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 Cylindrical Cuts (Posted on 2009-01-22)
 An object resembling that in the graphic, is carved from a cube of soft material such that its elevations, as seen from the top, front and side, are all that of the drawing at the top. Note that each curve is a segment of a quadrant circle centred on the opposite corner and having the radius of the side of the square. While the object may look like a stellated octahedron, each curved stellated extension is actually formed by six surfaces being parts of cylindrical forms (thus there are 48 actual surfaces which form the object). What percentage of the cube is wasted to create this object? "Impossible" Solid's Volume was a somewhat related problem.

 See The Solution Submitted by brianjn No Rating

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 Exact Surface Area Comment 5 of 5 |

After four or five failed attempts at calculating the surfaced area using double integrals and surface formulas, I found a successful method, using single variable calculus.  I thought about adding this as a problem in the queue (definitely D5), but I am probably the only person who would solve it anyways.

The surface of the object consists of 48 congruent segments.  With the cube at lattice points (0,0,0)-(1,1,1), one segment can be defined by the cylinder y^2+(1-z)^2=1 and the planes y=1/2, x=1/2, x=z, and x=y.

`             A(,B,C)    A    B            C+------------*----+     +----*-----------_+|             *   |     |   *           / ||              *  |     |  *          _/  ||              *  |  y  |  *         /    ||               * |     | *        _/     ||              _P |  a  | *       /       ||           __/  *|  x  |*      _/        ||         _/     *|  i  |*     /          ||      __/       *|  s  |*   _/           ||    _/           *     *   /             || __/             *     * _/              ||/  T(angle)      *     */                |R-----------------*     *-----------------+              D(,E,F)   D    E            F      z axis                  x axis`
`(Sorry about the crude graphics)`
`Point A is at (0,           1/2, 1-sqrt[3]/2)Point B is at (1-sqrt[3]/2, 1/2, 1-sqrt[3]/2)Point C is at (1/2,         1/2, 1-sqrt[3]/2)`
`Point D is at (0,           0,   0) [The origin]Point E is at (1-sqrt[3]/2, 0,   0)Point F is at (1/2,         0,   0)`

Point R is the cylinder's axis, (x,0,1)
Point P is a generic point on the cyninder (x,y,z).

ACDF is a segment of the cylinder and BCD is the segment whose area is in question.

BD is the intersection of plane x=z and the cylinder y^2+(1-z)^2=1.  You can also verify it is the intersection of the cylinders y^2+(1-z)^2=1 and y^2+(1-x)^2=1.

CD is the intersection of the cylinder y^2+(1-z)^2=1 and plane x=y.

In a yz cross section the length of an arc, such as arc PD, equals the measure of angle PRD in radians.  Then arc PD = angle T = arcsin(y) or arccos(1-z) or arctan(y/(1-z)).  An integrable function can be formed by expressing y or z as a function of x.

Expressing curve BD in terms of x: (x, sqrt[2x-x^2], x)
Curve CD: (x, x, 1-sqrt[1-x^2])

The area of BDC equals area(BDE) + area(BCEF) - area (CDF)

The arc function for BCEF is a constant pi/6 radians, which makes the area of BCEF equal
pi(sqrt[3]-1)/12

I will express the other arc functions in terms of arcsin(y).

The arc function for CDF is arcsin(x) over the range [0, 1/2], which makes integral:
Integ{0,1/2}[arcsin(x)]dx

The substitution x=sin t changes the integral to
Integ{0,pi/6}[t*cos(t)]dt

Then using integration by parts u=t and dv=cos(t) dt, the integral can be evaluated as
pi/12 + sqrt(3)/2 - 1

The arc function for BDE is arcsin(sqrt[2x-x^2]) over the range [0, 1-sqrt[3]/2], which makes integral:
Integ{0, 1-sqrt[3]/2}[arcsin(sqrt[2x-x^2])]dx

The substitution sin t = sqrt[2x-x^2] successfully simplifies the integral to:
Integ{0,pi/6}[t*sin(t)]dt

Then using integration by parts u=t and dv=sin(t) dt, the integral can be evaluated as
1/2 - pi*sqrt[3]/12

Now the area of the segment can be expressed as
(1/2 - pi*sqrt[3]/12) + (pi(sqrt[3]-1)/12) - (pi/12 + sqrt(3)/2 - 1)
= 3/2 - pi/6 - sqrt[3]/2

Since there are 48 segments, the total surface area (assuming a 1x1x1 cube) is
72 - 8*pi - 24*sqrt[3]

That surface area is 88.3 percent of the cube's surface area.

 Posted by Brian Smith on 2009-02-12 13:10:23

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