Single Variable Calculus Volume of a Sphere - Proof 1

Key Takeaways

hey everybody in this video we're going to be talking about how to prove the formulas for the volumes of a sphere and Cone using single variable calculus techniques as children we were given these formulas without any clear explanation as to how they were derived and since they are dried using calculus we couldn't hardly get an explanation back then so now we're going to look at what we can do to try to prove these formulas so starting with our sphere we're going to start with a technique taught in calculus 2 called the cylindrical shells technique the idea is that we're going to rotate a semicircle let's call this the right hand of a semicircle we all know the formula of a circle is given by x squared plus y squared equals r squared and yes the circle is centered at the origin to make things simpler for us where R is going to be the radius of the semicircle now we're going to have to rotate this circle around about the y-axis so that we're going to get a full sphere but the way to do it is we're going to form cylindrical shells so for example let's start with a small rectangle here at r it's going to be a small rectangle what it's going to look like when it's revolved around the y-axis is it's going to look like a shell essentially and then we're going to take the next bigger rectangle and each of these rectangles has infinitesimally small widths or we're going to call it DX because they're along the x-axis and the next one will have a bigger height the a smaller radius so it's going to look something like this and we're going to keep we're going to keep doing this we're going to keep taking bigger and bigger rectangles and we're eventually going to get if it looks more and more like this and as you can see as we take these beginning rectangles this shape is going to resemble a solid sphere and if we sum up all of these volumes of all these cylindrical shells so we're going to get the volume of this here now how do we do this so if you take one of these shells just pretend we take one of these rectangles and we unravel it it's going to look a little something like this size is of course exaggerated for you guys to see but this is essentially if we take one of these cylindrical shells and these cells do have they do have a width they have an infinitesimally small width called DX so it's as if we took one of these shells and we sliced it open right here and we just unfolded it and got this essentially block now if you think about it this is not a perfect rectangular prism but because the width is so small it it approximates the rectangular prism very closely so we have to think about this what was the radius of this shell it was some distance X right because it's X away from the center right here x so the circumference of the shell was 2 pi x what was the height well the height was this height right here half of it is y so this is 2y and what is the thickness it's the DX that we had here so to find the volume of this we just use length times width times height so essentially it's 2 pi x 2y DX and now we don't just want one of those we want to sum them up we want to sum all of those up as and we want to use DX because we're integrating with respect to X here we want to sum all them up I'm going to take this 2 pi right here and there's 2 and combine them to a 4 pi and move them outside the integral so we can clean this up a little bit um and I'm going to leave the X in here I'm going to leave the y in here okay um and essentially we're integrating these from when X is equal to zero to when X is equal to R and because this is the integral with respect to X we can't leave any y's in here we need to convert them and how do we convert them we're going to be using this formula that's right here so y equals it's going to be equal to radical r squared minus x squared we're going to go ahead and put that in so it's going to be x times the radical of r squared minus x squared DX now we need to use some U substitutions we're going to U equal r squared minus x squared so that negative d u over 2 equals and that's going to be equal to RX DX and our X DX quantity is right here comprised of this X and this DX so that when we do this going off from zero to R and it's going to be we're going to take the negative one half from the negative du over 2 and put it out here and um and then we're going to go ahead and do radical u d and these limits I should really change them at this point because they should be in respect to U so when I put U when we put 0 into our formula for U right here I want to put 0 into X we get r squared and we put R into X we get zero and integrating this we get two-thirds mute three over two going from r squared to zero so when we put 0 into here becomes zero when you put r squared into it becomes negative two-thirds R cubed so we get Negative two-thirds R cubed out here and don't forget this quantity out here the constant so that that's negative 2 pi because it's 4 pi times negative one half so it's negative two pi a negative and a negative right here become a basically positive and we get the formula multiplying all this together right here we get four pi over 3 R cubed which we know as the volume of our sphere