Supose we have a net given by y = 1 − x2 − z2,
where y ≥ 0, and this net is dipped in a river. If the river has velocity vector field given by
v = hx−y,z +y+4,z2i, determine the flow rate of water through the net in the positive
y-direction.
2
(Solution) Here’s a plot of the net in question, along with several vectors depicting the flow
of the river. Note that these velocity vectors have been given unit length for purposes of the
figure, but in fact have varying magnitudes.
Now we need a parametrization of this surface. Because the projection of this surface onto
the xz-plane is a unit circle, it’s very natural to define G(r,θ) = (x(r,θ),y(r,θ),z(r,θ)),
where
x(r,θ) = rcosθ, z(r,θ) = rsinθ, and then y(r,θ) = 1 −r2,
for 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. We can now compute our tangent vectors at each point (r,θ):
T(r,θ) = hcosθ,−2r,sinθi and T (r,θ) = h−rsinθ,0,rcosθi.
r θ
So we have a normal vector is given by
i j k
2 2
T(r,θ)×T (r,θ) = cosθ −2r sinθ = h−2r cosθ,−r,−2r sinθi.
r θ
−rsinθ 0 rcosθ
Notice that since −r < 0, this vector points in the negative y-direction, opposite our orien-
tation. So we choose the normal vector pointing in the opposite direction and have
N(r,θ) = h2r2cosθ,r,2r2sinθi.
Next, notice that
2 2 2 2
v(G(r,θ)) = hrcosθ −(1−r ),rsinθ+(1−r )+4,r sin θi.
3
So we have
v(G(r,θ))·N(r,θ) = 2r3cos2θ −2r2cosθ(1−r2)+r2sinθ+5r−r3+2r4sin3θ
2 2 4 3 2
=2r cosθ(r −1+rcosθ)+2r sin θ+r(5−r +rsinθ).
Finally, our flow rate across S is given by
ZZ v·dS=Z 2πZ 1v(G(r,θ))·N(r,θ)drdθ
S 0 0
Z 2π Z 1 2 2 4 3 2
= 0 0 2r cosθ(r −1+rcosθ)+2r sin θ+r(5−r +rsinθ)drdθ =5π.
This last integral isn’t hard, it’s just really ugly. Since this integral is not very nice, one
might expect a spherical parametrization such as G(θ,φ) = (cosθsinφ,cos2φ,sinθsinφ)
would work out better. I didn’t find the resulting integral to be any nicer. ♦
Line Integrals and Surface Integrals
We’ll finish by summarizing the various integrals we’ve considered in the last few sections.
Throughout, C is a curve parametrized by r(t) with t ∈ [a,b] and S is a surface parametrized
by G(u,v) with (u,v) ∈ D.
We write T for a unit tangent vector to a curve C, and we write n for a unit normal
vector to a curve C or a surface S. In both cases we assume that the vectors are positively
oriented. Given a parametrization G of S, we also write
N(u,v) = ±G ×G =±∂G×∂G.
u v ∂u ∂v
The ± is needed here because we want N(u,v) to be outward-pointing, and this may require
negating G ×G .
u v
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