Surface area of a sphere

In this video, I show how to find the area of any surface (yes, any surface) by calculating the surface area of a sphere in 3 dimensions. The calculation is absolutely beautiful, because there are 3 amazing simplifications that will happen. Enjoy!

[albertemcstein]

I really love your videos about different kinds of integrals! Thank you Peyam

[bulldawg]

As always, yet another EXCELLENT presentation simplifying a complex topic ... I've been struggling to get my mind around surface integrals, and this is one of the best videos on this subject ... Calculating the surface area of a sphere is big bonus ...

[dolev]

My new life motto:
@5:02 We're not scared - We're mathematicians!

[alexanderikejiofor]

Can you please make videos on Change of Basis? I find this topic very confusing, and your amazing at explaining? (i have an exam next week )

[tomatrix]

Wow, seeing that circle surface area derived logically is crazy! This is some powerful mind blowing stuff

[jacobahern]

I don't understand these videos yet, but I know I'm going to love them some day

[sodahead]

I did a simpler method. I used the surface area integral 2pi integral(r(x) * sqrt(1 + derivative(x)^2) from -r to r. The equation of a circle is x^2 + y^2 = r^2. In terms of x it is y = sqrt(r^2 - x^2) .When you do the derivative and the algebra, the sqrt(r^2 - x^2) cancels, leaving you with sqrt(r^2). Since sqrt(r^2) is a constant r, put it in the front so you get 2pir(integral(1)dx.) from -r to r. Then you solve the integral which is 2pir(x) from r to -r. Plugging in r and -r it is 2pi(r)^2 - (- 2pi(r)^2), add both and you get 4pi(r)^2

[MrRyanroberson]

I would like to ask a complicated question with a possibly trivial solution. What if you cut up a convex 3d volume by cutting up the surface and slicing to the origin (convex here being defined as "the surface crosses each Ray passing through the origin exactly once") and sorted the infinitesimally small pyramids in order from shortest around -x and longest around +x? Are there any guarantees about this shape? I don't think it would, but I also wouldn't be too surprised if it had to be a spheroid

[wompastompa]

I remember doing this exact problem in Calc. 3, man what a time.

[michaelempeigne]

I would like you to do a derivation of the spherical coordinate equations.

[jameswilson]

Brilliant! Your channel is better than bprp!

[shiina_mahiru_]

Why not take some common factors out of the determinant first, then you can find it much less pain and, when you find the magnitude, it's much easier?

[lucasfrykman]

Can't you just use arc length and rotate it 360?

[borg]

I'm Sorry, I'm just a physicist. But I'm also not scared..