Theory for computing areas of cone object?
827739
Not quite sure what you are asking for....\n\nThe surface area of a right cone (i.e. a cone with its vertex above the center of its base) is:\npi*R*sqrt(R^2+H^2)\n\nwhere R is the radius of the base and H is the height of the cone.
981062
The surface area of a sphere of radius 'r' is given by:\n\nsurface area = \n 4*pi*r^2\n \n\nNOTE: The value of p can never be known exactly, so surface areas of spheres cannot be calculated exactly. Common approximations for p are: 3.14, and 22/7. \n\nExample 1  if 'r' = 5 for a given sphere, and p = 3.14, then the surface area of the sphere is:\n\nsurface area = 4*pi*r^2\n\n         = 4 × 3.14 × 25\n\n         = 314
586782
The area of segment of a circle is given by\nA = r^2[theta - sin(theta)]/2\n\ntheta is given by\ntheta = 2 arcsin(c/[2r])\n\nwhere c is the chord length and r is the radius.  So,\ntheta = 2 arcsin(c/[2r])\n= 2 arcsin(20/[2(12)])\n= 1.97 radians\n\nSo the area then is\nA = r^2[theta - sin(theta)]/2\n= (12)^2[1.97 - sin(1.97)]/2\n= 75.5 m^2\n\nEDIT:  My answer is indeed correct.  greatindiangenius's angle below is wrong.  You can find that angle theta another way besides the way listed on the link below - if you imagine a triangle formed by the chord and two radii, and if you drop a height from the center to the chord, then you have two triangles with long leg 10 and hypotenuse 12.  The angle between the height and the radius is arcsin(10/12) = 56.44, so the entire angle is 112.88 degrees, or 1.97 radians, not 100.\n\nRecall that a _sector_ is the pie wedge shape of a circle, starting from the center.  But a _segment_ is only the part between the chord and the circle - the ice cream part of the cone, if you break the _sector_ down like that.  So the formula in my reference and my work is correct in that we are finding the area of the ice cream only, and not the ice cream and cone.\n\nWe can find the area of the segment another way: we can take the area of the sector and subtract the are of the triangle formed by the radii and chord.\n\nFor the area of the _sector_, recall that it's just a fraction of the entire area, based on that angle 112.88:\n\nArea of sector = (112.88/360)(pi*12^2)\n= 141.85 m^2\n\nFor the area of the triangle, we need the height (we already have the base, 20).  We can use the Pythagorean theorem:\n\nc^2 = a^2 + b^2\n12^2 = 10^2 + b^2\n144 = 100 + b^2\nb^2 = 44\nb = 2sqrt(11)\n~ 6.63 m\n\nArea of triangle = (1/2)bh\n= (1/2)(20)(6.63)\n= 66.3 m^2\n\nArea of segment = Area of sector - Area of triangle\n= 141.85 - 66.3\n= 75.5 m^2\n\nwhich is the same thing as above with the other formula.
37855
One possible three-dimensional cross section of a cone in the fourth dimension is a sphere. (Think of the analogy of a cross section of a cone in 3 dimensions being a circle. It can also be an ellipse or hyperbola.) There are some interesting illustrations of such cross sections (not only spheres, but some other shapes analogous to the hyperbola or ellipse) at the following link:\n\nhttp://www.well.com/~abs/SIGGRAPH96/4Dtess.html
380933
Surface areas and lateral areas are always fun to find.  Surface areas for right prisms and cylinders is always the sum of the lateral area and two times a base.  (Lateral area is the sum of all the surfaces that aren't bases.)\n\nIn the case of a cylinder, the lateral area is really a rectangle - imagine cutting open a roll of toilet paper.  The length is really the circumference of the top (base) and the height is, well, the height of the roll.  So the lateral area is 2*pi*r*h.\n\nThe surface area is the lateral area plus two bases.  A base in this case is given by pi*r^2.  So the total surface area is 2*pi*r^2 + 2*pi*r*h.\n\nThe idea behind lateral and surface areas will carry over when you start working on right prisms (if you're not already doing those).  Lateral area is, in the case of a prism, the perimeter of the base (similar to cylinders, except we used "circumference" in place of "perimeter") times the height of the prism.  And the surface area is the lateral area plus two bases - in this case, the bases vary (you may have squares, triangles, trapezoids, etc.), so your formula for the total surface area will vary as well.\n\nThis is probably more than you wanted to know, but it's always good to relate things you know to things that you're learning.
638243
1) think of  a series of disks stacked on top of eachother, so thier centers are aligned.  Each disk has a slightly smaller radius then the one under it.  If the disks are very thin (infinately so), and you stack enough of them so that the top disk has a radius of zero (a point), then you have a  cone. \n\n2) Take a circle, and remove some portion of it- like a pizza with one or more pieces removed. Take the remaining portion of the circle, and roll it, so that the straight edges adjoining the removed section are connected together along their entire lengths, and you have a cone.
837293
I haven't done one of these in quite a while. \n\nWhat would you differential element be?  How about r*(dtheta)*dr?\n\nthat sort of a ds*dr   where ds depends on the radius.  Try it! \n\nNote that without the sin theta you'd get the int rdthetadr with r 0 to R and theta 0 to 2pi  You'd get the correct area of a circle.\n\nTry the intergral, see if it makes sense.  Then sketch the figure and estimate it's area to compare.  My confidence is increasing as I write this.
241611
kindly refer to the formula..and do you math HW yourself...\nApart from that if Diameter=8 cm...radius got to be 4cm\n\nI dont know the Surface area formula for cone but should be easy to find it somewhere like google.
720350
Find the closest brick like shape lxbxh, also closest spherical shape by mean radius. Then by water displacement find the volume. Then find the surface area taking those regular shapes. Then make an avrage of the two. If the mango is long then take cylindrical shape intead of spherical.\n\nOtherwise put soft, wet tissue paper without much wrinkle like a paste/coating (you may tear some pieces and use small pcs). Then dry it. Peel off carefully. Place on a graph paper and measure. You may cut and paste.
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
747560
http://www.cplusplus.happycodings.com/Computer_Graphics/code22.html\nhttp://www.wol.net.pk/mtshome/cppComputerGraphics.html#Area%20Filling\n\nThis is C, but close enough:\nhttp://www.programmersheaven.com/download/15200/download.aspx
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
529200
Good that you care to ask. The answer to this should be evident to you by integration (hope you have been introduced to calculus? If not, you will be this year.). \n\nI assume that you don't have much of a background.\n \nOK, let the cone be such that its height is H, base radius R. \n\nWhat you need to do is consider an infinitesimally thin section of the cone (a diagram should be so helpful here, unfortunately not possible) at height h from the bottom and of height (along the axis) 'dh', such that the radius of the section is a uniform 'r'. \n\n(Note that you have a uniform radius r to the section because you have chosen a very thin section of the cone.)\n\nNow, the surface area of the section considered is \n2*pi*r*dh*cosec x. (x is the cone half-angle)\n\nThe cosec term comes from basic trigonommetry, when you realise that the section is tilted and not exactly parallel to the axis. \n\nThe surface area  of the entire cone, indeed the curved surface area, is a summation of all such infinitesimal sections, each of surface area 2*pi*r*dh* cosec x \n(where r is the radius of each section). \n\nThis above sum can be found by \nintegrating 2*pi*r*dh*cosec x (that is a word in the language of calculus for finding out the sum of many small parts) over the entire height of the cone. \n\nalso realise that r = h tan x. \n\nso an integration of 2*pi*htan x*cosecx*dh gives us,\n\nCSA = 2*pi*sec x* (H^2/2)  \n    = pi*R* H (sec x/tan x)\n    = pi*R* H cosec x\n    = pi*R*L\n\nHope this satisfies.
659736
Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!
701955
You could use graph paper to get a good approximation of the surface area. Using a large circle would increase the accuracy.\nCount the whole squares, then count the half squares. \n\nAlternatively would be to use graph paper, then precisely cut out the circle, then cut the graph paper. On another sheet of graph paper, try to fill squares with the incomplete pieces from the circle.\n\nModifying Mike's answer, using a sensitive microbalance, find an average mass of one graph square then measure the mass of the cut out circle.
568696
rare gem's answer is right..given the radius you'll just need to compute for the area using the formula right?\n\nArea=(radius^2)(pi)\n    = (5)(5)(pi)
311471
lateral surface area of a cone = pi*r*l\n\nwhere \n         pi = 3.14( a mathematical constant)\n         r = radius of base of cone\n         l = slant height of cone\n           =  square root of sum of squares of base radius  and height of cone\n         h = height of cone from centre of base circle to apex of cone\n          l = sq rt(r*r + h*h)