# Thread: Area of a circle vs area of a cone

1. ## Area of a circle vs area of a cone

Maybe this is a really stupid question, but its bugging me and seems very stupid

I understand that area of a circle is pi r^2 and area of a cone(excluding the base) is pi r s where s is the distance from the top of the cone to its base, and i understand how to get there.

But i was just thinking if that code was a retractable metal object made of very fine rings you could flatten it into a circle that should surely have the same area. but its only the case obviously if r = s then the areas are equal ie

pi r^2 = pi r s = pi r^2

Surely even by integration (not sure quite how) this would be the case? where you would integrate thin concentric rings...

Hope this isn't the dumbest question ever asked

2. Originally Posted by iva
Maybe this is a really stupid question, but its bugging me and seems very stupid

I understand that area of a circle is pi r^2 and area of a cone(excluding the base) is pi r s where s is the distance from the top of the cone to its base, and i understand how to get there.

But i was just thinking if that code was a retractable metal object made of very fine rings you could flatten it into a circle that should surely have the same area. but its only the case obviously if r = s then the areas are equal ie

pi r^2 = pi r s = pi r^2

Surely even by integration (not sure quite how) this would be the case? where you would integrate thin concentric rings...

Hope this isn't the dumbest question ever asked
No, the surface of each or the rings that you disect the cone into are not flat in the plane you wish to flatten the cone onto, so however narrow you make the rings the area will be different from the area of the disk.

This is the analogue of how long is a zig-zag lines with arbitrarily small zigs (and zags) so it is indistinguishable to the naked eye from a straight line. See >>Ask DrMath<< about this.

CB

3. Thanks for this i read the linked article so i guess then basically saying that however flat you take the circles to be ( and i thought integration could make it so flat that its got negligible height), there is still going to be a tiny height that will be enough to make the general area turn out to be more than the a circle with the circumference of its base

Makes sense now.. sort of