Film, Media & TV0 min ago
spheres and circles
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what is the proof that the area of circles is pi times r squared?
what is the proof that the volume of spheres is pi time 4/3 times r cubed?
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For more on marking an answer as the "Best Answer", please visit our FAQ.Now I have tried the problem, it is actually easier than it looks. In case you do not wish to solve it yourself, my derivation is as below. (Please imagine there to be line breaks where I have inserted < BR >. I have drawn some diagrams to help: http://images.dpchallenge.com/images_portfolio/16982/medium/ 114398.jpg and http://www.dpchallenge.com/image.php?IMAGE_ID=114399) <BR><BR>
First, imagine a square 8cm by 8cm [diagram 1]. It should be clear how the area of a square or rectangle (base x height) is arrived at, and that the area of this square is 64cm�. <BR><BR>
See how I have divided the square into 8 lines 1cm wide and 8cm long. This is just a simplified representation of how area is calculated, i.e. an infinite number of infinitely thin 8cm lines, all touching, and extending for 8cm. What if these 8 lines (which represent an infinite number) were in a more circular arrangement? Their area would still, obviously, add up to 64cm� [diagram 2]. <BR><BR>
I have shown this by turning each rectangle that makes up the square into a triangle of equal area, then joining the triangles back together. If there were more triangles you can see that the resultant shape would look more circular; if there were <I>infinite</I> triangles the shape <I>would</I> be a circle. <BR><BR>
[Continued]
The 8cm line that was the height of the square has now become the 8cm radius of my �circle�. Also, the 8cm line that was the <I>width</I> of the square has now become a <b>16cm</b> arc of the <I>circumference</I> of my circle. The doubling of this length comes from when the rectangles were converted to triangles (base of triangle doubles to keep area the same when opposite edge is pinched to a point). Now there is just one final step in calculating the area of a complete circle. <BR><BR>
Seeing as this collection of triangles with total area 8cm�8cm � in other words with an area of <I>r</I>� � forms a <I>sector</I> of the complete circle, all you need to do to find the whole area is multiply the area of the sector by the number of times the sector will fit into the whole, if that makes sense. The arc of the sector is of length <I>2r</I> [diagram 3], so logically, the number of times you have to multiply the sector by to get the area of the circle will be the same as the number of times you have to multiply <I>r</I> by to get the length of the circumference. This value is simply c/2r where c = circumference of circle. The common symbol for �c/2r� is lowercase pi, written �. Hence you get the equation A = �<I>r</I>�.
I also messed up the link for Diagram 3. It is http://images.dpchallenge.com/images_portfolio/16982/medium/ 114399.jpg.
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