Download 14.4 - Green`s Theorem two-dimensional curl dimensional

Andrew Rosen
Chapter 14 - Vector Calculus, Part II (LATEX)
14.4 - Green's Theorem
Green's Theorem - Circulation Form:
˛
−
→ −
F · d→
r =
˛
¨ f dx + g dy =
∂g
∂x
−
∂f
dA
∂y
R
Note: The partial derivatives that are subtracted in the double integrand are collectively known as the
two-dimensional curl
Note: From here on out, the positive direction is counterclockwise. If the curve is oriented clockwise, negate
the answer
→
−
If the two-dimensional curl of a vector eld, F = hf, gi, is zero then it is said to be irrotational
→
−
→
−
→
−
If the 2-D curl is zero, F is conservative ( F = ∇φ) because
¸→
− →
F · d−
r will also be zero
Area of a Plane Region by Line Integrals:
˛
˛
x dy = −
C
y dx =
C
1
˛
(x dy − y dx)
2
C
Green's Theorem - Flux Form:
˛
−
→
F · n̂ ds =
˛
¨ f dy − g dx =
∂f
∂x
+
∂g
∂y
dA
R
C
Note: The partial derivatives that are added in the double integrand are collectively known as the
dimensional divergence
two-
→
−
If the two-dimensional divergence of a vector eld, F = hf, gi, is zero then it is said to be source-free
14.5 - Divergence and Curl
∂g
∂h
∂f
−
→
−
→ −
→
div F = ∇ · F =
+
+
∂x
∂y
∂z
∂h
∂g ∂f
∂h ∂g
∂f
−
→
−
→ −
→
curl F = ∇ × F = h
−
,
−
,
−
i
∂y
∂z ∂z
∂x ∂x
∂y
Properties of a Conservative Vector Field:
→
−
→
−
1. There exists a potential function, φ, such that F = ∇φ
2.
´→
− →
F · d−
r = φ(B) − φ(A) for all points on a smooth, oriented curve, C , from A to B
C
3.
¸→
− →
F · d−
r = 0 on all simple, smooth, close oriented curves C
C
→
−
→
−
→
−
4. 1 ∇ × F = 0
Note: Rule #4 only applies if −→
F has continuous partial derivatives on all of R3 , so if there is a singularity anywhere then
the theorem cannot be used
1
1
→
−
→
−
→
−
→
−
Curl of a Gradient: If F is conservative then ∇ × ( ∇φ) = 0
→
−
→
−
→
−
Divergence of the Curl: ∇ · ( ∇ × F ) = 0
14.6 - Surface Integrals
A general description for a parametric surface is the following,
→
−
r (u, v) = hx(u, v), y(u, v), z(u, v)i
A general description for parameterizing a cylinder with its axis along the z axis with radius, r:
→
−
r (θ, z) = hr cos θ, r sin θ, zi
A general description for parameterizing a cone with its vertex at the origin with radius, r, and height, h:
rz
rz
→
−
r (θ, z) = h cos θ,
sin θ, zi
h
h
A general description for parameterizing a sphere with radius, ρ, centered at the origin2 :
→
−
r (φ, θ) = hρ sin φ cos θ, ρ sin φ sin θ, ρ cos φi
Surface Integral of Scalar Functions:
¨
¨
f (x, y, z) dS =
S
−
→
−
→
f (x(u, v), y(u, v), z(u, v)) | t u × t v | dA
R
Surface Integral of Vector Fields:
¨
F lux =
−
→
F · n̂ dS =
¨
S
−
→ −
→
−
→
F · ( t u × t v ) dA
R
Average value of a function on a surface:
˜
Average =
R
→
−
→
−
f (x(u, v), y(u, v), z(u, v)) | t u × t v | dA
˜ →
−
→
−
| t u × t v | dA
R
Finding the magnitude of the cross product of the two partial derivative vectors can be dicult here, but it will usually be
, which is exactly what is used for a spherical coordinate triple integral! This will be dA here. Even though the sphere
is parameterized with spherical coordinates, do not multiply in another factor of ρ2 sin φ. It's a regular, rectangular double
integral with bounds for φ and θ
2
ρ2 sin φ
2
Example for nding the equation of a tangent plane at a point on a surface:
→
−
Find the equation of the tangent plane at (1, 1, 3) of t (u, v) = hu2 , v 2 , u + 2vi
Step 1: Calculate the cross product of the partial derivatives
→
−
→
−
t u = h2u, 0, 1i and t v = h0, 2v, 2i
→
−
→
−
t u × t v = h2u, 0, 1i × h0, 2v, 2i = h−2v, −4u, 4uvi
Step 2: Solve for the normal vector by nding u, v and substituting into the cross product
x = 1 = u2 → u = ±1
y = 1 = v 2 → v = ±1
z = 3 = u + 2v ∴ u, v = 1
→
−
n (1, 1, 3) = h−2, −4, 4i
Step 3: Set up the tangent plane equation at that point
−2(x − 1) − 4(y − 1) + 4(z − 3) = 0 → −2x − 4y + 4z = 6
14.7 - Stokes' Theorem
→
−
Assuming that F is a vector eld show components have continuous rst partial derivatives on S ,
˛
−
→ −
F · d→
r =
¨
−
→ −
→
( ∇ × F ) · n̂ dS
S
Note: The normal vector for an explicitly dened surface z = g(x, y) is h−zx , −zy , 1i [Might help to solve
some problems]
Note: If the surface is in the xy plane, the unit normal vector is k̂
14.8 - Divergence Theorem
¨
−
→
F · n̂ dS =
S
˚
D
3
−
→ −
→
∇ · F dV