Hey guys, I have a question relating to integration which is puzzling me. I figured that I would try and derive the volume of a sphere myself. Having never done this before I started out with a strategy that to me seems perfectly logical. If it can be seen from the 10 second scribble in paint, I imagine the circle i've drawn to be a 3-d sphere. If i cut the sphere up in cross-sectional slices with cross sectional area πr^2 and integrate from 0 to r, taking the zero point to be the left most point on the circle, I would have found the volume of a hemisphere. Therefore I would need to multiply this integral by two to get the total volume. Like this: 2∫πr^2 dr (from 0 to r) = [2πr^3]/3 Obviously this is wrong so could someone point out the flaw in my approach as I am off by a factor of 2.
When you integrate from 0 to r, the radius of the circle is not equal r, it is equal to sqrt(1-r^2), so you have to take 2*∫π(1-r^2)dr = 2πr^3(1-1/3) = 4πr^3/3.
We've already established in this thread that the volume of a sphere is 4/3 * πr. This formula has been known for hundreds of years.