A right circular cylinder is inscribed in a sphere of radius r.?
969797
Let the radius of the sphere be R.\nLet x be the radius of the cylinder.\nThen height of half of the cylinder= sqrt(R^2-x^2)\n\nTotal surface area of cylinder = 2*pi*x^2 +2*pi*h, so\nS=2*pi*x^2 +2*pi*2*(R^2-x^2)^.5  [Eq I]\ndS/dx=4*pi*x + 4*pi*.5*(-2x)(R^2-x^2)^-.5\n=4*pi*x - 4*pi*x/(R^2-x^2)^.5\n=4*pi*x(R^2-x^2)-4*pi*x =0\nR^2-x^2-1=0\nx^2=R^2-1\nx=+ or - sqrt (R2-1)\nPlug these values of x into [EQ I] to get answer.
684992
That depends very much on the type of solid.  Mathematically you need to know a lot about its various dimensions and shape.  \n\nPrevious posters have given the answer for box-like shapes (length * width * height) and a cylinder (Pi*r^2*h).  A sphere is PI * 4/3 * (R ^ 3).  \n\nFor complex items you need calculus as well.\n\nIn physics, dunk it in a tank of water filled to the brim and measure the volume of water it displaces.  This works for any shape, provided it fits in the tank.
421815
Given that the volume of the cylindrical can is 100 cm^2...\n\n100 = (pi)r^2h\n\ndivide both sides by (pi)r^2 to isolate h\n\nh = 100/(pi)r^2\n\nSo, if the radius is 2 cm...\n\nh = 100/((pi)*4) - r^2 = 2^2 = 4\n\nh = 7.96 cm, rounded to first two decimal places, using 3.14 as the value for pi.
470432
The radius of a sphere centered at (0,0,0) is the square root of x^2 + y^2 + z^6\n\nr^2 = 36 + 9 + 49 = 94\nr = 9.7\n\nSo, the diameter is 19.4 (two times the radius)
1137727
I like the joke!  To play along, I define:\nr = metric r  and \nd = metric d\n\nV = (pi/3)*(d^2)*(3r - d)\n\nThis will work for the tank being any fraction of a sphere, not just 1/2
544480
36pi  \nThe length of the whole is measured down the side of the \nhole. So the bigger the sphere, the greater the diameter of the hole. No matter how big the sphere is, the remaining\nmaterial is the same. So assume the hole is infinitesimal\nin diameter. The remaining material is all the material\n which is a sphere with diameter 6 and radius 3 inches.\nSo V=4pi(3)^3/3=4pi3^2=36pi
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.
502090
Centre is 1.5,2 and the Radius is 3
475972
The volume of a cylinder is given by\n\nV = pi*r^2*h\n\nAnd the surface area of a cylinder is given by\n\nSA = 2*pi*r^2 + 2*pi*r*h\n= 2*pi*r(r + h)\n\nNow, we don't have enough information to specifically find either the radii or heights - we only have a volume.  But we can still work with that based on the fact that the cylinders are similar.\n\nIf the cylinders are similar, their measurements are proportional.  In other words, the radii are proportional and the heights are proportional (r1/r2 = h1/h2 and r1/h1 = r2/h2).  So let's just assume that the larger cylinder has a radius of 3 and a height of 3 (plug in to the volume formula and it works out to 27*pi cm^3, but for the sake of this argument, we'll ignore the pi for now).  For the smaller cylinder to be similar, it would have to have a radius of 2 and a height of 2 (again ignoring pi).\n\nThe surface area of the larger cylinder is then\n\nSA = 2*pi*r(r + h)\n= 2*pi*3(3 + 3)\n= 36*pi\n\nAnd the smaller one would have a surface area of\nSA = 2*pi*r(r + h)\n= 2*pi*2(2 + 2)\n= 16*pi\n\nSo your ratio would then be 36*pi / 16*pi = 9/4.  (If we went back and corrected out radii and heights for pi, they would have all cancelled out by this step.)\n\nNow, what if my initial assumption was wrong and the radii and heights weren't 2 and 3?  Well, we can still work with that, but we'd end up with the same ratio of 9/4.  If the smaller cylinder had a height of 4, then the radius would have to be sqrt(2) for the volume to work out to be 8.  So the ratio of radius to height would be sqrt(2)/4.  Because the two cylinders are similar, this ratio must hold for the larger one.\n\n(Crazy math warning!)\nSo r/h = sqrt(2)/4, so r = h*sqrt(2)/4.  We can find the radius and height of the larger cylinder now:\n\nV = pi*r^2*h\n27 = pi*(h*sqrt(2)/4)^2*h\n27 = pi*(h^2/8)*h\n27 = (pi*h^3)/8\n216 = pi*h^3   <-  we'll drop the pi at this point - same reason as before\n216 = h^3\nh = 6\n\nr = h*sqrt(2)/4\n= 6*sqrt(2)/4\n= 3*sqrt(2)/2\n\nNow let's compare surface areas of the two:\n\nSA(big) = 2*pi*r(r + h)\n= 2*pi*(3*sqrt(2)/2)(3*sqrt(2)/2 + 6)\n= (3*pi*sqrt(2))(3*sqrt(2)/2 + 6)\n= 9*pi + 18*pi*sqrt(2)\n= 9*pi(1 + 2sqrt(2))\n\nSA(small) = 2*pi*r(r + h)\n= 2*pi*(sqrt(2))(sqrt(2) + 4)\n= 4*pi + 8*pi*sqrt(2)\n= 4*pi(1 + 2sqrt(2))\n\nAnd if you took this ratio of big to small, you have 9*pi(1+2sqrt(2)) / 4*pi(1+2sqrt(2)) = 9/4!\n\nSo my original assumption is right, and the ratio turns out to be 9/4 for these two cylinders.\n\nNow I used cylinders for this (I have no idea why), but it turns out that it works for any solid.  There's a relationship among these measurements - if we compare cubes:\n\nRatio of a side = s1/s2\nRatio of area = (s1)^2/(s2)^2\nRatio of volume = (s1)^3/(s2)^3\n\nSo for our case, the ratio of the area can be found by taking the cube root of the ratio of volumes, and then squaring it - 9/4.
699573
Yes it is with radius of curvature = ∞\n\nHigher geometry takes this for granted
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.
358957
Volume = pi * r^2 * h\n\nr is 1/2 the radius of the drilled hole = (D'/2)\nh is equivalent to the diameter of the original cylinder = D
809750
Ratio of what?\nSurface area?\nVolume?\n\nI will go with volume.\n\nassume radius of sphere is = 1\n\nsphere = 4/3pi\ncylinder = 2pi\ncone = 2/3 pi\n\neat the pi's (get rid of them) and multiply what's left by 1.5\n\nSphere:Cylinder:Cone = 2:3:1\n\nI hope that helps
30747
2 cubes have 16 corners (since each cube has 8 corners)\n\na cylinder doesn't have any corners
1124334
two circles, in parallel planes with the line connecting their centers forming right-angles with each plane in all directions. ... and the walls comprising the set of line segments one of whose end points is on one circel and the other end point on the other circle withe the property that the segments at each point on circle(A) is the minimum distance to circle(B).\n\n.. or, since you use the word "cylinder" in the question:\n\na cylinder, each of whose end-faces is a circle.
918429
use the formulas your given\nto find the volume of the cylinder\nthen pretend the semi hemi on  the top is a sphere find the volume and multiply by 1/2\n\nadd two totals to find volume
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!