Calculus III (part 4): Vector Calculus
(by Evan Dummit, 2015, v. 2.87)
Contents
4
Vector Calculus
1
4.1
Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
4.2
Surfaces and Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
4.2.1
Parametric Surfaces
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
4.2.2
Surface Integrals
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
Vector Fields, Work, Circulation, Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
4.3.1
Circulation and Work Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
4.3.2
Flux Across a Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
4.3.3
Flux Across a Surface
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
4.4
Conservative Vector Fields, Path-Independence, and Potential Functions . . . . . . . . . . . . . . . .
16
4.5
Green's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
4.6
Stokes's Theorem and Gauss's Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
4.6.1
Stokes's Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
4.6.2
Gauss's Divergence Theorem
23
4.3
4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Vector Calculus
•
Our motivating problem for multivariable integration was to generalize the idea of integration to more complicated regions in space or more succinctly, to integrate a function over a region. We might also ask whether
there is a simple way to integrate a function over an arbitrary curve in the plane or in space, and whether
there is a way to integrate a function over an arbitrary surface in space.
•
The answer (as it always has been to this point) is yes: the generalization of single-variable integration to
arbitrary curves is called a line integral, and the generalization of double integration to arbitrary surfaces is
called a surface integral.
•
We will then discuss vector elds, which are vector-valued functions in 2-space and 3-space, and which also
provide a useful model for the ow of a uid through space. The principal applications of line and surface
integrals are to the calculation of the work done by a vector eld on a particle traveling through space, the
ux of a vector eld across a curve or through a surface, and the circulation of a vector eld along a curve.
•
Finally, we discuss several generalizations of the Fundamental Theorem of Calculus: the Fundamental Theorem of Calculus for line integrals, Green's Theorem, Gauss's Divergence Theorem, and Stokes's Theorem.
Collectively, these theorems unify all of the dierent notions of integration, as they each relate the integral of
a function on a region to the integral of an antiderivative of the function on the region's boundary.
4.1
•
Line Integrals
The motivating problem for our discussion of line integrals is: given a parametric curve
and a function
f (x, y),
r(t) = hx(t), y(t)i
z = f (x, y),
if we build a surface along the curve with height given by the function
how can we calculate the area of this surface? (This is a natural generalization of our typical single-variable
integration problem, in which we build the surface inside a plane, thus making it the area under a curve.)
1
◦
Here is an example (for visualization), with
◦
Another closely related question is:
f (x, y, z),
◦
given a parametric curve
how can we calculate the average value of
f (x, y, z)
for
0≤t≤
r(t) = hx(t), y(t), z(t)i
3
:
2
and a function
on the curve?
A third question: given a thin wire shaped along some curve
δ(x, y),
•
r(t) = t2 , t cos(2πt) , f (x, y) = t2 + 1,
r(t) = hx(t), y(t)i
with variable density
what is the wire's mass, and what are its moments about the coordinate axes?
As with all other types of integrals we have examined so far, we use Riemann sums to give the formal denition
of the line integral of a function
f (x, y)
on a plane curve
C.
(Also as before, we will use the formal denition
as infrequently as possible!)
◦
The idea is to approximate the curve with straight line segments, sum (over all the segments) the function
value times the length of the segment, and then take the limit as the segment lengths approach zero.
◦
(x0 , y0 ),
list of points
The norm of the partition
number
◦
C , a partition of Cpinto n pieces is a
nth segment having length ∆si = (∆xi )2 + (∆yi )2 .
among all of the segment lengths in P .
Denition: For a curve
with the
Denition: For
of
f (x, y)
on
D
f (x, y) a continuous function and P
corresponding to
P
to be RSP (f )
a partition of the curve
=
n
X
... ,
(xn , yn ) on C ,
P is the largest
C , we dene the Riemann
f (xk , yk ) ∆sk .
k=1
◦
f (x, y),
Denition: For a function
ˆ
we dene the line integral of
L such that, for every > 0, there exists
norm(P ) < δ , we have |RSP (f ) − L| < .
to be the value of
partition
◦
P
with
◦
L
a
f
f (x, y) ds,
C
(depending on ) such that for every
on the curve
δ>0
C,
denoted
f (x, y) is continuous and the curve C
Remark: It can be proven (with signicant eort) that, if
then a value of
sum
is smooth,
satisfying the hypotheses actually does exist.
Remark: The dierential
ds
in the denition of the line integral is the dierential of arclength, which
we discussed earlier in our study of vector-valued functions.
•
In exactly the same way, we can use Riemann sums to give a formal denition of the line integral along a
curve
•
C
in 3-space. (We simply add the appropriate
z -terms
to all the denitions.)
Like with the other types of integrals, line integrals have a number of formal properties which can be deduced
from the Riemann sum denition. Specically, for
D
an arbitrary constant and
f (x, y) and g(x, y) continuous
functions, the following properties hold:
ˆ
◦
D ds = D · Arclength(C)
Integral of constant:
.
C
ˆ
◦
ˆ
D f (x, y) ds = D ·
Constant multiple of a function:
f (x, y) ds
C
ˆ
◦
ˆ
f (x, y) ds +
Addition of functions:
C
ˆ
g(x, y) ds =
C
ˆ
◦
C
[f (x, y) + g(x, y)] ds
.
C
ˆ
f (x, y) ds −
Subtraction of functions:
.
C
ˆ
[f (x, y) − g(x, y)] ds
g(x, y) ds =
C
C
2
.
ˆ
◦
Nonnegativity: if
f (x, y) ≥ 0,
f (x, y) ds ≥ 0
then
.
C
◦
Union of curves: If
C1
and
C2
are curves such that
ˆ
C2
starts where
ˆ
f (x, y) ds +
by gluing the curves end-to-end, then
ends, and
ˆ
f (x, y) ds =
C1
◦
C1
C2
C
is the curve obtained
f (x, y) ds
.
C
Remark: These same properties also all hold for line integrals of a function
f (x, y, z)
in 3-space.
•
The key observation is that we can reduce calculations of line integrals to traditional single integrals:
•
Proposition (Line Integrals in the Plane):
ˆ
a ≤ t ≤ b,
ˆ
b
f (x, y) ds =
then
C
a
If the curve
ds
f (x(t), y(t))
dt,
dt
C
x = x(t), y = y(t) for
p
ds
=
x0 (t)2 + y 0 (t)2 is the derivative of
dt
can be parametrized as
where
arclength.
•
C
Proposition (Line Integrals in 3-Space): If the curve
ˆ
for
a ≤ t ≤ b,
ˆ
f (x, y, z) ds =
then
a
C
derivative of arclength.
◦
b
ds
f (x(t), y(t), z(t))
dt,
dt
´
f (x, y) ds
C
is also a Riemann sum
n
X
f (xk , yk )
∆sk
∆tk
∆tk
Thus, to evaluate the line integral of
f
C ;
´b
a
for the line
f (x(t), y(t))
ds
dt.
dt
s-coordinates to t-coordinates,
ds
dt.
ds =
dt
on the curve
f (xk , yk ) ∆sk
for the integral
Equivalently: we have made a substitution in the integral by changing from
where the dierential changes using the rule
n
X
k=1
k=1
•
where
The proof of both of these results is simply to observe that the Riemann sum
integral
◦
x = x(t), y = y(t), z = z(t)
p
ds
= x0 (t)2 + y 0 (t)2 + z 0 (t)2 is the
dt
can be parametrized as
namely the line integral
´
C
f (x, y, z) ds,
follow these
steps:
•
◦
Step 1: Parametrize the curve
◦
Step 2: Write
◦
Step 3:
◦
Step 4:
f
C
as a function of
t i.e., in the form r(t) = hx(t), y(t), z(t)i, for a ≤ t ≤ b.
t: f (x, y, z) = f (x(t), y(t), z(t)).
p
ds
Write ds =
dt = ||v(t)|| dt = x0 (t)2 + y 0 (t)2 + z 0 (t)2 dt.
dt
p
´b
Evaluate the resulting integral
f (x(t), y(t), z(t)) x0 (t)2 + y 0 (t)2 + z 0 (t)2 dt.
a
as a function of
To nd the average value of a function on a curve, we simply integrate the function over the curve, and then
divide by the curve's arclength.
•
Example: Integrate the function
(1, 0)
•
and ending at
f (x, y) = x2 + y
x2 + y 2 = 1,
starting at
(−1, 0).
◦
The unit circle is parametrized by
◦
We have
r(t) = hcos t, sin ti: the range we want is 0 ≤ t ≤ π .
p
f (x, y) = x2 + y = cos2 t + sin t, and we also have ds = (− sin t)2 + (cos t)2 = 1.
´π 2
´ π 1 + cos 2t
π
integral is therefore
cos t + sin t dt = 0
+ sin t dt =
+2 .
0
2
2
◦
The
Example: Find the average value of the function
to
along the top half of the unit circle
f (x, y, z) = x2 +y 2 +z 2
along the line segment from
(1, −1, 0)
(2, 2, 1).
◦
segment as
◦
◦
v = h2, 2, 1i − h1, −1, 0i = h1, 3, 1i.
hx, y, zi = h1, −1, 0i + t h1, 3, 1i for 0 ≤ t ≤ 1.
The direction vector for the line is
So the line segment is parametrized explicitly by
Now we set up the integral: the function is
Thus, we can parametrize the line
x = 1 + t, y = −1 + 3t, z = t
2
11t − 4t + 2.
3
2
for
2
0 ≤ t ≤ 1.
f (x, y, z) = x + y + z = (1 + t) + (−1 + 3t)2 + (t)2 =
2
2
√
ds √ 2
= 1 + 32 + 12 = 11.
dt
√
1
√
√
´1 2
11
11
11 3
2
◦ The integral of f is therefore 0 11t − 4t + 2 11dt = 11
=
t − 2t + 2t .
3
3
t=0
√
´1
´1√
◦ To compute the average value, we divide by the arclength, which is 0 1 ds = 0 11dt = 11.
x0 (t) = 1, y 0 (t) = 3,
◦
Since
◦
So the average value is
and
11
3
z 0 (t) = 1,
we also have
.
•
We also have formulas for the mass and moments of a wire of variable density:
•
Center of Mass and Moment Formulas (Thin Wire): Given a 1-dimensional wire of variable density
along a parametric curve
The total mass
◦
The
x-moment Myz
is given by
◦
The
y -moment Mxz
is given by
◦
The
z -moment Mxy
is given by
◦
The center of mass
is given by
(x̄, ȳ, z̄)
M=
has
´
δ(x, y, z) ds.
´
Myz = C x δ(x, y, z) ds.
´
Mxz = C y δ(x, y, z) ds.
´
Mxy = C z δ(x, y, z) ds.
Myz Mxz Mxy
,
,
.
coordinates
M
M
M
C
Note: For a wire in 2-space, the formulas are essentially the same (except without the
though the
x-moment
is denoted
My
and the
y -moment
is denoted
z -coordinate),
Mx .
Example: Find the total mass, and the center of mass, of a thin wire having the shape of the unit circle and
variable density
◦
◦
◦
◦
◦
•
M
δ(x, y, z)
in 3-space:
◦
◦
•
C
δ(x, y) = 2 + x.
ds p
x = cos t, y = sin t, so
= (− sin t)2 + (cos t)2 = 1.
dt
´
´ 2π
The total mass M is M =
δ(x, y) ds = 0 (2 + cos t) dt = 2π .
C
2π
´
´ 2π
1
1
= π.
The x-moment My is My =
x
δ(x,
y)
ds
=
t
+
sin(2t)
cos
t(2
+
cos
t)
dt
=
2
sin
t
+
C
0
2
4
t=0
2π
´
´ 2π
1
= 0.
The y -moment Mx is Mx =
y
δ(x,
y)
ds
=
cos(2t)
sin
t(2
+
cos
t)
dt
=
−2
cos
t
−
C
0
4
t=0
M x My
1
Therefore, the center of mass is
,
=
,0 .
M M
2
We can parametrize the unit circle with
We will also be interested in computing
line integrals involving the dierentials
ˆ
ds:
dx, dy ,
and
dz
rather than
f dx + g dy + h dz .
namely, expressions of the form
C
•
We evaluate such line integrals by making the appropriate substitutions: if
ˆ
C
is parametrized by
x = x(t),
y = y(t), z = z(t) for a ≤ t ≤ b, then the line integral
f dx + g dy + h dz is given by the single-variable
C
ˆ b
dy
dz
dx
+g
+h
dt.
integral
f
dt
dt
dt
a
´
• Example: Find C y dx + z dy + x2 dz , where C is the curve (x, y, z) = (t, t2 , t3 ) ranging from t = 0 to t = 1.
◦
Step 2: We have
x = t, y = t2 , z = t3 ,
◦
Step 3: We have
dx = dt, dy = 2t dt, dz = 3t2 dt.
◦
Step 4: The integral is therefore
so
f = t2 , g = t3 , h = t2 .
´1
´1 2
73
t · dt + 3t2 · 2t dt + t2 · 3t2 = 0 t2 + 6t3 + 3t4 dt =
0
30
4
.
4.2
•
Surfaces and Surface Integrals
We would now like to consider the problem of computing the integral of a function on a surface in 3-dimensional
space. In a similar way to how we computed line integrals using (single) integrals, we will be able to compute
surface integrals as double integrals.
•
There are essentially two ways to describe a surface in 3-space:
f (x, y, z) = c,
◦
or as a parametric surface
has the form
z = g(x, y) is simply a special case of the general
f (x, y, z) = c with f (x, y, z) = g(x, y) − z and c = 0.
In cases where the functions
the variables
s
and
t
x, y ,
z are suciently simple or nice,
x = x(s, t), y = y(s, t), z = z(s, t),
f (x, y, z) = c.
and
from the system
surface as an implicit surface
◦
for two parameters
Note that the explicit surface
g(x, y) − z = 0
◦
either as an implicit surface of the form
r(s, t) = hx(s, t), y(s, t), z(s, t)i
s
and
t.
implicit surface, since
it can be possible to eliminate
and obtain an equation for the
However, if we can nd a parametric description of a surface, it is often easier to work with it than with
an implicit description. For example, graphing a parametric surface requires only plugging in values and
plotting points, whereas graphing an implicit surface requires nding solutions to the implicit equation.
•
We will describe how to nd parametrizations of some common surfaces, and then give the denition of surface
integral and show how to compute them on both parametric and implicit surfaces.
4.2.1
•
Parametric Surfaces
If we graph a vector-valued function of two variables
r(s, t) = hx(s, t), y(s, t), z(s, t)i
as
s
and
t
vary, we will
obtain a surface in space (barring something strange happening).
•
Example: The surface
(x0 , y0 , z0 )
r(s, t) = hx0 , y0 , z0 i + t hv1 , v2 , v3 i + s hw1 , w2 , w3 i is the plane passing through the point
v = hv1 , v2 , v3 i and w = hw1 , w2 , w3 i, provided that v and w are
which contains the two vectors
not parallel.
◦
We could also describe the plane as an implicit surface of the form
is the normal vector to the plane and
◦
There are many ways to describe this plane as a parametric surface. For example, both of the parametriza-
r(s, t) = hs, t, 1 − s − ti
x + y + z = 1.
tions
•
ax+by +cz = d, where ha, b, ci = v×w
d = ax0 + by0 + cz0 .
and
r(s, t) = h−3 + s − 2t, 2 + t + 2s, 2 + t − 3si
describe the same plane
r and R, with r < R, the surface dened parametrically by
r(s, t) = hcos(t) · [R + r sin(s)], sin(t) · [R + r cos(s)], r sin(s)i, for 0 ≤ t ≤ 2π and 0 ≤ s ≤ 2π is a donut-
Example: For two positive radius parameters
shaped surface called a torus.
◦
It is the surface obtained by taking a circle of radius
in the
◦
•
Four tori, with respective parameters
(r, R)
Example: The surface dened parametrically by
4π
and
r
and moving its center along the circle
x2 + y 2 = R 2
xy -plane.
0 ≤ s ≤ 4π
equal to
(1, 5), (2, 5), (3, 5),
and
(4, 5),
are plotted below:
r(s, t) = hcos(s) + cos(t), s + t, sin(s) + sin(t)i,
is a helical ribbon:
5
for
0≤t≤
•
In general, it can be a somewhat involved problem to convert a geometric or verbal description of a surface
into a parametrization. (It is really more of an art form than a general procedure.)
◦
To parametrize parts of cylinders, cones, and spheres, it is almost always a very good idea to consider
whether cylindrical or spherical coordinates can be of assistance.
◦
•
Using translations and rescalings, we can also parametrize surfaces like ellipsoids.
There are many dierent ways to parametrize the same surface, and which description is best will depend on
what the parametrization will be used for.
√
x = s, y = t, z = s2 + t2
parametrization x = s cos t, y = s sin t, z = s.
◦
For example,
◦
If we want to describe the points lying over a rectangular region in the
parametrizes the cone
z =
p
x2 + y 2 ,
but so does the
xy -plane, the rst parametrization
is more useful, but if we want to describe the points on the cone up to a specic height, the second
parametrization is more useful.
•
Example: Parametrize the portion of the cylinder
◦
◦
In cylindrical coordinates, we know that
x2 + y 2 = 4
lying between the planes
x = r cos θ, y = r sin θ,
and
z = −2
and
z = 2.
z = z.
r = 2 in cylindrical coordinates, we see that a parametrization of
x = 2 cos t, y = 2 sin t, z = s, where 0 ≤ t ≤ 2π but with no restrictions on s. (Here
θ and s as z .)
Since the given cylinder has equation
the full cylinder is
we think of
◦
◦
as
To obtain just the portion with
and
and then restrict the ranges for
surface where
x = 2 cos t, y = 2 sin t, z = s,
where
x2 +y 2 = 4 lying between the planes z = y −2 and z = x+4.
s
t
and
x = 2 cos t, y = 2 sin t,
appropriately. In this case, we want the portion of the
y − 2 ≤ z ≤ x + 4.
It is straightforward to check that the two planes do not intersect inside the cylinder (since
inside the cylinder, while
◦
s.
Like in the previous example, we take the parametrization of the full cylinder as
z = s,
◦
we just restrict the range for
−2 ≤ s ≤ 2.
Example: Parametrize the portion of the cylinder
◦
•
−2 ≤ z ≤ 2
Thus the parametrization of the desired portion of the cylinder is
0 ≤ t ≤ 2π
•
t
So in this case, we take
y−2 ≤ 0
x + 4 ≥ 2).
0 ≤ t ≤ 2π
Example: Parametrize the sphere
and
2 sin t ≤ s ≤ 2 cos t + 4.
x2 + y 2 + z 2 = 9.
◦
In spherical coordinates, we know that
◦
The sphere has equation ρ = 9, so we can immediately see that x = 3 cos(t) sin(s), y = 3 sin(t) sin(s),
z = 3 cos(s), with 0 ≤ t ≤ 2π and 0 ≤ s ≤ π , will parametrize the sphere. (Here, we are thinking of t as
θ and s as ϕ.)
x = ρ cos(θ) sin(ϕ), y = ρ sin(θ) sin(ϕ), z = ρ cos(ϕ).
6
•
Example: Parametrize the sphere
◦
(x − 2)2 + (y + 1)2 + (z − 6)2 = 4.
It is not so easy to describe this sphere using spherical coordinates directly.
However, if we shift the
coordinates to center the sphere at the origin, we can easily write down the parametrization.
◦
By translating back, we can see that
with
•
0 ≤ t ≤ 2π
and
0 ≤ s ≤ π,
Example: Parametrize the ellipsoid
◦
x = 2 + 2 cos(t) sin(s), y = −1 + 2 sin(t) sin(s), z = 6 + 2 cos(s),
will parametrize the sphere.
x2
y2
z2
+
+
= 1.
4
9
16
It is again not so easy to write down the parametrization using any of our coordinate systems directly.
However, if we rescale the coordinates, then we can obtain the unit sphere whose parametrization we
know.
◦
By rescaling back, we can see that
and
•
0 ≤ s ≤ π,
x = 2 cos(t) sin(s), y = 3 sin(t) sin(s), z = 4 cos(s),
Example: Parametrize the portion of the cone
◦
with
0 ≤ t ≤ 2π
will parametrize this ellipsoid.
p
z = 3 x2 + y 2
that lies below the plane
z = 1 + x + y.
z = 3r and z = 2 + r cos θ + r sin θ. They are equal when 3r =
√
2
. (Note that sin θ + cos θ ≤
2, so the denominator is never
r=
3 − cos θ − sin θ
In cylindrical, the equations are
2 + r cos θ + r sin θ,
or
zero.)
4.2.2
•
◦
The full surface is parametrized by
◦
The portion under the plane
x = s cos(t), y = s sin(t), z = 3s.
2
corresponds to 0 ≤ s ≤
,
3 − cos t − sin t
with
0 ≤ t ≤ 2π .
Surface Integrals
The motivating problem for our discussion of surface integrals is:
hx(s, t), y(s, t), z(s, t)i
and a function
f (x, y, z),
given a parametric surface
r(s, t) =
we would like to integrate the function on that surface.
Like with line integrals, we have two natural applications: computing the average value of a function on the
surface, and analzying the physical properties of a thin surface with variable density.
◦
As with all the other types of integrals, the idea is to approximate the surface with small patches, sum
(over all the patches) the function value times the area of the patch, and then take the limit as the patch
sizes approach zero.
◦
Denition: For a parametric surface
S,
where the
partition
◦
P
Denition:
k th
S,
is the largest number among
f (x, y, z)
For
Riemann sum of
S into n pieces is a list of disjoint subregions inside
sk ≤ s ≤ s0k , tk ≤ t ≤ t0k , and has area ∆σk . The norm of the
the areas of the rectangles in P .
a partition of
subregion corresponds to
f (x, y, z)
a continuous function and
on
R
corresponding to
P
P
to be RSP (f )
=
n
X
Denition: For a function
be the value of
partition
◦
P
f (x, y, z),
we dene the surface integral of
L such that, for every > 0, there exists a δ > 0
norm(P ) < δ , we have |RSP (f ) − L| < .
f
on
¨
S,
denoted
to
with
Remark: It can be proven (with signicant eort) that, if
f (x, y, z)
is continuous, then a value of
As with all of the other types of integrals, surface integrals possess some formal properties:
¨
◦
f (x, y, z) dσ ,
S
(depending on ) such that for every
satisfying the hypotheses actually does exist.
•
we dene the
f (r(sk , tk )) ∆σk .
k=1
◦
S,
a partition of the surface
C dσ = C · Area(S)
Integral of constant:
.
S
7
L
¨
◦
¨
C f (x, y) dσ = C ·
Constant multiple of a function:
S
¨
¨
f (x, y) dσ +
Addition of functions:
g(x, y) dσ =
S
[f (x, y) + g(x, y)] dσ
S
¨
f (x, y) dσ −
Subtraction of functions:
.
S
¨
◦
.
S
¨
◦
f (x, y) dσ
¨
[f (x, y) − g(x, y)] dσ
g(x, y) dσ =
S
S
.
S
¨
◦
f (x, y) ≥ 0,
Nonnegativity: if
f (x, y) dσ ≥ 0
then
.
S
¨
◦
Union: If
S1
and
S2
¨
S , then
don't overlap and have union
f (x, y) dσ +
¨
f (x, y) dσ =
S1
S2
f (x, y) dσ
.
S
•
Like with line integrals, we can reduce calculations of surface integrals to traditional double integrals:
•
Proposition (Parametric Surface Integrals): If
f (x, y, z) is continuous on the surface S which is parametrized
r(s, t) = hx(s, t), y(s, t), z(s, t)i, where S is described by a region R in st-coordinates, then the surface
integral of f on S is
¨
¨
∂r ∂r dt ds.
f (x, y, z) dσ =
f (x(s, t), y(s, t), z(s, t)) ×
∂s
∂t S
R
as
◦
The key step is to recognize the Riemann sum for the surface integral as the Riemann sum for a particular
double integral.
◦
Ultimately, the dierential of surface area
a small patch in
st-coordinates:
changes slightly, the change in
◦
•
s
arises from computing the area of
changes slightly, the change in
is given by
r
is given by
∂r
.
∂t
dσ
is roughly equal to the area of this parallelogram, which is
There is also a formula for calculating the surface integral for an implicit surface of the form
Proposition (Implicit Surface Integrals):
g(x, y, z) = c, R
of f on S is
is the projection of
S
If
¨
=
f (x, y, z)
S
◦
is the gradient of
The statement that
never vertical above
region
◦
g
and
dierential
g(x, y, z) = c:
R
k = h0, 0, 1i.
(Thus,
||∇g||
dy dx
|∇g · k|
∇g · k = ∂g/∂z .)
∂g/∂z 6= 0 on R is equivalent to
R. In particular this implies that
saying that the tangent plane to
g(x, y, z) = c
is
the surface never doubles back on itself over the
R.
Thus for example, we could not use the method directly to compute a surface integral on the entire unit
sphere, because it has a vertical tangent plane above its projection
◦
t
f (x, y, z) is continuous on the implicit surface S dened by
xy -plane, and ∂g/∂z 6= 0 on R, then the surface integral
f (x, y, z) dσ
∇g
and when
into the
¨
where
∂r
,
∂s
S , sothe
∂r ∂r ∂s × ∂t .
These two vectors form a small parallelogram that closely approximates the surface
of surface area
•
r
when
∂r ∂r dt ds
dσ = ×
∂s
∂t x2 + y 2 ≤ 1
in the
xy -plane.
This formula can be derived from the parametric surface integral formula: after some simplication, it
is what one obtains by using the parametrization
via the relation
f (s, t, z(s, t)) = c.
8
r(s, t) = hs, t, z(s, t)i,
where
z(s, t)
is dened implicitly
•
Using these two results, we can reduce calculations of surface integrals to traditional double integrals: given
a description of the surface
◦
S,
we can convert it to a double integral using one of two methods:
For a parametric surface given in the form
∗
Find the bounds on
Step 1:
s
t
and
r(s, t) = hx(s, t), y(s, t), z(s, t)i:
which parametrize the portion of the surface that's being
integrated over.
◦
∗
Step 2: Express the function
∗
Step 3:
∗
Step 4:
f (x, y, z)
(s, t).
∂r ∂r Find the dierential of surface area dσ = ∂s × ∂t ds dt.
¨
∂r ∂r ds dt
Write down the integral
f (x(s, t), y(s, t), z(s, t)) ×
∂s
∂t S
For an implicit surface given in the form
∗
to be integrated in terms of
and evaluate.
g(x, y, z) = c:
Step 1: Sketch the surface, determine the shape of its projection
R into the xy -plane,
and make sure
that the surface does not cover any part of the projection more than once.
¨
∗
f (x, y, z)
Step 2: Evaluate the integral
R
||∇g||
dy dx
|∇g · k|
, where
∇g
is the gradient of
g
and
k=
h0, 0, 1i.
∗
x and y , so if the
g(x, y, z) = c to get rid
It is very important to note that the only variables allowed in the integral are
function being integrated has any
z
terms we must use the implicit equation
of them.
◦
projection into the
◦
•
z with x or with y ,
xz -plane or the yz -plane.
Note that, by swapping
Also note that for a surface of the form
Example: Integrate the function
for
0 ≤ t ≤ 2π
and
the implicit surface procedure can also be used with a
z = f (x, y),
either method will apply.
g(x, y, z) = z over the surface with parametrization r(s, t) = hsin(t), cos(t), s + ti
0 ≤ s ≤ π.
◦
We have an explicit parametrization of the surface, so we use the parametric formula.
◦
Step 2: On the surface, we have
z =s+t
so
g(x, y, z) = z = s + t.
i
j
k ∂r
∂r
∂r
∂r
0
1 =
◦ Step 3: We have
= h0, 0, 1i and
= hcos(t), − sin(t), 1i, so
×
= 0
∂s
∂t
∂s
∂t
cos(t) − sin(t) 1 ∂r ∂r = 1.
hsin(t), cos(t), 0i. Then ×
∂s
∂t ◦ Step 4: The integral is therefore given by
ˆ 2π ˆ π
ˆ 2π 2
s
π
(s + t) ds dt =
+ st dt
2
s=0
0
0
0
ˆ 2π 2
π
=
+ πt dt
2
0
2
π
π 2 2π
=
t+ t = 3π 3 .
2
2
t=0
•
To compute surface area, we can simply integrate the function 1 on the surface, in exactly the same way that
integrating 1 on a plane region gives its area or integrating 1 on a solid region gives its volume.
•
Example: Find the area of the portion of the surface
z = 2 − x2 − y 2
◦
We can rewrite the equation of the surface implicitly as
◦
Step 1: The projection of the surface into the
is the same as
2
2
x + y ≤ 2,
xy -plane
that lies above the
x2 + y 2 + z − 2 = 0 ,
so we use Method #2.
2
2
√ R on which 2 − x − y ≥ 0, which
2 centered at the origin. Since this
is the region
and this describes the disc of radius
surface is explicit we do not need to worry about having a vertical tangent plane.
9
xy -plane.
◦
therefore
◦
p
∇g = h2x, 2y, 1i so ||∇g|| = 4x2 + 4y 2 + 1 and |∇g · k| = 1. The desired integral is
˜ p
4x2 + 4y 2 + 1 dy dx, since to calculate surface area we simply integrate the function 1.
R
Step 2: We have
To evaluate this integral, we change to polar coordinates: both the region and the function to be inte-
√
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π ,
recall the area dierential in polar is r dr dθ .
´ 2π ´ √2 √
◦ Thus we obtain the polar integral 0 0
4r2 + 1 r dr dθ.
grated become simpler: the region is
◦
To evaluate this new integral, we make (another) substitution
ˆ
2π
ˆ
0
√
2
p
ˆ
4r2
+ 1 r dr dθ
2π
=
0
0
ˆ
2π
=
0
ˆ
=
0
•
2π
ˆ
√
and the function is
u = 4r2 + 1,
with
4r2 + 1.
We also
du = 8r dr:
9
1√
u du dθ
8
1
1 2 3/2 9
u
dθ
8 3
u=1
26
13π
dθ =
.
12
3
To nd the average value of a function, we integrate the function on the surface and then divide by the area.
• Example:
p
x2 + y 2
◦
◦
◦
◦
◦
the surface
S
given by the portion of the cone
z =
x = s cos(t),
0 ≤ t ≤ 2π .
dr
dr
= hcos(t), sin(t), 1i and
= h−s sin(t), s cos(t), 0i.
We then have r(s, t) = hs cos(t), s sin(t), si, so
ds
dt
i
j
k ∂r
∂r
sin(t)
1 = h−s cos(t), s sin(t), si, so the magnitude is given by
Then
×
= cos(t)
∂s
∂t
−s sin(t) s cos(t) 0 ∂r ∂r q
√
= s2 cos2 (t) + s2 sin2 (t) + s2 = s 2.
×
∂s
∂t
√
√
˜
´ 2π ´ 2
´ 2π 8 √
16π 2
.
We also have f (x, y, z) = z = s. So
z dσ = 0 0 s · s 2 ds dt = 0
2 dt =
S
3
3
√
˜
´ 2π ´ 2 √
´ 2π √
Also, the surface area is
1 dσ = 0 0 s 2 ds dt = 0 2 2 dt = 4π 2.
S
√
1 ˜
16π 2/3
4
√
Thus, the average value is
.
z
dσ
=
=
S
Area
3
4π 2
By using cylindrical coordinates we see that we can parametrize this portion of the cone as
y = s sin(t), z = s,
◦
f (x, y, z) = z on
x2 + y 2 = 4.
Find the average value of
which lies inside the cylinder
for
0≤s≤2
and
•
Like with double, triple, and line integrals, we have mass and moment formulas for surface integrals:
•
Center of Mass and Moment Formulas (Thin Surface): Given a surface
S
of variable density
δ(x, y, z)
in 3-
space:
4.3
•
◦
The total mass
◦
The
x-moment Myz
is given by
◦
The
y -moment Mxz
is given by
◦
The
z -moment Mxy
is given by
◦
The center of mass
M
is given by
(x̄, ȳ, z̄)
M=
has
˜
δ(x, y, z) dσ .
˜
Myz = S x δ(x, y, z) dσ .
˜
Mxz = S y δ(x, y, z) dσ .
˜
Mxy = S z δ(x, y, z) dσ .
Myz Mxz Mxy
,
,
.
coordinates
M
M
M
S
Vector Fields, Work, Circulation, Flux
Denition: A vector eld in the plane is a function
F(x, y) = hP (x, y), Q(x, y)i which associates a vector to
F(x, y, z) = hP (x, y, z), Q(x, y, z), R(x, y, z)i
each point in the plane. A vector eld in 3-space is a function
which associates a vector to each point in 3-space.
10
◦
One vector eld we have already encountered is the vector eld associated to the gradient of a function
f (x, y)
•
or
f (x, y, z):
for example, if
f (x, y) = x2 + xy ,
then
∇f (x, y) = h2x + y, xi.
To represent a vector eld visually, we choose some (nice) collection of points (generally in a grid) and draw the
vectors corresponding to those points as arrows pointing in the appropriate direction and with the appropriate
length.
◦
F(x, y) = hx, yi, G(x, y) = h−y, xi,
with −2 ≤ x ≤ 2, −2 ≤ y ≤ 2:
Example: The three vector elds
are plotted below on the region
2
2
2
1
1
1
0
0
0
-1
-1
-1
-2
-2
-2
◦
-1
0
1
-2
-2
-1
0
1
2
-2
-1
0
1
2
F(x, y, z) = hx, z − y, x + yi:
We can think of a vector eld as describing the ow of an incompressible uid through space: the vector
F(x, y)
•
H(x, y) = x + y 2 , 2 − 2xy
We can also produce these plots for 3-dimensional vector elds, but the diagrams tend to be quite
cluttered; here is such a diagram for
•
2
and
at any point
(x, y)
gives the direction and velocity of the uid's ow there.
In this context, if we have a particle that travels along some given path
r(t)
through the uid, we might like
to know how much work the uid does on the particle, or (essentially equivalently) how much the uid is
pushing the particle along its path. This is the central idea behind work integrals and circulation integrals.
◦
Intuitively, we see that the more the vector eld
F
aligns with the tangent vector
T
to the particle's
path, the more work it does.
◦
In the picture, a particle moving counterclockwise around the circle will be pushed along its path by the
vector eld:
2
1
0
-1
-2
-2
•
-1
0
1
2
Alternatively, if we have a particle traveling along a path, we could also ask: how much is the uid pushing
the particle o of the path? This is the central idea behind a ux integral.
◦
Another way of thinking about this is to imagine the path as being a thin membrane, and asking how
much uid is passing across the membrane.
11
◦
Here, we see that more uid is owing across the membrane if the vector eld
vector
N
F
aligns with the normal
to the particle's path:
2
1
0
-1
-2
-2
•
-1
0
1
2
We can also formulate these ideas in 3-dimensional space: the idea of circulation and work remain the same,
but the notion of ux requires a surface for the uid to ow across.
4.3.1
•
Circulation and Work Integrals
Denition: The (counterclockwise) circulation (or ow) of the vector eld
ˆ
Circulation
be
F · T ds
=
, where
T
F
along the curve
C
is dened to
is the unit tangent to the curve.
C
◦
f (t) = F(x(t), y(t), z(t))·
C,
nd the unit tangent vector T(t) to the curve, and then integrate the dot product F(x(t), y(t), z(t)) · T(t)
What this says is: the circulation is given by integrating the dot product function
T(t)
along the curve
C.
In order to evaluate the integral as written, we would need to parametrize
along the curve.
◦
F(x, y) = hP, Qi, where P and Q
r(t) = hx(t), y(t)i from t = a to t = b.
dx dy
dx
dy
hP, Qi ·
,
P
+Q
dt dt
dt
dt .
so F · T =
=
◦ Then
||v(t)||
||v(t)||
dy
dx
´
´ b P dt + Q dt
´b
dy
dx
◦ We can then write C F · T ds = a
· ||v(t)|| dt = a P
+Q
dt.
||v(t)||
dt
dt
ˆ b
dy
dx
+Q
dt ,
◦ Thus, the circulation integral can be written more explicitly as Circulation =
P
dt
dt
a
´
where P, Q have been rewritten as functions of t. Note that this expression is also equal to
P dx+Q dy .
C
We would like to see if there is a simpler way, so let us suppose that
y , and say C
dx dy
,
v(t)
dt dt
=
,
T(t) =
||v(t)||
||v(t)||
are functions of
◦
x
and
is parametrized by
Note: The denition also holds for a curve in 3-space, provided we simply add the corresponding
ˆ
if
F = hP, Q, Ri,
then
Circulation
=
a
◦
b
dx
dy
dz
P
+Q
+R
dt
dt
dt
dt
z -terms:
.
Terminology Note: Some authors reserve the term circulation for closed curves, and use ow to refer
to the general case. This terminology can be somewhat confusing given that there is also a ux integral,
and the words ux and ow (in the dictionary sense) are synonyms.
•
Example: Find the circulation of the vector eld
G(x, y) = h−y, xi
around a path that winds once counter-
clockwise around the unit circle.
◦
We can parametrize the path as
◦
Thus,
◦
So
x = cos t, y = sin t
0 ≤ t ≤ 2π .
dx
dy
= − sin t and
= cos t.
dt
dt
for
P = −y = − sin t and Q = x = cos t, and also
´b
´ 2π
´ 2π
dx
dy
P
Circulation =
+Q
dt = 0 ((− sin t)(− sin t) + (cos t)(cos t)) dt = 0 1 dt = 2π
a
dt
dt
12
.
•
Denition: The work performed on a particle by a vector eld
ˆ
Work
ˆ
F · dr =
=
C
F · T ds
F
as the particle travels along a curve
C
is
.
C
◦
Note that the work integral has the same form as the circulation integral.
◦
Notation: The vector dierential
dr
is dened as
dr = hdx, dyi
(or as
dr = hdx, dy, dzi
for a eld in
3-space).
ˆ
◦
Then
F · dr = P dx + Q dy ,
ˆ
F · dr =
C
•
C
ˆ
in the plane, or as
r(t) = t, t2 , 2t
from
t=0
b
P dx + Q dy + R dz =
to
ˆ
P dx + Q dy =
C
ˆ
C
a
Example: Find the work done by the vector eld
the path
ˆ
F · dr =
so the work integral is
a
b
dy
dx
+Q
dt
P
dt
dt
dy
dz
dx
+Q
+R
dt
P
dt
dt
dt
F(x, y, z) = h2x + z, yz, xyi
in 3-space.
on a particle traveling along
t = 1.
dx
dy
dz
P = 2x + z = 3t, Q = yz = 2t3 , and R = xy = t3 . Also,
= 1,
= 2t, and
= 2.
dt
dt
dt
´1
´b
´1
dy
dz
dx
+Q
+R
dt = 0 (3t)(1) + (2t3 )(2t) + (t3 )(2) dt = 0 (3t +
◦ Therefore, the work is a P
dt
dt
dt
14
.
4t4 + 2t3 ) =
5
◦
4.3.2
We have
Flux Across a Curve
ˆ
•
Denition: The ux of the vector eld
F
across the curve
C
is
Flux
F · N ds
=
, where
N
is the unit
C
normal to the curve.
◦
As with the circulation integral, we would like an easier way to evaluate the ux integral.
◦
If
◦
Note: The ux integral as dened here only makes sense for curves in the plane. In 3-dimensional space,
F(x, y) = hP, Qi
r(t)
= hx(t), y(t)i from t = a to t = b, after some algebra
1
dy
dx
we can calculate that N(t) =
,−
. (At the very least, it is easy to observe that this is a
||v(t)|| dt
dt
unit vector and it is orthogonal to T.)
dy
dx
dy
dx
hP, Qi ·
,−
P
−Q
dt
dt
dt
dt .
◦ Then F · N =
=
||v(t)||
||v(t)||
dy
dx
´
´ b P dt − Q dt
´b
dy
dx
◦ Plugging this in gives C F · N ds = a
· ||v(t)|| dt = a P
−Q
dt.
||v(t)||
dt
dt
ˆ
ˆ b
dy
dx
◦ Thus, the ux integral can be written more explicitly as Flux =
P dy − Q dx =
P
−Q
dt .
dt
dt
C
a
and
C
is parametrized by
the corresponding notion requires a surface integral, since a membrane will be a surface, rather than a
curve.
•
Example: Find the ux of the vector eld
G(x, y) = hx, yi
across a path that winds once counterclockwise
around the unit circle.
◦
We can parametrize the path as
◦
Thus,
P = x = cos t
and
x = cos t, y = sin t for 0 ≤ t ≤ 2π .
dx
dy
Q = y = sin t, and also
= − sin t and
= cos t.
dt
dt
13
◦
•
So Circulation
Example:
´b
a
dy
dx
−Q
dt
dt
P
dt =
´ 2π
0
((cos t)(cos t) − (sin t)(− sin t)) dt =
y) = h2x + y, 2y − xi,
F(x,
r(t) = t, t2 between (0, 0) and (2, 4).
For the vector eld
portion of the curve
◦
=
´ 2π
1 dt = 2π
0
.
nd the ux across, and circulation along, the
Here is a plot of the vector eld, along with the curve:
4
3
2
1
0
0.0
◦
0.5
1.0
1.5
2.0
From the picture, we would expect the circulation and ux to be roughly equal, since the vector eld
makes roughly a 45-degree angle with the path near the end.
y = t2 , so that P = 2x + y = 2t + t2 and Q = 2y − x = 2t2 − t.
Also, the start is t = 0 and the end is t = 2.
´
´b
´
dy
dx
◦ By denition we have Flux = C F · N ds = a P
−Q
dt and Circulation = C F · T ds =
dt
dt
´b
dx
dy
+Q
dt.
P
a
dt
dt
ˆ 2
ˆ 2
1 4 2 3
2
=
t + t + t2 ◦ Then, Flux =
(2t + t2 ) · 2t − (2t2 − t) · 1 dt =
2t3 + 2t2 + 2t dt =
2
3
t=0
0
0
52
.
3
´2
´2
1
2
◦ Also, Circulation = 0 (2t + t2 ) · 1 + (2t2 − t) · 2t dt = 0 4t3 − t2 + 2t dt = t4 − t3 + t2 =
3
t=0
52
.
3
4.3.3
•
◦
The parametrization given says
◦
Indeed, we see that the ux and circulation are roughly (and exactly) equal.
x=t
and
Flux Across a Surface
In 3-dimensional space, the corresponding notion of circulation along a curve remains the same. However,
in order to make sense of ux, we must instead talk about ux through a surface (rather than through a
curve):
¨
•
Denition: The (normal) ux of the vector eld
F across the surface S
is Flux across
F · n dσ
S=
, where
S
n
is the outward unit normal to the surface.
◦
Recall that the normal vector to a surface is orthogonal to the tangent plane (it is in fact the normal
vector to the tangent plane as we dened it earlier). If the surface is an implicit surface
then the normal vector
◦
n
is the gradient
Notation: When speaking of a unit normal to a surface we will use a lowercase
dierent from the unit normal
◦
Remark: The integral
˜
S
N
F · n dσ
to a curve (which is an uppercase
n,
to keep the notation
N).
computes the ux through the surface in the direction of the outward
normal. It is also possible to ask about ux in the direction of a particular unit vector
that case is
˜
F · u dσ ,
S
g(x, y, z) = c,
∇g .
u;
the integral in
instead. In general, when it is not specied what type of ux integral is meant,
the ux in the direction of the outward normal is intended.
14
◦
If
S
is parametrized by
r(s, t) = hx(s, t), y(s, t), z(s, t)i, then the two vectors
∂r
∂s
and
∂r
∂t
span the tangent
plane to the surface, and so a normal vector to the surface is given by the cross product
◦
The unit normal to the surface
∗
S
at
r(s, t)
is therefore
∂r ∂r
×
.
∂s
∂t
∂r ∂r
×
∂t .
n = ∂s
∂r × ∂r ∂s
∂t Important Warning: If we write the factors of the cross product in the opposite order, the cross
product vector is multiplied by
−1.
To remedy this ambiguity, one must specify which of these
two orientations is being asked for. One should always check to ensure that the normal vector is
pointing in the correct direction (typically, we intend for it to be pointing outward or upward.)
¨
◦
Plugging this into the surface integral formula yields
Flux across
F·
S=
S
vided that
∗
∂r ∂r
×
∂s
∂t
∂r ∂r
×
∂s
∂t
ds dt
, pro-
is the outward-pointing normal vector to the surface.
Observe that the ugly part of the surface-area dierential cancels out the normalization in the unit
normal vector. (Though the function to be integrated still involves both a dot product and a cross
product, so it's not entirely free sailing.)
¨
◦
We also have a formula for ux through an implicitly-dened surface: it says Flux across
S=
R
f (x, y, z) = c and R is the projection of S in the xy -plane.
• Example: Find the outward ux of the vector eld F = xz 2 , yz 2 , x3 ey through the portion of the cylinder
x2 + y 2 = 4 that lies between the planes z = −1 and z = 1.
where the surface
◦
S
F · ∇g
dy dx
|∇g · k|
is dened implicitly by
From cylindrical coordinates, we can parametrize the cylinder as
desired portion corresponds to
−1 ≤ s ≤ 1
and
r(s, t) = h2 cos t, 2 sin t, si,
where the
0 ≤ t ≤ 2π .
i
j
k ∂r
∂r
∂r ∂r = h−2 sin t, 2 cos t, 0i and
= h0, 0, 1i, so
×
= −2 sin t 2 cos t 0 = h2 cos t, 2 sin t, 0i.
∂t
∂s
∂t ∂s 0
0
1 ◦
Then
◦
This is indeed an outward-pointing normal vector since it is the vector pointing from
(0, 0, s) to the point
r(s, t) = (2 cos t, 2 sin t, s) on the surface.
∂r ∂r
×
= 2s2 cos t, 2s2 sin t, (2 cos t)3 e2 sin t · h2 cos t, 2 sin t, 0i = 4s2 cos2 t + 4s2 sin2 t =
◦ Then F ·
∂t
∂s
4s2 .
´ 2π ´ 1
´ 2π 8
16π
◦ The ux integral is thus 0 −1 4s2 ds dt = 0
dt =
.
3
3
•
Example: Find the outward ux of the vector eld
x2 + y 2 + z 2 = 4
that lies above the plane
F = hx − z, y, x + zi
◦
We use the ux across an implicit surface formula.
◦
On the sphere,
z=1
through the portion of the sphere
z = 1.
x2 + y 2 = 3, and as z increases to 2, the value of x2 + y 2
2
2
surface into the xy -plane is the region R : x + y ≤ 3.
corresponds to
to 0. Thus the projection of the
decreases
F · ∇g
2x2 − 2xz + 2y 2 + 2xz + 2z 2
4
.
=
=p
|∇g · k|
2z
4 − x2 − y 2
˜
4
◦ The ux integral is therefore given by R p
dy dx. We will evaluate this integral using polar
4 − x2 − y 2
◦
We have
∇g = h2x, 2y, 2zi,
so
coordinates.
◦
In polar coordinates, the region is
0≤r≤
√
3 and 0 ≤ θ ≤ 2π , so the integral is
15
´ 2π ´ √3
0
0
√
4
r dr dθ.
4 − r2
,
◦
Substituting
´ 2π
0
◦
u = 4 − r2
4 dθ = 8π
in the inner integral gives
´ 2π ´ √3
0
0
√
´ 2π ´ 0
4
2
r dr dθ = 0 1 − √ du dθ =
2
u
4−r
.
ρ centered at the origin, the outward
1
hx, y, zi.
unit normal vector is n =
ρ
˜ 1
˜ 1
˜
◦ The desired integral is therefore S hx, y, zi · hx − z, y, x + zi dσ = S (x2 + y 2 + z 2 ) dσ = S 2 dσ .
2
2
◦ This is twice the surface area of S , which we could compute (using a simpler surface integral) to be 4π ,
meaning that the desired ux is again 8π .
4.4
•
Alternatively, we could have observed that for a sphere of radius
Conservative Vector Fields, Path-Independence, and Potential Functions
F(x, y)
C2 between
the same two points, we might
´
F
·
dr
and
F
· dr.
C1
C2
2 the elds F(x, y) = hy, xi and G(x, y) = y , x evaluate the work integrals from (0, 0) to (1, 1)
3 2
7 10
three dierent paths C1 : (x, y) = (t, t), C2 : (x, y) = (t , t ), and C3 : (x, y) = (t , t ), for
If we have a vector eld
and two dierent paths
C´1
and
wonder if there is any relation between the work integrals
•
Example: For
along the the
0 ≤ t ≤ 1.
dx
dy
= 1, and
= 1.
F = ht, ti, G = t2 , t ,
dt
dt
´1
´1
´
´
5
∗ Then C1 F · dr = 0 [t · 1 + t · 1] dt = 1 , and C1 G · dr = 0 t2 · 1 + t · 1 dt =
.
6
dx
dy
◦ Along C2 we have F = t2 , t3 , G = t4 , t3 ,
= 3t2 , and
= 2t.
dt
dt
´
´1
´
´1
29
∗ Then C2 F · dr = 0 t2 · 3t2 + t3 · 2t dt = 1 , and C2 G · dr = 0 t4 · 3t2 + t3 · 2t dt =
.
35
dx
dy
◦ Along C3 we have F = t10 , t7 , G = t20 , t7 ,
= 7t6 , and
= 10t9 .
dt
dt
´1
´1
´
´
389
∗ Then C3 F·dr = 0 t10 · 7t6 + t7 · 10t9 dt = 1 , and C3 G·dr = 0 t30 · 7t6 + t7 · 10t9 dt =
459
◦
◦
•
Along
C1
we have
F,
Observe that for
all three paths give the same value, while for
G,
.
each path gives a dierent value.
F is conservative
on a region R if, for any two paths C1 and C2 (inside R) from P1
´
F
·
dr
=
F
·
dr
. In other words, F is conservative if any two paths yield the same
C2
C1
Denition: A vector eld
to
P2 ,
it is true that
´
work integral.
◦
Equivalent to the above denition is the following:
curve
◦
C
in
R,
¸
F · dr = 0.
C
F
is conservative on a region
if, for any closed
C1 and C2 are two paths from P1 to P2 , then we can
C
from
P1 to
´
´1
´ P2 and then following C2 from P2 back to P1 .
F
·
dr
=
F
·
dr
−
F · dr, and so the left-hand side is zero if and
C
C1
C2
These two statements are equivalent because, if
construct a closed path
C
by following
Then for this curve, we have
only if the right-hand side is zero.
◦
R
(A closed curve is one whose start and end points are the same.)
Notation: For a line integral around a closed curve, we often use the notation
¸
C
the circle being a
suggestive example of a closed curve.
•
The vector eld F is conservative on a
U , called a potential function for F, such that
´b
F · dr = U (b) − U (a) along any path from a to b.
a
Theorem (Fundamental Theorem of Calculus for Line Integrals):
simply-connected region
F = ∇U
◦
R
if and only if there exists a function
. If such a function
U
exists, then
Notice the similarity of the statement
´b
a
F · dr = U (b) − U (a)
to the Fundamental Theorem of Calculus,
which relates the integral of a derivative of a function to its values at the endpoints of a path.
16
◦
Technical Note:
The term simply-connected is a technical requirement needed for the proof of the
theorem: intuitively, a simply-connected region consists of a single piece that does not have any holes
in it. More rigorously, it means that the region is connected (contains only one piece) and that if we
take any closed loop in the region, we can shrink it to a point without leaving the region.
x2 + y 2 ≤ 4
is simply-connected, whereas the annulus
1 ≤ x2 + y 2 ≤ 4
◦
The full proof is not especially enlightening. We will instead show one direction of the proof.
◦
Proof (Reverse direction in 3 dimensions): Suppose that
∗
C is the path
dU
∂U dx ∂U dy ∂U dz
=
·
+
·
+
· .
dt
∂x dt
∂y dt
∂z dt
By the (multivariable) Chain Rule, if
a ≤ t ≤ b,
∗
F = ∇U =
then
with
The disc
is not.
∂U ∂U ∂U
,
,
∂x ∂y ∂z
.
x = x(t), y = y(t),
and
z = z(t)
for
Now we can write
ˆ
ˆ
b
dx dy dz
·
,
,
dt
dt dt dt
a
ˆ b
∂U dx ∂U dy ∂U dz
=
·
+
·
+
·
dt
∂x dt
∂y dt
∂z dt
a
ˆ b
dU
dt = U (r(b)) − U (r(a))
=
dt
a
F · dr
=
C
∂U ∂U ∂U
,
,
∂x ∂y ∂z
where we used the Fundamental Theorem of Calculus for the last step.
∗
Notice that this expression does not depend on
the two endpoints
r(b)
and
r(a).
C:
it only involves the potential function
U
Hence we see that the integral is independent of the path, so
and
F
is
conservative.
•
If we can see that a vector eld is conservative, then it is very easy to compute work integrals: we just need
to nd a potential function for the vector eld.
•
F(x, y) = h2x + y, xi
t = 1.
Example: Find the work done by the vector eld
r(t) = −2 cos(πet ), tan−1 (t)
from
t=0
to
on a particle traveling along the path
◦
If we try to set up the integral directly using the parametrization, it will be rather unpleasant.
◦
However, this vector eld is conservative: it is easily veried that for
U (x, y) = x2 + xy ,
then
∇U =
h2x + y, xi = F.
◦
By the Fundamental Theorem of Calculus for line integrals, the work done by the vector eld is then
simply the value of
◦
•
Since
U (r(1)) − U (r(0)).
r(1) = h2, π/4i
and
r(0) = h−2, 0i,
the work is
U (2, π/4) − U (−2, 0) =
π
2
.
An immediate question is whether there an easy way to determine whether a given vector eld is conservative.
There is, but to state the result we rst need to dene the curl of a vector eld:
If F = hP, Q, Ri then the curl of F is dened to be the vector eld curl F = ∇ × F =
• Denition:
i
j
k ∂/∂x ∂/∂y ∂/∂z = ∂R − ∂Q , ∂P − ∂R , ∂Q − ∂P = hRy − Qz , Pz − Rx , Qx − Py i.
∂y
∂z ∂z
∂x ∂x
∂y
P
Q
R ◦
Example: If
◦
If
F = 3x2 y, xyz, exy
F = hP, Qi
◦
F=∇×F=
xexy − xy, −yexy , yz − 3x2
is a vector eld in the plane then we dene the curl of
hP, Q, 0i:
then curl
namely,
∂Q ∂P
0, 0,
−
∂x
∂y
F
.
to be the curl of the vector eld
.
Since this vector only has one nonzero component, some authors dene the curl of a vector eld in the
plane to be the scalar quantity
∂Q
∂P
−
.
∂x
∂y
(We will not do this: for us, the curl of a vector eld will
always be a new vector eld.)
17
•
Theorem (Zero Curl Implies Conservative): A vector eld on a simply-connected region in the plane or in
F = hP, Qi is conservative
Py = Qx , and the vector eld F = hP, Q, Ri is
and only if Py = Qx , Pz = Rx , and Qz = Ry .
3-space is conservative if and only if its curl is zero. More explicitly, the vector eld
on a simply-connected region
R
in the plane if and only if
conservative on a simply-connected region
◦
D
in 3-space if
F = hP, Qi = ∇U
Py = Uxy = Uyx =
It is fairly easy to see why we need the equality of the derivatives of the components: if
then
P = Ux
and
Q = Uy ,
so by the equality of mixed partial derivatives, we see that
Qx .
◦
•
F = hP, Q, Ri follow in the same
Py = Uxy = Uyx = Qx , Pz = Uxz = Uzx = Rx ,
F = ∇U then P = Ux ,
Qz = Uyz = Uzy = Ry .
The three necessary equalities when
way: if
Q = Uy ,
and
and
R = Uz ,
so
The two theorems give us an eective procedure for determining whether a eld is conservative: we rst check
whether its curl is zero, and then (if it is) we can try to nd a potential function by computing antiderivatives.
•
Example: Determine whether
F(x, y) = x2 + y, x + y 2
is conservative. If it is, nd a potential function.
∂ 2
∂ x +y =1=
x + y 2 , so the eld is conservative .
∂y
∂x
◦ To nd a potential function U with ∇U = F, we need to nd U such that Ux = x2 + y and Uy = x + y 2 .
1
◦ Taking the antiderivative of Ux = x2 + y with respect to x yields U = x3 + xy + f (y), for some function
3
f (y).
1
◦ To nd f (y) we dierentiate: Uy = x + f 0 (y), so we get f 0 (y) = y 2 so f (y) = y 3 . (Plus an arbitrary
3
◦
For
F,
we see
constant, but we can ignore it.)
◦
•
Example: Determine whether
◦
•
Thus we see that a potential function for
For
G,
we see
F
is
U (x, y) =
G(x, y) = x + y 2 , x2 + y
.
is conservative. If it is, nd a potential function.
∂ ∂ 2
x + y 2 = 2y 6= 2x =
x +y ,
∂y
∂x
Example: Determine whether
1
1 3
x + xy + y 3
3
3
so the eld is
H(x, y, z) = hy + 2z, x + 3z, 2x + 3yi
not conservative .
is conservative. If it is, nd a potential
function.
◦
For
H,
we have
∂
[2x + 3y],
∂y
◦
and
∂
[x + 3z] = 3 =
∂z
conservative .
U
with
∇U = H,
we need to nd
U
such that
Ux = y + 2z , Uy = x + 3z ,
Uz = 2x + 3y .
Taking the antiderivative of
function
◦
so the eld is
To nd a potential function
and
◦
∂
∂
∂
∂
[y + 2z] = 1 =
[x + 3z],
[y + 2z] = 2 =
[2x + 3y],
∂y
∂x
∂z
∂x
To nd
Ux = y + 2z
with respect to
x
yields
U = xy + 2xz + f (y, z),
for some
f (y, z).
f (y, z) we dierentiate: x+fy = x+3z and 2x+fz = 2x+3y , so fy = 3z
f = 3yz + g(z), where g 0 (z) = 0.
and
fz = 3y .
Repeating
the process yields
◦
•
Thus we see that a potential function for
H
is
U (x, y, z) = xy + 2xz + 3yz
.
If we can nd a potential function for a conservative vector eld, then (as we saw above) it is very easy to
compute work integrals.
•
F = x3 + 4x3 sin y sin z + y 2 z, 2xyz + y + x4 cos y sin z, z 3 + x4 sin y cos z + xy 2 , nd the work
√
3
the eld F on a particle that travels along the curve C : r(t) = sin(πt), t t + 3, 2t + 2 for
Example: If
done by
0 ≤ t ≤ 1.
◦
In theory we could compute the work integral using the parametrization of the path, but this seems quite
unpleasant. Instead, we will check whether this vector eld is conservative: then determining the answer
only requires us to nd the potential function of the eld.
18
◦
◦
◦
We have
We have
Py = 4x3 cos y sin z + 2yz
3
Pz = 4x sin y cos z + y
Finally we have
2
and
and
Qx = 2yz + 4x3 cos y sin z
3
Rx = 4x sin y cos z + y
4
Qz = 2xy + x cos y cos z
and
2
so they are equal.
so they are also equal.
3
Ry = 4x cos y cos z + 2xy , and these are also equal.
Thus,
the eld is conservative.
◦
To nd a potential function
∗
We know
3
U
with
F = ∇U = hUx , Uy , Uz i:
3
Ux = x + 4x sin y sin z + y 2 z
so taking the antiderivative with respect to
1 4
x + x4 sin y sin z + xy 2 z + C(y, z).
4
∗ We then see Uy = x4 cos y sin z + 2xyz + Cy (y, z)
Cy = y .
must equal
Then taking the antiderivative with respect to
y
yields
x
yields
2xyz + y + x4 cos y sin z
1
C(y, z) = y 2 + D(z).
2
U =
so we see
1
1 4
x + x4 sin y sin z + xy 2 z + y 2 + D(z). Then Uz = x4 sin y cos z + xy 2 + D0 (z)
4
2
1 4
3
4
2
0
3
must equal z + x sin y cos z + xy so we see D (z) = z so we can take D(z) =
z .
4
1
1
1
◦ We conclude that a potential function for F is U (x, y, z) = x4 + x4 sin y sin z + xy 2 z + y 2 + z 4 .
4
2
4
◦ Then the desired work integral is equal to U (0, 2, 4) − U (0, 0, 2) = 62 .
∗
4.5
•
We now have
U=
Green's Theorem
Green's Theorem is a 2-dimensional version of the Fundamental Theorem of Calculus that relates a line
integral of a function around a closed curve
that is enclosed by the curve
•
Green's Theorem: If
C
C
to the double integral of a related function over the region
R is the region it
∂Q ∂P
−
dy dx .
∂x
∂y
is a simple closed rectiable curve oriented counterclockwise, and
ˆ
encloses, then for any dierentiable functions
¨ P dx + Q dy =
P (x, y) and Q(x, y),
C
◦
R
C.
Here is an example of a curve
C
and its corresponding region
R
R:
C
◦
Remark: The hypotheses about the curve (simple closed rectiable, oriented counterclockwise) are to
ensure the curve is nice enough for the theorem to hold. Simple means that the curve does not cross
itself, closed means that its starting point is the same as its ending point (e.g., a circle), rectiable
means piecewise-dierentiable (i.e., dierentiable except at a nite number of points), and oriented
counterclockwise means that
◦
C
runs around the boundary of
R
in the counterclockwise direction.
Proof: It suces to prove Green's Theorem for rectangular regions, as more complicated regions can
be built by gluing together simpler ones (in much the manner of a Riemann sum); the overlapping
boundary pieces on two rectangles which share a side will have opposite orientations hence they will
cancel each other out.
∗
a ≤ x ≤ b, c ≤ y ≤ d,
´
´
´
´
´
= C1 + C2 + C3 + C4 , where C1 ,
C
C2 , C3 , and C4 are the four sides of the rectangle (with the proper orientation), and the function to
For a rectangular region
we then have
P dx + Q dy .
´
´
´b
∗ Setting up parametrizations shows C1 [P dx + Q dy] + C3 [P dx + Q dy] = a [P (x, c) − P (x, d)] dx,
´
´
´d
and
[P dx + Q dy] + C4 [P dx + Q dy] = c [Q(b, y) − Q(a, y)] dy .
C2
be integrated on each curve is
19
∗
For the double integral we have
˜
R
−
´ b ´ d ∂P
´b
∂P
dy dx = a c −
dy dx = a [P (x, c) − P (x, d)] dx,
∂y
∂y
and
˜ ∂Q
´ c ´ b ∂Q
´d
dx dy = d a
dx dy = c [Q(b, y) − Q(a, y)] dy .
R ∂x
∂x
´
˜
∂Q ∂P
∗ Therefore, by comparing the expressions, we see that C [P dx + Q dy] = R
−
dy dx,
∂x
∂y
as
desired.
•
Green's Theorem can be seen as a generalization of the Fundamental Theorem of Calculus: both theorems
show that the integral of the derivative of a function (in an appropriate sense) on a region can be computed
using only the values of the function on the boundary of the region.
•
Green's Theorem can be used to convert line integrals into double integrals (which can often be easier to
evaluate if the curve is complicated).
•
Example: Evaluate the integral
having vertices
(0, 0), (1, 0),
¸
3x2 dx + 2xy dy ,
(1, 2).
C
and
where
C
is the counterclockwise boundary of the triangle
◦
We will evaluate the integral both as a line integral and using Green's Theorem.
◦
Green's Theorem says that
produces
◦
◦
˜
P dx + Q dy = R (Qx − Py ) dy dx, so setting P = 3x2
3x2 dx + 2xy dy = R 2y dy dx, where R is the interior of the triangle.
C
¸
´
C˜
A quick sketch shows that
∗
Thus, the integral is
R
0 ≤ x ≤ 1 and 0 ≤ y ≤ 2 − 2x.
´1
4
.
dx = 0 (2 − 2x)2 dx =
3
is dened by the inequalities
´ 1 ´ 2−2x
0
Q = 2xy
R:
To compute the double integral, we need to describe the region
∗
and
0
2y dy dx =
´1
2−2x
(y 2 )y=0
0
To compute the line integral, we need to parametrize each piece of the boundary.
∗
(0, 0) to (1, 0): This component is parametrized by x = t, y = 0 for 0 ≤ t ≤ 1.
´1
dy = 0, so the integral is 0 3t2 dt = 1.
∗ Component #2, joining (1, 0) to (1, 2): This component is parametrized by x = 1, y = t for 0 ≤ t ≤ 2.
´2
Then dx = 0 and dy = dt, so the integral is
2t dt = 4.
0
∗ Component #3, joining (1, 2) to (0, 0): This component is parametrized by x = 1−t, y = 2−2t for 0 ≤
´1
t ≤ 1. Then dx = −dt and dy = −2dt, so the integral is 0 3(1 − t)2 · (−dt) + 2(1 − t)(2 − 2t) · (−2dt) =
´1
11
[−11t + 22t − 11t2 ] dt = − .
0
3
11
4
∗ Thus, the value of the integral over the entire boundary is 1 + 4 −
=
.
3
3
Component #1, joining
Then
•
dx = dt
and
We frequently apply Green's Theorem to simplify the calculation of circulation and ux integrals: we can use
the theorem to give expressions for circulation and ux either as line integrals or as double integrals over a
region. Depending on the shape of the region and its boundary, and the nature of the eld
F,
either the line
integral or the double integral can be easier.
˛
•
Green's Theorem (Tangential form):
Circulation around
¨
F · T ds =
C=
C
◦
Recall that if
F = hP, Qi,
then curl
F=∇×F=
0, 0,
(curl F) · k dA
.
R
∂Q ∂P
−
∂x
∂y
and
(curl F) · k =
∂Q ∂P
−
.
∂x
∂y
The
curl measures how much the vector eld is rotating around a given point.
◦
Thus, if we write everything out in terms of vector eld components, the tangential form of Green's
Theorem reads
¸
C
P dx + Q dy =
˜
R
∂Q ∂P
−
∂x
∂y
dy dx,
which is just the statement we gave above.
˛
•
Green's Theorem (Normal form):
Flux across
C
20
¨
F · N ds =
C=
(div F) dA
R
.
◦
Here, if
F = hP, Qi
then div
F = ∇·F =
∂P
∂Q
+
.
∂x
∂y
This is called the divergence of
F
and measures
how much the vector eld is pushing inward or outward at the given point.
◦
Explicitly, the normal form of Green's Theorem reads
¸
we can recognize as the original statement of Green's Theorem except
place of
◦
˜
dy dx,
which
of
P
F
is modeling
and
P
in
Q.
There is a nice interpretation of the normal form of Green's Theorem:
population movement, and that
border
∂Q
∂P
+
R
∂x
∂y
with −Q in place
P dy − Q dx =
C
C,
the value
F·N
C
imagine that
is the border of a country taking up the region
R.
At a city along the
measures the immigration (in or out) to that city from across the border. At
a city inside the country, the value div F measures the net immigration (into or out of ) that city.
◦
The normal form of Green's Theorem then says: if we add up the net immigration along the border, this
equals the total population ow inside the country. (These two quantities are denitely equal, since they
both tally the net immigration into the country as a whole.)
•
Example: Find the outward ux through, and the (counterclockwise) circulation around, the square with
vertices
◦
(0, 0), (2, 0), (2, 2),
and
(0, 2),
for the vector eld
F(x, y) = x2 − 2xy, y 3 − x .
We could parametrize the boundary of this region and evaluate the line integrals to nd the ux and
circulation. However, this would be somewhat annoying because we'd have to do four line integrals each
time (one for each side of the square).
We can save a lot of eort by using Green's Theorem, which
applies because the boundary is a closed curve.
◦
Flux: Green's Theorem says that Flux across
C=
¸
F · N ds =
C
˜
R
∂P
∂Q
+
∂x
∂y
dy dx.
P = x2 − 2xy and Q = y 3 − x, and the region is 0 ≤ x ≤ 2 and 0 ≤ y ≤ 2.
∂Q
∂P
= 2x − 2y and
= 3y 2 , the ux is given by
since
∂x
∂y
ˆ 2ˆ 2
ˆ 2
2
2
2x − 2y + 3y dy dx =
2xy − y 2 + y 3 dx
∗
Here, we have
∗
Therefore,
0
0
y=0
0
ˆ
2
(4x + 4) dx = 16 .
=
0
◦
∗
•
C=
Circulation: Green's Theorem says that Circulation around
Since
∂Q
= −1
∂x
and
ˆ
∂P
= −2x,
∂y
ˆ
2
C
F · T ds =
˜
R
2
(−1 + 2x) dy dx =
the circulation is
0
0
´2
0
∂Q ∂P
−
∂x
∂y
dy dx.
(−2 + 4x) dx = 4 .
One of the many applications of Green's Theorem is to give various ways to compute the area of a planar
region using a line integral around its boundary. Specically, if
the region
R
(and
C
and
R
C
Area of
R=
˛
˛
−y dx =
x dy =
C
C
C
because by Green's Theorem, each of the line integrals is equal to
Example: Compute the area enclosed by the ellipse
◦
is the counterclockwise boundary curve of
satisfy the hypotheses of Green's Theorem), then
˛
•
¸
1
(x dy − y dx)
2
˜
1 dy dx, which
R
x = a cos t, y = b sin t, 0 ≤ t ≤ 2π .
Using the third formula, we compute
˛
A =
1
(x dy − y dx)
2
C
2π
ˆ
=
0
ˆ
=
0
2π
is the area of
1
[(a cos t)(b cos t) − (b sin t)(−a sin t)] dt
2
ab
dt = πab .
2
21
R.
4.6
•
Stokes's Theorem and Gauss's Divergence Theorem
Stokes's Theorem and Gauss's Divergence Theorem are generalizations of the two forms of Green's Theorem
to the 3-dimensional setting.
As with Green's Theorem, these theorems can be used in either direction,
depending on which integral is easier to set up and evaluate.
4.6.1
•
Stokes's Theorem
Stokes's Theorem: If
the surface
S,
C
is a simple closed rectiable curve in 3-space that is oriented counterclockwise around
then we have
˛
Circulation around
¨
F · T ds =
C=
C
where
◦
◦
T
is the unit tangent to the curve and
n
(curl F) · n dσ
S
is the unit normal to the surface.
Notice the similarity of the statement of Stokes's Theorem to the tangential form of Green's Theorem.
Important Note: The curve
C,
∗
C
must run counterclockwise around
S
in other words, when walking along
the surface should be on its left-hand side.
If one wishes to set up a problem where a curve runs clockwise around a surface, it is equivalent to
traversing the curve in the opposite direction, and so the integral will be scaled by
◦
−1.
Remark: The hypotheses about the curve (simple closed rectiable, oriented counterclockwise) are the
same as in Green's Theorem, and they ensure the curve is nice enough for the theorem to hold.
◦
◦
Recall curl
i
F = ∇×F = ∂/∂x
P
j
∂/∂y
Q
k
∂/∂z
R
= ∂R − ∂Q , ∂P − ∂R , ∂Q − ∂P if F = hP, Q, Ri.
∂y
∂z ∂z
∂x ∂x
∂y
(curl F) · n at (x, y, z)
(x, y, z) along the surface. Stokes's Theorem
Intuitively, if we think of a vector eld as modeling the ow of a uid, the quantity
measures how much the uid is circulating around the point
then says: we can measure how much the uid circulates around the whole surface by measuring how
much it circles around its boundary.
◦
The proof of Stokes's Theorem (which we omit) is essentially the same as the proof of Green's Theorem:
we can reduce to the case of showing the result for simple patches on the surface. Then, by parametrizing
the patches explicitly, we can show Stokes's Theorem is essentially the same as the tangential form of
Green's Theorem on each patch.
•
Stokes's Theorem generalizes the tangential form of Green's Theorem to cover 3-dimensional closed curves
and the surfaces they bound. Note that unlike in Green's Theorem, there are many possible surfaces that any
given curve can bound.
◦
x2 + y 2 = 1, z = 0 in the xy -plane bounds the upper portions (i.e.,
p where
2
2
2
x + y + z = 1, the paraboloid z = 2(1 − x − y ), and the cone z = 1 − x2 + y 2 ,
For example, the unit circle
z ≥ 0)
of the sphere
2
2
as pictured below:
22
•
Typically, we use Stokes's theorem when the line integral over the boundary is dicult, but there is a nice
surface available.
•
F(x, y, z) = y 2 z 3 , 2xyz 3 , 3xy 2 z 2 around the ellipse
2
2
2
2
2
2
ellipsoid x + 2y + 2z = 12 with the cone x + 2y = z .
Example: Find the circulation of the eld
intersection of the upper half of the
given by the
◦
Here is a picture of the surfaces and the ellipse:
◦
We could write down a parametrization for this ellipse with a little bit of eort: substituting the cone's
2
2
2
3z
√ = 12 hence z = 2. Then using the fact that x + 2y = 4
y = 2 sin(t) gives us a parametrization for the curve as r(t) =
equation into the sphere's equation gives
by x = 2 cos(t) and
√
is parametrized
2 cos(t), 2 sin(t), 2 . The resulting
circulation integral does not look so wonderful, although it is
possible to evaluate it.
◦
Another way is to try to use Stokes's Theorem. We have two obvious surfaces to choose from (ellipsoid
and cone); the curve runs counterclockwise around the top portion of
that.
◦
∗
So the curl of
F
j
∂/∂y
2xyz 3
is zero. Hence
(curl F) · n
C=
C
F · T ds =
so let's use
˜
(curl F) · n dσ .
S
k
∂/∂z = 6xyz 2 − 6xyz 2 , 3y 2 z 2 − 3y 2 z 2 , 2yz 3 − 2yz 3 = h0, 0, 0i.
3xy 2 z 2 Stokes's Theorem tells us that Circulation around
i
∗ We have curl F = ∂/∂x
y2 z3
¸
x2 + 2y 2 + 2z 2 = 12,
will also be zero, so we see that the circulation is
0
,
without even having to evaluate the surface integral.
•
Example: Find the ux of the curl
the part of the sphere
◦
2
2
S
2
curl(F)
x + y + z = 25
· n dσ ,
F = yzi − xzj + ex+y k, S is the surface
z = 3, and n is the outward normal.
where
below the plane
We will use Stokes' Theorem. In this case, we want
which is below the plane
◦
˜
S
to be the part of the sphere
x2 + y 2 + z 2 = 25
z = 3.
The boundary of this surface will be the intersection of the plane and the sphere:
(x, y, z) : x2 + y 2 = 16, z = 3 , which
r(t) = h4 cos(t), 4 sin(t), 3i for 0 ≤ t ≤ 2π .
∗ However: the surface S lies below the curve C , not above it:
curve is the set of points
which is
we see that the
is a circle that we can parametrize as
so, when viewed from below (which is
required because we are using the the outward normal), the curve runs clockwise around the surface.
∗
◦
C , which we can
t = 0.
´
F · dr = C P dx + Q dy + R dz .
C
In order to apply Stokes' Theorem, we need to reverse the orientation of the curve
do by interchanging the limits of integration: thus we start at
t = 2π
´
From Stokes's Theorem, the ux of the curl is given by the line integral
and end at
P = 12 sin(t), Q = −12 cos(t), and R = e4 cos(t)+4 sin(t) , and also dx = −4 sin(t) dt, dy =
4 cos(t) dt, and dz = 0 dt.
´
´0 ◦ We get C F · dr = 2π (12 sin(t)) · (−4 sin(t) dt) + (−12 cos(t)) · (4 cos(t) dt)) + e4 cos(t)+4 sin(t) · 0 dt =
´0
−48 dt = 96π .
2π
◦
4.6.2
•
We have
Gauss's Divergence Theorem
Gauss's Divergence Theorem: If
solid region
D,
and
F
S
is a closed (and bounded) piecewise-smooth surface which fully encloses a
is a continuously dierentiable vector eld, then we have
¨
Flux across
˚
F · n dσ =
S=
S
23
(div F) dV
D
where
◦
◦
n
is the outward unit normal to the surface.
Notice the similarity of the statement of the Divergence Theorem to the normal form of Green's Theorem.
x2 + y 2 + z 2 = 1, then D would be the unit ball
x + y + z ≤ 1. If S consists of the 6 faces of the unit cube, then D would be the interior of the cube.
∂P
∂Q ∂R
◦ Here, if F = hP, Q, Ri then div F = ∇ · F =
+
+
.
∂x
∂y
∂z
◦ Intuitively, if we think of a vector eld as modeling the ow of a uid, the divergence measures whether
To get an idea of the setup, if
2
2
S
is the unit sphere
2
there is a source or a sink at a given point (i.e., whether uid is owing inward toward that point,
or outward from that point). The ux through a surface measures how much uid is owing across the
surface. The Divergence Theorem then says: we can measure how much uid is owing in or out of a
solid region by measuring how much uid is owing across its boundary.
◦
The proof of the Divergence Theorem (which we omit) is essentially the same as the proof of Green's
Theorem: we reduce to the case of showing the result for rectangular boxes, and then parametrize the
boxes explicitly.
•
Typically, we want to use the Divergence Theorem whenever we're asked to nd the ux through a closed
surface, since it is almost always easier to evaluate the triple integral than the surface integral.
•
Example: Find the outward ux of the eld
by the planes
◦
F(x, y, z) = x3 − 3y, 2yz + 1, xyz
through the cube bounded
x = ±1, y = ±1, z = ±1.
We could do this directly by computing the ux across each of the six faces of the cube. This is not the
best idea, because it would require setting up six surface integrals.
◦
Instead, we use the Divergence Theorem: it says Flux across
∗
∗
∗
S=
We have div
1
ˆ
1
ˆ
ˆ
1
−1
1
ˆ
1
=
−1
−1
ˆ 1
−1
ˆ 1
V
(div F) dV .
−1
1
ˆ
=
−1
ˆ 1
2
x + y + z = 1,
and
n
dy dx
z=−1
−1
1
6x2 y + xy 2 dx
y=−1
12x2 dx = 8 .
=
−1
‚
F · n dσ , where F
S
is the outward normal.
Example: Compute the ux
1
3x2 z + z 2 + xyz 6x2 + 2xy dy dx
=
◦
˝
Thus the ux integral is
−1
2
F · n dσ =
V is dened by −1 ≤ x ≤ 1, −1 ≤ y ≤ 1, −1 ≤ z ≤ 1.
F = 3x2 + (2z) + (xy).
3x2 + 2z + xy dz dy dx
2
S
The solid region
ˆ
•
˜
We will use the Divergence Theorem. If
= 3x2 + 3y 2 + 3z 2 .
region enclosed by S is
3
= (x + yz)i + (y 3 + xz)j + (z 3 + xy)k, S
F = hP, Q, Ri
then div(F)
is the unit sphere
= Px + Qy + Rz ,
so here we have
div(F)
◦
◦
The
x2 + y 2 + z 2 ≤ 1 .
3x2 + 3y 2 + 3z 2 dz dy dx.
the unit ball
Thus the triple integral is
˝
x2 +y 2 +z 2 ≤1
◦
To evaluate this integral we switch to spherical coordinates: the region is bounded by the inequalities
0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π ,
◦
So we obtain
and
´ 2π ´ π ´ 1
0
0
0
0 ≤ θ ≤ 2π ,
the function is
3ρ2 · ρ2 sin(φ) dρ dφ dθ =
3ρ2 ,
and the dierential is
ρ2 sin(φ) dρ dφ dθ.
´ 2π ´ π 3
´ 2π 6
12π
sin(φ) dφ dθ = 0
dθ =
0
0 5
5
5
.
Well, you're at the end of my handout. Hope it was helpful.
Copyright notice: This material is copyright Evan Dummit, 2012-2015. You may not reproduce or distribute this
material without my express permission.
24