Vector Calculus
Dr. D. Sukumar
January 31, 2014
Green’s Theorem
Tangent form or Ciculation-Curl form
‰
c
Mdx + Ndy =
¨ R
∂M
∂N
−
∂x
∂y
dA
Green’s Theorem
Tangent form or Ciculation-Curl form
‰
Mdx + Ndy =
c
R
‰
◮
◮
◮
◮
¨ C
F · dr =
¨
R
∂M
∂N
−
∂x
∂y
dA
(∇ × F ) · k dA
C is a simple, closed, smooth curve in counterclockwise
direction
R is the region enclosed by C
dA is area element
dr is tangential length
Stokes Theorem
The circulation of F = Mi + Nj + Pk around the boundary C
of an oriented surface S in the direction counterclockwise with
respect to the surface’s unit normal vector n equals the
integral of ∇ × F · n over S
‰
¨
F · dr =
∇ × F · n dσ
C
◮
S
C is a simple,closed, smooth curve considered
counterclockwise direction
Stokes Theorem
The circulation of F = Mi + Nj + Pk around the boundary C
of an oriented surface S in the direction counterclockwise with
respect to the surface’s unit normal vector n equals the
integral of ∇ × F · n over S
‰
¨
F · dr =
∇ × F · n dσ
C
◮
◮
S
C is a simple,closed, smooth curve considered
counterclockwise direction
S is a surface (oriented) with boundary C
Stokes Theorem
The circulation of F = Mi + Nj + Pk around the boundary C
of an oriented surface S in the direction counterclockwise with
respect to the surface’s unit normal vector n equals the
integral of ∇ × F · n over S
‰
¨
F · dr =
∇ × F · n dσ
C
◮
◮
◮
S
C is a simple,closed, smooth curve considered
counterclockwise direction
S is a surface (oriented) with boundary C
d σ is surface area element
Stokes Theorem
The circulation of F = Mi + Nj + Pk around the boundary C
of an oriented surface S in the direction counterclockwise with
respect to the surface’s unit normal vector n equals the
integral of ∇ × F · n over S
‰
¨
F · dr =
∇ × F · n dσ
C
◮
◮
◮
◮
S
C is a simple,closed, smooth curve considered
counterclockwise direction
S is a surface (oriented) with boundary C
d σ is surface area element
dr is tangential length
Use Stoke’s Theorem to calculate the circulation of the Field
F = x 2 i + 2xj + z 2 k around the curve C : The ellipse
4x 2 + y 2 = 4 in the xy plane counter clockwise when viewed
from above.
Use Stoke’s Theorem to calculate the circulation of the Field
F = x 2 i + 2xj + z 2 k around the curve C : The ellipse
4x 2 + y 2 = 4 in the xy plane counter clockwise when viewed
from above.
The surface S is 4x 2 + y 2 = 4 ie f = 4x 2 + y 2 = 4
Use Stoke’s Theorem to calculate the circulation of the Field
F = x 2 i + 2xj + z 2 k around the curve C : The ellipse
4x 2 + y 2 = 4 in the xy plane counter clockwise when viewed
from above.
The surface S is 4x 2 + y 2 = 4 ie f = 4x 2 + y 2 = 4
i
j
k
∂ ∂ ∂
∇ × F = ∂x ∂y ∂z x 2 2x z 2 (∇ × F ) · k = 2
Use Stoke’s Theorem to calculate the circulation of the Field
F = x 2 i + 2xj + z 2 k around the curve C : The ellipse
4x 2 + y 2 = 4 in the xy plane counter clockwise when viewed
from above.
The surface S is 4x 2 + y 2 = 4 ie f = 4x 2 + y 2 = 4
i
j
k
∂ ∂ ∂
∇ × F = ∂x ∂y ∂z x 2 2x z 2 (∇ × F ) · k = 2
‰
Circulation
F · dr
c
Use Stoke’s Theorem to calculate the circulation of the Field
F = x 2 i + 2xj + z 2 k around the curve C : The ellipse
4x 2 + y 2 = 4 in the xy plane counter clockwise when viewed
from above.
The surface S is 4x 2 + y 2 = 4 ie f = 4x 2 + y 2 = 4
i
j
k
∂ ∂ ∂
∇ × F = ∂x ∂y ∂z x 2 2x z 2 (∇ × F ) · k = 2
‰
ˆ 2ˆ
Circulation
F · dr =
c
0
√
−
4−y 2
2
√
4−y 2
2
2dxdy = 2
ˆ
0
2p
4 − y 2 dy
Use Stoke’s Theorem to calculate the circulation of the Field
F = x 2 i + 2xj + z 2 k around the curve C : The ellipse
4x 2 + y 2 = 4 in the xy plane counter clockwise when viewed
from above.
The surface S is 4x 2 + y 2 = 4 ie f = 4x 2 + y 2 = 4
i
j
k
∂ ∂ ∂
∇ × F = ∂x ∂y ∂z x 2 2x z 2 (∇ × F ) · k = 2
‰
ˆ 2ˆ
Circulation
F · dr =
c
0
√
−
4−y 2
2
√
4−y 2
2
2dxdy = 2
p
4
y
y
4 − y 2 + sin−1
=2
2
2
2
ˆ
0
2
0
2p
4 − y 2 dy
Use Stoke’s Theorem to calculate the circulation of the Field
F = x 2 i + 2xj + z 2 k around the curve C : The ellipse
4x 2 + y 2 = 4 in the xy plane counter clockwise when viewed
from above.
The surface S is 4x 2 + y 2 = 4 ie f = 4x 2 + y 2 = 4
i
∂ j∂ k∂ ∇ × F = ∂x ∂y ∂z x 2 2x z 2 (∇ × F ) · k = 2
‰
ˆ 2ˆ
Circulation
F · dr =
c
0
√
−
4−y 2
2
√
4−y 2
2
2dxdy = 2
ˆ
0
2
p
4
y
y
−
1
4 − y 2 + sin
=2
2
2
2 0
π
− 0 = 2π
=4
2
2p
4 − y 2 dy
Exercise
Stoke’s Theorem
Use Stoke’s theorem to calculate the flux of the curl of the
field F across the surface S in the direction of the outward
unit normal n.
1. F = 2zi + 3xj + 5yk
12π
S : z + x2 + y2 = 4
2. F = 2zi + 3xj + 5yk
S : r(r , θ ) = (r cos θ )i + (r sin θ )j + (4 − r 2 )k
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π
3. F = x 2 yi + 2y 3 zj + 3zk
S : r(r , θ ) = (r cos θ )i + (r sin θ )j + rk
−π
0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π
4
Green’s Theorem
(Normal form or Flux-Divergence form)
˛
C
Mdy − Ndx =
¨ R
∂N
∂M
+
∂x
∂y
dA
Green’s Theorem
(Normal form or Flux-Divergence form)
˛
C
Mdy − Ndx =
˛
C
◮
◮
◮
◮
¨ ∂N
∂M
+
∂x
∂y
¨
∇ · F dA
F · n ds =
R
S
C is a simple, closed, smooth curve
R is the region enclosed by C
dA is area element
ds is length element
dA
˛
C
F · n ds =
¨
S
∇ · F dA
˛
C
F · n ds =
¨
¨
˚
S
◮
◮
◮
◮
F · n dσ =
S
D
∇ · F dA
∇ · F dV
S is a simple, closed, oriented surface.
D is solid regin bounded by S
d σ surface area element
dV is volume element
The Divergence Theorem
The flux of a vector field F = Mi + Nj + Pk across a closed
oriented surface S in the direction of the surface’s outward
unit normal field n equals the integral of ∇ · F (divergence of
F ) over the region D enclosed by the surface:
¨
˚
F · n dσ =
∇ · F dV .
S
D
F = yi + xyi − zk
D : The region inside the solid cylinder x 2 + y 2 ≤ 4 between
the plane z = 0 and the parabolaid z = x 2 + y 2
∇·F = 0+x −1 = x −1
˚
D
∇ · F dV
F = yi + xyi − zk
D : The region inside the solid cylinder x 2 + y 2 ≤ 4 between
the plane z = 0 and the parabolaid z = x 2 + y 2
˚
D
∇·F = 0+x −1 = x −1
ˆ 2 ˆ √4 − x 2 ˆ x 2 + y 2
(x − 1)dzdydx
∇ · F dV =
√
0
=
ˆ
0
2ˆ
− 4− x 2
√
4− x 2
0
(x
√
− 4− x 2
− 1)(x 2 + y 2 )dy dx
y 3 √ 4− x 2
] √
3 − 4− x 2
0
ˆ 2
p
p
2
(x − 1)(2x 2 4 − x 2 + (4 − x )2 4 − x 2 )dx
=
3
0
ˆ 2
p
1
=
(x − 1) 4 − x 2 [6x 2 + 2(16 − 8x + 8x 2 )]dx
3 0
ˆ
p
1 2
=
(x − 1) 4 − x 2 [8x 2 − 8x + 16]dx
3 0
= −16π
=
ˆ
2
(x − 1)[x 2 y +
Exercise
Divergence theorem
Use divergence theorem to calculate outward flux
1. F = (y − x )i + (z − y )j + (y − x )k
D :The cube bounded by the planes x ± 1, y ± 1 and
z ± 1.
−16
2
2. F = x i − 2xy j + 3xzk
D :The region cut from the first octant by the sphere
3π
x2 + y 2 + z2 = 4
◮
◮
F is conservative, F is irrotational=⇒ Ciruculation= 0
F is incompressible, ∇.F is 0 =⇒ Flux= 0
Fundamental Theorem of Calculus
ˆ
[a,b ]
df
dx = f (b ) − f (a)
dx
Fundamental Theorem of Calculus
ˆ
Let F = f (x )i
ˆ
[a,b ]
[a,b ]
df
dx = f (b ) − f (a)
dx
df
dx = f (b ) − f (a)
dx
= f (b )i · i + f (a )i · − i
Fundamental Theorem of Calculus
ˆ
Let F = f (x )i
ˆ
[a,b ]
[a,b ]
df
dx = f (b ) − f (a)
dx
df
dx = f (b ) − f (a)
dx
= f (b )i · i + f (a )i · − i
= F (b ) · n + F (a ) · n
Fundamental Theorem of Calculus
ˆ
[a,b ]
df
dx = f (b ) − f (a)
dx
Let F = f (x )i
ˆ
df
dx = f (b ) − f (a)
[a,b ] dx
= f (b )i · i + f (a )i · − i
= F (b ) · n + F (a ) · n
= total outward flux of F across the boundary
Fundamental Theorem of Calculus
ˆ
[a,b ]
df
dx = f (b ) − f (a)
dx
Let F = f (x )i
ˆ
df
dx = f (b ) − f (a)
[a,b ] dx
= f (b )i · i + f (a )i · − i
= F (b ) · n + F (a ) · n
= total outward flux of F across the boundary
ˆ
=
∇ · F dx
[a,b ]
Integral of the differential operator acting on a field over a
region equal the sum of (or integral of ) field components
appropriate to the operator on the boundary of the region