Area of Circular segment by integration

[MBuck2]

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,

[MikeArmstrong]

Something like the attached?

Mike

[AlanStevens]

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

Alan

[MikeArmstrong]

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

[MBuck2]

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

[MikeArmstrong · Accepted solution]

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.

Mike

[wayne]

Attached are two other ways to derive the formula,
1) using polar coordinates
2) using recangular coordinates.

Niether method uses double integration

[MikeArmstrong]

Very nice Wayne,

So I thiunk my initial answer was correct, was it?

Mike

[wayne]

Yes, I believe so, nice work to you too.

Wayne

[MBuck2]

Mike and others thank you. The reason I wasn't getting the incorrect symbolic result with Mikes' solution is that I forgot that Mike defined R as equal to 1 and if you have a unit circle you get pi/2 aka 1.571 which is correct for that case. However, R needs to be unassigned to generalise the symbolic result. Just wish that the symbolic engine didn't use such esoteric functions when a simple sin function is all that is needed.

Thanks again.

[MBuck2]

Mike

One final question on the subject. I understand the x-axis upper and lower limits of integration i.e. -R*Sin(pi/2) to R*Sin(pi/2) but why is the upper y limit of integration sqrt(R^2-x^2) and not just -R*cos(pi/2)? Because if we assume the circle radius is starting at point 0,0 and going in the negative y-axis direction then the way I see it is that the lower limit would be -R and the upper limit -R*cos(pi/2) for the segment. Could you please explain this?

Kind regards, Mark

[wayne]

This was Mike's solution, but I will answer anyway

You are integrating between two curves; think of a vertical strip, dx wide and a hieght equal to the the distance between the lower and upper boundary of the defining curves. The single integration method does this (except the area is rotated 90 degrees), and that is what the double integration in effect does also. The limites (y) for the first integration are different for each point x.

[MikeArmstrong]

This was Mike's solution, but I will answer anyway

You are integrating between two curves; think of a vertical strip, dx wide and a hieght equal to the the distance between the lower and upper boundary of the defining curves. The single integration method does this (except the area is rotated 90 degrees), and that is what the double integration in effect does also. The limites (y) for the first integration are different for each point x.

You beat me to it Wayne.

I have just noticed that I made a mistake in my worksheet. The (Beyer 1987) function should be like below (left).

Mike

[AlanStevens]

I took segment to mean the yellow part only. This might well not have been what was required!

Alan