@@BecauseMaths Depositing material one thin disk at a time, working from bottom to top, possibly using cylindrical coordinates, r, θ, and z, similar to integrating to find the volume of the sphere.
Always amazes me how I can understand derivations like these as soon as I look at them (I just skipped to the end and looked at the math and diagram) but idk if I’d have ever solved this on my own. I tried and all I realized was I couldn’t just integrate with discs without doing something clever before I went back to doing normal classwork. Don’t know if I’d have ever reasoned out this diagram with the right triangle. So ingenious to just integrate from 0 to r, and express each radius y in terms of r and x so that r can just be subbed for x, thereby giving a working formula.
Well, judging from your thumbnail it looks like you're sliding the sphere along x and doing the integral that way. It's much, MUCH simpler and more intuitive to do it integrated over spherical shells, from the center out to the radius. It's just integral from 0 to R of the surface area - 4*pi*r^2*dr.
I find it much easier to derive the volume first with a double integral of z=sqrt(R^2 - x^2 - y^2), double it, and then derive the surface area from differentiating it relative to the radius. I'd guess that deriving the surface area from first principals would require an integral similar to the arc length integral.
Bruh, I mean, good video, it was awesome, but didn’t you INTEGRATE the Superficial area of the sphere, instead of getting the DERIVATIVE of the sphere?
@naufarrelz.a7311 pretty sure they're just making a joke using wordplay about the video title saying "derive" and the term "derivative" but the video makes use of integration.
My favourite derivation of the volume of a sphere until now. This video is very intuitive and passionately explained. Thank you
❤️❤️❤️🙏
it would have been nice to see where A=πr² came from, which you assumed, and also to see how the Ancients did this without calculus.
What is the best way to make a solid sphere using a 3D Laser Printer?
Hi, Sir Roger, I have no experience with 3D laser printer yet…
@@BecauseMaths Depositing material one thin disk at a time, working from bottom to top, possibly using cylindrical coordinates, r, θ, and z, similar to integrating to find the volume of the sphere.
Always amazes me how I can understand derivations like these as soon as I look at them (I just skipped to the end and looked at the math and diagram) but idk if I’d have ever solved this on my own. I tried and all I realized was I couldn’t just integrate with discs without doing something clever before I went back to doing normal classwork. Don’t know if I’d have ever reasoned out this diagram with the right triangle. So ingenious to just integrate from 0 to r, and express each radius y in terms of r and x so that r can just be subbed for x, thereby giving a working formula.
❤️❤️❤️
Well, judging from your thumbnail it looks like you're sliding the sphere along x and doing the integral that way. It's much, MUCH simpler and more intuitive to do it integrated over spherical shells, from the center out to the radius. It's just integral from 0 to R of the surface area - 4*pi*r^2*dr.
I find it much easier to derive the volume first with a double integral of z=sqrt(R^2 - x^2 - y^2), double it, and then derive the surface area from differentiating it relative to the radius.
I'd guess that deriving the surface area from first principals would require an integral similar to the arc length integral.
Bruh, I mean, good video, it was awesome, but didn’t you INTEGRATE the Superficial area of the sphere, instead of getting the DERIVATIVE of the sphere?
wdym?
@naufarrelz.a7311 pretty sure they're just making a joke using wordplay about the video title saying "derive" and the term "derivative" but the video makes use of integration.