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Vector Calculus handout
The Fundamental Theorem of Line Integrals
Theorem 1 (The Fundamental Theorem of Line Integrals). Let C be a smooth curve given by a vector
function ~r(t), where a ≤ t ≤ b, and let f be a differentiable function of two or three variables whose
gradient vector ∇f is continuous on C. Then
Z
∇f · d~r = f (~r(b)) − f (~r(a)) .
C
Note that the answer doesn’t depend on the path, only the endpoints! So, for any two paths C1 and C2
having the same initial point and same terminal point,
Z
Z
∇f · d~r.
∇f · d~r =
C1
C1
This leads to. . .
R
Definition 1. Let F~ be a continuous vector field with domain D. The line integral C F~ · d~r is said to be
R
R
independent of path (or path independent if C1 F~ · d~r = C1 F~ · d~r for any two paths C1 and C2 in D that
have the same initial point and same terminal point.
Definition 2. A curve C is said to be closed if its initial point and terminal point are the same.
Theorem 2.
R
C
F~ · d~r is path independent in D if and only if
R
C
F~ · d~r = 0 for every closed path C in D.
Recall that a vector field is conservative if there exists a scalar function f such that ∇f = F~ .
Theorem
3. Let F~ be a continuous vector field on an open connected region. Then the line integral
R
~
F · d~r is path independent if and only if F~ is conservative.
C
R
Why do we care? If F~ is conservative, then that can make evaluating C F~ · d~r much easier (if C is closed
we get 0 immediately, if not then we could still choose a ‘nicer’ path). But it would be nice to have some
way to check to see if F~ is conservative without going back to the definition:
Theorem 4. Let F~ = P ı̂ + Q̂ be a vector field on an open, simply-connected region D such that P and
Q have continuous first-order partial derivatives. Then F~ is conservative if and only if
∂P
∂Q
=
∂y
∂x
on D.
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Curl and Divergence
For three-dimensional vector fields, we need another way to check for a conservative vector field. Recall
the “del” operator that we defined when we introduced the gradient:
∂
∂
∂
k̂
ı̂ +
̂ +
∂x
∂y
∂z
∂ ∂ ∂
=
,
,
∂x ∂y ∂z
∇=
We use this to define the curl and divergence of a three-dimensional vector field F~ = P ı̂ + Q̂ + R k̂:
curl F~ = ∇ × F~
ı̂
̂
∂
∂
= ∂x
∂y
P
Q
k̂ ∂ ∂z R
div F~ = ∇ · F~
Theorem 5. Let F~ = P ı̂ + Q̂ + R k̂ be a vector field on an open, simply-connected region D such that P ,
Q, and R have continuous first-order partial derivatives. Then F~ is conservative if and only if curl F~ = ~0
on D.
Example 1. Determine whether or not the following vector fields are conservative. If so, find f (x, y) such
that ∇f = F~ .
(a) F~ (x, y) = h y, x i
(b) F~ (x, y) = h y, 1 i
(c) F~ (x, y) = y 2 , 2xy
2
(d) F~ (x, y) = y 3 + 1, 3xy 2 + 1
(e) F~ (x, y) =
D
2
2
2xyex y , x2 ex
y
E
(f) F~ (x, y, z) = h sin(y), −x cos(y), 1 i
(g) F~ (x, y, z) = 2xy, x2 + 2yz, y 2
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We can combine these definitions and theorems as follows:
Theorem 6. Let F~ be a continuous vector field with continuous first partial derivatives in an open connected region D, and let C be a [piecewise] smooth curve in D given by ~r(t). The following are equivalent:
1. F~ is conservative.
2.
∂P
∂Q
=
(for F~ = P ı̂ + Q̂) or curl F~ = 0 (for F~ = P ı̂ + Q̂ + R k̂) on D .
∂y
∂x
3. There exists f such that ∇f = F~ (which then allows us to use the Fundamental Theorem).
R
4. C F~ · d~r is independent of path.
R
5. C F~ · d~r = 0 for every closed curve C in D.
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Example 2. Evaluate C F~ · d~r, where F~ = y 3 + 1, 3xy 2 + 1 and C is the semicircular path from (0, 0)
to (2, 0) given by ~r(t) = h 1 − cos(t), sin(t) i , 0 ≤ t ≤ π.
R
Example 3. Evaluate
F~ · d~
r, where F~ = y 2 , 2xy and C is the parabolic path from (4, 0) to (1, 3)
C
given by ~r(t) = 4 − t, 4t − t2 , 0 ≤ t ≤ 3.
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Example 4. Evaluate
R
C
F~ · d~r, where F~ = 2xy, x2 + 2yz, y 2 and C is the path given by
~r(t) = t2 , t cos(t), et sin(t) , 0 ≤ t ≤ 3π.
Green’s Theorem
Green’s Theorem relates a line integral around a simple closed curve C to a double integral over the region
in the plane D bounded by C:
Theorem 7 (Green’s Theorem). Let C be a positively oriented, piecewise-smooth, simple, closed curve in
R2 and let D be the region bounded by C, and let F~ = P ı̂ + Q̂ be a continuous vector field such that P
and Q have continuous partial derivatives on an open region containing D. Then
Z
Z
F~ · d~r =
P dx + Q dy
C
C
ZZ ∂Q ∂P
−
dA
=
∂x
∂y
ZDZ =
curl F~ · k̂ dA
D
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R
Example 5. Evaluate C x2 y 2 dx + xy dy where C is the positively oriented curve consisting of the arc of
the parabola y = x2 from (0, 0) to (1, 1) and the line segments from (1, 1) to (0, 1) and from (0, 1) to (0, 0).
Parametric Surfaces
Given a vector valued or parametric function in one variable, say t, we can trace out a curve in R2 or R3 :
~r(t) = x(t)ı̂ + y(t)̂[+z(t) k̂]. But with two parameters we can trace out a surface (in R3 ):
~r(u, v) = x(u, v)ı̂ + y(u, v)̂ + z(u, v) k̂.
Given a function z = f (x, y), one way to get a parameterization of the surface is to simply let u = x,
v = y, and then let z = f (u, v) (similar construction for surfaces defined by y = f (x, z) or x = f (y, z)).
Tangent Planes to Parametric Surfaces
Recall that to find the equation of a plane, we need to know a normal vector to the plane ~n = h a, b, c i
and a point in the plane (x0 , y0 , z0 ). Then the equation of the plane is
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0.
Now, given a parameterization ~r(u, v) = x(u, v)ı̂ + y(u, v)̂ + z(u, v) k̂ of a surface, to find an equation of
a tangent plane we still need a normal vector and a point. The point will either be given as (x0 , y0 , z0 ),
or we will be given (u0 , v0 ) which we’ll then plug into ~r to find the coordinates of the point. Our normal
vector is
~n = ~ru (u0 , v0 ) × ~rv (u0 , v0 )
(or, for this type of problem, we could also use ~ru (u0 , v0 ) × ~rv (u0 , v0 )) where
~ru =
∂y
∂z
∂x
∂y
∂z
∂x
ı̂ +
̂ +
k̂ and ~rv =
ı̂ +
̂ +
k̂
∂u
∂u
∂u
∂v
∂v
∂v
Example 6. Find the equation of the plane tangent to the surface
√
~r(u, v) = uı̂ + v̂ + uv k̂
at the point (1, 1).
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Example 7. Find the equation of the plane tangent to the surface
~r(u, v) = 3u cos(v)ı̂ + 3u sin(v)̂ + u2 k̂
at the point (0, 6, 4).
Surface Area of Parametric Surfaces
Recall that the area of a parallelogram with sides given by ~a and ~b is |~a × ~b|. This leads to the following:
Theorem 8. Let S is a smooth parametric surface given by ~r(u, v) = x(u, v)ı̂ + y(u, v)̂ + z(u, v) k̂,
(u, v) ∈ D, such that S is covered just once as (u, v) ranges throughout D. Then the surface area of S is
ZZ
A(S) =
|~ru × ~rv | dA
D
If we have a surface given by z = f (x, y) and view x and y as the parameters, then we have ~r(x, y) =
h x, y, f (x, y) i which means ~rx = h 1, 0, fx i and ~ry = h 0, 1, fy i, so
ı̂ ̂ k̂ ~ru × ~rv = 1 0 fx 0 1 fy = h −fx , fy , 1 i
and
|~ru × ~rv | =
Thus the surface area is
A(S) =
q
ZZ q
fx 2 + fy 2 + 1
fx 2 + fy 2 + 1 dA
D
(Similar formulae if the surface is given by y = f (x, z) or x = f (y, z).)
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Example 8. Find the surface area of
~r(u, v) = 2u cos(v)ı̂ + 2u sin(v)̂ + u2 k̂
for 0 ≤ u ≤ 1, 0 ≤ v ≤ 2π.
Example 9. Find the surface area of the portion of the hemisphere z =
the cylinder x2 + y 2 = 9.
8
p
25 − x2 − y 2 that lies inside
Surface Integrals
Previously, we had that the line integral of f (x, y, z) along a curve C given by ~r(t) = h x(t), y(t), z(t) i,
a ≤ t ≤ b, is
s 2 2
Z b
Z b
2
dx
dy
dz
or
f (x(t), y(t), z(t))
+
+
dt =
f (~r(t)) |~r 0(t)| dt
dt
dt
dt
a
a
Now we extend this definition to define the surface integral of f (x, y, z) over a surface S given by
~r(u, v) = x(u, v)ı̂ + y(u, v)̂ + z(u, v) k̂ where (u, v) ∈ D, as
ZZ
ZZ
f (x, y, z) dS =
f (~r(u, v)) |~ru × ~rv | dA
S
D
Example 10. Evaluate the surface integral
with 0 ≤ u ≤
π
2
RR
xy dS where S is given by ~r(u, v) = 2 cos(u)ı̂+2 sin(u)̂+v k̂
S
and 0 ≤ v ≤ 2.
If S is a surface given by a function z = g(x, y) then we can let x and y be the parameters and find |~rx ×~ry |
as we did previously to get
ZZ
ZZ
q
f (x, y, z) dS =
f (x, y, g(x, y)) (gx )2 + (gy )2 + 12 dA
S
D
(Similar formulae if the surface is given by y = g(x, z) or x = g(y, z).)
Example 11. Evaluate
RR
x2 + y 2 + z 2 dS where S is the portion of the plane z = x+2 that is bounded
S
by x = −1, x = 1, y = 0, and y = 2.
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Orientable Surfaces
Orientable surfaces have two sides, and we can distinguish between the two based on which way a normal
vector is pointing. For a closed surface (such as a sphere, cube, or any region E over which we evaluated
a triple integral) we can have a normal vector pointing outward from the surface (this is called positive
orientation) or it can point inward (this is called negative orientation). Some surfaces are non-orientable,
such as a Möbius strip.
Now that we have the notion of orientability, we can talk about. . .
Surface Integrals Over Oriented Surfaces
Let F~ = P ı̂ + Q̂ + R k̂ be a continuous vector field, defined on an oriented surface S with unit normal
vector ~n. The surface integral of F~ over S is
ZZ
ZZ
ZZ
~
~
F · dS =
F · ~n dS =
F~ · (~ru × ~rv ) dA
S
S
D
If S is given by the graph of a function z = g(x, y), then
RR
RR

ZZ
(−P gx − Q gy + R) dA
 F~ · h −gx , −gy , 1 i dA =
D
RRD
F~ · dS = RR

(P gx + Q gy − R) dA
 F~ · h gx , gy , −1 i dA =
S
D
(oriented upward)
(oriented downward)
D
where D is the projection of z = g(x, y) onto the xy-plane.
Example 12. Evaluate the surface integral
RR
F~ · dS, where F~ = h x, y, z i and S is the part of the
S
paraboloid z = 4 − x2 − y 2 that lies above the xy-plane, and has upward orientation.
Example 13. Evaluate the surface integral
RR
F~ · dS, where F~ = h 3z, −4, y i and S is the part of the
S
plane x + y + z = 1 in the first octant, with downward orientation.
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