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vector calculus integration identity problem

2

$begingroup$

This is a follow up from another post . I was using the integration symbol available in the Basic Math Assistance available in Wolfram Mathematica.

I am new to vector calculus operations. There is a known identity found in the textbooks

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

I have no idea how to do this type of integration. This is what I tried but return a dissaster

Integrate[s*(Dot[s, A]), s, 0, 4 [Pi]]

Also , without success

Integrate[Sin[[Theta]], 
 Cos[[Theta]]*(Dot[Sin[[Theta]], Cos[[Theta]], a1, 
 a2]), [Theta], 0, 4 [Pi]]

It is obviosu that I am doing something fundamentally not correct. I go to WM documentation on Vector Calculus but does not offer much substance or examples. How will you enter the equation above in order to return the identity in the right?

UPDATE 1

In respond to comment, here is a copy of the text. This is from page 10 Optical-Thermal Response of Laser-Irradiated Tissue ISBN 9789048188307

$$w$$ is the surface area of a sphere in solid angle steradian. s is the directional vector of a pencil of radiation located inside the sphere

vector-calculus

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.8k10311467

asked 2 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,058718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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♦
  2 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
  2 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♦
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 3 more comments

2

$begingroup$

This is a follow up from another post . I was using the integration symbol available in the Basic Math Assistance available in Wolfram Mathematica.

I am new to vector calculus operations. There is a known identity found in the textbooks

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

I have no idea how to do this type of integration. This is what I tried but return a dissaster

Integrate[s*(Dot[s, A]), s, 0, 4 [Pi]]

Also , without success

Integrate[Sin[[Theta]], 
 Cos[[Theta]]*(Dot[Sin[[Theta]], Cos[[Theta]], a1, 
 a2]), [Theta], 0, 4 [Pi]]

It is obviosu that I am doing something fundamentally not correct. I go to WM documentation on Vector Calculus but does not offer much substance or examples. How will you enter the equation above in order to return the identity in the right?

UPDATE 1

In respond to comment, here is a copy of the text. This is from page 10 Optical-Thermal Response of Laser-Irradiated Tissue ISBN 9789048188307

$$w$$ is the surface area of a sphere in solid angle steradian. s is the directional vector of a pencil of radiation located inside the sphere

vector-calculus

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.8k10311467

asked 2 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,058718

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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♦
  2 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
  2 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♦
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

 | 
show 3 more comments

$begingroup$

This is a follow up from another post . I was using the integration symbol available in the Basic Math Assistance available in Wolfram Mathematica.

I am new to vector calculus operations. There is a known identity found in the textbooks

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

I have no idea how to do this type of integration. This is what I tried but return a dissaster

Integrate[s*(Dot[s, A]), s, 0, 4 [Pi]]

Also , without success

Integrate[Sin[[Theta]], 
 Cos[[Theta]]*(Dot[Sin[[Theta]], Cos[[Theta]], a1, 
 a2]), [Theta], 0, 4 [Pi]]

It is obviosu that I am doing something fundamentally not correct. I go to WM documentation on Vector Calculus but does not offer much substance or examples. How will you enter the equation above in order to return the identity in the right?

UPDATE 1

In respond to comment, here is a copy of the text. This is from page 10 Optical-Thermal Response of Laser-Irradiated Tissue ISBN 9789048188307

$$w$$ is the surface area of a sphere in solid angle steradian. s is the directional vector of a pencil of radiation located inside the sphere

vector-calculus

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.8k10311467

asked 2 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,058718

$endgroup$

Integrate[s*(Dot[s, A]), s, 0, 4 [Pi]]

Integrate[Sin[[Theta]], 
 Cos[[Theta]]*(Dot[Sin[[Theta]], Cos[[Theta]], a1, 
 a2]), [Theta], 0, 4 [Pi]]

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.8k10311467

asked 2 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,058718

share|improve this question

edited 1 hour ago

J. M. is slightly pensive♦

98.8k10311467

asked 2 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,058718

edited 1 hour ago

J. M. is slightly pensive♦

98.8k10311467

edited 1 hour ago

J. M. is slightly pensive♦

98.8k10311467

asked 2 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,058718

asked 2 hours ago

Jose Enrique CalderonJose Enrique Calderon

1,058718

1 Answer
1

active

oldest

votes

2

$begingroup$

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 1 hour ago

Michael E2Michael E2

150k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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♦
  1 hour 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
  1 hour 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♦
  1 hour 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♦
  1 hour ago
  

  
  
  
  

 | 
show 2 more comments

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2

$begingroup$

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 1 hour ago

Michael E2Michael E2

150k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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♦
  1 hour 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
  1 hour 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♦
  1 hour 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♦
  1 hour ago
  

  
  
  
  

 | 
show 2 more comments

2

$begingroup$

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 1 hour ago

Michael E2Michael E2

150k12203482

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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♦
  1 hour 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
  1 hour 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♦
  1 hour 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♦
  1 hour 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 1 hour 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 1 hour ago

Michael E2Michael E2

150k12203482

answered 1 hour ago

Michael E2Michael E2

150k12203482

answered 1 hour ago

Michael E2Michael E2

150k12203482