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|Spherical shelter 3 (Posted on 2012-05-27)
A unit sphere is on a level surface in the rain. What is the volume of the largest convex solid that be placed under the sphere and not get rained on?
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Submitted by Jer
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For the convex shape to be maximum, Im supposing that it is bounded by a cylindrical surface (the outer edge of the dry space), the base plane and a tangent plane to the sphere, which cuts the base circle in a chord, with mid-point M say.
Let A be the acute angle which this plane makes with the horizontal, and let t be the perpendicular distance, MC, from the chord to the centre, C, of the base circle.
To find the volume of the shape, note that a typical horizontal section at height y above the base, will always be a segment of a circle, radius 1, with a chord at a distance x (say) from the vertical axis through the centre, O, of the sphere,
where dy/dx = tan(A).
Area of this segment = area sector area triangle
= arccos(x) - (1/2)*x*sqrt(1 x2)
Volume of shape = Integral(x=t..1) of [arccos(x) (1/2)*x*sqrt(1 x2)] dy
= Integral(x=t..1) of [arccos(x) (1/2)*x*sqrt(1 x2)]tan(A) dx
= tan(A)[x*arccos(x) sqrt(1 x2) + (1/6)(1 x2)3/2] between t & 1
V = (1/6)tan(A)[(t2 + 5)sqrt(1 t2) 6t*arccos(t)] (1)
If the plane touches the sphere at B, angle BOC = A, and OM bisects this angle
so that tan(A/2) = |CF|/|OC| = t and therefore tan(A) = 2t/(1 t2)
(1) now becomes V = (t/3)*[(t2 + 5)sqrt(1 t2) 6t*arccos(t)]/(1 t2)
Differentiating and simplifying eventually gives:
dV/dt = (1/3)[(5 2t4 +9t2)sqrt(1 t2) 12t*arccos(t)]/(1 t2)2
dV/dt = 0 then gives (5 2t4 +9t2)sqrt(1 t2) = 12t*arccos(t) (2)
Apart from the root t = 1 (V indeterminate), equation (2) cant be solved
explicitly for t. But the single root between 0 and 1 is shown numerically
to be t = 0.46506..., which gives a maximum volume of 0.3134..
Posted by Harry
on 2012-06-02 22:24:01
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