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Draw a circle with radius R, and add a line segment representing the jello line (length of 6, 1 unit above bottom of circle). Now draw two radii, one straight down, and another to one end of the line segment. This gives us a right triangle where
(R-1)^2 + 3^2 = R^2
Solving, R = 5 (the good old 3 4 5 triangle), and an equation for the circle is
x^2 + y^2 = 25.
Just as the volume of the sphere can be found by integrating an infinite series of flat discs:
Vsphere = integral(pi*x^2 dy) from -5 to 5
similarly, the volume of the depression can be found from
Vdep = integral(pi*x^2 dy) from -5 to -4 (or, by symmetry, from 4 to 5 to keep the minus signs at bay)
Vdep = integral[pi*(25 - y^2)]dy from 4 to 5
Vdep = pi*[25y - 1/3*y^3] from 4 to 5
Vdep = pi*[(125 - 1/3*125) - (100 - 1/3*64)]
Vdep = 14pi/3
Archimede's principle tells us that the weight of the floating ball = the weight of the displaced jello. Since the plastic is twice as dense as the jello, it follows that
Vplastic = 1/2*Vdep = 7pi/3
If the inside of the plastic shell has radius a,
Vplastic = 4pi/3*(5^3-a^3) = 7pi/3
simplifying,
a^3 = 125-7/4
a = 4.9766
Therefore, the average thickness is
THICKNESSave = 5 - 4.9766 = 0.0234cm
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