Solution to Problem #35



Matthew Brom of Walnut Grove High School MO found the volume. Philippe Fondanaiche of Paris (France), Lou Poulo of Andover MA, Alexey Vorobyov of Irvine CA, Vince Lynch of Doncaster (UK), Robin Stokes of the University of New England (Australia), Ignacio Larrosa Caņestro of Coruņa (Spain), and Sudipta Das of Jadavpur University, West Bengal (India) solved both problems. Here is Lou Poulo's solution:

Define regions 1 and 2 above and below the x-axis respectively, as shown in the diagrams. The area components A1 and A2 and volume components V1 and V2 can be found as follows:

Region 1, with theta from 0 - pi/3:

x = r cos(theta) - r/2 = r(cos(theta) - 1/2), ds = r d(theta), dy = cos(theta) ds = r cos(theta) d(theta).

dA1 = 2 pi x ds = pi r2 ( 2 cos(theta) - 1) d(theta)

dV1 = pi x2 dy = pi r3 ( cos(theta) - 1/2)2 cos(theta) d(theta).

Integrating over 0 - pi/3 gives:

A1 = pi r2 (sqrt(3) - pi/3)

V1 = pi r3 (3 sqrt(3)/8 - pi/6)

Region 2, with phi from 0 - pi/6:

x = r sin(phi), ds = r d(phi), dy = sin(phi) ds = r sin(phi) d(phi).

dA2 = 2 pi x ds = 2 pi (r sin(phi)) r d(phi) = 2 pi r2 sin(phi) d(phi),

dV2 = pi x2 dy = pi (r sin(phi))2 r sin(phi) d(phi) = pi r3 sin3(phi) d(phi).

Integrating over 0 - pi/6 gives:

A2 = pi r2 (2 - sqrt(3))

V2 = pi r3(2/3 - 3 sqrt(3)/8)

Combining for the totals,

Area = pi r2 (2 - pi/3),

Volume = pi r3 (2/3 - pi/6).




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