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Solutions: 1
The total volume of the hemisphere and cylinder is V, where
The surface area of the hemisphere is S_{h}, where
V/π = (2/3)R^{3} + R^{2}h [1]
The lateral surface area of the cylinder is S_{c}, where
S_{h}/π = 2R^{2} [2]
The surface area of the cylinder base is S_{b}, where
S_{c}/π = 2Rh [3]
The above four quantities are integers by convention. From [1], [2] and [3], we have
S_{b}/π = R^{2} [4]
so R is rational. By [4], R^{2} is an integer, and it follows that R is an integer. The equability requirement is
which reduces to
The numerator and denominator must have the same sign, so
[5]
2 < R < 9/2In the latter case, h = 2/3 by [5] but S_{c}/π = 2Rh = 16/3 is not an integer. However, if R = 3, then h = 3 and 2Rh = 18. Therefore, the only equable hemispherical cylinder has a radius and height of 3, and volume and surface area equal to 45 π. It is a solid of revolution of a non-equable closed shape consisting of aR = 3 or R = 4
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