Volume of this irregular object floating in deep space?

Question

A solid cube of radium is floating in deep space. Each edge of the cube is 1km in length. An astronaut is protected from its radiation if she remains at least 1km from the nearest spec of radium including the interior of the cube. What is the volume (in cubic km) of space that is forbidden to the astronaut?

We've concluded that:
-The shape can't be a sphere nor a cube
-The shape does not have vertices
-The corners are arced so that she remains exactly 1km from the radium
-The distance from arc to arc is 3km

We have some idea of what it looks like, but do not know how to calculate the volume. Any suggestions would be EXTREMELY helpful. :)

Answer 1

This is a very interesting question.

I believe that the region to be avoided is the union of a cube, cylinder and sphere.
The cubes will have a volume of 1 * 6
The sphere 1/6 pi * 8 = 4/3 pi(since there are 8 eights spheres one at each corner)
and the cylinder is pi/ 4 * 12 = 3pi

thus in total :6 + 4/3pi + 3pi

I am not sure of my answer though.
Note: I correct the sphere part of my answer, it should have been 8 * 1/8 th of spheres. Also, you say the radium is solid so the space inside cannot be occupied, else I would have added one to my answer.

Answer 2

The forbidden region is a 3x3x3 kilometer cube with all its edges filleted to a 1 kilometer radius.

To find the volume, break the shape into smaller simple shapes. If you dissect the shape on a grid with a 1 km spacing, you will end up with:
1 solid cube at the center
6 solid cubes at the faces
12 'edge' cubes with one edge rounded
8 'corner' cubes with three edges rounded

Obviously the volume of the center cube is
Vcenter = 1 km^3.

The 'face' cubes are also:
Vface = 1 km^3

The 'edge' cubes are really just a quarter of a cylinder. The volume of an 'edge' cube is:
Vedge = 1/4*(π*h*r^2)
Vedge = 1/4*(π*(1 km)*(1 km^2))
Vedge = π/4 km^3

The 'corner' cubes are really just an eight of a sphere. The volume of a 'corner' cube is:
Vcorner = 1/8*(4/3*π*r^3)
Vcorner = 1/8*(4/3*π*(1 km)^3)
Vcorner = π/6 km^3

So, the total volume is
Vtotal = 1*Vcenter + 6*Vface + 12*Vedge + 8*Vcorner
Vtotal = 1*(1 km^3) + 6*(1 km^3) + 12*(π/4 km^3) + 8*(π/6 km^3)
Vtotal = 1 km^3 + 6 km^3 + 3*π km^3 + 4/3*π km^3
Vtotal = 20.6136 km^3

Answer 3

The solid can be described as follows: it is a union of 7 congruent cubes (the initial one and other 6 built on its faces), 12 congruent cylindrical pieces (quarter of cylinder each), filling the spaces between every 2 of the 6 cubes above (they form 3 cylinders with radius 1 and height 1, volume π each) and 8 spherical octants, filling the remaining gaps, making a whole sphere with radius 1, volume 4π/3.

Assuming initial cube with edge length 1, the volume is (initial cube also taken into account, otherwise we'll have 6 instead of 7 below):
7 + 3π + 4π/3 = 7 + 13π/3

The shape is somewhat similar to one of the Archimedean solids - the Rombicuboctahedron (please follow the link below to see a picture):
http://en.wikipedia.org/wiki/Rhombicuboctahedron
but imagine spherical caps instead of the 8 triangles and 12 rounded cylindrical pieces instead of the 12 squares, sharing common side with the triangular faces, or, say it otherwise, imagine 8 curvilinear triangles (instead of equilateral on the picture), the sides of each are 3 arcs containing 90° in three 2 by 2 perpendicular planes, and 12 curvilinear rectangles instead of squares, sharing common sides with the above triangles.

Note: "The distance from arc to arc is 3km" is irrelevant.

Answer 4

The 6 flat sides each keep away 1 sq km of space.
The 8 vertices keep away a quarter sphere each of volume 1/3*pi giving 8/3 *pi
The 12 edges keep away each a quarter cylinder of volume pi/4 each.
Total gives 6+ 8/3 pi + 12*pi/4
=6 + 17/3*pi sq km