Multivariable Calculus Volume of a Sphere Proof - Double Integrals

Key Takeaways

hey everyone so in this video we're going to be using multivariable calculus techniques to find the formula for the volume of a sphere we'll be using double integrals in this video and triple integrals in the next video so let's just start by drawing our XYZ coordinate grid draw a small XYZ coordinate grid in our black marker and we're going to call this X Y Z now we're going to draw the top half of the sphere the hemisphere that will serve as our function so the equation for this hemisphere is given by Z = radical um a^ 2 - x^2 - y^ 2 where a let's define a a is going to be the radius of our circle of our sphere rather so now we're going to be using double integrals and we're going to be using instead of just typical double integrals we'll be using polar coordinates CU I make our calculations is a lot easier so we know that in polar coordinates we have function of x y r * Dr D Theta what are the limits for R going from 0 to a right from the inside to the outside and the limits for Theta is going to be going from the positive X AIS all the way around so basically zero to 2 pi um now what we want to do is f ofx y is here but we want to convert this to Something in terms of R and Theta so we know this this can be written as a^ 2 - x^2 + y^2 and we know that in polar coordinates r^ 2 = x^2 + y^2 so this again can be written as a^ 2us R 2 uh so good so now we have all the parts we need to solve this problem we're going to rewrite this 0 to a radal a^ 2 - R 2 * the r d r d Theta and we're going to be using U substitution to take our next step so U = a^2 - r² so that netive du / 2 = RDR and here's the RDR term and so we're I'm going to take the - 1/2 of the negative D over2 move to the very outside so we're going to have d 2 pi I'm going to leave these limits blank for just a second uh because I want to convert them in terms of U we're going to have radical U because this inside quantity we defined as U right here um let's move this up radical U and RDR became du and we have a d Theta so now the limits are going to in terms of U so when we plug in zero into this U equation it's going to be a s and when we plug in a into here we're going to get zero so now we're just going to integrate this as normal we're going to go over here and it's going to become one2 integral from 0 to 2 pi and it's we're going to evaluate um we're going to take the integral of radical U which is 2/3 U 3 / 2 evaluated from a S to 0o so this right here is going to give give us -2 integral from 0 to 2 pi um and when we put in 0 and a cub in here we get 2/3 a cubed D Theta we're going to take this constant -2/3 and move it to the outside which is going to combined with this 1/2 and became a 1/3 so 1/3 integral from 0 to 2 pi this a cub is in the inside D Theta and when we take this move up here I should use a biger paper next time uh we're going to have 1/3 and this when we take the integral of this since this is respect to Theta this is a constant so we're going to have a cub Theta 0 to 2 pi and when we do that we're going to have again moving up we're going to have 2 pi we're going have 2 pi a cub over 3 and remember this is the for uh this is this gives us the volume of this half of the sphere the top half so simply we'll just multiply this by two and we get 4 piun over 3 a cub which we know as the volume of our sphere