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Vector calculus integration identity problem

3

$begingroup$

This is a follow up from another post. I was using the integration symbols available in the Basic Math Assistant palette.

I am new to vector calculus operations. There is a known identity found in my textbook.

$$qquad int _4 pi hats (hatscdot A) d omega=frac4 pi3A$$

I have no idea how to do this type of integration. This is what I tried, but it returns a disaster:

Integrate[s*(Dot[s, A]), s, 0, 4 π]

Also without success:

Integrate[Sin[θ], Cos[θ]*(Dot[Sin[θ], Cos[θ], a1, a2]), θ, 0, 4 π]

It is obvious that I am doing something fundamentally not correct. I go to the documentation on Vector Calculus, but it does not offer much in substance or examples. How do you enter the integral expression shown above in order to return the identity in the right?

Update

In response to comments, here is a copy of the text. This is from page 10 of Optical-Thermal Response of Laser-Irradiated Tissue.

$omega$ is the surface area of a sphere in steradians. $hat s$ is the directional vector of a pencil of radiation located inside the sphere

symbolic vector-calculus

share|improve this question

edited 22 mins ago

J. M. is slightly pensive♦

98.8k10311467

asked 3 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,063718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are $s$ and $omega$ supposed to be? It might be helpful if you can give an example of the textbook with the formula.
  $endgroup$
  – J. M. is slightly pensive♦
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Here's my guess: With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] ] --- or this: With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] == 4 Pi/3 A ]
  $endgroup$
  – Michael E2
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Michael, yes, that does seem to be it. This is why people should always define what their variables mean in their formulae.
  $endgroup$
  – J. M. is slightly pensive♦
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Michael E2 please post it as an answear for upvote
  $endgroup$
  – Jose Enrique Calderon
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I've never seen this author's notation. My guess is that $int_4picdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
  $endgroup$
  – Michael E2
  2 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

3

$begingroup$

This is a follow up from another post. I was using the integration symbols available in the Basic Math Assistant palette.

I am new to vector calculus operations. There is a known identity found in my textbook.

$$qquad int _4 pi hats (hatscdot A) d omega=frac4 pi3A$$

I have no idea how to do this type of integration. This is what I tried, but it returns a disaster:

Integrate[s*(Dot[s, A]), s, 0, 4 π]

Also without success:

Integrate[Sin[θ], Cos[θ]*(Dot[Sin[θ], Cos[θ], a1, a2]), θ, 0, 4 π]

It is obvious that I am doing something fundamentally not correct. I go to the documentation on Vector Calculus, but it does not offer much in substance or examples. How do you enter the integral expression shown above in order to return the identity in the right?

Update

In response to comments, here is a copy of the text. This is from page 10 of Optical-Thermal Response of Laser-Irradiated Tissue.

$omega$ is the surface area of a sphere in steradians. $hat s$ is the directional vector of a pencil of radiation located inside the sphere

symbolic vector-calculus

share|improve this question

edited 22 mins ago

J. M. is slightly pensive♦

98.8k10311467

asked 3 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,063718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are $s$ and $omega$ supposed to be? It might be helpful if you can give an example of the textbook with the formula.
  $endgroup$
  – J. M. is slightly pensive♦
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Here's my guess: With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] ] --- or this: With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s [Element] Sphere[]] == 4 Pi/3 A ]
  $endgroup$
  – Michael E2
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Michael, yes, that does seem to be it. This is why people should always define what their variables mean in their formulae.
  $endgroup$
  – J. M. is slightly pensive♦
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Michael E2 please post it as an answear for upvote
  $endgroup$
  – Jose Enrique Calderon
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I've never seen this author's notation. My guess is that $int_4picdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
  $endgroup$
  – Michael E2
  2 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

This is a follow up from another post. I was using the integration symbols available in the Basic Math Assistant palette.

I am new to vector calculus operations. There is a known identity found in my textbook.

$$qquad int _4 pi hats (hatscdot A) d omega=frac4 pi3A$$

I have no idea how to do this type of integration. This is what I tried, but it returns a disaster:

Integrate[s*(Dot[s, A]), s, 0, 4 π]

Also without success:

Integrate[Sin[θ], Cos[θ]*(Dot[Sin[θ], Cos[θ], a1, a2]), θ, 0, 4 π]

It is obvious that I am doing something fundamentally not correct. I go to the documentation on Vector Calculus, but it does not offer much in substance or examples. How do you enter the integral expression shown above in order to return the identity in the right?

Update

In response to comments, here is a copy of the text. This is from page 10 of Optical-Thermal Response of Laser-Irradiated Tissue.

$omega$ is the surface area of a sphere in steradians. $hat s$ is the directional vector of a pencil of radiation located inside the sphere

symbolic vector-calculus

share|improve this question

edited 22 mins ago

J. M. is slightly pensive♦

98.8k10311467

asked 3 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,063718

$endgroup$

Integrate[s*(Dot[s, A]), s, 0, 4 π]

Integrate[Sin[θ], Cos[θ]*(Dot[Sin[θ], Cos[θ], a1, a2]), θ, 0, 4 π]

share|improve this question

edited 22 mins ago

J. M. is slightly pensive♦

98.8k10311467

asked 3 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,063718

share|improve this question

edited 22 mins ago

J. M. is slightly pensive♦

98.8k10311467

asked 3 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,063718

edited 22 mins ago

J. M. is slightly pensive♦

98.8k10311467

edited 22 mins ago

J. M. is slightly pensive♦

98.8k10311467

asked 3 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,063718

asked 3 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,063718

1 Answer
1

active

oldest

votes

3

$begingroup$

Here's my guess:

With[s = x, y, z,
 A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ] 
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)

--- or this:

With[s = x, y, z, A = A1, A2, A3,
 Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)

share|improve this answer

answered 2 hours ago

Michael E2Michael E2

150k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why it simply does not work with limits of integration s,0,4Pi
  $endgroup$
  – Jose Enrique Calderon
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
  $endgroup$
  – J. M. is slightly pensive♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
  $endgroup$
  – Jose Enrique Calderon
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Jose The syntax s, 0, 4 Pi already implies one-dimensional s from Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.
  $endgroup$
  – J. M. is slightly pensive♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can use RegionIntersection[] with Sphere[] and either ConicHullRegion[] or HalfSpace[].
  $endgroup$
  – J. M. is slightly pensive♦
  2 hours ago
  

  
  
  
  

 | 
show 2 more comments

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3

$begingroup$

Here's my guess:

With[s = x, y, z,
 A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ] 
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)

--- or this:

With[s = x, y, z, A = A1, A2, A3,
 Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)

share|improve this answer

answered 2 hours ago

Michael E2Michael E2

150k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why it simply does not work with limits of integration s,0,4Pi
  $endgroup$
  – Jose Enrique Calderon
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
  $endgroup$
  – J. M. is slightly pensive♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
  $endgroup$
  – Jose Enrique Calderon
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Jose The syntax s, 0, 4 Pi already implies one-dimensional s from Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.
  $endgroup$
  – J. M. is slightly pensive♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can use RegionIntersection[] with Sphere[] and either ConicHullRegion[] or HalfSpace[].
  $endgroup$
  – J. M. is slightly pensive♦
  2 hours ago
  

  
  
  
  

 | 
show 2 more comments

3

$begingroup$

Here's my guess:

With[s = x, y, z,
 A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ] 
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)

--- or this:

With[s = x, y, z, A = A1, A2, A3,
 Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)

share|improve this answer

answered 2 hours ago

Michael E2Michael E2

150k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why it simply does not work with limits of integration s,0,4Pi
  $endgroup$
  – Jose Enrique Calderon
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @Jose, the author was being lazy with the limits (basically, shorter than saying "integrate over the whole area of the unit sphere"). It is fine to be lazy in mathematics, but not so much when programming.
  $endgroup$
  – J. M. is slightly pensive♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.M. is slightly pensive Ok.. but why Mathematica function proposed in the answear does not work with With[s = x, y, z, A = A1, A2, A3, Integrate[s (s.A), s,0,4Pi] ]
  $endgroup$
  – Jose Enrique Calderon
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Jose The syntax s, 0, 4 Pi already implies one-dimensional s from Mathematica's view, while in the "abuse of notation" used in your reference, $hats$ is implied to be a vector.
  $endgroup$
  – J. M. is slightly pensive♦
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Jose: the easiest way is that you have to switch to spherical coordinates if you need to integrate across arbitrary angles. If you insist on keeping yourself to regions, you can use RegionIntersection[] with Sphere[] and either ConicHullRegion[] or HalfSpace[].
  $endgroup$
  – J. M. is slightly pensive♦
  2 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Here's my guess:

With[s = x, y, z,
 A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ] 
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)

--- or this:

With[s = x, y, z, A = A1, A2, A3,
 Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)

share|improve this answer

answered 2 hours ago

Michael E2Michael E2

150k12203482

$endgroup$

With[s = x, y, z,
 A = A1, A2, A3, Integrate[s (s.A), s ∈ Sphere[]] ] 
(* (4 A1 π)/3, (4 A2 π)/3, (4 A3 π)/3 *)

With[s = x, y, z, A = A1, A2, A3,
 Integrate[s (s.A), s ∈ Sphere[]] == 4 Pi/3 A ]
(* True *)

share|improve this answer

answered 2 hours ago

Michael E2Michael E2

150k12203482

answered 2 hours ago

Michael E2Michael E2

150k12203482

answered 2 hours ago

Michael E2Michael E2

150k12203482