# Vector calculus integration identity problem

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

UPDATE 2

Will like to share that a proof to this was included in a Mathematic post here

• What are $s$ and $\omega$ supposed to be? It might be helpful if you can give an example of the textbook with the formula. – J. M.'s technical difficulties Apr 1 '19 at 1:23
• 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 ] – Michael E2 Apr 1 '19 at 1:33
• Also to stave off possible questions about entry #5: Integrate[{x, y, z}, {x, y, z} ∈ RegionIntersection[Sphere[], HalfSpace[{0, 0, -1}, 0]]]. Entry #6 would instead use HalfSpace[{0, 0, 1}, 0]. – J. M.'s technical difficulties Apr 1 '19 at 2:24
• I've never seen this author's notation. My guess is that $\int_{4\pi}\cdots$ means the integral over the sphere of measure $4\pi$, i.e., the unit sphere. – Michael E2 Apr 1 '19 at 2:27
• I think the key point here is to understand the phrase solid angle: mathworld.wolfram.com/SolidAngle.html – xzczd Apr 25 '19 at 13:03

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  *)

• Why it simply does not work with limits of integration {s,0,4Pi} – Jose Enrique Calderon Apr 1 '19 at 2:15
• @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. – J. M.'s technical difficulties Apr 1 '19 at 2:17
• @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}] ] – Jose Enrique Calderon Apr 1 '19 at 2:20
• @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. – J. M.'s technical difficulties Apr 1 '19 at 2:23
• @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[]. – J. M.'s technical difficulties Apr 1 '19 at 2:31