What is the Volume of a cirlce with a 6" radius that has moved 10" vertically along an axis drawn in center?
96858
erm the volume of the circle would be the same, the volume of the cylinder would be (pi)r*r*h (r=radius, h=height)
141519
It is the length divided by the radius
783552
An annulus is just a ring shape.\nLet the radius of the inside ring be 'x' and the radius of the outside ring then becomes 'x+10'.\nThe area of the track is the area of the large circle minus the area of the small circle,\ni.e Pi[(x+10)^2-x^2] = 1000\nYou can whittle this down to x = 50/Pi - 5\nSo the smaller circle had radius 10.92 (to 2 dp) & the larger circle has radius 20.92 (to 2 dp)
565265
one can find the volume of a three-dimensional object. an oval (ellipse) only has an area. a three-dimensional oval will be an ellipsoid. \n\nthe volume formula is :(4/3)*(pi)*(r1)*(r2)*(r3)\nwhere \nwhere r1 = radius 1, r2 = radius 2, r3 = radius 3 \nthe radii of the three dimensions
365724
V = Pi*r^2 * h\n\n90 ft diameter and 6 ft deep\n\nor 45ft radius\n6ft high\n\nV = Pi * (45)^2 * 6\nV = 2025pi * 6\nV = 12150pi\n\nV = about 38170.35ft^3\n\nIn gallons, that would be about 285500 gallons
971773
Do you have the formula of the cillinder? Use it.\nThe weight (2 lbs.) is not important for areas and volumes.\nThe radius is the diameter divided by two.\n\na. 2 x π x radius x height\nb. π x radius x radius x height\n\nc. The formula for the volume of the cone is π x radius x radius x height / 3\n\nTo get your answer just divide the volume of the cillinder by the volume of the cone.
237156
Pi * r^2\n\n(Pi R Squared)\n\nwhere R is the radius (which is also equivalent to half the diameter)
26120
If the pressure's the same, then the change in volume would be "(3.61*A*X)/SQRT(Y) Where: A = Cross-sectional area of discharge\npipe in square inches\nX = Horizontal distances in inches\nY = Vertical distances in inches"
456802
Let x = length of the corner being cut out.\nThe volume is then\nV = x(12-2x)(8-2x)\nV = x(96 - 24x - 16x + 4x^2)\nV = 4x^3 - 40x^2 + 96x\nTake the derivative of V\nV' = 12x^2 - 80x + 96\nSet equal to 0 and solve for x\n12x^2 - 80x + 96 = 0\n3x^2 - 20x + 24 = 0\nx = (2sqrt(7) + 10)/3\n= 5.097 or 1.569\n12-2(1.569) = 8.861\n8-2(1.569) = 4.861\nThus, the dimensions are\n1.569 by 8.861 by 4.861
778715
increments of radius is 0.001 in/min\nincrement of area is 3.14(1.801^2-1.8^2)=0.0113 sq in\nhence increment of volume is 6*0.0113=0.0678 cu in/min\n\nMore scientific solution\nassume radius increments = dx\nthen volume increment V will be Height*area increment\nV = 6*3.14{(R+dx)^2-R^2} = 18.84(R^2+2Rdx+dx^2-R^2)\nV=18.84 (2Rdx)  when dx close to 0 then dx^2 can be omitted\nthen dV/dt = 37.68 R dx/dt = 37.68 * 1.8 * 0.001 =0.0678 cu in/min
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
745326
The volume of a cylinder is the area of the base (a circle) times\nthe height of the cylinder.  The pipe represents a cylinder.\nThe area of a circle is PI times the radius squared (PI X r^2)\n\ndiameter of pipe = 24 inches implies the radius = 12 inches implies radius = 1 foot.  Length of pipe is height of cylinder\nnew value of PI = 3.1416 exactly (?)\n\nVol of Cylinder = Area of Base                      X Height of Cylinder \nVol of Cylinder = Area of Circle                     X Height of Cylinder \nVol                   = PI(3.1416) X radius(1 ft.)^2 X height(8 ft.)\nVol                   = (3.1416 sq.ft.)                     X height(8 ft.)\nVolume            =  25.1328 cu.ft. exactly
173346
For a circular cone, it's V=(1/3)*pi*h*r^2. For a cone with a non-circular base, the volume is always 1/3 the volume of the corresponding cylinder. So if the area of the base is A, the volume of the cone will be (1/3)Ah. In this, h is the vertical height of the cone and r is the radius of the circle on the base.
676849
Cylinder = PI r^2 x 6 r = 6PI r^3\nBalls = 4/3 x PI r^3 x 3 = 4 PI r^3\nratio = 4 PI r^3 / 6 PI r^3 = 4/6\n\nRatio = 2/3 of the space is used by the balls.
586130
surface area = 3bh, where b is length of side, and h is the height dropped normally from center to any of the sides\n\nvolume = 6 times volumeof each triamgle, formed by all 3 diagonals
18353
There are many possibilities, but the one using whole numbers is a pyramid with a square base that is 3cm. on each side and has a height of 10cm.\n\nVolume = (length x width x height)/3
84844
Height = H = 10"\nRadius = r = 4"\n\nVOLUME (not area, volume is 3-dimensional) = (π/3)(r^2*H) = (160/3)*π\n\nSURFACE AREA = (πr)(sqrt(r^2+H^2)) = 4(sqrt(116))*pi