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How to find the volume of a sphere


Well, just how do we arrive at the formula for the volume of a sphere? We do know that any volume is measured in cubes of some sort and this is the same for a sphere. The formula (graphic from Wiki How) is:
Perhaps we can trace this formula back to Archimedes, who was born around 287 BC in the Greek city of Syracuse in Sicily. While around your age, he was sentarchimedes to Alexandria in Egypt to study mathematics with teachers who had learned from Euclid. Not only was this as a way to gain his education, but his father also felt it was better to send his son away from the warfare being waged around Syracuse.

Besides studying mathematics, Archimedes also was interesting in creating inventions to solve common problems. Archimedes had became a master at mathematics, especially geometry. He spent most of his time working on solving new problems – in fact, he often became so involved in his work that he forgot to eat, much like computer geeks today!

Some of the mathematical problems Archimedes solved concerned areas and volumes of geometric figures. He had to devise a better number system and a new way to determine the formulae for the areas and volumes of spheres, cylinders, parabolas, and other plane and solid figures. One of the methods he used was an early form of integration. This great discovery eventually led to the field of Calculus. To determine the area of sections bounded by geometric figures such as parabolas and ellipses, Archimedes broke the sections into an infinite number of rectangles and added the areas together. Of course, he may also have used the idea of putting a sphere in water and finding the volume of water it displaced in order to calculate this in terms of the diameter and radius – to arrive at the formula at the start of this post. We may never know, but the formula is widely used today and well known by every high school student by the time they enter their senior year in Mathematics.
(H3 footnote: Do cubes and squares have the same properties as spheres and circles? Check out this discussion, aimed at teachers)

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NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]

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