Can you help me with Volume of a cone?
85854
A cone holds exactly 1/3 as much as a cylinder with the same radius and height.  So multiply pi times radius squared times height, then divide by 3.
877943
volume of cyllinder = area of circle times height\n\n= Pi x radius x radius  x height\n\nremember its just the same as finding the volume of any prism.\nfind the area of the shape of the end then multiply by its length!!
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.
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)
887520
Volume = pi r^2 h \n\nMeasure a Coke can to get r and h and calculate the volume.  Plug it in for V in  the volume formula and solve it for h.\n\nThen take the surface area formula, SA = 2 pi r^2 + 2 pi r h,\nreplace the h with your answer from above, find the first derivative d(SA)/dr, and make it equal to zero to find the minimum.
934101
what's the height?  you need the height to finish this problem\nwhat type of shape is it?  a cylinder? cone?\nbe specific or no one can help you
1161095
The volume of a cone is 1/3(Area of Base)(height) = 1/3 π r2 h
543816
Yes it is !!\n\nFor proof of volume of cone, pls go to :\n\nhttp://mathforum.org/library/drmath/view/53646.html\n\n\nThe following site has got the proof for area of circle :-\n\n\nhttp://64.233.161.104/search?q=cache:FbBn10NweQkJ:www.artofproblemsolving.com/LaTeX/Examples/AreaOfACircle.pdf+area%2Bcircle%2Bproof&hl=en&gl=in&ct=clnk&cd=1
16584
You can find the volume using a triple integral
730124
1. volume = h*2*pi*r^2 =1000 so r^2= 1000/2/pi/h\n\nsurface area = 2*pi*(h+2*r^2)\nsurface area / volume = (h+2*r^2)/h/r^2 \n=1/r^2+2/h  = 2*pi*h/1000+2/h  which is a minimum when h is about 17.84  [(1000/pi)^0.5]
433823
Revolving about the x-axis yields a cone whose tip is at the origin and opens in the positive x direction.  Taking discrete x slices (dx) yields a disk whose volume is given by pi*(kx)^2 dx.  Integrating from 0 to h gives us (where "S" is used as the integration symbol):\n\nV = S pi*(kx)^2 dx\n= pi * k^2 S x^2 dx\n= (pi * k^2) * (x^3)/3\n\nEvaluating from 0 to h gives us\n= (pi * k^2) * (h^3)/3 - (pi * k^2) * (0^3)/3\n= (pi * k^2 * h^3)/3
499283
Hmm...\nThe formula for a cylinder volume is:\nArea * height\nThe area is a circle so it's pi * r * r\n\nVol = pi * r * r * h\n\nTherefore\n\nh = Vol / (pi * r * r)\nh = 121 / pi * r * r\nh = 38.515496228238671256069870736148 * r * r\nor h = 38.52 * (r square)
144053
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.
28152
Don't know what grade you're in, but looks like this is gonna involve some integration.\nLet the intersection point of the plane be z=h(=0.5 in this case).\n\nConsider a thin disc of thickness dz in the volume to be evaluated, at a distance z from the origin. Obviously, z>h. It would have a radius of \nr = sqrt(R^2 - z^2). R is the radius of the sphere, =1 in this case.\n\nHence, the volume of this disc would be\ndV = 2*pi*r^2*dz\n   = 2*pi*(R^2-z^2)*dz\n\nHence, for total volume, this is going to have to be integrated between z=h to z=R.\n\nI believe this works out to \nV = 2*pi*((2R^3/3)-R^2h+h^3/3) here. Put in the values mentioned above.\n\nWhew!
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
461296
Volume of a prism or cylinder = Area of the base (a circle in this case) * Height\n\nV = pi*r^2 * h,  (V=volume, r=radius=diameter/2, h=height)\nV = pi*0.75^2 * 9\nV = 15.9 ft^3
976121
Let us assume that we require a cone with d as bottom circle dia, h as vertical height which are the known factors. Let l be the slant height of the cone which is equal to Square root of (h^2+(d/2)^2). To arrive the cone we have to make a sector with l as radius and A as angle of the sector. \nl is already calculated. A is calculated as follows:\nA = pi()*d/l which is obtained in radians; \nTo convert it to degrees, multiply A with 180 and divide by pi();\nSo we get A in degrees and l, from which we can get a sector. By connecting the two slant heights together we can get a cone with dia d and height h.