Surface Integrals of Vector Fields: Definition
We learned how to find surface integrals of scalar fields. Let’s now learn
how to find surface integrals of vector fields.
Setup:
• Let F(x, y, z) be a vector field on R3 .
• Let S ⊂ R3 be an orientable surface with orientation n.
Def: The surface integral (or flux) of F(x, y, z) on an orientable surface
S with orientation n(x, y, z) is:
¨
¨
Flux =
F · dS =
F · n dS.
S
S
So: The surface integral of the vector field F is defined to be the surface
integral of the scalar field F · n.
Warning: The orientation of the surface S matters. Using the wrong orientation will often result in the wrong answer.
Surface Integrals of Vector Fields: Parametric Surfaces
Setup:
• Let F(x, y, z) be a vector field on R3 .
• Let S ⊂ R3 be an orientable surface with orientation n.
Suppose that S is a parametric surface r(u, v) = hx(u, v), y(u, v), z(u, v)i
with parameter domain D.
The orientation n is a normal vector field to S of unit length, so it must
be:
ru × rv
n=±
(1)
kru × rv k
The ± accounts for the fact that there are two possible orientations for S. In
practice, one has to figure out which sign (+ or −) is appropriate.
Recall also the formula
dS = kru × rv k du dv.
(2)
Putting the formulas (1) and (2) together, we observe that:
¨
¨
¨
ru × rv
(kru × rv k du dv)
F · dS =
F · n dS =
F· ±
kru × rv k
S
S
D
¨
=±
F · (ru × rv ) du dv.
D
Conclusion: If an orientable surface S is given as a parametric surface
r(u, v) with parameter domain D, then the surface integral of F across S is:
¨
¨
F · dS = ±
F · (ru × rv ) du dv.
(?)
S
D
¨
F · dS on parametric surfaces S, one typi-
Evaluation: To find the flux
S
cally needs to:
(1) Find a parametrization r of S.
(2) Find the parameter domain D.
(3) Compute ru × rv and determine + or − in Formula (?).
(4) Compute F · (ru × rv ) and apply Formula (?).
Surface Integrals of Vector Fields: Graphical Surfaces
The formula
¨
¨
F · dS = ±
S
F · (ru × rv ) du dv.
(?)
D
works for parametric surfaces r : D → R3 . In the special case where S is a
graphical surface z = g(x, y), the formula (?) simplifies. Let’s see how.
Setup:
• Let F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i be a vector field on R3 .
• Let S ⊂ R3 be a graphical surface z = g(x, y) over a region D in the
xy-plane, with a choice of orientation (either upwards or downwards).
In this case, we can choose our parametrization to be
r(x, y) = hx, y, g(x, y)i
so that
rx × ry = h−gx , −gy , 1i.
Therefore:
F · (rx × ry ) = hP, Q, Ri · h−gx , −gy , 1i
= −P gx − Qgy + R.
Finally, notice that rx × ry is always an upward normal vector field to S.
Plugging into (?), we have proved:
Conclusion: If S is given as a graphical surface z = g(x, y) over the region
D ⊂ R2 , then the surface integral of F = hP, Q, Ri across S is:
¨
¨
(−P gx − Qgy + R) dx dy.
(∗)
F · dS = ±
S
D
Here, we take the + sign if S is oriented upwards, and take the − sign if S
is oriented downwards.
¨
Evaluation: To find the flux
F · dS on graphical surfaces S:
S
(1) Choose + or − depending on orientation of S (upwards or downwards).
(2) Determine bounds for the region D in the xy-plane.
(3) Use the formula (∗).
Exercise: Surface Integrals of Vector Fields
1. Let S be the part of the plane 3x = z − 2 lying within the cylinder
x2 + y 2 = 4, oriented upward.
Find the flux of F(x, y, z) = h0, y, zi through S.
Solution: Our surface S = {3x = z − 2} can be expressed as a graphical
surface {z = 3x + 2}. Therefore, we can use formula (∗), meaning that the
flux of F across S is:
¨
¨
Flux =
F · dS = ±
(−P gx − Qgy + R) dx dy
S
¨ D
=
(−P gx − Qgy + R) dx dy
D
We chose the + sign because S is oriented upwards.
Our vector field is
F = hP, Q, Ri = h0, y, zi
Our surface is S = {z = 3x + 2}, so we have g(x, y) = 3x + 2, so that
gx = 3
gy = 0.
Therefore,
−P gx − Qgy + R = −0 · 3 − y · 0 + z = z
and hence:
¨
¨
F · dS =
Flux =
¨D
S
=
(−P gx − Qgy + R) dx dy
z dx dy
¨D
(3x + 2) dx dy
=
D
using the fact that z = 3x + 2 on S.
We need bounds for the region D, which is the projection of S into the
xy-plane. To see the situation clearly, draw a picture. The region D is the
disk x2 + y 2 ≤ 4, so we can express D in polar coordinates via
0≤r≤2
0 ≤ θ ≤ 2π.
Therefore:
¨
¨
F · dS =
Flux =
S
(3x + 2) dx dy
ˆ ˆ 2
=
(3r cos θ + 2) r dr dθ
ˆ0 2π ˆ0 2
=
3r2 cos θ + 2r dr dθ
ˆ0 2π 0
2
=
r3 cos θ + r2 0 dθ
ˆ0 2π
(8 cos θ + 4) dθ
=
D
2π
0
= 0 + 4(2π)
= 8π
♦
Exercise: Surface Integrals of Vector Fields
2. Find the flux of the vector field F(x, y, z) = hx, y, 3i out of the closed
surface bounded by z = x2 + y 2 and z = 4.
Solution: Let S be the closed surface in question. It has two pieces: the surface S1 consisting
of the plane z = 4, and the surface S2 consisting of the paraboloid z = x2 + y 2 . So, our flux
is:
‹
¨
¨
Flux =
F · dS =
F · dS +
F · dS.
S
S1
S2
We need to evaluate both surface integrals.
(1) The surface S1 is the graphical surface S1 = {z = 4}. Therefore,
¨
¨
F · dS = +
(−P gx − Qgy + R) dx dy,
S1
D1
where D1 is the projection of S1 into the xy-plane. We have chosen the + sign because S1
is oriented upwards.
Our vector field is
F = hP, Q, Ri = hx, y, 3i
Our surface S1 = {z = 4}, so g(x, y) = 4, so
gx = 0,
gy = 0.
Therefore,
−P gx − Qgy + R = −0 − 0 + 3 = 3,
so that
¨
¨
F · dS =
(−P gx − Qgy + R) dx dy
¨
S1
D1
=
3 dx dy
D1
= 3 Area(D1 )
Finally, note that D1 is the disk x2 + y 2 ≤ 4 (draw a picture to see this), so that:
¨
F · dS = 3 Area(D1 ) = 3(πr2 ) = 3(4π) = 12π.
S1
(2) The surface S2 is the graphical surface S2 = {z = x2 + y 2 }. Therefore,
¨
¨
F · dS = −
(−P gx − Qgy + R) dx dy
S2
D2
¨
=
(P gx + Qgy − R) dx dy
D2
where D2 is the projection of S2 into the xy-plane. We have chosen the − sign because S2
is oriented downwards.
Our vector field is
F = hP, Q, Ri = hx, y, 3i
Our surface S1 = {z = x2 + y 2 }, so g(x, y) = x2 + y 2 , so
gx = 2x,
gy = 2x.
Therefore,
P gx + Qgy − R = x(2x) + y(2y) − 3 = 2x2 + 2y 2 − 3,
so that
¨
¨
F · dS =
(P gx + Qgy − R) dx dy
¨
S2
D2
2x2 + 2y 2 − 3 dx dy.
=
D2
Finally, note that D2 is the disk x2 + y 2 ≤ 4 (draw a picture to see this). (Yes, in this
problem, D2 is the same as D1 . This does not always happen.) In polar coordinates, D2 has
bounds
0≤r≤2
Therefore:
0 ≤ θ ≤ 2π.
¨
ˆ
2π
ˆ
2
2r2 − 3 r dr dθ
F · dS =
S2
0
ˆ
0
2π
=
0
ˆ
2π
=
0
ˆ
ˆ
2
2r3 − 3r dr dθ
0
2
4
r2
r
−3
dθ
2
2 0
2π
(8 − 6) dθ
=
0
= 4π.
Conclusion: The flux is:
‹
¨
F · dS =
Flux =
S
¨
F · dS +
S1
F · dS
S2
= 12π + 4π
= 16π
♦
Outlook: This example was quite tedious. Later in the course, we’ll redo this exact problem
with the aid of a powerful theorem: the Divergence Theorem (in R3 ). We will see that the
Divergence Theorem will make the solution significantly easier.