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I got this problem from a journal and curious how they have calculated that.Previously I asked this problem and unfortunately did not get any answer. Hereby I am posting again and hopefully someone can answer as it is a popular platform to understand math. Consider I have a sphere that is equally divided into two different patch(gray and white). The total gray portion can be called as Pt The sphere can rotate and translate(up-down) at the air-water interface. enter image description here The angle α defines the position of the sphere (alpha increase means the sphere goes towards the air). The angle β defines the rotation of the sphere. So the surface area of any patch into any phase changes with the position and orientation(example: area of Pt inside air phase will change with α and β angle). Now the calculation of the surface area of Pt part of the sphere in air(Apt-A) is given by this equation.enter image description here

I understand the first integral limit in polar angle direction but did not find out how the limit comes for azimuth angle direction. Please keep in mind that the picture is only visual representation of a 3D sphere in 2D.

Intelligenti pauca
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T. an
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The white/grey interface lies on the plane $z/y=\tan\beta$. The $\phi$ integration is carried out along an arc of circle, whose endpoints are the intersections between the great circle on the above plane and plane $z=R\cos\theta$. Inserting that into the equation of the plane gives: $$ y={R\cos\theta\over\tan\beta}, \quad\text{that is:}\quad R\sin\theta\sin\phi={R\cos\theta\over\tan\beta}. $$ From the last equality it follows $$ \sin\phi={1\over\tan\beta\tan\theta}. $$

Intelligenti pauca
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  • @Arentino A great answer given by you. I was looking for this answer for last couple of months and many people gave it a try. Thanks. – T. an Feb 27 '19 at 18:28
  • Sir, I am not sure how the integral can be solved. I have tried in several ways but never reach the final result. Assuming your capabilities I think you can guide me that as well. – T. an May 05 '19 at 23:29
  • @Tanvir Integrate by parts and in the resulting integral make the change of variable $\sin\theta=u$. – Intelligenti pauca May 06 '19 at 13:50
  • Sir, did you check my this post?https://math.stackexchange.com/questions/3212640/calculate-double-integral-for-a-segment-of-a-sphere-unanswered – T. an May 07 '19 at 18:25
  • Apply what I wrote above to the last integral you get. – Intelligenti pauca May 07 '19 at 18:32
  • Ok sir. Is there anyplace where I can show you after solving(in case if I stuck somewhere). – T. an May 07 '19 at 18:35
  • Edit the question you linked above. – Intelligenti pauca May 07 '19 at 18:36
  • If I choose sinθ=u and rest is dv then the integral of dv is not going any where. I have tried with mathematica and it keep running for a long time and provide really long and unrealistic result. – T. an May 07 '19 at 22:05
  • I took another part as u and tried to calculate. I have created a new post related to that. Please have a look. https://math.stackexchange.com/questions/3218069/definite-double-integral – T. an May 08 '19 at 07:38