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If the sides of the cube were multiplied by 2, its surface area would be multiplied by the square of 2 and become 24 m 2. Its volume would be multiplied by the cube of 2 and become 8 m 3. The original cube (1 m sides) has a surface area to volume ratio of 6:1. The larger (2 m sides) cube has a surface area to volume ratio of (24/8) 3:1.
{{convert|123|cuyd|m3+board feet}} → 123 cubic yards (94 m 3; 40,000 board feet) The following converts a pressure to four output units. The precision is 1 (1 decimal place), and units are abbreviated and linked.
The surface-area-to-volume ratio has physical dimension inverse length (L −1) and is therefore expressed in units of inverse metre (m-1) or its prefixed unit multiples and submultiples. As an example, a cube with sides of length 1 cm will have a surface area of 6 cm 2 and a volume of 1 cm 3. The surface to volume ratio for this cube is thus
A cord is the amount of wood that, when "ranked and well stowed" (arranged so pieces are aligned, parallel, touching and compact), occupies a volume of 128 cubic feet (3.62 m 3). [46] This corresponds to a well-stacked woodpile, 4 feet deep by 4 feet high by 8 feet wide (122 cm × 122 cm × 244 cm) , or any other arrangement of linear ...
This formula holds whether or not the cylinder is a right cylinder. [7] This formula may be established by using Cavalieri's principle. A solid elliptic right cylinder with the semi-axes a and b for the base ellipse and height h. In more generality, by the same principle, the volume of any cylinder is the product of the area of a base and the ...
The generation of a bicylinder Calculating the volume of a bicylinder. A bicylinder generated by two cylinders with radius r has the volume =, and the surface area [1] [6] =.. The upper half of a bicylinder is the square case of a domical vault, a dome-shaped solid based on any convex polygon whose cross-sections are similar copies of the polygon, and analogous formulas calculating the volume ...
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Much more work is needed to find the volume if we use disc integration.First, we would need to solve = () for x.Next, because the volume is hollow in the middle, we would need two functions: one that defined an outer solid and one that defined the inner hollow.