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M
MBuck21-Visitor
1-Visitor
January 24, 2011
Solved

Area of Circular segment by integration

  • January 24, 2011
  • 2 replies
  • 11181 views

As you can probably detect my integration skills are not very sharp but what should be a simple problem has a certain subtly that escapes me. Can someone help.

Surprisingly I could not find an answer with a Google search. It seams the normal solution is to find the area of a circular sector and minus the triangular piece but in my view this is work around not a direct calculus solution..

Kind regards,

Best answer by MikeArmstrong

Thanks guys but I don't think either answer is what I am looking for? What I am trying to figure out is the general equation for the circular segment i.e. a symbolic result that is identical to Wikipedia. For example, how does Wikipedia arrive at the result R^2 / 2 (angle-sin(angle)) formula?

My example gave you the answer you asked for.

Also see below.

Clipboard01.jpg

Mike

2 replies

M
MikeArmstrong1-Visitor
1-Visitor
January 25, 2011

Something like the attached?

Mike

A
AlanStevens19-Tanzanite
19-Tanzanite
January 25, 2011

You might have confused theta and theta/2 - see attached for derivation.

Alan

M
MikeArmstrong1-Visitor
1-Visitor
January 25, 2011

You might have confused theta and theta/2 - see attached for derivation.

Are you sure?

Have a look at the link below.

http://en.wikipedia.org/wiki/Circular_segment

Mike

M
MBuck21-VisitorAuthor
1-Visitor
January 25, 2011

Thanks guys but I don't think either answer is what I am looking for? What I am trying to figure out is the general equation for the circular segment i.e. a symbolic result that is identical to Wikipedia. For example, how does Wikipedia arrive at the result R^2 / 2 (angle-sin(angle)) formula? For some reason the symbolic results even if one transferring Mikes' numerical result into a symbolic result in MathCad it still seems to produce the incorrect general symbolic result. But maybe further algebraic manipulations need to be done to get the Wikipedia answer?

Regards, Mark

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