Spheres and Cone Example
This example states that
the plane located y units above the "equator" intersects the sphere in a circle of area
the area of the plane's intersection with the part of the cylinder that is outside of the cone is also
Can someone double check that? I am trying to follow through the example and it seems that the area of the intersection circle in the sphere should be and the intersection ring of the inner cylinder outside of the ring should be , different from what is claimed. —Preceding unsigned comment added by 188.8.131.52 (talk • contribs)
- The article's statements are correct and yours are not. The circumference of the intersection of the plane with the sphere is 2π√(r2 − y2), i.e. it's 2π times the radius. The area is π times the square of the radius. Since the radius is √(r2 − y2), the area is π(r2 − y2).
- Remember that the area needs to be proportional to the squares of the distances involved, and that fails to happen in either of the two formulas you give.
- Similarly in the cone example, what you give is the difference between the circumferences rather than the difference between the areas of the two concentric circles. Michael Hardy (talk) 20:22, 28 May 2009 (UTC)
Missing working, missing triangle
I had to read the same proof several times until I realized that a whole lot of working is missing in the jump to Pythagoras, which is asserted but not demonstrated. It would help if it was shown on the diagram that it is the annulus we are interested in (it should be shaded) and similarly that the slice through the sphere (shaded the same way) has that same area -- we need the triangle from the centre of the sphere to the centre of the slice to a point on the slice's circumference. Then Pythagoras appears, and then even I can follow the rest of the proof. Chiswick Chap (talk) 08:44, 24 October 2013 (UTC)