Vector calculus integration identity problem
5
$begingroup$
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.
$$qquad int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$
I have no idea how to do this type of integration. This is what I tried, but 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
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked Apr 1 at 1:15
Jose Enrique CalderonJose Enrique Calderon
1,073719
$endgroup$
|
show 3 more comments
5
$begingroup$
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.
$$qquad int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$
I have no idea how to do this type of integration. This is what I tried, but 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
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked Apr 1 at 1:15
Jose Enrique CalderonJose Enrique Calderon
1,073719
$endgroup$
$begingroup$
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♦
Apr 1 at 1:23
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
Apr 1 at 1:33
$begingroup$
@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♦
Apr 1 at 1:37
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:10
2
$begingroup$
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
Apr 1 at 2:27
|
show 3 more comments
5
5
5
$begingroup$
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.
$$qquad int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$
I have no idea how to do this type of integration. This is what I tried, but 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
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked Apr 1 at 1:15
Jose Enrique CalderonJose Enrique Calderon
1,073719
$endgroup$
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.
$$qquad int _{4 pi }hat{s} (hat{s}cdot A) d omega=frac{4 pi}{3}A$$
I have no idea how to do this type of integration. This is what I tried, but 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
symbolic vector-calculus
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked Apr 1 at 1:15
Jose Enrique CalderonJose Enrique Calderon
1,073719
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
asked Apr 1 at 1:15
Jose Enrique CalderonJose Enrique Calderon
1,073719
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
edited 2 days ago
J. M. is slightly pensive♦
99k10311467
99k10311467
asked Apr 1 at 1:15
Jose Enrique CalderonJose Enrique Calderon
1,073719
asked Apr 1 at 1:15
Jose Enrique CalderonJose Enrique Calderon
1,073719
asked Apr 1 at 1:15
Jose Enrique CalderonJose Enrique Calderon
1,073719
1,073719
$begingroup$
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♦
Apr 1 at 1:23
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
Apr 1 at 1:33
$begingroup$
@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♦
Apr 1 at 1:37
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:10
2
$begingroup$
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
Apr 1 at 2:27
|
show 3 more comments
$begingroup$
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♦
Apr 1 at 1:23
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
Apr 1 at 1:33
$begingroup$
@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♦
Apr 1 at 1:37
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:10
2
$begingroup$
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
Apr 1 at 2:27
$begingroup$
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♦
Apr 1 at 1:23
$begingroup$
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♦
Apr 1 at 1:23
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
Apr 1 at 1:33
$begingroup$
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
Apr 1 at 1:33
$begingroup$
@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♦
Apr 1 at 1:37
$begingroup$
@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♦
Apr 1 at 1:37
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:10
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:10
2
2
$begingroup$
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
Apr 1 at 2:27
$begingroup$
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
Apr 1 at 2:27
|
show 3 more comments
1 Answer
1
active
oldest
votes
5
$begingroup$
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 *)
answered Apr 1 at 2:13
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
Why it simply does not work with limits of integration {s,0,4Pi}
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:15
3
$begingroup$
@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♦
Apr 1 at 2:17
$begingroup$
@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
Apr 1 at 2:20
2
$begingroup$
@Jose The syntax{s, 0, 4 Pi}already implies one-dimensionalsfrom 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♦
Apr 1 at 2:23
3
$begingroup$
@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 useRegionIntersectionwithSphereand eitherConicHullRegionorHalfSpace.
$endgroup$
– J. M. is slightly pensive♦
Apr 1 at 2:31
|
show 2 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
5
$begingroup$
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 *)
answered Apr 1 at 2:13
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
Why it simply does not work with limits of integration {s,0,4Pi}
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:15
3
$begingroup$
@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♦
Apr 1 at 2:17
$begingroup$
@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
Apr 1 at 2:20
2
$begingroup$
@Jose The syntax{s, 0, 4 Pi}already implies one-dimensionalsfrom 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♦
Apr 1 at 2:23
3
$begingroup$
@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 useRegionIntersectionwithSphereand eitherConicHullRegionorHalfSpace.
$endgroup$
– J. M. is slightly pensive♦
Apr 1 at 2:31
|
show 2 more comments
5
$begingroup$
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 *)
answered Apr 1 at 2:13
Michael E2Michael E2
150k12203482
$endgroup$
$begingroup$
Why it simply does not work with limits of integration {s,0,4Pi}
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:15
3
$begingroup$
@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♦
Apr 1 at 2:17
$begingroup$
@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
Apr 1 at 2:20
2
$begingroup$
@Jose The syntax{s, 0, 4 Pi}already implies one-dimensionalsfrom 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♦
Apr 1 at 2:23
3
$begingroup$
@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 useRegionIntersectionwithSphereand eitherConicHullRegionorHalfSpace.
$endgroup$
– J. M. is slightly pensive♦
Apr 1 at 2:31
|
show 2 more comments
5
5
5
$begingroup$
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 *)
answered Apr 1 at 2:13
Michael E2Michael E2
150k12203482
$endgroup$
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 *)
answered Apr 1 at 2:13
Michael E2Michael E2
150k12203482
answered Apr 1 at 2:13
Michael E2Michael E2
150k12203482
answered Apr 1 at 2:13
Michael E2Michael E2
150k12203482
answered Apr 1 at 2:13
Michael E2Michael E2
150k12203482
150k12203482
$begingroup$
Why it simply does not work with limits of integration {s,0,4Pi}
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:15
3
$begingroup$
@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♦
Apr 1 at 2:17
$begingroup$
@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
Apr 1 at 2:20
2
$begingroup$
@Jose The syntax{s, 0, 4 Pi}already implies one-dimensionalsfrom 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♦
Apr 1 at 2:23
3
$begingroup$
@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 useRegionIntersectionwithSphereand eitherConicHullRegionorHalfSpace.
$endgroup$
– J. M. is slightly pensive♦
Apr 1 at 2:31
|
show 2 more comments
$begingroup$
Why it simply does not work with limits of integration {s,0,4Pi}
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:15
3
$begingroup$
@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♦
Apr 1 at 2:17
$begingroup$
@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
Apr 1 at 2:20
2
$begingroup$
@Jose The syntax{s, 0, 4 Pi}already implies one-dimensionalsfrom 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♦
Apr 1 at 2:23
3
$begingroup$
@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 useRegionIntersectionwithSphereand eitherConicHullRegionorHalfSpace.
$endgroup$
– J. M. is slightly pensive♦
Apr 1 at 2:31
$begingroup$
Why it simply does not work with limits of integration {s,0,4Pi}
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:15
$begingroup$
Why it simply does not work with limits of integration {s,0,4Pi}
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:15
3
3
$begingroup$
@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♦
Apr 1 at 2:17
$begingroup$
@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♦
Apr 1 at 2:17
$begingroup$
@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
Apr 1 at 2:20
$begingroup$
@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
Apr 1 at 2:20
2
2
$begingroup$
@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♦
Apr 1 at 2:23
$begingroup$
@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♦
Apr 1 at 2:23
3
3
$begingroup$
@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♦
Apr 1 at 2:31
$begingroup$
@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♦
Apr 1 at 2:31
|
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$begingroup$
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♦
Apr 1 at 1:23
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
Apr 1 at 1:33
$begingroup$
@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♦
Apr 1 at 1:37
$begingroup$
@Michael E2 please post it as an answear for upvote
$endgroup$
– Jose Enrique Calderon
Apr 1 at 2:10
2
$begingroup$
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
Apr 1 at 2:27