# A question regarding the geometry of spheres and circles.

All these facts came in my mind while the study of fluid mechanics where I got to know that fluids always have a tendency to decrease their surface area.

Assume that a closed irregular shaped loop of thread is put on the surface of a ring having a film of soap solution on it,now prick a hole in the middle by my observation I came to know that after pricking the hole the force of surface tension pulls the thread which leads to the formation of a circle.If fluids want to get the minimum area possible then the loop formed must have the maximum surface area possible and by observation loop is a circle.

It's my appeal to mathematicians that please give me the rigid proof regarding this observation that circle had the largest area possible formed by a closed loop.

A small drop of liquid takes nearly a spherical shape.Because of gravitational force between the particles there is some deviation from the spherical shape but for small drops it can be neglected.

The formation of spherical shape must be due to the property of fluid to decrease its surface area.Please give here a proof regarding the fact that for a given volume sphere assumes the smallest surface area possible.

Edit:- Give a rigid proof to the concerned problems do not apply induction or arguments.If a proof by geometry or high order calculus exists then only give the answer.Well I know the whole physics of this problem and basically the mathematics is complex that's why I am concerned to it.

• google.com/url?sa=t&source=web&rct=j&url=https://… Mar 6 '19 at 14:11
• Thanks for the link what about the first part of my concern?As well ast the answers to the link does not give the complete proof,give a rigid proof.
– user628320
Mar 6 '19 at 14:23
• I heard about a thing called isoperimetric inequality look it up on the net it will be helpful Mar 6 '19 at 17:19
• Thanks Aditya for recommending but I know that isoperimetric inequality is based on the principal of least action of physics.Do you have any proof by calculus or pure geometry.
– user628320
Mar 6 '19 at 18:39
• cut-the-knot.org/do_you_know/isoperimetric.shtml
– user
Mar 7 '19 at 21:53

Take a line $$AB$$ dividing the closed curve in two equal area parts, forget one half (or always take a mirror image), and pick any other point $$C$$ on the curve. Let $$\theta$$ be the angle $$ACB$$. The area of $$ABC$$ is maximal when $$\theta = 90^\circ$$; otherwise we can always find another closed curve with larger area by hinging $$AC$$ and $$BC$$ about $$C$$ until $$\theta=90^\circ$$. The perimeter of the curve remains the same but the subtended area is larger. The only curve for which all points subtend $$90^\circ$$ to $$AB$$ is a circle.