Centers of adajacent faves of a unit cube are joined to form a regular octahedron. What is the volume?
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Consider the unit cube to becentered on the point (0,0,0).  The centers of the faces are at:\n(+0.5,0,0) (-0.5,0,0) (0,+0.5,0) (0,-0.5,0) (0,0,+0.5) and (0,0,-0.5).  I trust that is obvious...\n\nNow, consider only the top half of this: it is a square pyramid; the base has sides SquareRoot(0.5*0.5+0.5*0.5), or sides SquareRoot(2)/2  Again, simple.  That gives the square forming the base of this pyramid an area of 1/2.  The pyramid has altitude 1/2, and the volume of the pyramid is (1/3)*base*height.  That comes to(1/3)*(1/2)*(1/2)= 1/12.\n\nNow, this is just the volume of the top half of our octahedron; the lower pyramid has exactly the same volume--again, I trust that is painfully obvious.  Therefore, the total volume of the octahedron is 1/12 + 1/12, or 1/6\n\nThe trick here is breaking the octahedron into two parts, whose volume is easily computed.  From there, the problem is comparatively straight forward!
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polyhedron (pŏl'ēhē'drən) , closed solid bounded by plane faces; each face of a polyhedron is a polygon. A cube is a polyhedron bounded by six polygons (in this case squares) meeting at right angles. Although regular polygons are possible for any number of sides, there are only five possible regular polyhedrons, having congruent faces, each a regular polygon and meeting at equal angles. The five regular polyhedrons are also known as the Platonic solids, although they were known to the Greeks before the time of Plato. They are the tetrahedron, bounded by four equilateral triangles; the hexahedron, or cube, bounded by six squares; the octahedron, bounded by eight equilateral triangles; the dodecahedron, bounded by twelve regular pentagons; and the icosahedron, bounded by twenty equilateral triangles. The 18th-century Swiss mathematician Leonhard Euler showed that for any simple polyhedron, i.e., a polyhedron containing no holes, the sum of the number of vertices V and the number of faces F is equal to the number of edges E plus 2, or V+F=E+2
840147
Volume = (4/3) pi * radius^3\n\nRadius = Diameter / 2\n\nCristi has the right answer. Different people will use a comma in place of a decimal. It all depends on how you were taught. That's why it seems like the answer was a few orders of magnitude off when it was actually exactly right.\n\nTherefore, the answer is 113.097 cubic feet.
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The volume of a pyramid is equal to one-third the base area multiplied by the height.\n\n\nImagine a step-pyramid made up of small square prisms with dimensions\na-by-a-by-b, arranged in square layers. If the layers are k-by-k for\nk = 1, 2, 3, ..., n, then the total volume of the step-pyramid is:\n\n   V = a^2*b*[1^2 + 2^2 + 3^2 + ... + n^2],\n     = a^2*b*n*(n + 1)*(2*n + 1)/6 (which can be proved by induction),\n     = a^2*b*n^3*(1/3 + 1/[2*n] + 1/[6*n^2]).\n\nNow, the height of the step-pyramid is h = b*n, and the area of the \nbase is:\n\n   B = a^2*n^2, \n\nso the volume is:\n\n   V = h*B*(1/3 + 1/[2*n] + 1/n^2).\n\nNow let n -> infinity, so the steps get smaller and smaller, and the\nvolume of the step-pyramid approaches the volume of the inscribed \npyramid. Then the right-hand side approaches h*B/3, as the volume V \napproaches the volume of the pyramid.\n\nThis same proof works for a pyramid with any polygonal base. For a \ncone, take a pyramid with a regular m-gon for a base, and let \nm -> infinity.\n\nThe base approaches a circle, and the pyramid approaches a cone, and \nthe formula V = h*B/3 still holds in the limit.\n\nThere is another proof, which cuts a rectangular solid into six pieces \nof equal volume, each of which is a triangular pyramid (or irregular\ntetrahedron). This proves that the volume of each is 1/6th the volume \nof the starting rectangular solid, and each has base of area 1/2 the \narea of one of the faces of the solid, and height equal to the \ndimension of the solid perpendicular to that face
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cube = a 3  \n\nrectangular prism = a b c  \n\nirregular prism = b h  \n\ncylinder = b h = pi r 2 h  \n\npyramid = (1/3) b h  \n\ncone = (1/3) b h = 1/3 pi r 2 h  \n\nsphere = (4/3) pi r 3  \n\nellipsoid = (4/3) pi r1 r2 r3
189752
sphere, cube, pyramid, cilinder
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Most crystals form three basic shapes: simple cubic, body-centered cubic, and face-centered cubic.  Simple cubic has molecules at each corner of a cube, body-centered cubic has molecules at each corner of a cube and one in the center, and face-centered cubic has molecules at each corner of a cube and one in the center of each face.
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Volume = (area of the base) x (height)
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291.13 L is the correct volume.\n\nCheck your algebra when you are solving for V1.
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6\nSee http://en.wikipedia.org/wiki/Tilings_of_regular_polygons
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Within your 3x3x3 cube, we have ten classes of rectangular prisms, including the two boundary cases of the whole cube and of the single cublet:\n1x1x1, 1x1x2, 1x2x2, 1x1x3, 1x2x3, 2x2x2, 2x2x3, 1x3x3, 2x3x3, 3x3x3.\n\nFor each prism class, we have to count all possible arrangements within the main cube. The contributes to this number are from translations, then from rotations (but only for non-symmetric prisms). If I'm not wrong, they are:\n1x1x1: 27\n1x1x2: 54\n1x2x2: 12\n1x1x3: 27\n1x2x3: 18\n2x2x2: 8\n1x3x3: 9\n2x2x3: 12\n2x3x3: 6\n3x3x3: 1\n\nTotal=174 prisms\n\nAs per your later addition, if we want to discard cubic prisms, we have: 174-27-8-1 = 138 non-cubic prisms.
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let "a" be the side of the cube the volume of the cube will be a^3\nso V = (-9kx^3)^3\nV = - 729 k^3 x^9
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There are different formulas for different shapes. This page has a list of them: http://www.math.com/tables/geometry/volumes.htm
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The volume of a sphere with radius r is (4/3)pi * r^3. Your radius is 7yds, so the volume is (4/3) pi (7yd)^3 = (4/3)(343)(pi) yd^3 = 457 1/3 (pi) yd^3
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1.) The cylinder with the lowest surface area per unit volume will be cubelike. \n\nSurface area for a cylinder is 2πr^2 + 2πrh, where the 2πr^2 is for the base, and 2πrh is for the tube itself.\n\nOn a similar cube, the base would be 2 sides (top and bottom), and the tube would be remaining 4 sides. Thus, for a cubelike cylinder, 2πrh will equal 2 * 2πr^2 (since the tube accounts for twice the area as the base).\n\n2πrh = 2 * 2πr^2\nh = 2r (h/r = 2 for your ratio)\n\nV = πr^2h = 2πr^3\n1000 = 2πr^3\n500 / π = r^3 = 159.15\nr = 5.419 cm, so h = 10.838 cm\n\nSA = 2πr^2 + 2πrh = 2πr^2 + 4πr^2 = 6πr^2\n6πr^2 = 553.52 cm^2 \n\nCheck:\nr = 5.418\nV = πr^2h = 1000\nh = 1000 / π * (5.418 ^2) = 10.844\nSA = 2πr^2 + 2πrh = 2π (5.418 ^2) + 2π (5.418 * 10.844) = 553.60\n\nr = 5.42\nV = πr^2h = 1000\nh = 1000 / π * (5.42 ^2) = 10.836\nSA = 2πr^2 + 2πrh = 2π (5.42 ^2) + 2π (5.42 * 10.836) = 553.56\n\nWithout the diagram, it'll be hard to answer 2 and 3.
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An octahedron is a regular solid with eight identical faces. Octahedral coordination of a metal occurs when six bonded atoms of coordinating ligands occupy sites that would correspond to the six vertices (corners) of a regular octahedron surrounding the metal atom.
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All diagonals of a perfect cuboid are integers\n\nAssume the cuboid has height h, width w and depth d and h>w>d.\n\nThe lengths of the face diagonals are:\n\n    * fd12 = h2 + w2\n    * fd22 = h2 + d2\n    * fd32 = w2 + d2\n\nThe length of the space diagonal is:\n\n    * sd2 = h2 + w2 + d2\n\nIf h,w and d are all even, then every fd and the sd are divisible by 2, and so we can divide h,w and d by 2.\n\nThe sum of 2 odd integers squared is never a square, and so the only possiblity is for exactly one of h,w,d to be odd.\n\nIn turn, this implies that two of the fdi are odd, the other even. Consider fd1 to be even.\n\nThe aim of this proof is to demonstrate that any such fd1 must contain an infinite power of 2 as a factor.\n\nIf we expand the fd equations, using a,b,c as parameters (h=2a, w=2b, d=2c+1), then we arrive at:\n\n4[a2 + b2 + c2 + c] + 1 = sd2\n\na and b must be of the same parity, as 4k+1 is only square for even k.\n\nHowever a2 + b2 = fd12/4\n\nAnd so fd12/4 must be even, i.e. h=4a, w=4b.\n\nIf we repeat the process with the new parameterization, then we arrive at:\n\n4[8a2 + 8b2 + c2 + c] + 1 = sd2\n\nBut now fd12/16 must be even, i.e. h=8a, w=8b.\n\nThis is obviously absurd.