A hollow plastic ball floats in a tub of jello (the flavor is irrelevant, but I’ll tell you: it’s key lime). The jello sets, and when the ball is removed, the saucer-shaped depression in the jello measures 6cm across and 1cm deep. If the density of the plastic is twice that of jello, what is the average thickness of the plastic shell?

The first thing to do is to find the radius of the ball. A right triangle can be formed from bottommost point in the depression to the point 1 cm above that, at the center of the circle made at top-of-jello level, along a radius of that circle to some point on that circle, and then straight back to the bottom point. It has legs of 1 cm and 3 cm, and so the hypotenuse, extending from the bottom of the depression to the rim is √10 in length

The center of the whole ball lies along a perpendicular bisector of that hypotenuse where it meets a vertical line above the center of the depression. That line, an extension of the 1-cm side of the first rt triangle, forms a radius of the sphere and is the hypotenuse of a larger rt triangle similar to the first as they share a common angle at the bottom of the depression. The shorter side of the new triangle is half the hypotenuse of the old: (√10)/2. As it is similar, the ratio of the shorter leg to the hypotenuse is still √10, so the hypotenuse is ((√10/2)(√10) = 5, which is the outer radius of the ball, which we'll call R.

At this point we can either use a pre-built formula for the volume of the cap of a sphere (which is what the depression is) or use calculus directly. The formula is (1/3)πh²(3R-h), where h is the 1-cm height of the cap. This comes out to 14π/3.

To be continued due to lack of space...

Posted by Charlie on 2003-04-04 10:52:10