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



$$int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$



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



enter image description here






vector-calculus







share|improve this question











$endgroup$












  • $begingroup$
    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♦
    1 hour 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
    1 hour 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♦
    1 hour 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 $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
    $endgroup$
    – Michael E2
    51 mins ago


















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



$$int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$



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



enter image description here






vector-calculus







share|improve this question











$endgroup$












  • $begingroup$
    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♦
    1 hour 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
    1 hour 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♦
    1 hour 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 $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
    $endgroup$
    – Michael E2
    51 mins ago
















2












2








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



$$int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$



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



enter image description here






vector-calculus







share|improve this question











$endgroup$




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



$$int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$



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



enter image description here






vector-calculus




vector-calculus






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 46 mins ago









J. M. is slightly pensive♦

98.8k10311467




98.8k10311467










asked 2 hours ago









Jose Enrique CalderonJose Enrique Calderon

1,058718




1,058718












  • $begingroup$
    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♦
    1 hour 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
    1 hour 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♦
    1 hour 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 $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
    $endgroup$
    – Michael E2
    51 mins ago




















  • $begingroup$
    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♦
    1 hour 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
    1 hour 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♦
    1 hour 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 $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
    $endgroup$
    – Michael E2
    51 mins ago


















$begingroup$
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♦
1 hour ago




$begingroup$
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♦
1 hour ago




2




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
1 hour ago






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












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






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






1




1




$begingroup$
I've never seen this author's notation. My guess is that $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
$endgroup$
– Michael E2
51 mins ago






$begingroup$
I've never seen this author's notation. My guess is that $int_{4pi}cdots$ means the integral over the sphere of measure $4pi$, i.e., the unit sphere.
$endgroup$
– Michael E2
51 mins ago












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









$endgroup$













  • $begingroup$
    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
    58 mins 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    56 mins 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♦
    47 mins ago












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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









$endgroup$













  • $begingroup$
    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
    58 mins 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    56 mins 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♦
    47 mins ago
















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









$endgroup$













  • $begingroup$
    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
    58 mins 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    56 mins 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♦
    47 mins ago














2












2








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









$endgroup$



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












share|improve this answer



share|improve this answer










answered 1 hour ago









Michael E2Michael E2

150k12203482




150k12203482












  • $begingroup$
    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
    58 mins 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    56 mins 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♦
    47 mins ago


















  • $begingroup$
    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
    58 mins 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, $hat{s}$ is implied to be a vector.
    $endgroup$
    – J. M. is slightly pensive♦
    56 mins 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♦
    47 mins ago
















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




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




1




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






$begingroup$
@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
58 mins 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
58 mins ago






1




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, $hat{s}$ is implied to be a vector.
$endgroup$
– J. M. is slightly pensive♦
56 mins ago






$begingroup$
@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, $hat{s}$ is implied to be a vector.
$endgroup$
– J. M. is slightly pensive♦
56 mins ago






1




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♦
47 mins ago




$begingroup$
@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♦
47 mins ago


















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