Multivariable Calculus Volume of a Sphere Proof - Triple Integrals

Key Takeaways

hey everyone in this video we're going to be using triple integrals to find the formula for the volume of a sphere so triple integrals are another multivariable calculus technique and in this one we're going to be using three integrals and we're going to using a full sphere to find the volume so again like in the last video when we used um polar coordinates instead of regular double integral we're going to be using spherical coordinates here instead of regular triple integral because it'll make our calculations a lot simpler so we're going to start with the form for spherical coordinates which is R3 and and our three limits so D row which is that what's side called d f or Fe and D Theta and of course we have to incre uh we have to include the interior term which is Row 2 s of five so now what are our limits for row row is going from 0 to a what are our limits for what are limits for uh F or F it's going from zero to Pi because it's going uh V is always measured from the positive z- axis to the negative z- axis so it's going the full the full angle from 0 to Pi and Theta we know goes from the it's measured from the positive x axis and it goes all the way around again so again that's 0 to 2 pi so now we're going to go ahead and try to solve this integral as it is because there's no F of XY in here because with triple integrals it's already assumed to be a volume that's what it means it's taking little individual cubes of this uh solid sphere whereas when using double integrals it was a hollow sphere and we're adding up all these little cubes over all this big region and when we add up those cubes we're going to be getting the volume so let's start so the first integral the inside integral is with respect to row so it's going to be 1/3 row cubed s of F and that's evaluated from what from 0 to a so it's going to be 1/3 a cubed s of F and now we have two more integrals to evaluate 0 to Pi 0 to 2 pi d f d Theta this 1/3 a cub we can move to the outside because it's a constant remember our radius a is a constant so we're going to move it to the outside so we don't have to deal with it all the way in there and we're going to integral from 0 to 2 pi integral from 0 to Pi S 5 d d Theta so now sin ofi uh when you take an integral that it's just going to be negative cosine of pi evaluated from what 0 to Pi so negative move the negative outside cosine of I is 1 minus um cosine of 0 is 1 so it's going to be negative quantity 1 minus one which is going to come out to a 2 again this two is a constant we're going to move it to the outside of the integral with a 1/3 a cub so it's going to be 2/3 a cubed integral of from 0 to 2 pi and just D Theta so now when we integrate this we're going to have Theta because this is actually just an imaginary one inside here and that comes out to Theta you value from 0 to 2 pi and that comes out to 2 pi and this 2 piun ultip our 2/3 a cub is 2/3 a cub * 2 piun which comes out to 43 Pi a cubed this is by far the simplest most elegant way to solve for this volume uh because it takes it uses the full advantage of how to find the volume it takes those little tiny cubes and just approximates them and finds the whole volume rather than revolving some area or using some uh region underneath the solid so now we know where that Elementary formula for the volume of sphere comes from so hopefully we all learn something thanks for watching