How do I find the volume of air that surrounds 3 balls each w/a radius of 5 cm in a cylindrical can?
147975
First, find the volume of the can: [pi]r-squared x the height of the can. If the 3 balls in it each have a radius of 5 cm, then they are each 10 cm tall, and the can is 30 cm tall. So the volume of the can = [pi]x25x30 = 750[pi] cubic cm.\n\nNext, the volume of each ball is 4/3 x [pi]r-cubed, or 166.67[pi] cubic cm. Three of them would be 500[pi] cubic cm.\n\nLast, subtract the volume of the balls from the volume of the cylinder to get 250[pi] cubic cm, or about 785.4 cubic cm.
856026
Use the gas law\nPV=nRT
819179
1)  the answer is h/r =2 or h = 2r\n\nThis is a classic optimization problem. You want the minimum surface area(SA) for a cylindrical shaped tin for a fixed Volume(V). This ratio of h/r that minimizes SA is independent of the volume though of course h and r are not independent of the volume.\n\n(1) SA = 2pi r^2 + 2pi r h\n\nV = pi r^2 h (for simplicity set V = 1)\n\nso (2) h = 1/(pi r^2)\n\nSubstituting this value for h into (1) gives:\n\nSA = 2pi r^2 + 2/r\n\nd/dr SA = 4pi r - 2/r^2\n\nd/dr SA = 0 when 2/r^2 = 4pi r\n\nthat is: (3) r = 1/(2 pi r^2)\n\nTo get the value of the ratio h/r divide equation (2) by equation (3), which yields a value of 2 to 1.\n\nYou are given the volume of the tins as 1 liter\n\n1000 = pi r^2 h,  but h=2r\n\n1000= 2 pi r^3 so r = (1000/2pi)^1/3 = 12.62 cm approx\nand h= 25.24 cm\n\n2) The answer is h/r = 8/pi\n\nIn this case your you have waste metal.\n\n2r is the diameter of the top of the tin. If the top and bottom are cut out of the square then the square will have to have an area of (2r)^2 = 4r^2. So the area A of material required for the tin is now: \n\n(1)   A = 2(4 r^2) +  2 pi r h  since the top and bottom are cut from squares.The procedure is the same in determining h/r, just a different function for the area of aluminum required\n\nAgain use V = 1 and so h = 1/(pi r^2)\n\nSubstituting this value of h in the above equation gives:\n\nA = 8r^2 + 2/r \n\ndA /dr = 16r -2/r^2\n\ndA/dr =0 when 16r = 2/r^2\n\nr^3 = 1/8 so r = 1/2 and h = 1/(pi r^2) = 4/pi\n\nSo h/r = 8/pi\n\nNow that you have the ratio h/r, you can determine the values of h and r when the volume of the can = 1000 cm^3\n\nNotice the value of h/r in this case is a little higher than in the no aluminum waste case. Of course, that makes sense\n\n3) This is also true for the equilateral triangle and hexagon\nYou need to determine the area of the equilateral triangle and hexagon having a maximum inscribed circle of radius r. The inscribed cirlcle will touch the midpoints of the edges of the regular polygon. We saw for the square this was 4 r^2. The general formula for the area of a regular polygon having N sides, in terms of r (its inradius, i.e. apothem) is:\n\nArea = r^2 x N x tan(pi / N)\n\nFor equilateral triangle N =3. For a hexagon, N=6. Plug these values in this formula and it will give you the area in terms of r When you calculate it, make sure your calculator is using radian measure, not degrees. So the area of aluminum required for the for the can if the top and bottom are cut from a regular polygon having N sides is\n\nA = 2 r^2 x N x tan(pi/N) + 2/r\n\nDetermine d A/dr. Set it equal to 0 and solve for r, and then solve for h.\n\nHere's a link about the Area of regular polygons.\n\nhttp://www.mathopenref.com/polygonregulararea.html
728936
1. obtain formulas for surface area and volume of cylinder\n2. let volume equal 1000 and solve this volume equation for h\n3. substitute the expression from previous step into surface area    equation to get equation with only r's\n4. differentiate surface area formula with respect to r set this equal to 0 and solve for r i got   r=10/3rd root(pi)\n5. substitute value for r into volume equation solve for h  i got h=10/(pi)^(2/3)\n\nincidentally, i used 2nd derivative test to conclude r is minimum\n\n\n\ni'll get back to u with more
482843
Boyle's law\n\nP1xV1=P2xV2\n\n(.75 atm)x(2 L) = (2 atm)x(X)\n\n1.5=2x\n\ndivide both sides by 2\n\n.75 L would be the volume of helium if the container was at a pressure of 2 atm.
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.
501128
Inside or outside volume?\n\nInside:  Can't use Fluid => Gas or liquid.  Use solid like sand to fill the vase.  Takeout the sand and measure the volume either by weight/density or use another standard measuring device.\n\nOutside: Trace the curve, find out how many continuous sections are there that are connected at discontinuous boundaries, find volumes of each (use cylinder or cone volume formula) and add up the volumes.
760290
Popcorn ball. . . . \n\nGoogle search:  volume sphere (should have used "volume sphere," but it turned out not to matter)\n\nhttp://www.aaamath.com/geo79-volume-sphere.html\n\n4/3 (pi) r^3\n\n4/3 times pi times (6 to the 3rd power) \n\nequals 4/3 times pi times 216. . . and work from there.\n\nGoogle search:  volume cylinder\n\nhttp://www.321know.com/exp79_x5.htm\n\n(area of the base) * height =(pi r squared times height)\n\nPi times (16 squared) times 275. . . work from there.
29575
well, the formula for the volume of a cone is 1/3 * base * height.\n\nthe trick to doing this is to find the volume of the large cone, then subtract the volume of the small cone.  what's leftover will be the tumbler-shaped volume.  plug in numbers for the height and base for both the large and small cones, and then subtract. voila!
329292
V=(4/3)*pi*r^3\n\nwhere r is the radius of the sphere and V is volume.
852456
Total surface area of a tin can = A = 2pr^2 + 2prh\nVolume of a tin can = V = pr^2h = 1000 cm^3\nTherefore, h = 1000 / pr^2\n\nSubstituting this into the area equation gives :\nA =  2pr^2 + 2pr * 1000 / pr^2\n= 2pr^2 + 2000 / r\n\nNow take the derivative of A with respect to r :\ndA / dr = 4pr - 2000 / r^2\n\nSet this equal to zero to find the minimum :\n4pr - 2000 / r^2 = 0\nThis can be rearranged to show that : 1000 / pr^2 = 2r, which equals h, as shown above.\nContinuing to find r, we get : r = (500 / p) ^ (1/3), or approx. 5.42 cm.\n\nNow h = 2r = 2 * (500 / p) ^ (1/3), or approx. 10.84 cm.\n\nThus, ratio is h / r = 2r / r = 2\n\nI'm a bit confused about the second question, but I think what you want is a square sheet of aluminium, such that all 3 pieces of a can will fit, with the minimum of wastage. This would be a square sheet with the dimensions : (perimeter of can) by (height + 2 * radius).\n\nThus, Perimeter = Height + 2 * Radius, because the sides of a square are equal.\nTherefore, 2*pi*r = h + 2r,\nso, h = 2r(pi - 1)\n\nBut, V = pi*r^2*h = 1000, so h = 1000 / (pi*r^2)\n\nEquating the 2 values for h gives :\n2r(pi - 1) = 1000 / (pi*r^2)\nfrom which we get : r  = {500 / [pi(pi - 1)]} ^ (1/3)\n\nSo now we know that h = 2r(pi - 1), so that after substituting for r, we get : \nh = 2 * (pi - 1) ^ (2/3) * (500 / pi) ^ (1/3)\nAnd we find that the ratio h / r = 2r(pi - 1) / r = 2(pi - 1)\n\nWell, that was a bit of speculation, but even if I've got it wrong, then perhaps you can adapt it to your needs.\n\nEdit\nCan't help feeling I've misunderstood the second question the first time around, but on re-reading it many times, I still don't understand it.
471793
Well here's some extra info you need to solve the problem:\n\n1 cubic inch = 0.00432900431 US gallons\n\nSo one can has a volume of:\n\nV=(pi)(5)(3/2)^2=35.34291 cubic inches\n\nThey filled 86400 cans which has a volume of:\n\nVtotal = V*86400 = 35.34291*86400 = 3053627.424 cubic inches\n\nThen you need to convert this to gallons so:\n\nVgallons = 0.00432900431 * 3053627.424 = 13219.1662 Gallons
540563
weigh the graduated cylinder full with water.\ninvert the cylinder (still full) in a tub of water.\nblow bubbles into the cyliner to find the volume of the air you blew in.\nremove and reweigh the cyliner.\n\nit may be a little tricky, but it's not too bad.
796040
material needed=2pir(h+r)\nvolume=pir^2h\npir^2h=1000cc (given)\nh=1000/(pir^2)\nsubstituing h in the equation 2pir(h+r)\nmaterial=2pir(1000/pir^2+r)\n=2pir*1000/pir^2+2pir^2\ndA/dt=-2000/r^2+4pir\nsetting this to zero\n-2000+4pir^3=0\n4pir^3=2000\nr^3=2000/4pi\nr=cube root of 500/3.14\n=5.4 approx\nsubstituting\nh=1000/(3.14*5.4*5.4)\n=10.9 aapprox\nso the radius is 5.4 cm approx\nand the height 10.9 approx
350090
Use the formula PV=rnT\n\np= pressure in atm\nV= volume in L or dm^3\nr=number of moles\nn= constant: 0.0821 atm x L/ moles x K (This constant is used to balance out units)\nt= temperature in K\n\nSo for your problem:\n2.3 atm x 120 L = r x (0.0821 atm x L/ moles x K) x 340 K\n\nthe constant should allow you to be left with moles as your unit\nusing algebra, r is left, which also stands for the number of moles (meaning the problem coordinates with your question)\n\nusing this setup, i got 9.89 moles
864957
diameter is 70 means radius=35\nlenth=130\nso if it is the volume4 u want\n\nitssimply 2*3.14*35*130\n\nif not so question is in complete
587954
Measurements (In distance I'm assuming) would give you volume.\n\nMass / Volume = Density\n\nIf you have volume & density, you can rewrite the equation:\n\nMass = Volume * Density