The parameter a for the closed curve #r = f(theta;a)# is rational. For example, a is rational in #r=a(1+cos theta)#. How do you prove that its length, area enclosed and the volume of the solid of revolution, about an axis, are all transcendental?

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Answer 1 · 2016-08-19T13:06:56

The #pi#-implicit values are #pi#-involved. They are obtained by mathematical operations on #pi#. Hence, they are all transcendental.

The trigonometric functions are all #pi#-related. It is just a matter of

convenience that we convert the transcendental #pi# into #180^o#.

It is impossible to graduate, in mathematical exactitude, #90^o# on

the x-axis for y = sin x, for the amplitude y = 1., on the same scale.

Also, this is the problem in evaluating orbital characteristics of

planets.

None of them, like eccentricity, period, speeds, ..., could be

expressed in mathematical exactitude, in the form of finite-sd digital

strings. All we use, like 1 year = 365.256363004... days, are finite-sd

approximations only.

It is important that, as arc of a unit circle, #pi# is length of the semi-

circle. #pi# is the area of the circle. As an angle, it is dimensionless.

A ticklish issue here is the marking of the edge between irrational

and transcendental numbers, to identify indubitably which is which.. .