find the volume of sphere whose surface area is 64 pi cm2?
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
208438
There is a delightful article by the late Dr. John Kenny on that, back a few years ago, looking at the formulae for surface area and volumes of spheres; I'll see if I can find it for you--it's in my files SOMEWHERE--and give you the reference, and the formulae he generated. It turns out that they're not precisely the same as what you've got--but not terribly far from it.\n\nAs I said, I'll see if I can find it, and I'll get you the reference and the formulae.\n\nAbout 9pm CST...\n\nOK, here is the formulae, done the best I can on this thing...\n\n\nFor a sphere of dimension 2*k, radius R\n\nVolume: (pi^k)*(R^(2*k))/k!\n\nFor a sphere of dimension 2*k + 1 and radius R\n\nVolume:\n(2^(2*k+1))*k!*(pi^k)* (R^(2*k+1))/(2*k+1)!\n\n\nTo get the surface areas for each of these, all you do is differentiate with respect to R...\n\nAs I said, it was a little more complex than you thought--the surface areas come out a little bit differently for odd and even numbered dimensions.  I do not recall why, frankly.
885732
well...lets call your known volume V.  so...what kind of shape are we dealing with...eh, lets claim we are dealing with a sphere...we know the volume of the sphere is determined by v=(4/3)*pi*r^3 so we could solve for r (the radius).\n\nso....\n\nV=(4/3)*pi*r^3\nr=((3V)/(4pi))^(1/3)\n\nwe can now substitute that r into our surface area of a sphere equation....which is SA=4*pi*r^2, so we would have\n\nSA=4*pi*((3V)/(4pi))^(2/3)\n\na similar series of actions would occur for any other shape\n\nhope this helps,\n\nMatttlocke
288086
I assume this is a question from a calculus class.  The only way I know of solving this is using derivatives.\n\nsurface area = 2Pi*r^2 + 2Pi*r*h\n\nV = Pi*(r^2)*h\nV = 400Pi\nThus, 400Pi = Pi*(r^2)*h  Solve this equation for h by dividing both sides by Pi*r^2\n400PI/Pi*r^2 = h  Simplifying will give you\nh = 400/r^2\nNow replace h with 400/r^2 in the equation \nsurface area = 2Pi*r^2 + 2Pi*r*h\nLet S stand for surface area\nS = 2Pi*r^2 + 2Pi*r*(400/r^2)  Simplify\nS = 2Pi*r^2 + 800Pi/r\nNow take the derivative.\nS' = 4Pi*r - 800Pi/r^2\nSet this equal to zero and solve for r.\nI get that r = the cube root of 200 which is 5.84803547643\nNow you should test to see whether this is a max or min.  It should be a min; although, I did not check it.\nAfter you check it plug this value of r = 5.84803547643 into the equation h = 400/r^2\nh = 400/(5.84803547643)^2 \n= 11.6960709529\nThus,\nr = 5.8480354764cm and h = 11.6960709529cm
919539
Area =pi (Radius)^2 = 22/7(3)^2 =22(9)/7 = 28.29 ft^2 Ans.
820928
I am guessing that you want to know how big one of the sections are, so that's what I will try to determine.\n\nThe first step is to find the surface area of the hemisphere with a radius of 18 meters:\nThe surface area of a sphere, with r as the radius, is: A = (4)(π)(r²). Since this is a hemisphere, we can divide the expression equal to the surface area by 2, leaving us: A = (2)(π)(r²).\nA = 2(π)(18²) = 2(π)(324) = 648(π)\n\nSince we know that the surface area of the hemisphere is 648π m², we can divide this by 12 to find out the surface area of one of the sections:\n648π / 12 = 54π meters² ≈ 169.65 meters²
572270
triangle area = 1/2(height * width)\n\ncircle area = pi * radius^2
653555
Not considering the spheric harmonic of high orders I would have to tell you that my physics spirit tell the surface area  is what counts and there you could have k0.2Pi.R=2Pi <=>k0=1/R\nk1=1/(2R), k2=1/(distance of the 3 points more away from each other)........................
976222
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.
761667
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)
1175861
Area of Semicircle = (PI * R * R)/2\n\nAns. = 2.262 sq. cm
1068729
i think n=6..\nyes this can be done..if you know the formula..\nof volume and surface arae of a hypersphere and also the form of gamma fucntion..\nhope you can do it yourself..i saw these two links and based on the formula given for volume and surface ares and the gamma fucntion expression its evident that n=6.\nHope you can figure it out yourself.\nvolume= pi^3/6 (radius =1 here)
975805
1. A sphere with a diameter of 16 mm has the same surface area as the total surface area of a right cylinder with the base diameter equal to the sphere diameter. how high is the cylinder?\nA (sphere)=4*pi*r^2=4*pi*8^2=256*pi mm^2\nA cylinder=2*pi*r^2+2*pi*r*h==128*pi+16*pi*h\nA sphere=A cyl\n256*pi=128*pi+16*pi*h\n128*pi=16*pi*h\n8=h\nheight is 8 mm\n\n2. find the volume of a frustum of a right pyramid whose lower base is a square with a side 5 in. , whose upper base is a square with a side 3 in. , and whose altitude is 12 in.?\n\nB1 area=25in^2\nB2 area=9in^2\naltitude=3\nV=1/3(3)(25+15+9)=49in^2\n\n3. The figure formed by the intersection of a solid with a plane parallel to the base of the solid is congruent to the base if the solid is a........? \n\nprism
466950
Volume = Area x height\nArea = r x r x pi\npi is approximated by the fraction 22/7\nTherefore Volume of cylinder = 4 x4 x 22/7 x7 = 352 units^3
529963
The volume of a sphere is (4/3) pi r^3\n\nIf r is in meters than the volume is in cubic meters.\n\nFor r = 123,234,000 meters, the volume is \n\n2, 495, 343, 701, 035, 872, 000, 000, 000 pi\n\nor a little over\n\n7, 839, 353, 439, 355, 860, 807, 159, 721. 15 cubic meters\n\n(where the spaces keep Yahoo Answers from truncating the number and replacing the chopped off part with "...")\n\ni.e., approx.\n\n2.494E+24 pi or 7.839E+24 cubic meters
42062
Surface Area of a Sphere = 4 pi r^2\n\nCompute it for 95mm and 130mm and take the difference.
901347
You could use the formula for surface area of a sphere to solve this.\n\nUsing google, I found the diameter of the earth is 12756.3 km, which is equal to 41,851,378 ft.  So the radius is 20,925,689 ft.\n\nThe formula for surface area of a sphere is 4pi r².\n\nSo the square feet covering the earth is 4pi(20,925,689)² square feet.  \n\nThis is approx. equal to 5.5026 * 10^15 square feet.\n\n~ ♥ ~