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pikapika!
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The forumula for 2¶r can intergrated to make ¶r^2 (at least I think). So can anyone derive the volume of a sphere
4/3¶r^3?
4/3¶r^3?
The forumula for [itex]2\pir[/itex] can intergrated to make [itex]\pir^2[/itex] (at least I think). So can anyone derive the volume of a sphere
[itex]4/3\pir^3[/itex]?
It is certainly true that the formula for the circumference of a circle can be integrated to give the formula for area of a circle but that has nothing to do with "deriving" the area formula- unless you mean "differentiating" it!
It is true that the formula for the surface area of a sphere, [itex]4\pir^2[/itex] can be integrated to give the volume of the ball but, again, that is not "deriving the formula".
The formula for calculating the volume of a sphere is V = (4/3)πr³, where V is the volume and r is the radius of the sphere.
The formula for the volume of a sphere can be derived by using the formula for the volume of a cylinder and substituting the radius of the sphere for the height of the cylinder, then integrating to get the volume of the sphere.
The formula (4/3)πr³ is derived from the fact that the surface area of a sphere is 4πr². By integrating the surface area over the radius, we get the volume formula (4/3)πr³.
Yes, the formula for the volume of a sphere can be used for any size sphere as long as the radius is known.
The units for the volume of a sphere will depend on the units used for the radius. If the radius is in meters, the volume will be in cubic meters. If the radius is in inches, the volume will be in cubic inches.