A hemispherical dome has a height of 21 meters. Find the total volume enclosed in the hemisphere?
598222
19396 cubic meters\n\nVolume of a sphere v=4/3*Pi*r^3\nthen divide by 2 to get volume of hemisphere.
1000601
The method to finnd the volume of sphereis computed by The mathematician, Archimedes.\nThe formula for find the volume =4/3 pi  r cube\nie, pi = 3.14\n     r is the radius of the sphere.\nFor eg, r=1.2 m\n v =4/3 pi*(1.2*1.2*1.2) \n    =4/3*3.14*1728\n     =7.2346 cu.m\n     1 cu.m = 1000 litres\nV   =7.2346*1000\n       =7234.6\n\nDon't u understand.
625137
Volume is L * W * H ... if the Volumen is 1000^3 and it's a cube, then L, W and H all = 1000cm\n\nDuh.
612367
Volume of cylinder = Area of the base X height = πr2h.
176440
2 meters.\n\nLet s be the length of the side. Since the glass top is a square, the area is s^2.\n\ns^2 = 4\ns = +/- square root of 4\ns = 2 or -2\n\nDrop the negative 2 because a side length can't be negative.\n\nBy the way, in the future, you may want to put math problems in the Science/Mathematics category to get better answers.
218464
If by area you mean "Surface area" then\nSA(sphere) = 4(pi)(r)^2\nand\nV(sphere) = 4(pi)(r)^3/3\n\nSince SA = 196 cm^2, then\n\n196cm^n = 4(pi)r^2\nr^2 = 49/pi\nr = 7/sqrt(pi) or approx. 3.949 cm\n\nfor the volume,\nV = 4(pi)(r)^3/3\nV = 4(pi)(7/pi^0.5)^3/3\nV = (4pi/3)(343/pi^1.5)\nV = (1372/(3pi^.5)) or approx. 258.02 cm^3\n\n^_^ Hope this helps ^_^
679609
Surface Area=sqrt{3}a^2\n    Volume =(1/12)sqrt(2)a^3\nhttp://en.wikipedia.org/wiki/Tetrahedron
877943
volume of cyllinder = area of circle times height\n\n= Pi x radius x radius  x height\n\nremember its just the same as finding the volume of any prism.\nfind the area of the shape of the end then multiply by its length!!
101953
There's a simple way to go about this. The answer is reasonably accurate too.\n\nNow, we know that the formula for the volume of a sphere is 4/3 times pi times the cube of earth's radius.\n\nGiven the value of pi as 3.14 and the earth's radius as 6400 km, and also assuming that the earth is a perfect sphere, we get the volume as 1.09806622 * 10^12 cubic kilometers.
1175686
This can be solved in several different ways.  One of course is by calculus, by developing a formula for the surface area of a "slice" of the remainder of the sphere and integrating it about a rotated axis, or something similar.  Another is by digging for the rather obscure formula for the "volume of the endcaps of a sphere" and subtracting two of those and the volume of a cylinder from the volume of the original sphere (and discovering that many of the terms will cancel out nicely).\n\nThe easiest method, however, is via the use of simple logic.  Unfortunately, this method also requires one more "given" in order for it to be valid, though this given is not an unreasonable one:  There must be a single, real answer (or more specifically, a unique answer on the real number line).  Such an assumption would be folly in a practical (real-world) context, since without some form of proof, one should never immediately assume that this is the case.  Perhaps there are multiple answers, or there is no answer at all?  But as luck would have it, there actually is one answer, and it may be derived logically due to its unique existence.  (*thumbs nose at any philosophers who may be reading*)\n\nSolution:\n\nLet R be the radius of the cylinder drilled into the sphere.  If and only if there is a single, real answer, then that answer would still be the same no matter the value of R... even if R=0.  Thus, the answer is the volume of a sphere of radius 3 (diameter 6), or 36*pi cubic units.\n\n(Yes, the volume is always the same, whether the original sphere were the size of a basketball, or the size of a planet.)\n\nIt's a surprising answer, along the lines of the more familiar puzzle: "If the Earth were a perfect sphere and a band were placed snugly about its equator, and then one more meter were added to the length of the band and it were stretched so that all points along it were equidistant from the surface of the Earth, how far above the Earth would the band rise?"  Many are surprised to find that the size of the Earth plays absolutely no part in the equation, and that the "rise" is exactly the same if the sphere were basketball-sized or, umm... Earth-sized (with a circumference greater than 40,000,000 meters).  It rises exactly the radius of a circle with a 1-meter circumference: 1 / (2*pi) meters.
192399
In the front?  You have to consider the whole cube to find volume.  Volume = base area * height  You could find the surface area of one side, but you'd need to know the side length.
549577
The formula for surface area of a sphere is \nA = 4*pi(r^2)\n64 pi = 4*pi(r^2)\nwhich means r^2 = 16 or r = 4\nVolume of a sphere is \nV = (4/3)*pi (r^3)\nV = (4/3)*pi (4^3)\nV = (4/3)*pi (64)\nV = (256/3)*pi
430210
Volume of a shphere  = 4/3 * pi * R^3\n\nwhere pi = 3.142 (approx)\nand R is the radius
174332
The surface area of a sphere is A=4*pi*r^2 where r is the radius of the sphere. The *volume* is (4/3)*pi*r^3.
492961
Yes 152 cm is correct.\n\nTo solve the problem factor 2002 = 2x7x11x13\n\nThe closer the box is to a perfect cube the lower the ratio of the surface area to the volume.\n\nThis occurs when the sides are  11, 13 and 2x7=14\n\nIn the cube, each edge appears 4 times and so the sum of the edges will be 4 times (11+13+14) = 4x38 = 152 cm
875857
find what?...the dimensions of the box with the most volume?\n\nArea = w^2 + 4wh = 1100 cm^2\nh = (1100 - w^2 ) / 4w\n\nVolume = h w^2 =\nV = w^2 * (1100 - w^2 ) / 4w \nV = w (1100 - w^2 ) / 4\nV = 275w - (w^3)/4\ndV/dw = 275 - (3/4)w^2\nlet dV/dw = 0\n(3/4)w^2 = 275\n\nw = 19.15 cm\nh = 9.57 cm
952353
The curved surface area is 2*pi*r*h\n\nor 2*3.142*27*12 sq cms\n\nand 2*3.142*17*35 sq mms\n\nyou can do the rest