6. Integral Theorems of Vector Calculus

Question 1

Integral Definition of Divergence

Equation

Answer 1

|∇ . |F = (𝛿V->0) lim 1/𝛿V ∯|F . d|S

-where 𝛿S is the surface enclosing the volume 𝛿V with outward normal

Question 2

Integral Definition of Divergence

Derivation

Answer 2

-take 𝛿V to be the volume of a cuboid of sides 𝛿x 𝛿y 𝛿z centred on the point |x=(x,y,z)
-on face S1 (left face in x direction) |n=(1,0,0) so:
|F.|n = F1(x + 𝛿x/2, y, z)
-on face S2 (right face in x direction, |n=(-1,0,0) so:
|F.|n = -F1(x - 𝛿x/2, y, z)
-hence:
(S1) ∬|F.|n dS = F1(x+𝛿x/2, y, z) 𝛿y 𝛿z
(S2) ∬|F.|n dS = -F1(x-𝛿x/2, y, z) 𝛿y 𝛿z
-Thus:
(S1∪S2)∬|F.|n dS = [F1(x+𝛿x/2, y, z)-F1(x-𝛿x/2,y,z) / 𝛿x] 𝛿x 𝛿y 𝛿z
-so:
(𝛿x->0) lim ∬|F.|n dS = ∂F1/∂x 𝛿V
-simirlarly for the other four faces, so:
(𝛿x,𝛿y,𝛿z ->0) lim ∬|F.|n dS =
(∂F1/∂x + ∂F2/∂y + ∂F3/∂z)𝛿V
-so:
|∇ . |F = (𝛿V->0) lim 1/𝛿V ∯|F . d|S

Question 3

Integral Definition of Divergence

Description

Answer 3

-the divergence of a vector field |F is equal to the flux of |F out of 𝛿V divided by the volume of 𝛿V

Question 4

Divergence Theorem

Equation

Answer 4

∯|F . d|S = ∭ (|∇ . |F) dV

Question 5

Divergence Theorem

Description

Answer 5

-if V is a volume enclosed by a surface S and |F is a continuously differentiable vector field in V, then the divergence theorem states that the flux through that closed surface S is equal to the sum of the divergence at every point within the volume V.

Question 6

Divergence Theorem

Derivation

Question 7

Applications of Divergence Theorem

Answer 7

• divergence theorem allows surface integrals (which are often difficult) to be replaced by volume integrals, which are easier
• REMEMBER - we must check every time that the surface is closed, as divergence theorem only applies to closed surfaces

Question 8

Green’s First Identity

Equation

Answer 8

∭[ (|∇f) . (|∇g) + ∇²g ] dV = ∯ f|∇g . d|S

Question 9

Green’s First Identity

Derivation

Answer 9

-start with a vector field of the form |F = f |∇g, where f and g are scalar fields
-find the divergence by converting to index notation, rearranging (use product rule) and converting back to vector notation:
|∇ . |F = (|∇f) . (|∇g) + ∇²g
-hence the divergence theorem states that:
∭[ (|∇f) . (|∇g) + ∇²g ] dV = ∯ f|∇g . d|S

Question 10

Green’s Second Identity

Equation

Answer 10

∭[ f ∇²g - g ∇²f ] dV = ∯ ( f |∇g - g |∇f ) . d|S

Question 11

Green’s First Identity

Description

Answer 11

-the divergence theorem applied to a vector field of the form |F = f |∇g , where f and g are scalar fields

Question 12

Another Integral Theorem

Description

Answer 12

-the divergence theorem applied to a vector field of the form |F = |a f , where f is a scalar field and |a is a constant vector

Question 13

Another Integral Theorem

Derivation

Answer 13

-start with a vector field of the form |F = |a f, where f is a scalar field and |a is a constant vector
-find the divergence by converting to index notation, switching order of variables and converting back to vector notation:
|∇ . |F = |a . (|∇f)
-substitute into the divergence theorem, and cancel |a . from both sides since |a can be taken as any vector if we take |e1, |e2 and |e3 it is clear all components are equal so both vectors must be equal:
∭ |∇f dV = ∯ f d|S

Question 14

Another Integral Theorem

Equation

Answer 14

∭ |∇f dV = ∯ f d|S

Question 15

Green’s Second Identity

Description

Answer 15

-the divergence theorem applied to a vector field of the form, |F = f |∇g - g |∇f , where f and g are both scalar fields

Question 16

Green’s Second Identity

Derivation

Answer 16

-start with a vector field of the form, |F = f |∇g - g |∇f , where f and g are both scalar fields
-find the divergence by using the divergence identity from the derivation of Green’s First Identity, cancel terms:
|∇ . |F = f ∇²g - g ∇²f
-substitute into the divergence theorem:
∭[ f ∇²g - g ∇²f ] dV = ∯ ( f |∇g - g |∇f ) . d|S

Question 17

Gauss’s Law of Electrostatics

Integral Form

Answer 17

∯ |E . ^n dA = Q/εo

Question 18

Gauss’s Law of Electrostatics

Differential Form

Answer 18

|∇ . |E = ρ / εo

-i.e. charge density makes an electric field diverge

Question 19

Gauss’s Law of Electrostatics

Description

Answer 19

-Gauss’s law of electrostatics states that the flux of electric field |E through a closed surface S is proportional to the enclosed charge Q

Question 20

Integral Definition of Curl

Symbol Equation

Answer 20

|n . (|∇ x |F) = (δS->0) lim 1/δS ∮|F . d|x

where δS ia an element of surface with normal |n and δ| is the closed curve forming the boundary to δS

Question 21

Integral Definition of Curl

Word Equation

Answer 21

(curl |F) . |n = circulation of |F around δC / δS

Question 22

Integral Definition of Curl

Derivation

Question 23

Stokes’s Theorem

Equation

Answer 23

-let C be a closed curve and S ANY surface bounded by C, if |F is a continuously differentiable field on S then:
∬ (|∇ x |F) . d|S = ∮|F . d|x

Question 24

Stokes’s Theorem

Derivation

Question 25

Applications of Stokes’s Theorem

Show that every irrotational field is conservative

Answer 25

-let |∇x|F = 0 everywhere, i.e. let |F be an irrotational field
-let C be a closed curve and S be any surface bounded by C, then by Stokes’s Theorem:
∮|F . d|x = ∬ (|∇ x |F) . d|S
-and since |∇x|F = 0 everywhere:
∮|F . d|x = 0
-hence by definition, |F is conservative

Question 26

Green’s Theorem

Equation

Answer 26

∮ p(x,y)dx + q(x,y)dy = ∬ (∂q/∂x - ∂p/∂y) dxdy

Question 27

Green’s Theorem

Derivation

Answer 27

-Let |F = ( p(x,y) , q(x,y) , 0 ) , i.e. a vector field in the x-y plane and dedpendent only on x and y
-the curl of |F is:
( 0 , 0 , ∂q/∂x - ∂p/∂y )
-consider any surface S lying in the x-y plane with a perimeter described in an anticlockwise direction
-S has normal ^n = (0,0,1)
-so d|S = (0,0,1) dxdy
-apply Stoke’s Theorem:
∮ p(x,y)dx + q(x,y)dy = ∬ (∂q/∂x - ∂p/∂y) dxdy

Question 28

Other Vector Identities

Answer 28

∬ |∇f x d|S = - ∮ f d|x