What's the formula for area of a graduated cylinder?  Please help me.?
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I am assuming you mean something like a round flower pot, where the base is smaller than the top.  You are saying the bottom has a diameter of 17 inches and the top has a diameter of 34 inches, with a height of 34 inches (not the slant).\n\nIs this correct?  I've also heard this called a truncated cone.  The formulas for volume is given below:\n\nHeight: h\nRadius of bases: r, R\nVolume: V\n\nV = Pi(R^2+rR+r^2)h/3\n\nSo plugging in your values (don't forget to cut the diameters in half to get radii):\n\nh = 34\nr = 8.5\nR = 17\n\nV = Pi(17^2+(8.5)(17)+8.5^2) 34 / 3\nV = Pi(289 + 144.5 + 72.25) 34/3\nV = Pi(505.75)34/3\nV = Pi(505.74)(10.333)\nV = Pi(5731.833)\nV = 18007 cubic inches
667893
Pi*r^2*h in the first answer is the volume of a cylinder and is rather wrong!\n\nthe surface area of a cone is pi*r*l where l is the slant height of the cone. you need to calculate this from the height and the radius (not the diamteter) by pythagoras.\n\ntaking l=5.707506\nsurface area is 27.344231865 sq metres\nconvert units to cm^2 =273442.31865 sq cm
856302
Surface Area = (2 •  p   • r²) + (2 • p  • r • height)\n\nWhere (2 • p • r² ) is the surface area of the "ends" and\n(2 • p • r • height) is the lateral area (the area of the "side").\n\nVolume   =   p • r² • height   =   ¼ • p • d² • height\n\np = Pi
508516
Surface Area Formulas\nIn general, the surface area is the sum of all the areas of all the shapes that cover the surface of the object.\n\nCube | Rectangular Prism | Prism | Sphere | Cylinder | Units\n\n\nNote: "ab" means "a" multiplied by "b". "a2" means "a squared", which is the same as "a" times "a".\nBe careful!! Units count. Use the same units for all measurements. Examples\n  Surface Area of a Cube = 6 a 2  \n\n\n (a is the length of the side of each edge of the cube) \n\nIn words, the surface area of a cube is the area of the six squares that cover it. The area of one of them is a*a, or a 2 . Since these are all the same, you can multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared. \n\n \n\nSurface Area of a Rectangular Prism = 2ab + 2bc + 2ac  \n\n (a, b, and c are the lengths of the 3 sides) \n\nIn words, the surface area of a rectangular prism is the are of the six rectangles that cover it. But we don't have to figure out all six because we know that the top and bottom are the same, the front and back are the same, and the left and right sides are the same. \n\nThe area of the top and bottom (side lengths a and c) = a*c. Since there are two of them, you get 2ac. The front and back have side lengths of b and c. The area of one of them is b*c, and there are two of them, so the surface area of those two is 2bc. The left and right side have side lengths of a and b, so the surface area of one of them is a*b. Again, there are two of them, so their combined surface area is 2ab. \n\n  Surface Area of Any Prism \n\n\n  (b is the shape of the ends) \n\nSurface Area = Lateral area + Area of two ends \n\n(Lateral area) = (perimeter of shape b) * L \n\nSurface Area = (perimeter of shape b) * L+ 2*(Area of shape b) \n\n  Surface Area of a Sphere = 4 pi r 2  \n\n\n (r is radius of circle) \n\n  Surface Area of a Cylinder = 2 pi r 2 + 2 pi r h \n\n\n (h is the height of the cylinder, r is the radius of the top) \n\nSurface Area = Areas of top and bottom +Area of the side \n\nSurface Area = 2(Area of top) + (perimeter of top)* height \n\nSurface Area = 2(pi r 2) + (2 pi r)* h \n\nIn words, the easiest way is to think of a can. The surface area is the areas of all the parts needed to cover the can. That's the top, the bottom, and the paper label that wraps around the middle. \n\nYou can find the area of the top (or the bottom). That's the formula for area of a circle (pi r2). Since there is both a top and a bottom, that gets multiplied by two. \n\nThe side is like the label of the can. If you peel it off and lay it flat it will be a rectangle. The area of a rectangle is the product of the two sides. One side is the height of the can, the other side is the perimeter of the circle, since the label wraps once around the can. So the area of the rectangle is (2 pi r)* h. \n\nAdd those two parts together and you have the formula for the surface area of a cylinder. \n\nSurface Area = 2(pi r 2) + (2 pi r)* h \n\n  \n\n\nTip! Don't forget the units.  \n\n\nThese equations will give you correct answers if you keep the units straight. For example - to find the surface area of a cube with sides of 5 inches, the equation is: \n\nSurface Area = 6*(5 inches)2 \n\n= 6*(25 square inches) \n\n= 150 sq. inches
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cylinder\n1.volume=pir^2h\n2.LSA=2pirh\n3.TSA=2pir(h+r)\n\nsphere\n1.volume=(4/3)pir^3\n2.area=4pir^2\n\nhemisphere\n1.volume=(2/3)pir^3\n2.basearea=pir^2\n3.LSA=2pir^2\n4.TSA=3pir^2\n\ncone\n1.volume=(1/3)pir^2h\n2.LSA=pirl\n3.TSA=pir(l+r)\n\ncube\n1.volume=a^3\n2.surface area=6a^2\n\ncuboid\n1.volume=lbh\n2.surface area=2(lb+lh+bh)
220777
use the following formula .. \n\ntotatal surface area = \n[(H/D)*4(pie)(0.5D*0.5D)] + [ (H-t)/(D-2t)*4(pie)*(0.5D-t)(0.5D-t)] + [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]\n\nthis is considering D = tank diametere is from outer surface and not the cavity diameter \n\n\nouter surface area of cover = (H/D)*4*(pie)(0.5D*0.5D)\ninner surface area of cover = [(H-t)/(D-2t)*4*(pie)*(0.5D-t)(0.5D-t)] \n\nthe base line are = [(pie)(0.5D*0.5D)-(pie)(0.5D-t)(0.5D-t)]
919521
Imagine the center of the triangle. Draw a line segment from it to a vertex, and then one from the center to a midpoint of a side. This makes a 30-60-90 triangle, so the line segment is (p/2) sqrt(3). This is a radius of the cylinder, so we multiply it by 2 to get the diameter.\n\nd = p sqrt(3)\n\nor, as the question wants it\np = d/sqrt(3) = d sqrt(3)/3
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The volume of the sphere is 4/3*pi*r^3\n\nThe volume of the cylinder is pi*r^2*h\n\nSince the cylinder's h is the same as 2r, then the volume becomes pi*r^2*2r which is the same as 2*pi*r^3.\n\nThe fraction of the space taken up by the sphere is the same as:\n\n4/3*pi*r^3\n-------------\n2*pi*r^3\n\nwhich is the same as:\n\n4/3\n----\n2\n\nwhich reduces down to 2/3
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Area A = area of two ends  +  Curved surfaces.\n  A  =  2π r²  +  2π r h\n\nVolume V = Area of end  x  Heigh.\n  V  = π r²  *  h\n  V  = π r² h\n  h  =  V / π r²\n Now substitute for h into the area formula.\n\n  A  =  2π r²  +  2π r h\n  A  =  2π r²  +  2π r (V / π r²)\n  A  =  2π r²  +  2V / r\nThe volume of the can is 1000cm^3, so substitue in.\n  A  =  2π r²  +  2πV / r\n  A  =  2π r²  +  2000 / r\n\nTake first derivide and let it equat to zero (zero for max. or min. pt.).\n  A  =  2π r²  +  2000 / r\n  δA/δr  =  4π r  -  2000 / r²\n  4π r  -  2000 / r² = 0\n  4π r  =  2000 / r²\n  4π r^3  =  2000\n  r^3  =  2000/ 4π\n  r  =  (2000/ 4π)^1/3    (that is, cubed root).\n  r  =  5∙419 260 701\n  r  ≈  5∙42 cm.\n\nLet's use the r value to calculate h.\n  V  = π r²  *  h\n  1000  = π (5∙419 260 701)²  *  h\n  h  =  1000 /π (5∙419 260 701)²\n  h  =  10∙838 5214\n  h  ≈  10 51 cm.\n\nFor ratio:\n h/r = 10∙838 5214 / 5∙419 260 701\n h/r = 2/1   That is, a two to one ratio.\n\nFor queation two, follow the same proceedure after you set up your formulas.\nLike wise for question three.\n\nWith question two it's not clear what value you are taken for the length of the side of the square edge.  Because you are still using r, I'm assuming you are taken the largest size of a square that fits into the circle or radius r.  Based on this assumsion, the calculations are are follows:\n\nTo identify the length x of one side of the square of radius r:\nDraw two diameters of the circle at 90º to each other.  Where these diameters make contact with the circumference of the circle will be the corner points of the square.  The area of this square now consists of four triangles.  These triangles will have two sides of length r with the enclosed angle of 90º.  So the length of x will be:\n\n x² = r² + r²\n x² = 2 r²\n x = √(2 r²)   That is, length of one side of the square.\n\nArea A = area of two square ends + parameter of square * heigh.\nA = (2r² * 2) + [(√2)r * 4 * h]\nA =  4r²  +  4√(2)r h\n\nVolume V = square area x heigh.\nV =  2 r² * h\nV =  2 r² h\nh = V / 2 r²\nNow substitute for h into the area formula.\n\nA =  4r²  +  4√(2)r h\nA =  4r²  +  4√(2)r V / 2 r²\nA =  4r²  +  2√(2) V/r\nA =  4r²  +  2000√(2) r^-1     (Volume is 1000cm^3).\n\nTake first derivide and let it equat to zero (zero for max. or min. pt.).\nA =  4r²  +  2000√(2) r^-1\nδA/δr = 8r  -  2000√(2) r^-2\n 8r  -  2000√(2) r^-2  =  0\n 8r  =  2000√(2) r^-2\n 8r  =  [2000√(2)] /r²\n 8r^3  =  2000√(2)\n r^3  =  [2000√(2)]/ 8\n r  =  ([2000√(2)]/ 8)^1/3             (that is, cubed root).\n r  =  (353∙55 33 906...)^1/3\n r  =  7∙071 0678 12\n r  ≈  7∙07 cm.\n\nLet's use the r value to calculate h.\nV = 2r² * h\n1000 = 2r² * h\n1000 = 2 (50) * h\n1000 = 100h\n h = 10 cm.\n\nFor ratio:\nh/r = 10 / 7∙071 0678 12\nh/r = 1∙845 270 149 / 1\n h : r  ≈ 1∙85 : 1\nThese ratios do not take into account the scrap metal from the circle when the square is cut out of it.
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C=2(Pi)r or diameter times Pi\nI don't know how to solve the problem cause not enough info is given. Does it say what C equals?\n\nv = lwh . H stands for the height. Does it say what v equals?\n\na= 1/2h(b1+b2) b1 is base 1. Does it say waht the area is?\n\nProvide more info please. :)
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Think of the first as the area of a flat round plate sweeping out the inside of the figure you get from the rotation.  What will be its area? πr² where the r is the y value of √(9 + X²).  That changing area will sweep out the volume from whatever the limits are (you don't say), so it's the integral of π(√(9 + X²))², or π(9 + X²).\n\nIn the second case, the thing sweeping out the volume is the surface area of a cylinder as its radius grows.  Thats the circumfrence of the base times the height.  The circumfrence is 2πr where r is the X value and the height is the difference between Y = x and Y = X/2.  So it's the integral from 0 to 2 of:\n\n   2πX(X-X/2).\n\nThinking of an area sweeping thru the volume helps.
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graduated cylinder or callibrated pipette are the answers to the test questions in my classroom.  And in scientific situations, volume is measured in liters/litres- except when measured in cc's (cubic centimenters)
224253
You know that the area of a circle is given by:\nA = (pi)(r²)\n\nYou also know that the radius is half the diameter, so\nr = d/2\n\nYou may substitute this to the area formula:\nA = pi * (d/2)²\nA = pi * d²/4\nA = (pi)(d²)/4\n\n^_^
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Example: Find the length of a helix which winds three times around a cylinder of height 12" and radius 3"\n\nAnswer: divide the cylinder into three sections. The helix goes once around a cylinder of height 4" and radius 3"\n\nThe surface of the cylinder can be "peeled" into a rectangle 4" by 2 * pi * 3" = 18.850". The helix is a diagonal of that rectangle. So its length is sqt(4^2 + 18.850^2) = 19.27"\n\nIt goes around the cylinder three times and its total length is therefore 57.81"
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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
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Take the formula for the surface area(S) and solve for r\nS=4*pi*r^2\nr^2 = S/(4pi)\nr =squareroot(S/4pi) or (S/4pi)^(1/2)\n\nNow take the formula for the volume\n4/3*pi*r^3\nand substitute the formula for r into the equation\n4/3*pi*(squareroot(S/4pi))^3\n4/3*pi(S/4pi)^(3/2)
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Volume, V = 2πrh...\n\nSince there is only one equation, but two unknowns...\n\nyou must supply the value of one unknown (you must decide on a pre-determined value. You either decide the value for the radius of the cylinder or its height) to solve for the other dimension.