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 Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora My Apologies Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Richard Ruquist on June 1, 2003 02:28:08 UTC

My math is incorrect. Your corrections are correct.

Except, using your correct math, the volume is pi /5 and abounded calculation of the surface area is between pi and 2.5 times pi.

Both surface area and volume are finite. Nelson is still wrong and your intuitive remarks some years ago are still correct as previously stated.

However, I have been wrong mathematically in my two last posts.

The key calculation is of the surface area which is the integral over 2*pi/x^^3 times sqrt(1+9/x^^2).

Upper and lower bounds can be obtained by evaluating the integral using either the first term and then the second term in the sq root.

Using the first term we integrate 2*pi/x^^3 from 1 to infinity to obtain a=pi

Using the second term we integrate 6*pi/x^^5 from 1 to infinity t obtain 6*pi/4. The upper bound is then the summ of both calculations.

The answer lies between pi and 2.5 pi and is finite.

My apologies for screwing up calculus that I taught to Lexington High School students.

Even warmer regards than you have extended me,

yanniru