Vector Calculus

Dr. D. Sukumar

January 31, 2014

Green’s TheoremTangent form or Ciculation-Curl form

‰

c

Mdx + Ndy =

¨

R

(

∂N

∂x− ∂M

∂y

)

dA

Green’s TheoremTangent form or Ciculation-Curl form

‰

c

Mdx + Ndy =

¨

R

(

∂N

∂x− ∂M

∂y

)

dA

‰

C

F · dr =¨

R

(∇× F ) · k dA

◮ C is a simple, closed, smooth curve in counterclockwisedirection

◮ R is the region enclosed by C

◮ dA is area element

◮ dr is tangential length

Stokes Theorem

The circulation of F = M i+ Nj+ Pk around the boundary C

of an oriented surface S in the direction counterclockwise withrespect to the surface’s unit normal vector n equals theintegral of ∇× F · n over S

‰

C

F · dr =¨

S

∇× F · n dσ

◮ C is a simple,closed, smooth curve consideredcounterclockwise direction

Stokes Theorem

The circulation of F = M i+ Nj+ Pk around the boundary C

of an oriented surface S in the direction counterclockwise withrespect to the surface’s unit normal vector n equals theintegral of ∇× F · n over S

‰

C

F · dr =¨

S

∇× F · n dσ

◮ C is a simple,closed, smooth curve consideredcounterclockwise direction

◮ S is a surface (oriented) with boundary C

Stokes Theorem

The circulation of F = M i+ Nj+ Pk around the boundary C

of an oriented surface S in the direction counterclockwise withrespect to the surface’s unit normal vector n equals theintegral of ∇× F · n over S

‰

C

F · dr =¨

S

∇× F · n dσ

◮ C is a simple,closed, smooth curve consideredcounterclockwise direction

◮ S is a surface (oriented) with boundary C

◮ dσ is surface area element

Stokes Theorem

The circulation of F = M i+ Nj+ Pk around the boundary C

of an oriented surface S in the direction counterclockwise withrespect to the surface’s unit normal vector n equals theintegral of ∇× F · n over S

‰

C

F · dr =¨

S

∇× F · n dσ

◮ C is a simple,closed, smooth curve consideredcounterclockwise direction

◮ S is a surface (oriented) with boundary C

◮ dσ is surface area element

◮ dr is tangential length

Use Stoke’s Theorem to calculate the circulation of the FieldF = x2i + 2xj + z2k around the curve C : The ellipse4x2 + y2 = 4 in the xy plane counter clockwise when viewedfrom above.

Use Stoke’s Theorem to calculate the circulation of the FieldF = x2i + 2xj + z2k around the curve C : The ellipse4x2 + y2 = 4 in the xy plane counter clockwise when viewedfrom above.The surface S is 4x2 + y2 = 4 ie f = 4x2 + y2 = 4

Use Stoke’s Theorem to calculate the circulation of the FieldF = x2i + 2xj + z2k around the curve C : The ellipse4x2 + y2 = 4 in the xy plane counter clockwise when viewedfrom above.The surface S is 4x2 + y2 = 4 ie f = 4x2 + y2 = 4

∇× F =

∣

∣

∣

∣

∣

∣

i j k∂

∂x∂

∂y∂

∂z

x2 2x z2

∣

∣

∣

∣

∣

∣

(∇× F ) · k = 2

Use Stoke’s Theorem to calculate the circulation of the FieldF = x2i + 2xj + z2k around the curve C : The ellipse4x2 + y2 = 4 in the xy plane counter clockwise when viewedfrom above.The surface S is 4x2 + y2 = 4 ie f = 4x2 + y2 = 4

∇× F =

∣

∣

∣

∣

∣

∣

i j k∂

∂x∂

∂y∂

∂z

x2 2x z2

∣

∣

∣

∣

∣

∣

(∇× F ) · k = 2

Circulation

‰

c

F · dr

Use Stoke’s Theorem to calculate the circulation of the FieldF = x2i + 2xj + z2k around the curve C : The ellipse4x2 + y2 = 4 in the xy plane counter clockwise when viewedfrom above.The surface S is 4x2 + y2 = 4 ie f = 4x2 + y2 = 4

∇× F =

∣

∣

∣

∣

∣

∣

i j k∂

∂x∂

∂y∂

∂z

x2 2x z2

∣

∣

∣

∣

∣

∣

(∇× F ) · k = 2

Circulation

‰

c

F · dr =ˆ 2

0

ˆ

√4−y2

2

−√

4−y2

2

2dxdy = 2

ˆ 2

0

√

4− y2dy

Use Stoke’s Theorem to calculate the circulation of the FieldF = x2i + 2xj + z2k around the curve C : The ellipse4x2 + y2 = 4 in the xy plane counter clockwise when viewedfrom above.The surface S is 4x2 + y2 = 4 ie f = 4x2 + y2 = 4

∇× F =

∣

∣

∣

∣

∣

∣

i j k∂

∂x∂

∂y∂

∂z

x2 2x z2

∣

∣

∣

∣

∣

∣

(∇× F ) · k = 2

Circulation

‰

c

F · dr =ˆ 2

0

ˆ

√4−y2

2

−√

4−y2

2

2dxdy = 2

ˆ 2

0

√

4− y2dy

=2

(

y

2

√

4− y2 +4

2sin−1 y

2

)2

0

Use Stoke’s Theorem to calculate the circulation of the FieldF = x2i + 2xj + z2k around the curve C : The ellipse4x2 + y2 = 4 in the xy plane counter clockwise when viewedfrom above.The surface S is 4x2 + y2 = 4 ie f = 4x2 + y2 = 4

∇× F =

∣

∣

∣

∣

∣

∣

i j k∂

∂x∂

∂y∂

∂z

x2 2x z2

∣

∣

∣

∣

∣

∣

(∇× F ) · k = 2

Circulation

‰

c

F · dr =ˆ 2

0

ˆ

√4−y2

2

−√

4−y2

2

2dxdy = 2

ˆ 2

0

√

4− y2dy

=2

(

y

2

√

4− y2 +4

2sin−1 y

2

)2

0

=4(

π

2− 0

)

= 2π

ExerciseStoke’s Theorem

Use Stoke’s theorem to calculate the flux of the curl of thefield F across the surface S in the direction of the outwardunit normal n.

1. F = 2zi + 3xj + 5ykS : z + x2 + y2 = 4 12π

2. F = 2zi + 3xj + 5ykS : r(r , θ) = (r cos θ)i + (r sin θ)j + (4− r2)k0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π

3. F = x2yi + 2y3zj + 3zkS : r(r , θ) = (r cos θ)i + (r sin θ)j + rk

0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π−π

4

Green’s Theorem(Normal form or Flux-Divergence form)

˛

C

Mdy − Ndx =

¨

R

(

∂M

∂x+

∂N

∂y

)

dA

Green’s Theorem(Normal form or Flux-Divergence form)

˛

C

Mdy − Ndx =

¨

R

(

∂M

∂x+

∂N

∂y

)

dA

˛

C

F · n ds =¨

S

∇ · F dA

◮ C is a simple, closed, smooth curve

◮ R is the region enclosed by C

◮ dA is area element

◮ ds is length element

˛

C

F · n ds =¨

S

∇ · F dA

˛

C

F · n ds =¨

S

∇ · F dA

¨

S

F · n dσ =

˚

D

∇ · F dV

◮ S is a simple, closed, oriented surface.

◮ D is solid regin bounded by S

◮ dσ surface area element

◮ dV is volume element

The Divergence Theorem

The flux of a vector field F = M i+Nj+ Pk across a closedoriented surface S in the direction of the surface’s outwardunit normal field n equals the integral of ∇ · F (divergence ofF ) over the region D enclosed by the surface:

¨

S

F · n dσ =

˚

D

∇ · F dV .

F = yi + xyi − zkD : The region inside the solid cylinder x2 + y2 ≤ 4 betweenthe plane z = 0 and the parabolaid z = x2 + y2

∇ · F = 0+ x − 1 = x − 1˚

D

∇ · F dV

F = yi + xyi − zkD : The region inside the solid cylinder x2 + y2 ≤ 4 betweenthe plane z = 0 and the parabolaid z = x2 + y2

∇ · F = 0+ x − 1 = x − 1˚

D

∇ · F dV =

ˆ

2

0

ˆ

√4−x2

−√4−x2

ˆ x2+y2

0

(x − 1)dzdydx

=

ˆ

2

0

ˆ

√4−x2

−√4−x2

(x − 1)(x2 + y2)dydx

=

ˆ

2

0

(x − 1)[x2y +y3

3]√4−x2

−√4−x2

=

ˆ

2

0

(x − 1)(2x2√

4− x2 +2

3(4− x)2

√

4− x2)dx

=1

3

ˆ

2

0

(x − 1)√

4− x2[6x2 + 2(16− 8x + 8x2)]dx

=1

3

ˆ

2

0

(x − 1)√

4− x2[8x2 − 8x + 16]dx

= −16π

ExerciseDivergence theorem

Use divergence theorem to calculate outward flux

1. F = (y − x)i+ (z − y )j+ (y − x)kD :The cube bounded by the planes x ± 1, y ± 1 andz ± 1. −16

2. F = x2i− 2xy j+ 3xzkD :The region cut from the first octant by the spherex2 + y2 + z2 = 4 3π

◮ F is conservative, F is irrotational=⇒ Ciruculation= 0

◮ F is incompressible, ∇.F is 0 =⇒ Flux= 0

Fundamental Theorem of Calculus

ˆ

[a,b]

df

dxdx = f (b)− f (a)

Fundamental Theorem of Calculus

ˆ

[a,b]

df

dxdx = f (b)− f (a)

Let F = f (x)i

ˆ

[a,b]

df

dxdx = f (b)− f (a)

= f (b)i · i + f (a)i · −i

Fundamental Theorem of Calculus

ˆ

[a,b]

df

dxdx = f (b)− f (a)

Let F = f (x)i

ˆ

[a,b]

df

dxdx = f (b)− f (a)

= f (b)i · i + f (a)i · −i

= F (b) · n+ F (a) · n

Fundamental Theorem of Calculus

ˆ

[a,b]

df

dxdx = f (b)− f (a)

Let F = f (x)i

ˆ

[a,b]

df

dxdx = f (b)− f (a)

= 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

dxdx = f (b)− f (a)

Let F = f (x)i

ˆ

[a,b]

df

dxdx = f (b)− f (a)

= f (b)i · i + f (a)i · −i

= F (b) · n+ F (a) · n= total outward flux of F across the boundary

=

ˆ

[a,b]∇ · Fdx

Integral of the differential operator acting on a field over aregion equal the sum of (or integral of ) field componentsappropriate to the operator on the boundary of the region